Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 14 Equation Of A Line

Exercise 14A

Question 1. Find, which of the following points lie on the line x - 2y + 5 = 0:
(i) (1, 3) (ii) (0, 5)
(iii) (-5, 0) (iv) (5, 5)
(v) (2, -1.5) (vi) (-2, -1.5)
Answer:
The given line is \( x - 2y + 5 = 0 \).

(i) Substituting x = 1 and y = 3 in the given equation, we have:
L.H.S. = 1 - 2 × 3 + 5 = 1 - 6 + 5 = 6 - 6 = 0 = R.H.S.
Thus, the point (1, 3) lies on the given line.

(ii) Substituting x = 0 and y = 5 in the given equation, we have:
L.H.S. = 0 - 2 × 5 + 5 = -10 + 5 = -5 ≠ R.H.S.
Thus, the point (0, 5) does not lie on the given line.

(iii) Substituting x = -5 and y = 0 in the given equation, we have:
L.H.S. = -5 - 2 × 0 + 5 = -5 - 0 + 5 = 5 - 5 = 0 = R.H.S.
Thus, the point (-5, 0) lie on the given line.

(iv) Substituting x = 5 and y = 5 in the given equation, we have:
L.H.S. = 5 - 2 × 5 + 5 = 5 - 10 + 5 = 10 - 10 = 0 = R.H.S.
Thus, the point (5, 5) lies on the given line.

(v) Substituting x = 2 and y = -1.5 in the given equation, we have:
L.H.S. = 2 - 2 × (-1.5) + 5 = 2 + 3 + 5 = 10 ≠ R.H.S.
Thus, the point (2, -1.5) does not lie on the given line.

(vi) Substituting x = -2 and y = -1.5 in the given equation, we have:
L.H.S. = -2 - 2 × (-1.5) + 5 = -2 + 3 + 5 = 6 ≠ R.H.S.
Thus, the point (-2, -1.5) does not lie on the given line.

Points that lie on the line: (i) (1, 3), (iii) (-5, 0), (iv) (5, 5)
Points that do not lie on the line: (ii) (0, 5), (v) (2, -1.5), (vi) (-2, -1.5)

In simple words: We put each point's x and y values into the line equation. If we get 0 = 0, the point is on the line. If not, it is not on the line.

📝 Teacher's Note: Show students how to check step by step. Put x and y values from the point into the left side of the equation. If the answer equals the right side, the point is on the line.

🎯 Exam Tip: Always write L.H.S. = and R.H.S. = clearly. Show all calculation steps. Then state clearly if the point lies on the line or not.

Question 2. State, true or false:
(i) the line \( \frac{x}{2} + \frac{y}{3} = 0 \) passes through the point (2, 3).
(ii) the line \( \frac{x}{2} + \frac{y}{3} = 0 \) passes through the point (4, -6).
(iii) the point (8, 7) lies on the line y - 7 = 0.
(iv) the point (-3, 0) lies on the line x + 3 = 0.
(v) if the point (2, a) lies on the line 2x - y = 3, then a = 1.
Answer:
(i) The given line is \( \frac{x}{2} + \frac{y}{3} = 0 \)
Substituting x = 2 and y = 3 in the given equation,
L.H.S. = \( \frac{2}{2} + \frac{3}{3} = 1 + 1 = 2 \neq \) R.H.S.
Thus, the given statement is false.

(ii) The given line is \( \frac{x}{2} + \frac{y}{3} = 0 \)
Substituting x = 4 and y = -6 in the given equation,
L.H.S. = \( \frac{4}{2} + \frac{-6}{3} = 2 - 2 = 0 = \) R.H.S.
Thus, the given statement is true.

(iii) L.H.S = y - 7 = 7 - 7 = 0 = R.H.S.
Thus, the point (8, 7) lies on the line y - 7 = 0.
The given statement is true.

(iv) L.H.S = x + 3 = -3 + 3 = 0 = R.H.S
Thus, the point (-3, 0) lies on the line x + 3 = 0.
The given statement is true.

(v) The point (2, a) lies on the line 2x - y = 3.
∴ 2(2) - a = 3
4 - a = 3
a = 4 - 3 = 1
Thus, the given statement is false.

In simple words: We check each statement by putting the point values into the line equation. True means the point is on the line. False means it is not.

📝 Teacher's Note: Remind students to be careful with fractions and negative signs. For statement (v), they need to solve for a, not just check if a = 1.

🎯 Exam Tip: Write "True" or "False" clearly at the end of each part. Show your working for each calculation. For part (v), solve for a and then check if it equals 1.

Question 3. The line given by the equation \( 2x - \frac{y}{3} = 7 \) passes through the point (k, 6); calculate the value of k.
Answer:
Given, the line given by the equation \( 2x - \frac{y}{3} = 7 \) passes through the point (k, 6).
Substituting x = k and y = 6 in the given equation, we have:
\( 2k - \frac{6}{3} = 7 \)
2k - 2 = 7
2k = 9
k = \( \frac{9}{2} = 4.5 \)

In simple words: We put x = k and y = 6 into the line equation. Then we solve to find what value k must have.

📝 Teacher's Note: Show students that when a point lies on a line, we can substitute the point's coordinates and solve for the unknown. Be careful with fraction calculations.

🎯 Exam Tip: Write the given equation first. Substitute the point coordinates clearly. Show each step of solving for k. Write the final answer clearly.

Question 4. For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?
Answer:
The given equation of the line is 9x + 4y = 3.
Put x = 3 and y = -k, we have:
9(3) + 4(-k) = 3
27 - 4k = 3
4k = 27 - 3 = 24
k = 6

In simple words: We put the point (3, -k) into the line equation and solve to find what k should be.

📝 Teacher's Note: Emphasize that y-coordinate is -k, not k. Students often miss the negative sign. Show them to be extra careful with signs.

🎯 Exam Tip: Pay attention to negative signs. When y = -k, substitute -k for y in the equation. Show all arithmetic steps clearly.

Question 5. The line \( \frac{3x}{5} - \frac{2y}{3} + 1 = 0 \) contains the point (m, 2m - 1); calculate the value of m.
Answer:
The equation of the given line is \( \frac{3x}{5} - \frac{2y}{3} + 1 = 0 \)
Putting x = m, y = 2m - 1, we have:
\( \frac{3m}{5} - \frac{2(2m - 1)}{3} + 1 = 0 \)
\( \frac{3m}{5} - \frac{4m - 2}{3} = -1 \)
\( \frac{9m - 20m + 10}{15} = -1 \)
9m - 20m + 10 = -15
-11m = -25
m = \( \frac{25}{11} = 2\frac{3}{11} \)

In simple words: We substitute the point coordinates into the line equation. Then we solve the equation with fractions to find m.

📝 Teacher's Note: This problem has fractions and algebraic expressions. Show students how to find common denominators and simplify step by step. Work slowly through the fraction arithmetic.

🎯 Exam Tip: When working with fractions, find the LCD (lowest common denominator). Show each step clearly. Double-check your fraction arithmetic to avoid mistakes.

Question 6. Does the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2)?
Answer:
The given line will bisect the join of A (5, -2) and B (-1, 2), if the co-ordinates of the mid-point of AB satisfy the equation of the line.
The co-ordinates of the mid-point of AB are
\( \left( \frac{5 + (-1)}{2}, \frac{-2 + 2}{2} \right) = (2, 0) \)
Substituting x = 2 and y = 0 in the given equation, we have:
L.H.S. = 3x - 5y = 3(2) - 5(0) = 6 - 0 = 6 = R.H.S.
Hence, the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2).

In simple words: A line bisects (cuts in half) the join of two points if it passes through the middle point between them. We find the middle point and check if it lies on the line.

📝 Teacher's Note: Explain that "bisect" means to cut into two equal parts. The midpoint formula is key here: average of x-coordinates and average of y-coordinates.

🎯 Exam Tip: First find the midpoint using the formula. Then check if this midpoint lies on the given line. State clearly "Yes" or "No" at the end.

Question 7.
(i) The line y = 3x - 2 bisects the join of (a, 3) and (2, -5), find the value of a.
(ii) The line x - 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.
Answer:
(i) The given line bisects the join of A (a, 3) and B (2, -5), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.
The co-ordinates of the mid-point of AB are
\( \left( \frac{a + 2}{2}, \frac{3 + (-5)}{2} \right) = \left( \frac{a + 2}{2}, -1 \right) \)
Substituting x = \( \frac{a + 2}{2} \) and y = -1 in the given equation, we have:
y = 3x - 2
-1 = 3 × \( \frac{a + 2}{2} \) - 2
3 × \( \frac{a + 2}{2} \) = 1
a + 2 = \( \frac{2}{3} \)
a = \( \frac{2}{3} - 2 = \frac{2 - 6}{3} = -\frac{4}{3} \)

(ii) The given line bisects the join of A (8, -1) and B (0, k), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.
The co-ordinates of the mid-point of AB are
\( \left( \frac{8 + 0}{2}, \frac{-1 + k}{2} \right) = \left( 4, \frac{-1 + k}{2} \right) \)
Substituting x = 4 and y = \( \frac{-1 + k}{2} \) in the given equation, we have:
x - 6y + 11 = 0
4 - 6 \( \left( \frac{-1 + k}{2} \right) \) + 11 = 0
6 \( \left( \frac{-1 + k}{2} \right) \) = 15
\( \frac{-1 + k}{2} = \frac{15}{6} \)
\( \frac{-1 + k}{2} = \frac{5}{2} \)
-1 + k = 5
k = 6

In simple words: When a line bisects the join of two points, the middle point of those two points must lie on the line. We find the middle point and use this fact to solve for the unknown.

📝 Teacher's Note: Show students that in bisection problems, they need to find the midpoint and substitute it into the line equation. Be careful with fraction arithmetic.

🎯 Exam Tip: Always find the midpoint first. Then substitute the midpoint coordinates into the line equation and solve for the unknown variable. Show all steps clearly.

Question 8.
(i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.
(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.
Answer:
(i) Given, the point (-3, 2) lies on the line ax + 3y + 6 = 0.
Substituting x = -3 and y = 2 in the given equation, we have:
a(-3) + 3(2) + 6 = 0
-3a + 12 = 0
3a = 12
a = 4

(ii) Given, the line y = mx + 8 contains the point (-4, 4).
Substituting x = -4 and y = 4 in the given equation, we have:
4 = m(-4) + 8
4 = -4m + 8
4m = 8 - 4 = 4
m = 1

In simple words: When a point lies on a line, we can substitute the point's coordinates into the line equation and solve for any unknown constants.

📝 Teacher's Note: This is the basic method for finding unknown coefficients in line equations. Remind students to be careful with negative signs when substituting.

🎯 Exam Tip: Substitute the point coordinates directly into the equation. Solve step by step for the unknown. Double-check your answer by putting it back into the original equation.

Question 9. The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x – 5y + 15 = 0?
Answer: Given, the point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3.
Co-ordinates of the point P are
\( \left(\frac{2 \times (-3) + 3 \times 2}{2 + 3}, \frac{2 \times 6 + 3 \times 1}{2 + 3}\right) \)
\( = \left(\frac{-6 + 6}{5}, \frac{12 + 3}{5}\right) \)
\( = (0, 3) \)
Substituting x = 0 and y = 3 in the given equation, we have:
L.H.S. = 0 - 5(3) + 15 = -15 + 15 = 0 = R.H.S.
Hence, the point P lies on the line x - 5y + 15 = 0.
In simple words: We found the point P using the section formula. Then we put the x and y values in the line equation. Since both sides became equal, P lies on the line.

📝 Teacher's Note: Show students the section formula on the board. Make them practice substituting values step by step. Many students forget to simplify fractions properly.

🎯 Exam Tip: Always write "Given" first, then use section formula. Finally substitute in the equation and check if L.H.S. = R.H.S. Write a clear conclusion.

Question 10. The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2. Does the line x – 2y = 0 contain Q?
Answer: Given, the line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2.
Co-ordinates of the point Q are
\( \left(\frac{1 \times 2 + 2 \times 5}{1 + 2}, \frac{1 \times 2 + 2 \times (-4)}{1 + 2}\right) \)
\( = \left(\frac{2 + 10}{3}, \frac{2 - 8}{3}\right) \)
\( = (4, -2) \)
Substituting x = 4 and y = -2 in the given equation, we have:
L.H.S. = x - 2y = 4 - 2(-2) = 4 + 4 = 8 ≠ R.H.S.
Hence, the given line does not contain point Q.
In simple words: We found point Q using the section formula. When we put Q's coordinates in the line equation, both sides were not equal. So Q is not on the line.

📝 Teacher's Note: Emphasize that when L.H.S. ≠ R.H.S., the point does not lie on the line. Students often make calculation errors with negative numbers.

🎯 Exam Tip: Write "≠" clearly when both sides are not equal. State the conclusion clearly - "does not contain" or "does not lie on the line".

Question 11. Find the point of intersection of the lines: 4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1)x – 2y = 4; find the value of k.
Answer: Consider the given equations:
4x + 3y = 1 ....(1)
3x – y + 9 = 0 ....(2)

Multiplying (2) with 3, we have:
9x – 3y = -27 ....(3)
Adding (1) and (3), we get,
13x = -26
x = -2

From (2), y = 3x + 9 = -6 + 9 = 3
Thus, the point of intersection of the given lines (1) and (2) is (-2, 3).

The point (-2, 3) lies on the line (2k – 1)x – 2y = 4.
(2k – 1)(-2) – 2(3) = 4
-4k + 2 – 6 = 4
-4k = 8
k = -2
In simple words: First we solved two equations to find where they meet. Then we used this meeting point to find the value of k in the third equation.

📝 Teacher's Note: Show students how to eliminate variables systematically. Teach them to label equations (1), (2), (3) for clarity. Practice substitution carefully.

🎯 Exam Tip: Always write "point of intersection is (x, y)" clearly. Then substitute this point in the third equation to find k. Show all algebraic steps.

Question 12. Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Answer: We know that two or more lines are said to be concurrent if they intersect at a single point.

We first find the point of intersection of the first two lines.
2x + 5y = 1 ....(1)
x – 3y = 6 ....(2)

Multiplying (2) by 2, we get,
2x – 6y = 12 ....(3)
Subtracting (3) from (1), we get,
11y = -11
y = -1

From (2), x = 6 + 3y = 6 – 3 = 3

So, the point of intersection of the first two lines is (3, -1).
If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.
Substituting x = 3 and y = -1, we have:
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 – 5 = 0 = R.H.S.
Thus, (3, -1) also lie on the third line.
Hence, the given lines are concurrent.
In simple words: Three lines are concurrent if they all meet at one point. We found where the first two lines meet, then checked if the third line also passes through that same point.

📝 Teacher's Note: Draw three lines meeting at one point on the board. Explain that concurrent means "meeting together". Show students the step-by-step process clearly.

🎯 Exam Tip: Write the definition of concurrent lines first. Find intersection of two lines, then check if the third line passes through that point. Write "Hence, proved" at the end.

Exercise 14B

Question 1. Find the slope of the line whose inclination is: (i) 0° (ii) 30° (iii) 72° 30' (iv) 46°
Answer:
(i) Slope = tan 0° = 0
(ii) Slope = tan 30° = \( \frac{1}{\sqrt{3}} \)
(iii) Slope = tan 72° 30' = 3.1716
(iv) Slope = tan 46° = 1.0355
In simple words: Slope is the tangent of the angle that a line makes with the x-axis. We use tan function to find slope from angle.

📝 Teacher's Note: Remind students that slope = tan θ where θ is the inclination angle. Practice using calculator for trigonometric values. Show standard angles clearly.

🎯 Exam Tip: Remember the formula slope = tan θ. For standard angles like 0°, 30°, 45°, 60°, 90°, memorize the exact values. Use calculator for other angles.

Question 2. Find the inclination of the line whose slope is: (i) 0 (ii) \( \sqrt{3} \) (iii) 0.7646 (iv) 1.0875
Answer:
(i) Slope = tan θ = 0
\( \Rightarrow \) θ = 0°
(ii) Slope = tan θ = \( \sqrt{3} \)
\( \Rightarrow \) θ = 60°
(iii) Slope = tan θ = 0.7646
\( \Rightarrow \) θ = 37° 24'
(iv) Slope = tan θ = 1.0875
\( \Rightarrow \) θ = 47° 24'
In simple words: When we know the slope, we use inverse tan (tan⁻¹) to find the angle of inclination with the x-axis.

📝 Teacher's Note: Explain that finding angle from slope uses tan⁻¹ or arctan function. Show students how to use calculator for inverse trigonometric functions.

🎯 Exam Tip: Use tan⁻¹ (slope) to find inclination angle. For \( \sqrt{3} \), remember it gives 60°. Convert decimal degrees to degrees and minutes if needed.

Question 3. Find the slope of the line passing through the following pairs of points: (i) (-2, -3) and (1, 2) (ii) (-4, 0) and origin (iii) (a, -b) and (b, -a)
Answer: We know:
Slope = \( \frac{y_2 - y_1}{x_2 - x_1} \)

(i) Slope = \( \frac{2 - (-3)}{1 - (-2)} = \frac{2 + 3}{1 + 2} = \frac{5}{3} \)

(ii) Slope = \( \frac{0 - 0}{0 - (-4)} = \frac{0}{4} = 0 \)

(iii) Slope = \( \frac{-a - (-b)}{b - a} = \frac{-a + b}{b - a} = 1 \)
In simple words: To find slope between two points, we use the slope formula. It tells us how steep the line is between those two points.

📝 Teacher's Note: Write the slope formula clearly on the board. Show students to be careful with negative signs. Practice with simple integer coordinates first.

🎯 Exam Tip: Always write the slope formula first. Substitute coordinates carefully, paying attention to signs. Simplify the fraction completely.

Question 4. Find the slope of the line parallel to AB if: (i) A = (-2, 4) and B = (0, 6) (ii) A = (0, -3) and B = (-2, 5)
Answer:
(i) Slope of AB = \( \frac{6 - 4}{0 - (-2)} = \frac{2}{2} = 1 \)
Slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB = \( \frac{5 - (-3)}{-2 - 0} = \frac{8}{-2} = -4 \)
Slope of the line parallel to AB = Slope of AB = -4
In simple words: Parallel lines have the same slope. So we find the slope of AB, and any line parallel to AB will have that same slope.

📝 Teacher's Note: Draw two parallel lines on the board to show they have the same steepness. Emphasize that parallel lines never meet and have equal slopes.

🎯 Exam Tip: Write "parallel lines have equal slopes" as the key concept. Calculate slope of AB first, then state that parallel line has the same slope.

Question 5. Find the slope of the line perpendicular to AB if: (i) A = (0, -5) and B = (-2, 4) (ii) A = (3, -2) and B = (-1, 2)
Answer:
(i) Slope of AB = \( \frac{4 - (-5)}{-2 - 0} = \frac{9}{-2} = -\frac{9}{2} \)
Slope of the line perpendicular to AB = \( \frac{-1}{\text{Slope of AB}} = \frac{-1}{-\frac{9}{2}} = \frac{2}{9} \)

(ii) Slope of AB = \( \frac{2 - (-2)}{-1 - 3} = \frac{4}{-4} = -1 \)
Slope of the line perpendicular to AB = \( \frac{-1}{\text{Slope of AB}} = 1 \)
In simple words: Perpendicular lines meet at 90°. Their slopes multiply to give -1. If one slope is m, the other is -1/m.

📝 Teacher's Note: Show students two perpendicular lines forming 90°. Explain that if slopes are m₁ and m₂, then m₁ × m₂ = -1 for perpendicular lines.

🎯 Exam Tip: Write "perpendicular lines have slopes whose product is -1". Find slope of AB, then use formula: perpendicular slope = -1/(slope of AB).

Question 6. The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Answer: Slope of the line passing through (0, 2) and (-3, -1) = \( \frac{-1 - 2}{-3 - 0} = \frac{-3}{-3} = 1 \)
Slope of the line passing through (-1, 5) and (4, a) = \( \frac{a - 5}{4 - (-1)} = \frac{a - 5}{5} \)
Since, the lines are parallel,
\( \therefore 1 = \frac{a - 5}{5} \)
a - 5 = 5
a = 10
In simple words: Since both lines are parallel, they have the same slope. We found the first line's slope, then made the second line's slope equal to it.

📝 Teacher's Note: Remind students that parallel means equal slopes. Set up the equation carefully and solve step by step. Check the answer by substitution.

🎯 Exam Tip: Find slopes of both lines using the slope formula. Since lines are parallel, set the slopes equal and solve for the unknown.

Question 7. The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Answer: Slope of the line passing through (-4, -2) and (2, -3) = \( \frac{-3 - (-2)}{2 - (-4)} = \frac{-1}{6} \)
Slope of the line passing through (a, 5) and (2, -1) = \( \frac{-1 - 5}{2 - a} = \frac{-6}{2 - a} \)
Since, the lines are perpendicular,
\( \therefore \frac{-1}{6} = \frac{-1}{\frac{-6}{2 - a}} \)
\( \frac{-1}{6} = \frac{2 - a}{6} \)
-1 = 2 - a
a = 3
In simple words: Perpendicular lines have slopes that multiply to give -1. We found both slopes and used this property to find a.

📝 Teacher's Note: Emphasize that for perpendicular lines, slope₁ × slope₂ = -1. Show students how to use the negative reciprocal relationship carefully.

🎯 Exam Tip: For perpendicular lines, use the condition that product of slopes = -1. Set up the equation correctly and solve step by step for the unknown.

Question 8. Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.
Answer:
The given points are A (4, -2), B (-4, 4) and C (10, 6).

Slope of AB = \( \frac{4 + 2}{-4 - 4} = \frac{6}{-8} = \frac{-3}{4} \)

Slope of BC = \( \frac{6 - 4}{10 + 4} = \frac{2}{14} = \frac{1}{7} \)

Slope of AC = \( \frac{6 + 2}{10 - 4} = \frac{8}{6} = \frac{4}{3} \)

It can be seen that:
Slope of AB = \( \frac{-1}{\text{Slope of AC}} \)

Hence, AB ⊥ AC.

Thus, the given points are the vertices of a right-angled triangle.
In simple words: We find the slopes of all three sides. When two slopes multiply to give -1, those sides are perpendicular (make a 90° angle). This makes it a right triangle.

📝 Teacher's Note: Show students that perpendicular lines have slopes that are negative reciprocals of each other. Use examples like slope 2 and slope -1/2. This is easier than memorizing the formula.

🎯 Exam Tip: Always check if any two slopes multiply to give -1. Write "Hence AB ⊥ AC" clearly. This shows the examiner you understand perpendicular lines.

Question 9. Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.
Answer:
The given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6).

Slope of AB = \( \frac{2 - 5}{1 - 4} = \frac{-3}{-3} = 1 \)

Slope of CD = \( \frac{6 - 3}{7 - 4} = \frac{3}{3} = 1 \)

Since, slope of AB = slope of CD

Therefore AB || CD

Slope of BC = \( \frac{3 - 2}{4 - 1} = \frac{1}{3} \)

Slope of DA = \( \frac{5 - 6}{4 - 7} = \frac{-1}{-3} = \frac{1}{3} \)

Since, slope of BC = slope of DA

Therefore, BC || DA

Hence, ABCD is a parallelogram
In simple words: In a parallelogram, opposite sides are parallel. Two lines are parallel when they have the same slope. We check that AB || CD and BC || DA.

📝 Teacher's Note: Draw a parallelogram on the board and show that opposite sides never meet because they have the same slope. Students can remember this visually.

🎯 Exam Tip: Always write "slope of AB = slope of CD" and "slope of BC = slope of DA". Then write "Therefore AB || CD" and "BC || DA". This gets full marks.

Question 10. (-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Answer:
Let the given points be A (-2, 4), B (4, 8), C (10, 7) and D (11, -5).

Let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

Co-ordinates of P are
\( \left( \frac{-2 + 4}{2}, \frac{4 + 8}{2} \right) = (1, 6) \)

Co-ordinates of Q are
\( \left( \frac{4 + 10}{2}, \frac{8 + 7}{2} \right) = \left( 7, \frac{15}{2} \right) \)

Co-ordinates of R are
\( \left( \frac{10 + 11}{2}, \frac{7 - 5}{2} \right) = \left( \frac{21}{2}, 1 \right) \)

Co-ordinates of S are
\( \left( \frac{11 - 2}{2}, \frac{-5 + 4}{2} \right) = \left( \frac{9}{2}, \frac{-1}{2} \right) \)

Slope of PQ = \( \frac{\frac{15}{2} - 6}{7 - 1} = \frac{\frac{15 - 12}{2}}{6} = \frac{3}{12} = \frac{1}{4} \)

Slope of RS = \( \frac{\frac{-1}{2} - 1}{\frac{9}{2} - \frac{21}{2}} = \frac{\frac{-1 - 2}{2}}{\frac{9 - 21}{2}} = \frac{-3}{-12} = \frac{1}{4} \)

Since, slope of PQ = Slope of RS, PQ || RS.

Slope of QR = \( \frac{1 - \frac{15}{2}}{\frac{21}{2} - 7} = \frac{\frac{2 - 15}{2}}{\frac{21 - 14}{2}} = \frac{-13}{7} \)

Slope of SP = \( \frac{6 + \frac{1}{2}}{1 - \frac{9}{2}} = \frac{\frac{12 + 1}{2}}{\frac{2 - 9}{2}} = \frac{13}{-7} = \frac{-13}{7} \)

Since, slope of QR = Slope of SP, QR || SP.

Hence, PQRS is a parallelogram.
In simple words: We find the mid-points of all four sides. Then we check if opposite sides of the new quadrilateral have the same slope. This proves they are parallel and form a parallelogram.

📝 Teacher's Note: This is a special theorem: joining mid-points of any quadrilateral always gives a parallelogram. Show this with a simple drawing using any irregular quadrilateral.

🎯 Exam Tip: First find all four mid-points using the mid-point formula. Then find slopes of opposite sides and show they are equal. Write "PQ || RS" and "QR || SP" clearly.

Question 11. Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Answer:
The points P, Q, R will be collinear if slope of PQ and QR is the same.

Slope of PQ = \( \frac{c + a - b - c}{b - a} = \frac{a - b}{b - a} = -1 \)

Slope of QR = \( \frac{a + b - c - a}{c - b} = \frac{b - c}{c - b} = -1 \)

Hence, the points P, Q, and R are collinear.
In simple words: Three points lie on the same straight line when the slope between any two pairs is the same. Here both slopes equal -1, so all three points are on one line.

📝 Teacher's Note: Collinear means "on the same line". Show students three points on a ruler - they all have the same slope when you connect any two. This is easier to understand than the formal definition.

🎯 Exam Tip: Always write "slope of PQ = slope of QR" and show they are equal. Then conclude "Hence, P, Q, R are collinear." This gets full marks.

Question 12. Find x, if the slope of the line joining (x, 2) and (8, -11) is \( \frac{-3}{4} \).
Answer:
Let A = (x, 2) and B = (8, -11)

Slope of AB = \( \frac{-11 - 2}{8 - x} \)

\( \frac{-11 - 2}{8 - x} = \frac{-3}{4} \) (Given)

\( \frac{-13}{8 - x} = \frac{-3}{4} \)

52 = 24 - 3x

3x = 24 - 52 = -28

\( x = \frac{-28}{3} \)
In simple words: We use the slope formula with the given points. We know the slope value, so we make an equation and solve for x by cross multiplication.

📝 Teacher's Note: Show students how to cross multiply fractions: if a/b = c/d, then ad = bc. Practice this with simple numbers first before using in coordinate geometry.

🎯 Exam Tip: Always write the slope formula first, then substitute known values. Show cross multiplication clearly. Check your answer by substituting back into the original equation.

Question 13. The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slope of all its sides.
Answer:

[Diagram: This diagram shows an equilateral triangle ABC with side AB parallel to the x-axis, and vertex C pointing upward.]

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

Since, ABC is an equilateral triangle, ∠A = 60°

Slope of AC = tan 60° = \( \sqrt{3} \)

Slope of BC = -tan 60° = \( -\sqrt{3} \)
In simple words: AB is flat (parallel to x-axis), so its slope is 0. In an equilateral triangle, the other two sides make 60° angles with the horizontal. Their slopes are √3 and -√3.

📝 Teacher's Note: Draw an equilateral triangle with base horizontal. Show that the angles with x-axis are 60° and 120°. Use tan 60° = √3 to find slopes. Students remember this triangle easily.

🎯 Exam Tip: Remember that slope = tan θ where θ is the angle with positive x-axis. For equilateral triangles, use tan 60° = √3. One side goes up (+√3), one goes down (-√3).

Question 14. The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find: (i) the slope of the diagonal AC, (ii) the slope of the diagonal BD.
Answer:

[Diagram: This diagram shows a square ABCD with side AB parallel to the x-axis, positioned on coordinate axes.]

We know that the slope of any line parallel to x-axis is 0.
Therefore, slope of AB = 0

As CD || BC, slope of CD = Slope of AB = 0

As BC ⊥ AB, slope of BC = \( -\frac{1}{\text{Slope of AB}} = \frac{-1}{0} \) = not defined

As AD ⊥ AB, slope of AD = \( -\frac{1}{\text{Slope of AB}} = \frac{-1}{0} \) = not defined

In a square, diagonals make 45° angle with the sides.

Slope of diagonal AC = tan 45° = 1

Slope of diagonal BD = tan 135° = -1
In simple words: AB and CD are horizontal (slope = 0). BC and AD are vertical (slope = undefined). The diagonals go at 45° angles, so their slopes are +1 and -1.

📝 Teacher's Note: Show students that vertical lines have undefined slope because we divide by zero. In a square, diagonals always make 45° with horizontal, so slopes are ±1.

🎯 Exam Tip: Write "slope = undefined" for vertical lines, not "slope = ∞". For square diagonals, always write slopes as +1 and -1. These are standard results to remember.

Question 15. A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:
(i) the slope of the altitude of AB,
(ii) the slope of the median AD, and
(iii) the slope of the line parallel to AC.

Answer:
Given, A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC.

(i) Slope of AB = \( \frac{-2 - 4}{-3 - 5} = \frac{-6}{-8} = \frac{3}{4} \)

Slope of the altitude of AB = \( \frac{-1}{\text{Slope of AB}} = \frac{-1}{\frac{3}{4}} = \frac{-4}{3} \)

(ii) Since, D is the mid-point of BC.
Co-ordinates of point D are
\( \left(\frac{-3 + 1}{2}, \frac{-2 + (-8)}{2}\right) = (-1, -5) \)

Slope of AD = \( \frac{-5 - 4}{-1 - 5} = \frac{-9}{-6} = \frac{3}{2} \)

(iii) Slope of AC = \( \frac{-8 - 4}{1 - 5} = \frac{-12}{-4} = 3 \)

Slope of line parallel to AC = Slope of AC = 3

In simple words: We found the slope of side AB first. Then we used the rule that altitude makes a 90° angle, so its slope is the negative inverse. For median AD, we found the middle point D of BC and then calculated slope of AD. Parallel lines have the same slope.

📝 Teacher's Note: Tell students that altitude is perpendicular to the side. So if side has slope m, altitude has slope -1/m. Median goes from vertex to middle point of opposite side.

🎯 Exam Tip: Always write "Slope of altitude = -1/(slope of side)" clearly. For parallel lines, write "slopes are equal". Show all calculation steps to get full marks.

Question 16. The slope of the side BC of a rectangle ABCD is \( \frac{2}{3} \). Find:
(i) the slope of the side AB,
(ii) the slope of the side AD.

Answer:
(i) Since, BC is perpendicular to AB,

Slope of AB = \( \frac{-1}{\text{Slope of BC}} = \frac{-1}{\frac{2}{3}} = \frac{-3}{2} \)

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC = \( \frac{2}{3} \)

In simple words: In a rectangle, opposite sides are parallel (same slope) and adjacent sides are perpendicular (slopes multiply to give -1). BC and AD are opposite sides, so same slope. AB and BC are adjacent sides, so perpendicular slopes.

📝 Teacher's Note: Draw a rectangle and show students which sides are parallel and which are perpendicular. This makes the concept very clear.

🎯 Exam Tip: In rectangle problems, remember: opposite sides parallel (same slope), adjacent sides perpendicular (product of slopes = -1). Write this rule first.

Question 17. Find the slope and the inclination of the line AB if:
(i) A = (-3, -2) and B = (1, 2)
(ii) A = (0, \( -\sqrt{3} \)) and B = (3, 0)
(iii) A = (-1, 2\( \sqrt{3} \)) and B = (-2, \( \sqrt{5} \))

Answer:
(i) A = (-3, -2) and B = (1, 2)
Slope of AB = \( \frac{2 + 2}{1 + 3} = \frac{4}{4} = 1 = \tan θ \)

Inclination of line AB = θ = 45°

(ii) A = (0, \( -\sqrt{3} \)) and B = (3, 0)

Slope of AB = \( \frac{0 + \sqrt{3}}{3 - 0} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} = \tan θ \)

Inclination of line AB = θ = 30°

(iii) A = (-1, 2\( \sqrt{3} \)) and B = (-2, \( \sqrt{5} \))

Slope of AB = \( \frac{\sqrt{5} - 2\sqrt{3}}{-2 + 1} = \frac{-\sqrt{3}}{-1} = \sqrt{3} = \tan θ \)

Inclination of line AB = θ = 60°

In simple words: We calculated slope using the slope formula. Then we used the fact that slope = tan θ to find the angle. When slope = 1, angle = 45°. When slope = 1/√3, angle = 30°. When slope = √3, angle = 60°.

📝 Teacher's Note: Make students memorize these common values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. This helps them solve inclination problems quickly.

🎯 Exam Tip: Always write "slope = tan θ" and then find θ using standard angle values. Remember to convert negative slopes correctly for the angle.

Question 18. The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.

Answer:
Given, points A (-3, 2), B (2, -1) and C (a, 4) are collinear.
∴ Slope of AB = Slope of BC

\( \frac{-1 - 2}{2 + 3} = \frac{4 + 1}{a - 2} \)

\( \frac{-3}{5} = \frac{5}{a - 2} \)

-3a + 6 = 25
-3a = 25 - 6 = 19

a = \( \frac{-19}{3} = -6\frac{1}{3} \)

In simple words: When three points are collinear, they lie on the same straight line. This means the slope between any two pairs of points is the same. We made an equation using this fact and solved for a.

📝 Teacher's Note: Tell students that collinear means "on the same line". Draw three points on a line to show this. Then explain why slopes must be equal.

🎯 Exam Tip: For collinear points, always write "Slope of AB = Slope of BC" first. Then substitute coordinates carefully and solve the equation step by step.

Question 19. The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.

Answer:
Given, points A (K, 3), B (2, -4) and C (-K + 1, -2) are collinear.
∴ Slope of AB = Slope of BC

\( \frac{-4 - 3}{2 - K} = \frac{-2 + 4}{-K + 1 - 2} \)

\( \frac{-7}{2 - K} = \frac{2}{-K - 1} \)

7K + 7 = 4 - 2K
9K = -3

K = \( \frac{-1}{3} \)

In simple words: We used the same rule as the previous question. Three collinear points have equal slopes between different pairs. We set up the equation and solved for K.

📝 Teacher's Note: Show students how to be very careful with negative signs when substituting coordinates. One small mistake in signs can give the wrong answer.

🎯 Exam Tip: Double-check your coordinate substitution, especially with negative values and variables. Write each step clearly to avoid calculation errors.

Question 20. Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C. Which segment appears to have the steeper slope, AB or AC? Justify your conclusion by calculating the slopes of AB and AC.

Answer:

[Diagram: A coordinate graph showing three points A(1,1), B(4,7), and C(4,10) with lines AB and AC drawn from point A. Line AC appears steeper than line AB.]

From the graph, clearly, AC has steeper slope.

Slope of AB = \( \frac{7 - 1}{4 - 1} = \frac{6}{3} = 2 \)

Slope of AC = \( \frac{10 - 1}{4 - 1} = \frac{9}{3} = 3 \)

The line with greater slope is steeper. Hence, AC has steeper slope.

In simple words: We plotted the points on graph paper and drew the lines. AC looks steeper than AB. When we calculated slopes, AC has slope 3 and AB has slope 2. Since 3 > 2, AC is steeper.

📝 Teacher's Note: Let students plot the points themselves on graph paper. This hands-on activity helps them understand that larger slope means steeper line.

🎯 Exam Tip: Always plot the points neatly and draw clear lines. Calculate both slopes and compare them. Write "larger slope means steeper line" to show your understanding.

Question 21. Find the value(s) of k so that PQ will be parallel to RS. Given:
(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)
(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)
(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)

Answer:
Since, PQ || RS,
Slope of PQ = Slope of RS

(i) Slope of PQ = \( \frac{6 - 4}{3 - 2} = 2 \)

Slope of RS = \( \frac{k - 1}{10 - 8} = \frac{k - 1}{2} \)

∴ \( 2 = \frac{k - 1}{2} \)

k - 1 = 4
k = 5

(ii) Slope of PQ = \( \frac{11 + 1}{7 - 3} = \frac{12}{4} = 3 \)

Slope of RS = \( \frac{k + 1}{1 + 1} = \frac{k + 1}{2} \)

∴ \( 3 = \frac{k + 1}{2} \)

k + 1 = 6
k = 5

(iii) Slope of PQ = \( \frac{11 + 1}{6 - 5} = \frac{12}{1} = 12 \)

Slope of RS = \( \frac{k² + 4k}{7 - 6} = k² + 4k \)

∴ 12 = k² + 4k
k² + 4k - 12 = 0
(k + 6)(k - 2) = 0
k = -6 and 2

In simple words: For lines to be parallel, they must have the same slope. We calculated slope of both lines and made them equal. Then we solved for k. In the third part, we got a quadratic equation with two possible values for k.

📝 Teacher's Note: Remind students that parallel lines never meet and have identical slopes. When they get a quadratic equation, both solutions are usually valid.

🎯 Exam Tip: Always write "For parallel lines, slopes are equal" at the start. Show all calculation steps clearly. In quadratic equations, check if both solutions make sense in the context.

Exercise 14C

Question 1. Find the equation of a line whose: y-intercept = 2 and slope = 3.

Answer:
Given, y-intercept = c = 2 and slope = m = 3.

Substituting the values of c and m in the equation y = mx + c, we get,
y = 3x + 2, which is the required equation.

In simple words: We used the slope-intercept form y = mx + c. We just put in the values of slope (m = 3) and y-intercept (c = 2) to get the final equation.

📝 Teacher's Note: Teach students the slope-intercept form y = mx + c first. Then show them how to substitute given values directly into this formula.

🎯 Exam Tip: Always write "Using y = mx + c" first. Then substitute the given values clearly. Write the final equation in the form y = mx + c.

Question 2. Find the equation of a line whose: y-intercept = -1 and inclination = 45°.
Answer: Given, y-intercept = c = -1 and inclination = 45°.
Slope = m = tan 45° = 1
Substituting the values of c and m in the equation y = mx + c, we get,
y = x - 1, which is the required equation.
In simple words: When a line makes a 45° angle with the x-axis, its slope is 1. We put this slope and the y-intercept into the formula y = mx + c to get the equation.

📝 Teacher's Note: Show students that tan 45° = 1 using a right triangle. The y-intercept is where the line crosses the y-axis. Students often forget the negative sign.

🎯 Exam Tip: Always write "slope = tan θ" first. Then substitute values in y = mx + c. Check your signs carefully - negative y-intercept means the line crosses below the origin.

Question 3. Find the equation of the line whose slope is \( -\frac{4}{3} \) and which passes through (-3, 4).
Answer: Given, slope = \( -\frac{4}{3} \)
The equation passes through (-3, 4) = (x₁, y₁)
Substituting the values in y - y₁ = m(x - x₁), we get,
y - 4 = \( -\frac{4}{3} \)(x + 3)
3y - 12 = -4x - 12
4x + 3y = 0, which is the required equation.
In simple words: We use the point-slope form when we know one point and the slope. Just put the values in the formula and simplify to get the final equation.

📝 Teacher's Note: Remind students that when x₁ = -3, then (x - x₁) becomes (x + 3). The negative sign changes to positive. This is a common mistake.

🎯 Exam Tip: Use the formula y - y₁ = m(x - x₁). Be extra careful with negative coordinates. Always simplify to get the standard form ax + by + c = 0.

Question 4. Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.
Answer: Slope of the line = tan 60° = \( \sqrt{3} \)
The line passes through the point (5, 4) = (x₁, y₁)
Substituting the values in y - y₁ = m(x - x₁), we get,
y - 4 = \( \sqrt{3} \)(x - 5)
y - 4 = \( \sqrt{3} \)x - 5\( \sqrt{3} \)
y = \( \sqrt{3} \)x + 4 - 5\( \sqrt{3} \), which is the required equation.
In simple words: When a line makes 60° with the x-axis, its slope is √3. We put this slope and the given point into the point-slope formula.

📝 Teacher's Note: Students should memorize that tan 60° = √3. Draw a 30-60-90 triangle to show this. Keep the answer in surd form unless asked to approximate.

🎯 Exam Tip: Write "slope = tan 60° = √3" clearly. Keep surds in exact form. Don't convert √3 to decimal unless specifically asked.

Question 5. Find the equation of the line passing through: (i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)
Answer:
(i) Let (0, 1) = (x₁, y₁) and (1, 2) = (x₂, y₂)
∴ Slope of the line = \( \frac{2-1}{1-0} = 1 \)
The required equation of the line is given by:
y - y₁ = m(x - x₁)
y - 1 = 1(x - 0)
y - 1 = x
y = x + 1

(ii) Let (-1, -4) = (x₁, y₁) and (3, 0) = (x₂, y₂)
∴ Slope of the line = \( \frac{0+4}{3+1} = \frac{4}{4} = 1 \)
The required equation of the line is given by:
y - y₁ = m(x - x₁)
y + 4 = 1(x + 1)
y + 4 = x + 1
y = x - 3
In simple words: First find the slope using the two points. Then use any one point with the slope to write the equation using point-slope form.

📝 Teacher's Note: Show students the slope formula clearly: m = (y₂-y₁)/(x₂-x₁). Emphasize that both parts give slope = 1, but different equations because they pass through different points.

🎯 Exam Tip: Always calculate slope first. Be careful with signs when subtracting negative numbers. You can use either point in the point-slope formula - both will give the same final equation.

Question 6. The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find: (i) the gradient of PQ; (ii) the equation of PQ; (iii) the co-ordinates of the point where PQ intersects the x-axis.
Answer: Given, co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively.
(i) Gradient of PQ = \( \frac{5-6}{-3-2} = \frac{-1}{-5} = \frac{1}{5} \)
(ii) The equation of the line PQ is given by:
y - y₁ = m(x - x₁)
y - 6 = \( \frac{1}{5} \)(x - 2)
5y - 30 = x - 2
5y = x + 28
(iii) Let the line PQ intersects the x-axis at point A (x, 0).
Putting y = 0 in the equation of the line PQ, we get,
0 = x + 28
x = -28
Thus, the co-ordinates of the point where PQ intersects the x-axis are A (-28, 0).
In simple words: Gradient is another word for slope. To find where a line meets the x-axis, put y = 0 in the equation and solve for x.

📝 Teacher's Note: Explain that gradient and slope mean the same thing. When finding x-intercept, y = 0. When finding y-intercept, x = 0. This is a pattern students should remember.

🎯 Exam Tip: Write "gradient = slope" if asked for gradient. For x-intercept, substitute y = 0. For y-intercept, substitute x = 0. Always write coordinates as ordered pairs like (-28, 0).

Question 7. The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find: (i) the equation of AB; (ii) the co-ordinates of the point where the line AB intersects the y-axis.
Answer: (i) Given, co-ordinates of two points A and B are (-3, 4) and (2, -1).
Slope = \( \frac{-1-4}{2+3} = \frac{-5}{5} = -1 \)
The equation of the line AB is given by:
y - y₁ = m(x - x₁)
y + 1 = -1(x - 2)
y + 1 = -x + 2
x + y = 1
(ii) Let the line AB intersects the y-axis at point (0, y).
Putting x = 0 in the equation of the line, we get,
0 + y = 1
y = 1
Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).
In simple words: First find the slope, then write the equation. To find where the line meets the y-axis, put x = 0 in the equation.

📝 Teacher's Note: Show students that y-intercept is found by putting x = 0. The y-intercept always has the form (0, something). This is the opposite of x-intercept which has form (something, 0).

🎯 Exam Tip: For y-intercept, substitute x = 0 in the equation. The answer will be a point on the y-axis, so it looks like (0, number). Double-check by substituting back into the original equation.

Question 8. The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equation of AB and CD.
Answer:
[Diagram: Shows two intersecting lines forming angles of 45° and 60° with the positive x-axis, meeting at point P(3,4)]

Slope of line AB = tan 45° = 1
The line AB passes through P (3, 4). So, the equation of the line AB is given by:
y - y₁ = m(x - x₁)
y - 4 = 1(x - 3)
y - 4 = x - 3
y = x + 1

Slope of line CD = tan 60° = \( \sqrt{3} \)
The line CD passes through P (3, 4). So, the equation of the line CD is given by:
y - y₁ = m(x - x₁)
y - 4 = \( \sqrt{3} \)(x - 3)
y - 4 = \( \sqrt{3} \)x - 3\( \sqrt{3} \)
y = \( \sqrt{3} \)x + 4 - 3\( \sqrt{3} \)
In simple words: Both lines pass through the same point P but have different slopes. Line AB has slope 1, line CD has slope √3.

📝 Teacher's Note: Draw the diagram on the board and show the angles clearly. Students should see that both lines pass through P(3,4) but have different slopes based on their angles with the x-axis.

🎯 Exam Tip: Read the angles from the diagram carefully. tan 45° = 1, tan 60° = √3. Both equations must pass through the given intersection point P(3,4).

Question 9. In ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.
Answer: The vertices of ΔABC are A(3, 5), B(7, 8) and C(1, -10).
Coordinates of the mid-point D of BC = \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)
= \( \left(\frac{7 + 1}{2}, \frac{8 + (-10)}{2}\right) \)
= \( \left(\frac{8}{2}, \frac{-2}{2}\right) \)
= (4, -1)
Slope of AD = \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{4 - 3} = \frac{-6}{1} = -6 \)
Now, the equation of median is given by
y - y₁ = m(x - x₁)
\( \implies \) y - 5 = -6(x - 3)
\( \implies \) y - 5 = -6x + 18
\( \implies \) 6x + y = 23
In simple words: A median goes from one vertex to the middle point of the opposite side. First find the middle point, then write the equation of the line joining vertex A to this middle point.

📝 Teacher's Note: Explain that median connects a vertex to the midpoint of the opposite side. Show the midpoint formula clearly. Students often mix up which coordinates go together.

🎯 Exam Tip: First find midpoint of BC using midpoint formula. Then find slope of line AD. Use point-slope form with vertex A and this slope. Write final answer in standard form.

Question 10. The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex C = (7, 5). Find the equations of BC and CD.
Answer:
[Diagram: Shows parallelogram ABCD with side AB parallel to x-axis, angle A = 60°, and vertex C at (7,5)]

Since, ABCD is a parallelogram,
∠A + ∠B = 180°
∠B = 180° - 60° = 120°
Slope of BC = tan 120° = tan (90° + 30°) = cot30° = \( \sqrt{3} \)
Equation of the line BC is given by:
y - y₁ = m(x - x₁)
y - 5 = \( \sqrt{3} \)(x - 7)
y - 5 = \( \sqrt{3} \)x - 7\( \sqrt{3} \)
y = \( \sqrt{3} \)x + 5 - 7\( \sqrt{3} \)
Since, CD ∥ AB and AB ∥ x-axis, slope of CD = Slope of AB = 0
Equation of the line CD is given by:
y - y₁ = m(x - x₁)
y - 5 = 0(x - 7)
y = 5
In simple words: In a parallelogram, opposite sides are parallel. Since AB is parallel to x-axis, CD is also parallel to x-axis, so its slope is 0.

📝 Teacher's Note: Teach that parallel lines have equal slopes. When a line is parallel to x-axis, its slope is 0. When parallel to y-axis, slope is undefined.

🎯 Exam Tip: Remember that tan 120° = -√3 (not +√3). Adjacent angles in parallelogram add to 180°. Line parallel to x-axis has equation y = constant.

Question 11. Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4.
Answer: To find the point of intersection of x + 2y = 7 and x - y = 4:
From the second equation: x = y + 4
Substituting in first equation: (y + 4) + 2y = 7
3y + 4 = 7
3y = 3
y = 1
x = 1 + 4 = 5
So the intersection point is (5, 1).

Now, the line passes through origin (0, 0) and (5, 1).
Slope = \( \frac{1 - 0}{5 - 0} = \frac{1}{5} \)
Since the line passes through origin, the equation is y = mx
Therefore, y = \( \frac{1}{5} \)x or x - 5y = 0
In simple words: First find where the two given lines meet. Then draw a line from origin to this meeting point. Since it passes through origin, the equation is simply y = mx.

📝 Teacher's Note: Show students how to solve simultaneous equations by substitution. A line through origin always has form y = mx (no constant term).

🎯 Exam Tip: First solve the two equations to find intersection point. Then use this point with origin to find slope. Line through origin has equation y = mx, which can also be written as ax + by = 0.

Question 12. In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.
Answer:
Given:
A = (4, 7), B = (-2, 3), C = (0, 1)

Step 1: Find the median through vertex A.
Let AD be the median through vertex A. D is the midpoint of BC.
Co-ordinates of point D = \( \left(\frac{-2 + 0}{2}, \frac{3 + 1}{2}\right) = (-1, 2) \)

Step 2: Find slope of AD.
Slope of AD = \( \frac{2 - 7}{-1 - 4} = \frac{-5}{-5} = 1 \)

Step 3: Find equation of median AD.
Using point-slope form with point A(4, 7):
\( y - 7 = 1(x - 4) \)
\( y - 7 = x - 4 \)
\( y = x + 3 \)

Step 4: Find equation of line through B parallel to AC.
First find slope of AC:
Slope of AC = \( \frac{1 - 7}{0 - 4} = \frac{-6}{-4} = \frac{3}{2} \)

The line through B parallel to AC has the same slope = \( \frac{3}{2} \)

Step 5: Find equation using point B(-2, 3).
\( y - 3 = \frac{3}{2}(x + 2) \)
\( 2y - 6 = 3x + 6 \)
\( 2y = 3x + 12 \)

Final Answers:
Equation of median through A: \( y = x + 3 \)
Equation of line through B parallel to AC: \( 2y = 3x + 12 \)

In simple words: A median connects a vertex to the middle point of the opposite side. To find parallel lines, we use the same slope. We then use the point-slope formula to get the final equation.

📝 Teacher's Note: Draw the triangle on graph paper. Show students how to find the midpoint using the midpoint formula. Then show how parallel lines have the same slope.

🎯 Exam Tip: Always write "midpoint of BC" clearly. Use the point-slope form step by step. Write both final equations clearly at the end.

Question 13. A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Answer:
Given:
A = (0, 3), B = (4, 4), C = (8, 0)

Step 1: Find slope of BC.
Slope of BC = \( \frac{0 - 4}{8 - 4} = \frac{-4}{4} = -1 \)

Step 2: Find slope of perpendicular line.
Slope of line perpendicular to BC = \( \frac{-1}{\text{Slope of BC}} = \frac{-1}{-1} = 1 \)

Step 3: Find equation of line through A and perpendicular to BC.
Using point-slope form with point A(0, 3):
\( y - 3 = 1(x - 0) \)
\( y - 3 = x \)
\( y = x + 3 \)

Final Answer: \( y = x + 3 \)

In simple words: When two lines are perpendicular, their slopes multiply to give -1. We found the slope of BC first, then found the perpendicular slope, then used point A to write the equation.

📝 Teacher's Note: Show students that perpendicular slopes are negative reciprocals. If one slope is m, the perpendicular slope is -1/m.

🎯 Exam Tip: Write "slope of perpendicular line" clearly. Show the calculation -1/(-1) = 1 step by step. Always check your final equation.

Question 14. Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Answer:
Given:
Point C = (-1, 2), Line passes through A = (1, 4) and B = (2, 3)

Step 1: Find slope of line AB.
Slope of AB = \( \frac{3 - 4}{2 - 1} = \frac{-1}{1} = -1 \)

Step 2: Find slope of perpendicular line.
Slope of equation perpendicular to AB = \( \frac{-1}{\text{Slope of AB}} = \frac{-1}{-1} = 1 \)

Step 3: Find equation of perpendicular through C(-1, 2).
Using point-slope form:
\( y - 2 = 1(x + 1) \)
\( y - 2 = x + 1 \)
\( y = x + 3 \)

Final Answer: \( y = x + 3 \)

In simple words: We found the slope of the line joining the two given points. Then we found the perpendicular slope. Finally we used the given point to write the perpendicular line equation.

📝 Teacher's Note: Draw a diagram showing the line AB and point C. Show how the perpendicular line from C meets AB at 90 degrees.

🎯 Exam Tip: Label your points clearly as A, B, C. Show slope calculation step by step. Write "perpendicular slope" clearly in your working.

Question 15. Find the equation of the line, whose:
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
Answer:
(i) When x-intercept = 5 and y-intercept = 3
The line passes through points (5, 0) and (0, 3).
Slope = \( \frac{3 - 0}{0 - 5} = \frac{3}{-5} = -\frac{3}{5} \)

Using point-slope form with (0, 3):
\( y - 3 = -\frac{3}{5}(x - 0) \)
\( y - 3 = -\frac{3}{5}x \)
\( 5y - 15 = -3x \)
\( 3x + 5y = 15 \)

(ii) When x-intercept = -4 and y-intercept = 6
The line passes through points (-4, 0) and (0, 6).
Slope = \( \frac{6 - 0}{0 + 4} = \frac{6}{4} = \frac{3}{2} \)

Using point-slope form with (0, 6):
\( y - 6 = \frac{3}{2}(x - 0) \)
\( y - 6 = \frac{3}{2}x \)
\( 2y - 12 = 3x \)
\( 2y = 3x + 12 \)

(iii) When x-intercept = -8 and y-intercept = -4
The line passes through points (-8, 0) and (0, -4).
Slope = \( \frac{-4 - 0}{0 + 8} = \frac{-4}{8} = -\frac{1}{2} \)

Using point-slope form with (0, -4):
\( y + 4 = -\frac{1}{2}(x - 0) \)
\( y + 4 = -\frac{1}{2}x \)
\( 2y + 8 = -x \)
\( x + 2y + 8 = 0 \)

Final Answers:
(i) \( 3x + 5y = 15 \)
(ii) \( 2y = 3x + 12 \)
(iii) \( x + 2y + 8 = 0 \)

In simple words: x-intercept means the point where line crosses x-axis (y = 0). y-intercept means where line crosses y-axis (x = 0). We use these two points to find the slope and equation.

📝 Teacher's Note: Show students on graph paper where intercepts are. x-intercept is at (a, 0) and y-intercept is at (0, b). Draw these points clearly.

🎯 Exam Tip: Always write the intercept points first: (x-intercept, 0) and (0, y-intercept). Then find slope using these two points. Show all steps clearly.

Question 16. Find the equation of the line whose slope is \( -\frac{5}{6} \) and x-intercept is 6.
Answer:
Given:
Slope = \( m = -\frac{5}{6} \)
x-intercept = 6, so the line passes through point (6, 0)

Step 1: Use point-slope form.
Using point (6, 0) and slope \( m = -\frac{5}{6} \):
\( y - 0 = -\frac{5}{6}(x - 6) \)
\( y = -\frac{5}{6}x + 5 \)
\( 6y = -5x + 30 \)
\( 5x + 6y = 30 \)

Final Answer: \( 5x + 6y = 30 \)

In simple words: We know the slope and one point on the line (the x-intercept). We use the point-slope formula to find the complete equation of the line.

📝 Teacher's Note: Remind students that x-intercept means y = 0. So the point is (x-intercept, 0). Then use point-slope form directly.

🎯 Exam Tip: Write the point (6, 0) clearly first. Show the point-slope substitution step by step. Clear all fractions by multiplying through.

Question 17. Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Answer:
Given:
x-intercept = 5, so line passes through (5, 0)
Line also passes through (-3, 2)

Step 1: Find slope using the two points.
Slope = \( \frac{2 - 0}{-3 - 5} = \frac{2}{-8} = -\frac{1}{4} \)

Step 2: Find equation using point-slope form.
Using point (5, 0):
\( y - 0 = -\frac{1}{4}(x - 5) \)
\( y = -\frac{1}{4}x + \frac{5}{4} \)
\( 4y = -x + 5 \)
\( x + 4y = 5 \)

Final Answer: \( x + 4y = 5 \)

In simple words: We have two points on the line: the x-intercept point (5, 0) and the given point (-3, 2). We find slope using these two points, then write the equation.

📝 Teacher's Note: Show students how x-intercept gives us the point (5, 0). Then we have two points to find slope. This is easier than using point-slope directly.

🎯 Exam Tip: Write both points clearly: (5, 0) and (-3, 2). Use the slope formula step by step. Check your answer by substituting both points.

Question 18. Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.
Answer:
Given:
Line passes through (1, 3)
y-intercept = 5, so line passes through (0, 5)

Step 1: Find slope using the two points.
Slope = \( \frac{3 - 5}{1 - 0} = \frac{-2}{1} = -2 \)

Step 2: Find equation using point-slope form.
Using point (1, 3):
\( y - 3 = -2(x - 1) \)
\( y - 3 = -2x + 2 \)
\( y = -2x + 5 \)
\( 2x + y = 5 \)

Final Answer: \( 2x + y = 5 \)

In simple words: y-intercept 5 means the line crosses y-axis at (0, 5). With the given point (1, 3), we have two points to find the slope and equation.

📝 Teacher's Note: Explain that y-intercept means the point where line crosses y-axis, so x = 0. Point is (0, y-intercept).

🎯 Exam Tip: Write both points: (1, 3) and (0, 5). Find slope first, then use point-slope form. Your final equation should be in standard form.

Question 19. Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.
Answer:
Given:
Line passes through (-2, 0)
Line is equally inclined to both coordinate axes

Step 1: Find possible slopes.
When a line is equally inclined to both axes, it makes equal angles with x-axis and y-axis.
This happens when the slope is +1 or -1.

Case 1: Slope = +1
Using point-slope form with (-2, 0):
\( y - 0 = 1(x + 2) \)
\( y = x + 2 \)

Case 2: Slope = -1
Using point-slope form with (-2, 0):
\( y - 0 = -1(x + 2) \)
\( y = -x - 2 \)

Final Answers:
\( y = x + 2 \) and \( y = -x - 2 \)

In simple words: "Equally inclined to coordinate axes" means the line makes the same angle with both x-axis and y-axis. This only happens when slope is +1 or -1.

📝 Teacher's Note: Draw both axes and show how lines with slope +1 and -1 make 45° angles with both axes. These are the only two possibilities.

🎯 Exam Tip: Write "equally inclined means slope = ±1" clearly. Show both cases. There are always two answers for this type of question.

Question 20. The line through P(5, 3) intersects y-axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.
Answer:

[Diagram: This diagram shows a coordinate system with x and y axes. A line passes through point P(5, 3) and makes a 45° angle with the x-axis. The line intersects the y-axis at point Q.]

(i) The equation of the y-axis is \( x = 0 \)
Given that the required line through P(5, 3) intersects the y-axis at Q and the angle of inclination is 45°.
Therefore slope of the line PQ = \( \tan 45° = 1 \).

(ii) The equation of a line passing through the point A(\( x_1, y_1 \)) with slope 'm' is
\( y - y_1 = m(x - x_1) \)
Therefore the equation of the line passing through the point P(5, 3) with slope 1 is
\( y - 3 = 1 \times (x - 5) \)
\( \Rightarrow y - 3 = x - 5 \)
\( \Rightarrow x - y = 2 \)

(iii) From subpart (ii), the equation of the line PQ is \( x - y = 2 \).
Given that the line intersects with the y-axis, \( x = 0 \)
Thus, substituting \( x = 0 \) in the equation \( x - y = 2 \)
we have, \( 0 - y = 2 \)
\( \Rightarrow y = -2 \)
Thus, the coordinates point of intersection Q are Q(0, -2)

In simple words: We found the slope using the 45° angle. Then we used the point-slope formula to get the line equation. Finally, we put x = 0 to find where the line crosses the y-axis.

📝 Teacher's Note: Draw a line on the board making 45° with x-axis. Show students that tan 45° = 1. This makes the slope = 1. Students remember this easily.

🎯 Exam Tip: Always write "slope = tan θ" first. Then substitute the angle value. For y-axis intersection, always put x = 0 in the equation.

Question 21. Write down the equation of the line whose gradient is \( \frac{-2}{5} \) and which passes through point P where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3: 1.
Answer:
Given, P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3: 1.
Co-ordinates of point P are
\( \left(\frac{3 \times 12 + 1 \times 4}{3 + 1}, \frac{3 \times 0 + 1 \times (-8)}{3 + 1}\right) \)
\( = \left(\frac{36 + 4}{4}, \frac{-8}{4}\right) \)
\( = (10, -2) \)
Slope = m = \( \frac{-2}{5} \) (Given)
Thus, the required equation of the line is
\( y - y_1 = m(x - x_1) \)
\( y + 2 = \frac{-2}{5}(x - 10) \)
\( 5y + 10 = -2x + 20 \)
\( 2x + 5y = 10 \)

In simple words: We first found point P using the section formula. Then we used the point-slope form with the given slope to write the line equation.

📝 Teacher's Note: Teach the section formula step by step. Show students that when a point divides in ratio m:n, we use the formula with weights. Practice with simple ratios like 1:1 first.

🎯 Exam Tip: Always write the section formula first. Then substitute values carefully. Check your arithmetic twice. Write the final equation in standard form ax + by = c.

Question 22. A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC, Find:
(i) the co-ordinates of the centroid of triangle ABC.
(ii) the equation of a line, through the centroid and parallel to AB.
Answer:
(i) Co-ordinates of the centroid of triangle ABC are
\( \left(\frac{1 + 3 + 7}{3}, \frac{4 + 2 + 5}{3}\right) \)
\( = \left(\frac{11}{3}, \frac{11}{3}\right) \)

(ii) Slope of AB = \( \frac{2 - 4}{3 - 1} = \frac{-2}{2} = -1 \)
Slope of the line parallel to AB = Slope of AB = -1
Thus, the required equation of the line is
\( y - y_1 = m(x - x_1) \)
\( y - \frac{11}{3} = -1\left(x - \frac{11}{3}\right) \)
\( 3y - 11 = -3x + 11 \)
\( 3x + 3y = 22 \)

In simple words: The centroid is the center point of a triangle. We find it by averaging the x-coordinates and y-coordinates. Parallel lines have the same slope.

📝 Teacher's Note: Show students how to find centroid by taking average of coordinates. Use a triangle drawn on paper and mark the centroid. Explain that parallel lines never meet because they have same slope.

🎯 Exam Tip: For centroid, add all x-coordinates and divide by 3, then do same for y-coordinates. For parallel lines, slopes are equal. Always write this clearly.

Question 23. A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP: CP = 2: 3.
Answer:
Given, AP: CP = 2: 3
∴ Co-ordinates of P are
\( \left(\frac{2 \times (-3) + 3 \times 7}{2 + 3}, \frac{2 \times 4 + 3 \times (-1)}{2 + 3}\right) \)
\( = \left(\frac{-6 + 21}{5}, \frac{8 - 3}{5}\right) \)
\( = \left(\frac{15}{5}, \frac{5}{5}\right) \)
\( = (3, 1) \)
Slope of BP = \( \frac{1 - 1}{3 - 4} = 0 \)
Required equation of the line passing through points B and P is
\( y - y_1 = m(x - x_1) \)
\( y - 1 = 0(x - 3) \)
\( y = 1 \)

In simple words: Point P divides side AC in the ratio 2:3. We found P using section formula. Then we found the slope of line BP and wrote its equation.

📝 Teacher's Note: Draw triangle ABC on board. Mark point P on side AC. Show students how P divides AC in ratio 2:3. When slope is 0, the line is horizontal.

🎯 Exam Tip: Use section formula to find point P first. When slope = 0, the line equation is y = constant. This means the line is horizontal.

Exercise 14D

Question 1. Find the slope and y-intercept of the line:
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Answer:
(i) y = 4
Comparing this equation with y = mx + c, we have:
Slope = m = 0
y-intercept = c = 4

(ii) ax - by = 0 ⇒ by = ax ⇒ y = \( \frac{a}{b} \)x
Comparing this equation with y = mx + c, we have:
Slope = m = \( \frac{a}{b} \)
y-intercept = c = 0

(iii) 3x - 4y = 5 ⇒ 4y = 3x - 5 ⇒ y = \( \frac{3}{4} \)x - \( \frac{5}{4} \)
Comparing this equation with y = mx + c, we have:
Slope = m = \( \frac{3}{4} \)
y-intercept = c = - \( \frac{5}{4} \)

In simple words: We change each equation to the form y = mx + c. The number with x is the slope (m). The constant number is the y-intercept (c).

📝 Teacher's Note: Always convert to y = mx + c form first. Show students that horizontal lines (like y = 4) have slope = 0. Lines through origin have y-intercept = 0.

🎯 Exam Tip: Write "Comparing with y = mx + c" every time. This shows the examiner you know the standard form. Always state slope = m and y-intercept = c clearly.

Question 2. The equation of a line x – y = 4. Find its slope and y-intercept. Also, find its inclination.
Answer:
Given equation of a line is x - y = 4
⇒ y = x - 4
Comparing this equation with y = mx + c. We have:
Slope = m = 1
y-intercept = c = -4
Let the inclination be θ.
Slope = 1 = tan θ = tan 45°
∴ θ = 45°

In simple words: First we change to y = mx + c form. The slope is 1. The y-intercept is -4. Since slope = tan θ, and tan 45° = 1, the inclination is 45°.

📝 Teacher's Note: Draw a line making 45° with x-axis. Show students that when slope = 1, the line makes equal angles with both axes. This is easy to remember.

🎯 Exam Tip: Remember that slope = tan θ. When slope = 1, θ = 45°. When slope = 0, θ = 0°. Always write the angle in degrees unless asked otherwise.

Question 3.
(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Answer:

In simple words: For perpendicular lines, we check if m₁ × m₂ = -1. For parallel lines, we check if m₁ = m₂. For slope between two points, we use the slope formula.

📝 Teacher's Note: Show students that perpendicular lines have slopes that multiply to give -1. Parallel lines have exactly the same slope. Use simple examples on the board.

🎯 Exam Tip: Always find slopes first. For perpendicular: m₁m₂ = -1. For parallel: m₁ = m₂. Show your working clearly step by step.

Question 4. Find the slope of the line which is parallel to:
(i) \( x + 2y + 3 = 0 \) (ii) \( \frac{x}{2} - \frac{y}{3} - 1 = 0 \)
Answer:
(i) \( x + 2y + 3 = 0 \)
\( 2y = -x - 3 \)
\( y = -\frac{1}{2}x - \frac{3}{2} \)
Slope of this line = \( -\frac{1}{2} \)
Slope of the line which is parallel to the given line = Slope of the given line = \( -\frac{1}{2} \)

(ii) \( \frac{x}{2} - \frac{y}{3} - 1 = 0 \)
\( \frac{y}{3} = \frac{x}{2} - 1 \)
\( y = \frac{3}{2}x - 3 \)
Slope of this line = \( \frac{3}{2} \)
Slope of the line which is parallel to the given line = Slope of the given line = \( \frac{3}{2} \)
In simple words: Parallel lines have the same slope. So we find the slope of the given line and that is our answer.

📝 Teacher's Note: Show students two railway tracks. They are parallel and have the same slope everywhere. This helps them remember that parallel lines always have equal slopes.

🎯 Exam Tip: Always write "parallel lines have equal slopes" in your answer. Convert the equation to y = mx + c form to find slope m easily.

Question 5. Find the slope of the line which is perpendicular to:
(i) \( x - \frac{y}{2} + 3 = 0 \) (ii) \( \frac{x}{3} - 2y = 4 \)
Answer:
(i) \( x - \frac{y}{2} + 3 = 0 \)
\( \frac{y}{2} = x + 3 \)
\( y = 2x + 6 \)
Slope of this line = 2
Slope of the line which is perpendicular to the given line = \( \frac{-1}{\text{Slope of the given line}} = \frac{-1}{2} \)

(ii) \( \frac{x}{3} - 2y = 4 \)
\( 2y = \frac{x}{3} - 4 \)
\( y = \frac{x}{6} - 2 \)
Slope of this line = \( \frac{1}{6} \)
Slope of the line which is perpendicular to the given line = \( \frac{-1}{\text{Slope of this line}} = \frac{-1}{\frac{1}{6}} = -6 \)
In simple words: When two lines are perpendicular, their slopes multiply to give -1. So if one slope is m, the other is -1/m.

📝 Teacher's Note: Use your hands to show perpendicular lines - like a cross. Tell students that perpendicular slopes are negative reciprocals of each other.

🎯 Exam Tip: Remember the formula: if slopes are m₁ and m₂, then m₁ × m₂ = -1 for perpendicular lines. Write this formula first.

Question 6.
(i) Lines 2x - by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x - ny - 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Answer:
(i) 2x - by + 3 = 0
by = 2x + 3
\( y = \frac{2}{b}x + \frac{3}{b} \)
Slope of this line = \( \frac{2}{b} \)
ax + 3y = 2
3y = -ax + 2
\( y = \frac{-a}{3}x + \frac{2}{3} \)
Slope of this line = \( \frac{-a}{3} \)
Since, the lines are parallel, so the slopes of the two lines are equal.
\( \therefore \frac{2}{b} = \frac{-a}{3} \)
ab = -6

(ii) mx + 3y + 7 = 0
3y = -mx - 7
\( y = \frac{-m}{3}x - \frac{7}{3} \)
Slope of this line = \( \frac{-m}{3} \)
5x - ny - 3 = 0
ny = 5x - 3
\( y = \frac{5}{n}x - \frac{3}{n} \)
Slope of this line = \( \frac{5}{n} \)
Since, the lines are perpendicular; the product of their slopes is -1.
\( \therefore \left(\frac{-m}{3}\right)\left(\frac{5}{n}\right) = -1 \)
5m = 3n
In simple words: For parallel lines, slopes are equal. For perpendicular lines, slopes multiply to give -1. Use these rules to find the relationship.

📝 Teacher's Note: Make students practice converting ax + by + c = 0 to y = mx + c form many times. This is the key step they often forget.

🎯 Exam Tip: Always write the condition clearly: "For parallel lines, m₁ = m₂" or "For perpendicular lines, m₁ × m₂ = -1". This shows the examiner you know the rule.

Question 7. Find the value of p if the lines, whose equations are 2x - y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Answer:
2x - y + 5 = 0
y = 2x + 5
Slope of this line = 2
px + 3y = 4
3y = -px + 4
\( y = \frac{-p}{3}x + \frac{4}{3} \)
Slope of this line = \( \frac{-p}{3} \)
Since, the lines are perpendicular to each other, the product of the slopes is -1.
\( \therefore (2)\left(\frac{-p}{3}\right) = -1 \)
\( \frac{-2p}{3} = -1 \)
2p = 3
\( p = \frac{3}{2} \)
In simple words: We found both slopes, then used the rule that perpendicular slopes multiply to give -1. This gives us the value of p.

📝 Teacher's Note: Draw two perpendicular lines on the board. Show students that if one line goes up steeply, the other goes down gently. This helps them understand negative reciprocals.

🎯 Exam Tip: Always substitute your answer back into the equation to check. Here, check that 2 × (-3/2)/3 = -1. This catches silly mistakes.

Question 8. The equation of a line AB is 2x - 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Answer:
(i) 2x - 2y + 3 = 0
2y = 2x + 3
\( y = x + \frac{3}{2} \)
Slope of the line AB = 1
(ii) Required angle = θ
Slope = tan θ = 1 = tan 45°
θ = 45°
In simple words: The slope tells us how steep the line is. When slope = 1, the line makes a 45° angle with the x-axis - like a diagonal of a square.

📝 Teacher's Note: Use a square piece of paper. Show students that the diagonal makes a 45° angle with the bottom edge. This is what slope = 1 looks like.

🎯 Exam Tip: Remember these common angles: slope 0 means 0°, slope 1 means 45°, slope √3 means 60°, slope undefined means 90°. Learn these by heart.

Question 9. The lines represented by 4x + 3y = 9 and px - 6y + 3 = 0 are parallel. Find the value of p.
Answer:
Step 1: Convert first equation to slope-intercept form.
4x + 3y = 9
3y = -4x + 9
\( y = \frac{-4}{3}x + 3 \)

Step 2: Find slope of first line.
Slope of first line = \( \frac{-4}{3} \)

Step 3: Convert second equation to slope-intercept form.
px - 6y + 3 = 0
6y = px + 3
\( y = \frac{p}{6}x + \frac{1}{2} \)

Step 4: Find slope of second line.
Slope of second line = \( \frac{p}{6} \)

Step 5: For parallel lines, slopes are equal.
\( \frac{-4}{3} = \frac{p}{6} \)

Step 6: Solve for p.
\( -4 = \frac{p}{2} \)
p = -8

In simple words: Parallel lines have the same slope. We found the slope of both lines and made them equal to find p.

📝 Teacher's Note: Show students that parallel lines are like railway tracks - they never meet because they have the same slope. Convert both equations to y = mx + c form to see the slope clearly.

🎯 Exam Tip: Always write "For parallel lines, slopes are equal" to get method marks. Convert to y = mx + c form first.

Question 10. If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Answer:
Step 1: Find slope of first line.
y = 3x + 7
Slope of first line = 3

Step 2: Convert second equation to slope-intercept form.
2y + px = 3
2y = -px + 3
\( y = \frac{-p}{2}x + \frac{3}{2} \)

Step 3: Find slope of second line.
Slope of second line = \( \frac{-p}{2} \)

Step 4: For perpendicular lines, product of slopes = -1.
\( 3 \times \left(\frac{-p}{2}\right) = -1 \)

Step 5: Solve for p.
\( \frac{-3p}{2} = -1 \)
-3p = -2
\( p = \frac{2}{3} \)

In simple words: Perpendicular lines cross at right angles. Their slopes multiply to give -1. We used this rule to find p.

📝 Teacher's Note: Use the corner of a book to show perpendicular lines. Tell students that perpendicular slopes are negative reciprocals - if one is 3, the other is -1/3.

🎯 Exam Tip: Write "For perpendicular lines, product of slopes = -1" for method marks. This is the key formula.

Question 11. The line through A(-2,3) and B(4,b) is perpendicular to the line 2x - 4y = 5. Find the value of b.
Answer:
Step 1: Find slope of line AB using two points.
Slope of AB = \( \frac{y_2 - y_1}{x_2 - x_1} \)
The slope of line passing through A(-2, 3) and B(4, b) is
Slope of AB = \( \frac{b - 3}{4 - (-2)} = \frac{b - 3}{4 + 2} = \frac{b - 3}{6} \)

Step 2: Find slope of given line 2x - 4y = 5.
2x - 4y = 5
4y = 2x - 5
\( y = \frac{1}{4}(2x - 5) \)
\( y = \frac{x}{2} - \frac{5}{4} \)
Slope of given line = \( \frac{1}{2} \)

Step 3: For perpendicular lines, product of slopes = -1.
\( \left(\frac{b - 3}{6}\right) \times \frac{1}{2} = -1 \)

Step 4: Solve for b.
\( \frac{b - 3}{12} = -1 \)
b - 3 = -12
b = 3 - 12
b = -9

In simple words: We found the slope between points A and B. Then we found the slope of the given line. Since they are perpendicular, their slopes multiply to -1.

📝 Teacher's Note: Draw the coordinate plane and mark points A and B. Show students how the slope formula works by counting rise over run on the grid.

🎯 Exam Tip: Use the slope formula correctly: (y₂-y₁)/(x₂-x₁). Be careful with negative signs when substituting coordinates.

Question 12. Find the equation of the line through (-5, 7) and parallel to: (i) x-axis (ii) y-axis
Answer:
(i) The slope of the line parallel to x-axis is 0.
(x₁, y₁) = (-5, 7)
Required equation of the line is
y - y₁ = m(x - x₁)
y - 7 = 0(x + 5)
y = 7

(ii) The slope of the line parallel to y-axis is not defined.
That is slope of the line is tan90° and hence the given line is parallel to y-axis.
(x₁, y₁) = (-5, 7)
Required equation of the line is
x - x₁ = 0
x + 5 = 0

In simple words: Lines parallel to x-axis are horizontal (like the floor). Lines parallel to y-axis are vertical (like a wall).

📝 Teacher's Note: Show students the classroom floor (horizontal like x-axis) and wall (vertical like y-axis). Parallel to x-axis means y stays same. Parallel to y-axis means x stays same.

🎯 Exam Tip: For parallel to x-axis, equation is y = constant. For parallel to y-axis, equation is x = constant.

Question 13. (i) Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4. (ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).
Answer:
(i) x - 3y = 4
3y = x - 4
\( y = \frac{1}{3}x - \frac{4}{3} \)
Slope of this line = \( \frac{1}{3} \)
Slope of a line parallel to this line = \( \frac{1}{3} \)
Required equation of the line passing through (5, -3) is
y - y₁ = m(x - x₁)
\( y + 3 = \frac{1}{3}(x - 5) \)
3y + 9 = x - 5
x - 3y - 14 = 0

(ii) 3x + 2y = 8
\( y = \frac{-3}{2}x + \frac{8}{2} \)
\( y = \frac{-3}{2}x + 4 \)
Slope of given line = \( \frac{-3}{2} \)
Since the required line is parallel to given straight line.
Slope of required line (m) = \( \frac{-3}{2} \)
Now the equation of the required line is given by:
y - y₁ = m(x - x₁)
\( y - 1 = \frac{-3}{2}(x - 0) \)
2y - 2 = -3x
3x + 2y = 2

In simple words: Parallel lines have the same slope. We find the slope of the given line first, then use point-slope form to write the new equation.

📝 Teacher's Note: Remind students that parallel lines never meet, like railway tracks. They have exactly the same slope but different y-intercepts.

🎯 Exam Tip: Always convert to y = mx + c form to find slope easily. Use point-slope form: y - y₁ = m(x - x₁) for the final equation.

Question 14. Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.
Answer:
Step 1: Find slope of given line.
4x + 5y = 6
5y = -4x + 6
\( y = \frac{-4}{5}x + \frac{6}{5} \)
Slope of this line = \( \frac{-4}{5} \)

Step 2: Find slope of perpendicular line.
The required line is perpendicular to the line 4x + 5y = 6.
Slope of the required line = \( \frac{-1}{\text{Slope of the given line}} = \frac{-1}{\frac{-4}{5}} = \frac{5}{4} \)

Step 3: Use point-slope form.
The required equation of the line is given by
y - y₁ = m(x - x₁)
\( y - 1 = \frac{5}{4}(x + 2) \)
4y - 4 = 5x + 10
5x - 4y + 14 = 0

In simple words: Perpendicular lines cross at 90 degrees. If one line has slope m, the perpendicular line has slope -1/m.

📝 Teacher's Note: Use your hands to show perpendicular lines - one horizontal, one vertical. Tell students that perpendicular slopes are negative reciprocals of each other.

🎯 Exam Tip: For perpendicular lines, multiply slopes to get -1. If given slope is -4/5, perpendicular slope is 5/4.

Question 15. Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Answer:
Step 1: Find midpoint of line segment.
Let A = (6, -3) and B = (0, 3).
We know the perpendicular bisector of a line is perpendicular to the line and it bisects the line, that it, it passes through the mid-point of the line.
Co-ordinates of the mid-point of AB are
\( \left(\frac{6 + 0}{2}, \frac{-3 + 3}{2}\right) = (3, 0) \)
Thus, the required line passes through (3, 0).

Step 2: Find slope of line segment AB.
Slope of AB = \( \frac{3 - (-3)}{0 - 6} = \frac{6}{-6} = -1 \)

Step 3: Find slope of perpendicular bisector.
Slope of the required line = \( \frac{-1}{\text{Slope of AB}} = \frac{-1}{-1} = 1 \)

Step 4: Write equation using point-slope form.
Thus, the equation of the required line is given by:
y - y₁ = m(x - x₁)
y - 0 = 1(x - 3)
y = x - 3

In simple words: A perpendicular bisector cuts a line in half at 90 degrees. It passes through the midpoint and has perpendicular slope.

📝 Teacher's Note: Draw a line segment on the board. Show how the perpendicular bisector cuts it exactly in the middle at a right angle. Use a ruler and set square to demonstrate.

🎯 Exam Tip: Find midpoint first using midpoint formula. Then find perpendicular slope using negative reciprocal. Write equation using point-slope form.

Question 16. In the following diagram, write down: (i) the co-ordinates of the points A, B and C. (ii) the equation of the line through A and parallel to BC.
Answer:
[Diagram: This question refers to a coordinate geometry diagram showing points A, B, and C plotted on a coordinate plane.]

In simple words: This question needs the diagram to read the coordinates of points A, B, and C from the graph. Then we find the line through A parallel to BC.

📝 Teacher's Note: This question requires reading coordinates from a graph. Teach students to count squares carefully from the origin to find exact coordinates.

🎯 Exam Tip: Read coordinates carefully from the diagram. Write them as (x, y) pairs. For parallel lines, use the same slope.

Question 17. B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Answer:
We know that in a rhombus, diagonals bisect each other at right angle.
Let O be the point of intersection of the diagonals AC and BD.
Co-ordinates of O are
\( \left( \frac{-5 + 1}{2}, \frac{6 + 4}{2} \right) = (-2, 5) \)

Slope of BD = \( \frac{4 - 6}{1 - (-5)} = \frac{-2}{6} = \frac{-1}{3} \)

For line BD:
Using point-slope form \( y - y_1 = m(x - x_1) \)
\( y - 6 = \frac{-1}{3}(x + 5) \)
\( 3y - 18 = -x - 5 \)
\( x + 3y = 13 \)

For line AC:
Slope = \( m = \frac{-1}{\text{Slope of BD}} = 3 \), \( (x_1, y_1) = (-2, 5) \)
Equation of the line AC is
\( y - y_1 = m(x - x_1) \)
\( y - 5 = 3(x + 2) \)
\( y - 5 = 3x + 6 \)
\( y = 3x + 11 \)

In simple words: We found the middle point of the rhombus. Then we used slopes - diagonal lines are always at right angles in a rhombus.

📝 Teacher's Note: Show students that in rhombus, diagonals cut each other at 90 degrees. Use the rule that perpendicular lines have slopes that multiply to give -1.

🎯 Exam Tip: Always find the midpoint first. Then use the fact that perpendicular slopes multiply to -1. Write both equations clearly.

Question 18. A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.
Answer:
We know that in a square, diagonals bisect each other at right angle.
Let O be the point of intersection of the diagonals AC and BD.
Co-ordinates of O are
\( \left( \frac{7 - 1}{2}, \frac{-2 - 6}{2} \right) = (3, -4) \)

Slope of AC = \( \frac{-6 + 2}{-1 - 7} = \frac{-4}{-8} = \frac{1}{2} \)

For line AC:
Slope = \( m = \frac{1}{2} \), \( (x_1, y_1) = (7, -2) \)
Equation of the line AC is
\( y - y_1 = m(x - x_1) \)
\( y + 2 = \frac{1}{2}(x - 7) \)
\( 2y + 4 = x - 7 \)
\( 2y = x - 11 \)

For line BD:
Slope = \( m = \frac{-1}{\text{Slope of AC}} = \frac{-1}{\frac{1}{2}} = -2 \), \( (x_1, y_1) = (3, -4) \)
Equation of the line BD is
\( y - y_1 = m(x - x_1) \)
\( y + 4 = -2(x - 3) \)
\( y + 4 = -2x + 6 \)
\( 2x + y = 2 \)

In simple words: In a square, diagonals cross at right angles. We use the midpoint and perpendicular slope rule to find both diagonal equations.

📝 Teacher's Note: Emphasize that square and rhombus both have perpendicular diagonals. Students often forget to find midpoint first.

🎯 Exam Tip: Write "diagonals bisect at right angles" to show you understand the property. Check that slopes multiply to -1.

Question 19. A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Answer:
(i) We know the median through A will pass through the mid-point of BC. Let AD be the median through A.
Co-ordinates of the mid-point of BC, i.e., D are
\( \left( \frac{2 - 2}{2}, \frac{2 + 4}{2} \right) = (0, 3) \)

Slope of AD = \( \frac{3 + 5}{0 - 1} = -8 \)
Equation of the median AD is
\( y - 3 = -8(x - 0) \)
\( 8x + y = 3 \)

(ii) Let BE be the altitude of the triangle through B.
Slope of AC = \( \frac{4 + 5}{-2 - 1} = \frac{9}{-3} = -3 \)
∴ Slope of BE = \( \frac{-1}{\text{Slope of AC}} = \frac{1}{3} \)
Equation of altitude BE is
\( y - 2 = \frac{1}{3}(x - 2) \)
\( 3y - 6 = x - 2 \)
\( 3y = x + 4 \)

(iii) Slope of AB = \( \frac{2 + 5}{2 - 1} = 7 \)
Slope of the line parallel to AB = Slope of AB = 7
So, the equation of the line passing through C and parallel to AB is
\( y - 4 = 7(x + 2) \)
\( y - 4 = 7x + 14 \)
\( y = 7x + 18 \)

In simple words: Median goes to midpoint of opposite side. Altitude is perpendicular to opposite side. Parallel lines have same slope.

📝 Teacher's Note: Draw triangle on board. Show median connects vertex to midpoint. Altitude is like dropping a line straight down to opposite side.

🎯 Exam Tip: For median, find midpoint first. For altitude, use perpendicular slope rule. For parallel lines, copy the same slope.

Question 20.
(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.
Answer:
(i) 2y = 3x + 5
⇒ y = \( \frac{3}{2}x + \frac{5}{2} \)
Slope of this line = \( \frac{3}{2} \)
Slope of the line AB = \( \frac{-1}{\frac{3}{2}} = \frac{-2}{3} \)

\( (x_1, y_1) = (3, 2) \)
The required equation of the line AB is
\( y - y_1 = m(x - x_1) \)
\( y - 2 = \frac{-2}{3}(x - 3) \)
\( 3y - 6 = -2x + 6 \)
\( 2x + 3y = 12 \)

(ii) For the point A (the point on x-axis), the value of y = 0.
∴ \( 2x + 3y = 12 ⇒ 2x = 12 ⇒ x = 6 \)
Co-ordinates of point A are (6, 0).
For the point B (the point on y-axis), the value of x = 0.
∴ \( 2x + 3y = 12 ⇒ 3y = 12 ⇒ y = 4 \)
Co-ordinates of point B are (0, 4).

Area of △OAB = \( \frac{1}{2} × OA × OB = \frac{1}{2} × 6 × 4 = 12 \) sq units

In simple words: We made a line perpendicular to the given line. Then we found where it crosses the x and y axes. Triangle area is base times height divided by 2.

📝 Teacher's Note: Show students that perpendicular slopes multiply to -1. To find axis intercepts, put y=0 for x-axis and x=0 for y-axis.

🎯 Exam Tip: First find perpendicular slope. Then write line equation. For intercepts, substitute 0 for one coordinate. Area formula: ½ × base × height.

Question 21. The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A. Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.
Answer:
For the point A (the point on x-axis), the value of y = 0.
∴ \( 4x - 3y + 12 = 0 ⇒ 4x = -12 ⇒ x = -3 \)
Co-ordinates of point A are (-3, 0).
Here, \( (x_1, y_1) = (-3, 0) \)
The given line is 4x - 3y + 12 = 0
3y = 4x + 12
\( y = \frac{4}{3}x + 4 \)

Slope of this line = \( \frac{4}{3} \)
∴ Slope of a line perpendicular to the given line = \( \frac{-1}{\frac{4}{3}} = \frac{-3}{4} \)

Required equation of the line passing through A is
\( y - y_1 = m(x - x_1) \)
\( y - 0 = \frac{-3}{4}(x + 3) \)
\( 4y = -3x - 9 \)
\( 3x + 4y + 9 = 0 \)

In simple words: We found where the given line crosses x-axis by putting y = 0. Then we made a perpendicular line through that point.

📝 Teacher's Note: Remind students that x-axis means y = 0. Show how to rearrange line equation to find slope easily.

🎯 Exam Tip: For x-axis intercept, put y = 0. For perpendicular slope, flip the fraction and change sign. Write final answer in standard form.

Question 22. The point P is the foot of perpendicular from A (-5, 7) to the line whose equation is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP
(ii) the co-ordinates of P
Answer:
(i) The given equation is
2x - 3y + 18 = 0
3y = 2x + 18
\( y = \frac{2}{3}x + 6 \)

Slope of this line = \( \frac{2}{3} \)
Slope of a line perpendicular to this line = \( \frac{-1}{\frac{2}{3}} = \frac{-3}{2} \)

\( (x_1, y_1) = (-5, 7) \)
The required equation of the line AP is given by
\( y - y_1 = m(x - x_1) \)
\( y - 7 = \frac{-3}{2}(x + 5) \)
\( 2y - 14 = -3x - 15 \)
\( 3x + 2y + 1 = 0 \)

(ii) P is the foot of perpendicular from point A.
So P is the point of intersection of the lines 2x - 3y + 18 = 0 and 3x + 2y + 1 = 0.
2x - 3y + 18 = 0 ⇒ 4x - 6y + 36 = 0
3x + 2y + 1 = 0 ⇒ 9x + 6y + 3 = 0
Adding the two equations, we get,
13x + 39 = 0
x = -3
∴ 3y = 2x + 18 = -6 + 18 = 12
y = 4
Thus, the co-ordinates of the point P are (-3, 4).

In simple words: We drew a line from A perpendicular to the given line. The foot of perpendicular is where these two lines meet.

📝 Teacher's Note: Explain that "foot of perpendicular" means the point where the perpendicular line touches the original line. Like dropping a stone straight down.

🎯 Exam Tip: First find perpendicular slope. Then find intersection point by solving both equations together. This gives coordinates of foot of perpendicular.

Question 23. The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC. If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.
Answer:
For the line AB:
Slope of AB = \( m = \frac{2 - 0}{2 - 4} = \frac{2}{-2} = -1 \)

Using point (4, 0):
Equation of line AB is:
\( y - 0 = -1(x - 4) \)
\( y = -x + 4 \)
\( x + y = 4 \) ... (1)

For the line BC:
Slope of BC = \( m = \frac{6 - 2}{0 - 2} = \frac{4}{-2} = -2 \)

Using point (2, 2):
Equation of line BC is:
\( y - 2 = -2(x - 2) \)
\( y - 2 = -2x + 4 \)
\( 2x + y = 6 \) ... (2)

Finding point P (where AB cuts y-axis):
When AB cuts the y-axis, x = 0
Putting x = 0 in equation (1): \( 0 + y = 4 \)
\( y = 4 \)
Thus, coordinates of point P are (0, 4).

Finding point Q (where BC cuts x-axis):
When BC cuts the x-axis, y = 0
Putting y = 0 in equation (2): \( 2x + 0 = 6 \)
\( 2x = 6 \)
\( x = 3 \)
Thus, coordinates of point Q are (3, 0).

In simple words: We found slopes of both lines using two points. Then we made equations. To find where lines cut axes, we put x = 0 for y-axis and y = 0 for x-axis.

📝 Teacher's Note: Draw the points on graph paper first. Students can see where lines cross the axes. This makes finding P and Q very clear.

🎯 Exam Tip: Always write the slope formula first. Then use point-slope form. Show each step clearly to get full marks.

Question 24. Match the equations A, B, C and D with lines L₁, L₂, L₃ and L₄, whose graphs are roughly drawn in the given diagram.
A = y = 2x; B = y - 2x + 2 = 0;
C = 3x + 2y = 6; D = y = 2
Answer:
Putting x = 0 and y = 0 in equation y = 2x:
LHS = 0 and RHS = 0
Thus, line y = 2x passes through origin.
Hence, A = L₃

Putting x = 0 in y - 2x + 2 = 0, we get y = -2
Putting y = 0 in y - 2x + 2 = 0, we get x = 1
So, x-intercept = 1 and y-intercept = -2
So, x-intercept is positive and y-intercept is negative.
Hence, B = L₄

Putting x = 0 in 3x + 2y = 6, we get y = 3
Putting y = 0 in 3x + 2y = 6, we get x = 2
So, both x-intercept and y-intercept are positive.
Hence, C = L₂

The slope of line y = 2 is 0.
So, line y = 2 is parallel to x-axis.
Hence, D = L₁

In simple words: We check where each line crosses the x and y axes. Then we match with the diagram. Line y = 2 is horizontal (flat).

📝 Teacher's Note: Show students that y = 2x goes through (0,0). Lines with positive intercepts are in the first quadrant. Horizontal lines have slope zero.

🎯 Exam Tip: Always find where lines cross both axes. Check if intercepts are positive or negative. This helps match with the graph correctly.

Question 25. Find the value of a for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.
Answer:
If 3 points are collinear, the slope between any 2 points is the same.
Thus, for A(a, 3), B(2, 1) and C(5, a) to be collinear, the slope between A and B and between B and C should be the same.

\( \frac{1 - 3}{2 - a} = \frac{a - 1}{5 - 2} \)

\( \frac{-2}{2 - a} = \frac{a - 1}{3} \)

\( \frac{-2}{a - 2} = \frac{a - 1}{3} \)

\( 6 = (a - 2)(a - 1) \)
\( a² - 3a + 2 = 6 \)
\( a² - 3a - 4 = 0 \)
\( a = -1 \text{ or } 4 \)

Thus, slope can be:
\( \frac{2}{a - 2} = \frac{2}{-1 - 2} = -\frac{2}{3} \) OR \( \frac{2}{a - 2} = \frac{2}{4 - 2} = 1 \)

Thus, equation of line can be:
\( \frac{y - 1}{x - 2} = -\frac{2}{3} \)
\( 3y + 2x = 5 \)
or
\( \frac{y - 1}{x - 2} = 1 \)
\( y - x = -1 \)
\( x - y = 1 \)

In simple words: Collinear means all points lie on the same straight line. So slope between any two pairs must be equal. We solve to find the value of a.

📝 Teacher's Note: Use three points on a straight line to explain collinear. Show that slope between A-B equals slope between B-C when points are on same line.

🎯 Exam Tip: Write "slope between A-B = slope between B-C" first. Then substitute coordinates and solve the equation. Show both values of a clearly.

Exercise 14E

Question 1. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7.
Answer:
Using section formula, the coordinates of point P are:
\( \left( \frac{3 \times 16 + 5 \times 8}{3 + 5}, \frac{3 \times (-8) + 5 \times 0}{3 + 5} \right) \)

\( = \left( \frac{48 + 40}{8}, \frac{-24 + 0}{8} \right) \)

\( = \left( \frac{88}{8}, \frac{-24}{8} \right) = (11, -3) \)

3x + 5y = 7
\( y = -\frac{3}{5}x + \frac{7}{5} \)

Slope of this line = \( -\frac{3}{5} \)

As the required line is parallel to line 3x + 5y = 7,
Slope of required line = Slope of given line = \( -\frac{3}{5} \)

Thus, equation of required line is:
y - y₁ = m(x - x₁)
\( y + 3 = -\frac{3}{5}(x - 11) \)

5y + 15 = -3x + 33
3x + 5y = 18

In simple words: We use section formula to find where P divides the line. Parallel lines have the same slope. We use point-slope form to get the equation.

📝 Teacher's Note: Show students the section formula on board. Remind them that parallel lines never meet and have same slope. Draw two parallel lines to show this.

🎯 Exam Tip: Write section formula clearly first. Then find slope of given line. Use "parallel lines have same slope" and apply point-slope form for full marks.

Question 2. The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x - 3y + 4 = 0.
Answer:
Using section formula, the coordinates of point P are:
\( \left( \frac{1 \times (-2) + 3 \times 3}{1 + 3}, \frac{1 \times 1 + 3 \times (-4)}{1 + 3} \right) \)

\( = \left( \frac{-2 + 9}{4}, \frac{1 - 12}{4} \right) = \left( \frac{7}{4}, -\frac{11}{4} \right) \)

The equation of given line is:
5x - 3y + 4 = 0
\( y = \frac{5}{3}x + \frac{4}{3} \)

Slope of this line = \( \frac{5}{3} \)

Since required line is perpendicular to given line,
Slope of required line = \( -\frac{1}{\frac{5}{3}} = -\frac{3}{5} \)

Thus, equation of required line is:
y - y₁ = m(x - x₁)
\( y + \frac{11}{4} = -\frac{3}{5}\left(x - \frac{7}{4}\right) \)

\( \frac{4y + 11}{4} = -\frac{3}{5}\left(\frac{4x - 7}{4}\right) \)

20y + 55 = -12x + 21
12x + 20y + 34 = 0
6x + 10y + 17 = 0

In simple words: We find point P using section formula. Perpendicular lines have slopes that multiply to give -1. We use this to find the new slope.

📝 Teacher's Note: Show students that perpendicular lines meet at 90 degrees. If one slope is m, the other is -1/m. Draw perpendicular lines on board to show this.

🎯 Exam Tip: Write section formula first. Then write "for perpendicular lines, m₁ × m₂ = -1". This shows you know the concept and gets you marks.

A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M.

[Diagram: This diagram shows a coordinate plane with a straight line passing through points P(-1, 4) and Q(5, -2). The line intersects the y-axis at point B and the x-axis at point A. Point M is marked as the midpoint of segment AB.]

Answer:

(i) Finding the equation of the line:
Slope of PQ = \( \frac{-2 - 4}{5 - (-1)} = \frac{-6}{6} = -1 \)

Equation of the line PQ is given by
y - y₁ = m(x - x₁)
y - 4 = -1(x + 1)
y - 4 = -x - 1
x + y = 3

(ii) Finding the co-ordinates of A and B:
For point A (on x-axis), y = 0.
Putting y = 0 in the equation of PQ, we get,
x = 3
Thus, the co-ordinates of point A are (3, 0).

For point B (on y-axis), x = 0.
Putting x = 0 in the equation of PQ, we get,
y = 3
Thus, the co-ordinates of point B are (0, 3).

(iii) Finding the co-ordinates of M:
M is the mid-point of AB.
So, the co-ordinates of point M are
\( \left( \frac{3 + 0}{2}, \frac{0 + 3}{2} \right) = \left( \frac{3}{2}, \frac{3}{2} \right) \)

In simple words: We found the slope of the line using two given points. Then we wrote the equation. To find where it meets the axes, we put y = 0 for x-axis and x = 0 for y-axis. The midpoint is just the average of the two endpoints.

📝 Teacher's Note: Show students that slope is "rise over run" - how much y changes when x changes by 1. Use the midpoint formula as finding the middle point between two places on a map.

🎯 Exam Tip: Always write the slope formula first. Then substitute values carefully. For axis intercepts, remember: x-axis means y = 0, y-axis means x = 0.

Additional Problem: A = (1, 5) and C = (-3, -1). We know that in a rhombus, diagonals bisect each other at right angle. Let O be the point of intersection of the diagonals AC and BD. Co-ordinates of O are

Answer:

\( \left( \frac{1 - 3}{2}, \frac{5 - 1}{2} \right) = (-1, 2) \)

Slope of AC = \( \frac{-1 - 5}{-3 - 1} = \frac{-6}{-4} = \frac{3}{2} \)

For line AC:
Slope = m = \( \frac{3}{2} \), (x₁, y₁) = (1, 5)

Equation of the line AC is
y - y₁ = m(x - x₁)
y - 5 = \( \frac{3}{2} \)(x - 1)
2y - 10 = 3x - 3
3x - 2y + 7 = 0

For line BD:
Slope = m = \( \frac{-1}{\text{Slope of AC}} = \frac{-2}{3} \), (x₁, y₁) = (-1, 2)

Equation of the line BD is
y - y₁ = m(x - x₁)
y - 2 = \( \frac{-2}{3} \)(x + 1)
3y - 6 = -2x - 2
2x + 3y = 4

In simple words: In a rhombus, the diagonals cut each other at right angles. So if one diagonal has slope 3/2, the other has slope -2/3 (negative reciprocal). They meet at the center point.

📝 Teacher's Note: Explain that perpendicular lines have slopes that multiply to give -1. Like 3/2 × (-2/3) = -1. This is a key property students must remember.

🎯 Exam Tip: For perpendicular lines, write "slope₁ × slope₂ = -1". This shows you know the property. Always find the intersection point by solving the two equations together.

Using distance formula, we have:
AB = \( \sqrt{(6 - 3)^2 + (-2 - 2)^2} = \sqrt{9 + 16} = 5 \)
BC = \( \sqrt{(2 - 6)^2 + (-5 + 2)^2} = \sqrt{16 + 9} = 5 \)
Thus, AC = BC

Also, Slope of AB = \( \frac{-2 - 2}{6 - 3} = \frac{-4}{3} \)
Slope of BC = \( \frac{-5 + 2}{2 - 6} = \frac{-3}{-4} = \frac{3}{4} \)
Slope of AB × Slope of BC = -1
Thus, AB ⊥ BC

Hence, A, B, C can be the vertices of a square.

(i) Finding the fourth vertex D:
Slope of AB = \( \frac{-2 - 2}{6 - 3} = \frac{-4}{3} \) = Slope of CD

Equation of the line CD is
y - y₁ = m[x - x₁]
⇒ y + 5 = \( \frac{-4}{3} \)(x - 2)
⇒ 3y + 15 = -4x + 8
⇒ 4x + 3y = -7 ...(1)

Slope of BC = \( \frac{-5 + 2}{2 - 6} = \frac{-3}{-4} = \frac{3}{4} \) = Slope of AD

Equation of the line AD is
y - y₁ = m[x - x₁]
⇒ y - 2 = \( \frac{3}{4} \)(x - 3)
⇒ 4y - 8 = 3x - 9
⇒ 3x - 4y = 1 ...(2)

Now, D is the point of intersection of CD and AD.
(1) ⇒ 16x + 12y = -28
(2) ⇒ 9x - 12y = 3
Adding the above two equations we get,
25x = -25
⇒ x = -1
So, 4y = 3x - 1 = -3 - 1 = -4
⇒ y = -1
Thus, the co-ordinates of point D are (-1, -1).

(ii) Finding other properties:
The equation of line AD is found in part (i)
It is 3x - 4y = 1 or 4y = 3x - 1.

Slope of BD = \( \frac{-1 + 2}{-1 - 6} = \frac{1}{-7} = \frac{-1}{7} \)

The equation of diagonal BD is
y - y₁ = m[x - x₁]
⇒ y + 1 = \( \frac{-1}{7} \)(x + 1)
⇒ 7y + 7 = -x - 1
⇒ x + 7y + 8 = 0

In simple words: We used the distance formula to check if sides are equal. Then we checked if adjacent sides are perpendicular (slopes multiply to -1). For a square, opposite sides are parallel, so they have the same slope.

📝 Teacher's Note: Remind students that in a square: all sides are equal, all angles are 90°, and opposite sides are parallel. Use these three properties to solve such problems step by step.

🎯 Exam Tip: First prove it's a square by showing equal sides and right angles. Then use the property that opposite sides are parallel (same slope) to find the fourth vertex. Show all working clearly.

Finding intersection points:

The given line is
x = 3y + 2 ...(1)
3y = x - 2
y = \( \frac{1}{3} \)x - \( \frac{2}{3} \)

Slope of this line is \( \frac{1}{3} \).

The required line intersects the given line at right angle.
∴ Slope of the required line = \( \frac{-1}{\frac{1}{3}} = -3 \)

The required line passes through (0, 0) = (x₁, y₁)
The equation of the required line is
y - y₁ = m(x - x₁)
y - 0 = -3(x - 0)
3x + y = 0 ...(2)

Point X is the intersection of the lines (1) and (2).
Using (1) in (2), we get,
9y + 6 + y = 0
y = \( \frac{-6}{10} = \frac{-3}{5} \)

∴ x = 3y + 2 = \( \frac{-9}{5} + 2 = \frac{1}{5} \)

Thus, the co-ordinates of the point X are \( \left( \frac{1}{5}, \frac{-3}{5} \right) \).

In simple words: When two lines meet at right angles, their slopes multiply to give -1. We found the slope of the perpendicular line, then wrote its equation. The intersection point is where both equations are satisfied.

📝 Teacher's Note: Use the example of crossing roads - they meet at right angles. If one road goes up gently, the perpendicular road goes down steeply. That's what negative reciprocal means.

🎯 Exam Tip: Always write "perpendicular lines have slopes m₁ × m₂ = -1". Find the perpendicular slope first, then write the line equation. Solve the two equations together to find intersection.

Another geometric problem:

[Diagram: This diagram shows a coordinate plane with points A, B, and P marked, where P(3, 2) is shown and line segments are drawn to form a triangle.]

Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).
Since, P is the mid-point of AB, we have:

\( \left( \frac{x + 0}{2}, \frac{0 + y}{2} \right) = (3, 2) \)

\( \left( \frac{x}{2}, \frac{y}{2} \right) = (3, 2) \)

x = 6, y = 4
Thus, A = (6, 0) and B = (0, 4)

Slope of line AB = \( \frac{4 - 0}{0 - 6} = \frac{4}{-6} = \frac{-2}{3} \)

Let (x₁, y₁) = (6, 0)
The required equation of the line AB is given by
y - y₁ = m(x - x₁)
y - 0 = \( \frac{-2}{3} \)(x - 6)
3y = -2x + 12
2x + 3y = 12

In simple words: If P is the midpoint of AB, then P is exactly halfway between A and B. We used the midpoint formula backwards to find A and B from the given midpoint.

📝 Teacher's Note: Show students that if they know the midpoint, they can work backwards to find the endpoints. It's like knowing the center of a see-saw and finding where the ends are.

🎯 Exam Tip: When given a midpoint, use the midpoint formula: midpoint = ((x₁+x₂)/2, (y₁+y₂)/2). Work backwards to find the unknown points. Always verify your answer by checking the midpoint again.

Question 3. A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Answer:
Part 1: Finding coordinates of point P
When a line meets the y-axis, x = 0.
Substituting x = 0 in 5x + 3y + 15 = 0:
5(0) + 3y + 15 = 0
3y + 15 = 0
3y = -15
y = -5
Therefore, P = (0, -5)

Part 2: Finding perpendicular line
Given line: x – 3y + 4 = 0
Rewriting: x = 3y – 4
Slope of given line = \( \frac{1}{3} \)
Slope of perpendicular line = \( -3 \)

Using point-slope form with P(0, -5):
y - (-5) = -3(x - 0)
y + 5 = -3x
3x + y + 5 = 0

Final answers:
P = (0, -5)
Required line: 3x + y + 5 = 0
In simple words: To find where a line meets the y-axis, put x = 0. For perpendicular lines, multiply their slopes to get -1.

📝 Teacher's Note: Show students that y-axis means x = 0. Also teach them that perpendicular slopes multiply to give -1.

🎯 Exam Tip: Always write coordinates in brackets like (0, -5). Show all steps clearly to get full marks.

Question 4. Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Answer:
For two lines to be perpendicular, the product of their slopes = -1.

Line 1: kx – 5y + 4 = 0
Rewriting: 5y = kx + 4
y = \( \frac{k}{5}x + \frac{4}{5} \)
Slope \( m_1 = \frac{k}{5} \)

Line 2: 5x – 2y + 5 = 0
Rewriting: 2y = 5x + 5
y = \( \frac{5}{2}x + \frac{5}{2} \)
Slope \( m_2 = \frac{5}{2} \)

For perpendicular lines: \( m_1 \times m_2 = -1 \)
\( \frac{k}{5} \times \frac{5}{2} = -1 \)
\( \frac{k}{2} = -1 \)
k = -2

Answer: k = -2
In simple words: When two lines are perpendicular, their slopes multiply to give -1. We use this rule to find k.

📝 Teacher's Note: Teach students to write y = mx + c form first. This makes finding slopes easy.

🎯 Exam Tip: Always write "For perpendicular lines, m₁ × m₂ = -1" to show you know the rule.

Question 5. (1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Answer:
Step 1: Find equation of diagonal AC
A = (1, 5), C = (-3, -1)
Slope of AC = \( \frac{-1-5}{-3-1} = \frac{-6}{-4} = \frac{3}{2} \)

Using point-slope form with point A(1, 5):
y - 5 = \( \frac{3}{2} \)(x - 1)
2(y - 5) = 3(x - 1)
2y - 10 = 3x - 3
3x - 2y + 7 = 0

Step 2: Find equation of diagonal BD
In a rhombus, diagonals are perpendicular and bisect each other.
Slope of BD = \( -\frac{2}{3} \) (perpendicular to AC)

Midpoint of AC = \( \left(\frac{1+(-3)}{2}, \frac{5+(-1)}{2}\right) = (-1, 2) \)
BD passes through this midpoint.

Using point-slope form:
y - 2 = \( -\frac{2}{3} \)(x - (-1))
y - 2 = \( -\frac{2}{3} \)(x + 1)
3(y - 2) = -2(x + 1)
3y - 6 = -2x - 2
2x + 3y - 4 = 0

Final answers:
Diagonal AC: 3x - 2y + 7 = 0
Diagonal BD: 2x + 3y - 4 = 0
In simple words: In a rhombus, diagonals cut each other at right angles. We use this property to find both diagonal equations.

📝 Teacher's Note: Remind students that rhombus diagonals are always perpendicular and bisect each other. This is the key property.

🎯 Exam Tip: Write "diagonals of rhombus are perpendicular" to show you know the property. Find midpoint correctly.

Question 7. Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Answer:
Showing ABC can be vertices of a square:
AB = \( \sqrt{(6-3)^2 + (-2-2)^2} = \sqrt{9+16} = 5 \)
BC = \( \sqrt{(2-6)^2 + (-5-(-2))^2} = \sqrt{16+9} = 5 \)
AC = \( \sqrt{(2-3)^2 + (-5-2)^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2} \)

Since AB = BC and AC = \( AB\sqrt{2} \), ABC forms a right-angled triangle with equal sides.
Also, AB ⊥ BC (can be verified using slopes).
Therefore, ABC can be three vertices of a square.

(i) Finding fourth vertex D:
In square ABCD, \( \vec{AD} = \vec{BC} \)
\( \vec{BC} = (2-6, -5-(-2)) = (-4, -3) \)
If D = (x, y), then \( \vec{AD} = (x-3, y-2) \)
So: x - 3 = -4 and y - 2 = -3
x = -1, y = -1
Therefore, D = (-1, -1)

(ii) Finding equations without using D coordinates:
Slope of AB = \( \frac{-2-2}{6-3} = \frac{-4}{3} \)
Since ABCD is a square, AD ⊥ AB
Slope of AD = \( \frac{3}{4} \)

Equation of side AD:
y - 2 = \( \frac{3}{4} \)(x - 3)
4y - 8 = 3x - 9
3x - 4y - 1 = 0

For diagonal BD, B = (6, -2) and we need to find where it intersects AD.
Using properties of square: BD passes through center and is perpendicular to AC.
After calculation: x + 7y + 8 = 0

Final answers:
(i) D = (-1, -1)
(ii) Side AD: 3x - 4y - 1 = 0, Diagonal BD: x + 7y + 8 = 0
In simple words: In a square, all sides are equal and all angles are 90°. We use these facts to find the missing vertex and equations.

📝 Teacher's Note: Show students how to use distance formula and slope formula. Square properties are very important here.

🎯 Exam Tip: Always verify your square by checking equal sides and right angles. Show all distance calculations clearly.

Question 8. A line through origin meets the line x = 3y + 2 at right angles at point X. Find the coordinates of X.
Answer:
Step 1: Find slope of given line
x = 3y + 2
Rewriting: y = \( \frac{x-2}{3} \)
Slope of given line = \( \frac{1}{3} \)

Step 2: Find slope of perpendicular line
Slope of line through origin = -3 (perpendicular to \( \frac{1}{3} \))

Step 3: Find equation of line through origin
Since line passes through (0,0) with slope -3:
y = -3x

Step 4: Find intersection point X
Solving: x = 3y + 2 and y = -3x
Substituting y = -3x in first equation:
x = 3(-3x) + 2
x = -9x + 2
10x = 2
x = \( \frac{1}{5} \)

y = -3 × \( \frac{1}{5} = -\frac{3}{5} \)

Therefore, X = \( \left(\frac{1}{5}, -\frac{3}{5}\right) \)
In simple words: We find where two perpendicular lines meet. One goes through origin, the other is given. We solve their equations together.

📝 Teacher's Note: Show students that lines through origin have equation y = mx. Perpendicular slopes multiply to give -1.

🎯 Exam Tip: Write coordinates as fractions clearly. Show the perpendicular slope calculation step by step.

Question 9. A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Answer:
Let the line meet x-axis at A(a, 0) and y-axis at B(0, b), where a > 0 and b > 0.

Since (3, 2) bisects the line segment AB:
Midpoint of AB = \( \left(\frac{a+0}{2}, \frac{0+b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \)

Given that this midpoint is (3, 2):
\( \frac{a}{2} = 3 \) ⟹ a = 6
\( \frac{b}{2} = 2 \) ⟹ b = 4

So the line passes through A(6, 0) and B(0, 4).

Using intercept form of line:
\( \frac{x}{a} + \frac{y}{b} = 1 \)
\( \frac{x}{6} + \frac{y}{4} = 1 \)

Multiplying by 12:
2x + 3y = 12
2x + 3y - 12 = 0

Answer: 2x + 3y - 12 = 0
In simple words: The given point is the middle of the line segment between the axes. We use midpoint formula to find where the line cuts the axes.

📝 Teacher's Note: Draw a diagram showing the line cutting positive x and y axes. Mark the midpoint clearly.

🎯 Exam Tip: Use intercept form of line equation. Remember that midpoint formula is \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).

Question 10. Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.
Answer:
Step 1: Find intersection point
7x + 6y = 71 ...(1)
5x - 8y = -23 ...(2)

Multiply (1) by 4 and (2) by 3:
28x + 24y = 284 ...(3)
15x - 24y = -69 ...(4)

Adding (3) and (4):
43x = 215
x = 5

From (2): 8y = 5x + 23 = 25 + 23 = 48
y = 6

Intersection point = (5, 6)

Step 2: Find slope of perpendicular line
4x - 2y = 1
2y = 4x - 1
y = 2x - \( \frac{1}{2} \)
Slope = 2

Slope of perpendicular line = \( -\frac{1}{2} \)

Step 3: Find required equation
Using point-slope form with (5, 6):
y - 6 = \( -\frac{1}{2} \)(x - 5)
2(y - 6) = -(x - 5)
2y - 12 = -x + 5
x + 2y = 17

Answer: x + 2y = 17
In simple words: First find where the two given lines meet. Then draw a perpendicular line through that meeting point.

📝 Teacher's Note: Show students how to solve two equations together. Elimination method works well here.

🎯 Exam Tip: Always find the intersection point first. Then use perpendicular slope rule (multiply slopes = -1).

Question 11. Find the equation of the line which is perpendicular to the line \( \frac{x}{a} - \frac{y}{b} = 1 \) at the point where this line meets y-axis.
Answer:
Step 1: Find where given line meets y-axis
When line meets y-axis, x = 0
Putting x = 0 in \( \frac{x}{a} - \frac{y}{b} = 1 \):
\( 0 - \frac{y}{b} = 1 \)
\( \frac{y}{b} = -1 \)
y = -b

So the line meets y-axis at P(0, -b)

Step 2: Find slope of given line
\( \frac{x}{a} - \frac{y}{b} = 1 \)
Rewriting: \( \frac{y}{b} = \frac{x}{a} - 1 \)
y = \( \frac{b}{a}x - b \)
Slope = \( \frac{b}{a} \)

Step 3: Find slope of perpendicular line
Slope of perpendicular line = \( -\frac{a}{b} \)

Step 4: Find equation of required line
Using point-slope form with P(0, -b):
y - (-b) = \( -\frac{a}{b} \)(x - 0)
y + b = \( -\frac{a}{b}x \)
by + b² = -ax
ax + by + b² = 0

Answer: ax + by + b² = 0
In simple words: First find where the line cuts y-axis by putting x = 0. Then draw perpendicular line through that point.

📝 Teacher's Note: Teach students that y-axis means x = 0. Also show how to find slope from standard form of line equation.

🎯 Exam Tip: Write the final answer in standard form ax + by + c = 0. Show all slope calculations clearly.

Question 12. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of triangle OAB through vertex O.
(ii) the equation of altitude of triangle OAB through vertex B.
Answer:
(i) Median through O:
Let the median through O meets AB at D. So D is the midpoint of AB.
Coordinates of D = \( \left(\frac{3+(-5)}{2}, \frac{5+(-3)}{2}\right) = (-1, 1) \)

Slope of OD = \( \frac{1-0}{-1-0} = -1 \)

Since line passes through O(0, 0):
y - 0 = -1(x - 0)
y = -x
x + y = 0

(ii) Altitude through B:
The altitude through B is perpendicular to OA.
Slope of OA = \( \frac{5-0}{3-0} = \frac{5}{3} \)

Slope of altitude = \( -\frac{3}{5} \) (perpendicular to OA)

Using point-slope form with B(-5, -3):
y - (-3) = \( -\frac{3}{5} \)(x - (-5))
y + 3 = \( -\frac{3}{5} \)(x + 5)
5(y + 3) = -3(x + 5)
5y + 15 = -3x - 15
3x + 5y + 30 = 0

Final answers:
(i) Median: x + y = 0
(ii) Altitude: 3x + 5y + 30 = 0
In simple words: Median joins a vertex to the middle of the opposite side. Altitude is perpendicular from a vertex to the opposite side.

📝 Teacher's Note: Draw the triangle clearly. Show students where median and altitude are. Median goes to midpoint, altitude is perpendicular.

🎯 Exam Tip: For median, find midpoint first. For altitude, use perpendicular slope rule. Label your answers clearly.

Question 13. Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does the line 3x = y + 1 bisect the line segment joining the two given points?
Answer:
Part 1: Checking perpendicularity
Let A = (-2, 3) and B = (4, 1)
Slope of AB = \( \frac{1-3}{4-(-2)} = \frac{-2}{6} = -\frac{1}{3} \)

Given line: 3x = y + 1
Rewriting: y = 3x - 1
Slope of given line = 3

Product of slopes = \( -\frac{1}{3} × 3 = -1 \)

Since product = -1, the lines are perpendicular.

Part 2: Checking if line bisects AB
Midpoint of AB = \( \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2) \)

Checking if (1, 2) lies on 3x = y + 1:
3(1) = 2 + 1
3 = 3 ✓

Yes, the midpoint lies on the given line.

Final answers:
Yes, the lines are perpendicular.
Yes, the line 3x = y + 1 bisects the line segment AB.
In simple words: Two lines are perpendicular if their slopes multiply to give -1. A line bisects a segment if it passes through the midpoint.

📝 Teacher's Note: Show students both conditions: perpendicular (slopes multiply to -1) and bisection (passes through midpoint).

🎯 Exam Tip: Calculate slopes first, then check if product = -1. For bisection, find midpoint and check if it satisfies the line equation.

Question 14. Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).
Answer:
x cos 30° + y sin 30° = 2

Substituting values: cos 30° = \( \frac{\sqrt{3}}{2} \) and sin 30° = \( \frac{1}{2} \)

x × \( \frac{\sqrt{3}}{2} \) + y × \( \frac{1}{2} \) = 2

\( \frac{\sqrt{3}x}{2} + \frac{y}{2} = 2 \)

Multiplying by 2:
\( \sqrt{3}x + y = 4 \)

y = \( -\sqrt{3}x + 4 \)

Slope of given line = \( -\sqrt{3} \)

Since parallel lines have equal slopes, slope of required line = \( -\sqrt{3} \)

Using point-slope form with (4, 3):
y - 3 = \( -\sqrt{3} \)(x - 4)
y - 3 = \( -\sqrt{3}x + 4\sqrt{3} \)
y = \( -\sqrt{3}x + 4\sqrt{3} + 3 \)

\( \sqrt{3}x + y = 4\sqrt{3} + 3 \)

Answer: \( \sqrt{3}x + y = 4\sqrt{3} + 3 \)
In simple words: Parallel lines have the same slope. We find the slope of the given line, then write the equation of the new line with the same slope.

📝 Teacher's Note: Remind students that cos 30° = √3/2 and sin 30° = 1/2. These are important values to remember.

🎯 Exam Tip: Write the trigonometric values clearly. Show that parallel lines have equal slopes. Use point-slope form correctly.

Question 15. Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Answer:
Given line: (k - 2)x + (k + 3)y - 5 = 0
Rewriting: (k + 3)y = -(k - 2)x + 5
y = \( -\frac{k-2}{k+3}x + \frac{5}{k+3} \)

Slope of given line = \( m_1 = -\frac{k-2}{k+3} \)

Reference line: 2x - y + 7 = 0
Rewriting: y = 2x + 7
Slope = \( m_2 = 2 \)

(i) For perpendicular lines:
\( m_1 × m_2 = -1 \)
\( -\frac{k-2}{k+3} × 2 = -1 \)
\( \frac{2(k-2)}{k+3} = 1 \)
2(k - 2) = k + 3
2k - 4 = k + 3
k = 7

(ii) For parallel lines:
\( m_1 = m_2 \)
\( -\frac{k-2}{k+3} = 2 \)
-(k - 2) = 2(k + 3)
-k + 2 = 2k + 6
-3k = 4
k = \( -\frac{4}{3} \)

Final answers:
(i) k = 7 (for perpendicular)
(ii) k = \( -\frac{4}{3} \) (for parallel)
In simple words: For perpendicular lines, slopes multiply to -1. For parallel lines, slopes are equal.

📝 Teacher's Note: Show students how to extract slope from general form ax + by + c = 0. The slope is -a/b.

🎯 Exam Tip: Write the slope conditions clearly: parallel means equal slopes, perpendicular means product of slopes = -1.

Question 16. The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.
Answer:
Step 1: Find the slope of BC.
Slope of BC = \( \frac{7 + 2}{11 + 1} = \frac{9}{12} = \frac{3}{4} \)

Step 2: Write the equation of line BC.
Using point-slope form with point C(11, 7):
\( y - 7 = \frac{3}{4}(x - 11) \)
\( y + 2 = \frac{3}{4}(x + 1) \)
\( 4y + 8 = 3x + 3 \)
\( 3x - 4y = 5 \)...(1)

(i) Step 3: Find the slope of line perpendicular to BC.
Slope of perpendicular line = \( -\frac{1}{\frac{3}{4}} = -\frac{4}{3} \)

Step 4: Write equation of line through A(0, 5) perpendicular to BC.
\( y - 5 = -\frac{4}{3}(x - 0) \)
\( 3y - 15 = -4x \)
\( 4x + 3y = 15 \)...(2)

(ii) Step 5: Find intersection point of lines (1) and (2).
From equation (1): \( 3x - 4y = 5 \)
\( \implies 9x - 12y = 15 \)
From equation (2): \( 4x + 3y = 15 \)
\( \implies 16x + 12y = 60 \)

Adding the equations:
\( 25x = 75 \)
\( x = 3 \)

Substituting in equation (2):
\( 4(3) + 3y = 15 \)
\( 12 + 3y = 15 \)
\( y = 1 \)

Therefore:
Equation of BC: \( 3x - 4y = 5 \)
(i) Equation of perpendicular through A: \( 4x + 3y = 15 \)
(ii) Point of intersection: (3, 1)

In simple words: We found the slope of BC first, then used it to get the perpendicular slope. The perpendicular line passes through A and meets BC at point (3, 1).

📝 Teacher's Note: Show students that perpendicular slopes multiply to give -1. When slope is 3/4, perpendicular slope is -4/3. This rule helps solve such problems quickly.

🎯 Exam Tip: Always write the final equations clearly. Show each step of finding intersection by solving simultaneous equations. Check your answer by substituting back.

Question 17. From the given figure, find:
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC.
Answer:
(i) From the graph:
A = (2, 3), B = (-1, 2), C = (3, 0)

(ii) Step 1: Find slope of BC.
Slope of BC = \( \frac{0 - 2}{3 + 1} = \frac{-2}{4} = -\frac{1}{2} \)

Step 2: Find equation of line through A parallel to BC.
Slope of required line = Slope of BC = \( -\frac{1}{2} \)

Using point-slope form with A(2, 3):
\( y - 3 = -\frac{1}{2}(x - 2) \)
\( 2y - 6 = -x + 2 \)
\( x + 2y = 8 \)

In simple words: We read the coordinates from the graph, then used the fact that parallel lines have the same slope to write the equation.

[Diagram: This diagram shows a coordinate plane with triangle ABC plotted. Point A is at (2,3), B is at (-1,2), and C is at (3,0).]

📝 Teacher's Note: Teach students to read coordinates carefully from graphs. Count squares step by step. Parallel lines always have equal slopes - this is the key rule.

🎯 Exam Tip: Write coordinates in brackets like (2, 3). For parallel lines, write "slope = slope of BC" to show you know the rule. Always simplify final equation.

Question 18. P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
Answer:
Step 1: Find the midpoint of PQ (opposite side to R).
Midpoint X = \( \left(\frac{3 + 7}{2}, \frac{4 + (-2)}{2}\right) = (5, 1) \)

Step 2: Find slope of median RX.
Slope of RX = \( \frac{1 + 1}{5 + 2} = \frac{2}{7} \)

Step 3: Write equation of median RX.
Using point-slope form with R(-2, -1):
\( y - (-1) = \frac{2}{7}(x - (-2)) \)
\( y + 1 = \frac{2}{7}(x + 2) \)
\( 7y + 7 = 2x + 4 \)
\( 7y = 2x - 3 \)

In simple words: A median connects a vertex to the midpoint of the opposite side. We found midpoint of PQ, then drew line from R to this midpoint.

📝 Teacher's Note: Draw a triangle on board and show that median divides opposite side into two equal parts. Students often forget to find midpoint first.

🎯 Exam Tip: Always write "midpoint of opposite side" to show you understand what a median is. Use midpoint formula correctly: \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)

Question 19. A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.
Answer:
Step 1: Find coordinates of midpoint P of AB.
P = \( \left(\frac{8 + (-4)}{2}, \frac{-6 + 2}{2}\right) = (2, -2) \)

Step 2: Find coordinates of midpoint Q of AC.
Q = \( \left(\frac{8 + 0}{2}, \frac{-6 + (-10)}{2}\right) = (4, -8) \)

Step 3: Find slope of PQ.
Slope of PQ = \( \frac{-8 + 2}{4 - 2} = \frac{-6}{2} = -3 \)

Step 4: Find slope of BC.
Slope of BC = \( \frac{-10 - 2}{0 + 4} = \frac{-12}{4} = -3 \)

Since slope of PQ = slope of BC, therefore PQ || BC.

Step 5: Check slopes of other sides.
Slope of PB = \( \frac{-2 - 2}{2 + 4} = \frac{-4}{6} = -\frac{2}{3} \)

Slope of QC = \( \frac{-8 + 10}{4 - 0} = \frac{2}{4} = \frac{1}{2} \)

Since PB is not parallel to QC, quadrilateral PBCQ is a trapezium.

In simple words: When we join midpoints of two sides of a triangle to the third side, we get a trapezium. One pair of opposite sides is parallel.

📝 Teacher's Note: This proves the midpoint theorem. Show students that the line joining midpoints of two sides is parallel to the third side. A trapezium has one pair of parallel sides.

🎯 Exam Tip: Calculate both slopes clearly and show they are equal. Write "Therefore PQ || BC" to prove parallelism. Name the quadrilateral as "trapezium" - this is the special name asked for.

Question 20. A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
(i) the co-ordinates of A and B.
(ii) the equation of line through P and perpendicular to AB.
Answer:
(i) Step 1: Let A(x, 0) be on x-axis and B(0, y) be on y-axis.
Given P(-4, -2) divides AB in ratio 1:2.

Using section formula:
\( (-4, -2) = \left(\frac{1 \times 0 + 2 \times x}{1 + 2}, \frac{1 \times y + 2 \times 0}{1 + 2}\right) \)

\( (-4, -2) = \left(\frac{2x}{3}, \frac{y}{3}\right) \)

Step 2: Equate coordinates.
\( -4 = \frac{2x}{3} \)
\( \implies x = -6 \)

\( -2 = \frac{y}{3} \)
\( \implies y = -6 \)

Therefore, A(-6, 0) and B(0, -6).

(ii) Step 3: Find slope of AB.
Slope of AB = \( \frac{-6 - 0}{0 + 6} = \frac{-6}{6} = -1 \)

Step 4: Find slope of perpendicular line.
Slope of perpendicular = \( -\frac{1}{-1} = 1 \)

Step 5: Write equation of line through P(-4, -2) with slope 1.
\( y - (-2) = 1(x - (-4)) \)
\( y + 2 = x + 4 \)
\( y = x + 2 \)

In simple words: We used section formula to find where line AB meets the axes, then found the perpendicular line through the given point.

📝 Teacher's Note: Section formula is key here. If point divides line in ratio m:n, use coordinates carefully. Remember perpendicular slopes multiply to give -1.

🎯 Exam Tip: Write section formula clearly. Show substitution step by step. For perpendicular lines, always check that slopes multiply to -1. Write final equation in simplest form.

Question 21. A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis. Find the equation of the line.
Answer:
Step 1: Identify the points.
Line intersects x-axis at A(-2, 0).
y-intercept = 3, so line passes through B(0, 3).

Step 2: Find slope of line AB.
Slope = \( \frac{3 - 0}{0 + 2} = \frac{3}{2} \)

Step 3: Write equation using point-slope form.
Using point A(-2, 0):
\( y - 0 = \frac{3}{2}(x + 2) \)
\( y = \frac{3}{2}(x + 2) \)
\( 2y = 3x + 6 \)
\( 3x - 2y + 6 = 0 \)

In simple words: The line crosses x-axis at (-2, 0) and y-axis at (0, 3). We joined these two points to get the line equation.

📝 Teacher's Note: Intercept means where line cuts the axis. x-intercept is point (a, 0), y-intercept is point (0, b). Join these points to get line equation.

🎯 Exam Tip: Write intercept points clearly as (a, 0) and (0, b). Find slope using these two points. Convert final equation to standard form ax + by + c = 0.

Question 22. Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.
Answer:
Step 1: Identify the points.
Line passes through A(2, 3).
x-intercept = 4, so line passes through B(4, 0).

Step 2: Find slope of line AB.
Slope = \( \frac{0 - 3}{4 - 2} = \frac{-3}{2} \)

Step 3: Write equation using point-slope form.
Using point B(4, 0):
\( y - 0 = \frac{-3}{2}(x - 4) \)
\( y = \frac{-3}{2}(x - 4) \)
\( 2y = -3x + 12 \)
\( 3x + 2y = 12 \)

In simple words: We know the line passes through (2, 3) and cuts x-axis at (4, 0). We used these two points to find the equation.

📝 Teacher's Note: Students often confuse x-intercept with y-intercept. x-intercept means point (a, 0) where line crosses x-axis. Use any two points to find line equation.

🎯 Exam Tip: Write the intercept point correctly as (4, 0). Calculate slope carefully with correct signs. Always check your final equation by substituting both points.

Question 23. The given figure (not drawn to scale) shows two straight lines AB and CD. If equation of the line AB is: y = x + 1 and equation of line CD is: y = √3 x - 1. Write down the inclination of lines AB and CD; also, find the angle θ between AB and CD.
Answer:
Step 1: Find inclination of line AB.
For line AB: y = x + 1
Slope = 1 = tan α
Therefore, α = 45°
Inclination of AB = 45°

Step 2: Find inclination of line CD.
For line CD: y = √3 x - 1
Slope = √3 = tan β
Therefore, β = 60°
Inclination of CD = 60°

Step 3: Find angle between the lines.
Angle θ between two lines with slopes m₁ and m₂ is:
\( \tan θ = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \)

\( \tan θ = \left|\frac{1 - \sqrt{3}}{1 + 1 \times \sqrt{3}}\right| = \left|\frac{1 - \sqrt{3}}{1 + \sqrt{3}}\right| \)

Rationalizing:
\( \tan θ = \left|\frac{(1 - \sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}\right| = \left|\frac{(1 - \sqrt{3})^2}{1 - 3}\right| = \left|\frac{1 - 2\sqrt{3} + 3}{-2}\right| \)

\( \tan θ = \left|\frac{4 - 2\sqrt{3}}{-2}\right| = \left|\frac{2\sqrt{3} - 4}{2}\right| = |\sqrt{3} - 2| = 2 - \sqrt{3} \)

Since \( \tan 15° = 2 - \sqrt{3} \), therefore θ = 15°

In simple words: Inclination is the angle a line makes with x-axis. We used slopes to find these angles, then used the formula to find angle between two lines.

[Diagram: This diagram shows two intersecting lines AB and CD on a coordinate plane, with angles marked to show their inclinations with the x-axis.]

📝 Teacher's Note: Inclination is always measured from positive x-axis. tan 45° = 1 and tan 60° = √3 are important values. The angle between lines formula needs careful calculation.

🎯 Exam Tip: Write "slope = tan(inclination)" clearly. Remember key values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Use absolute value in angle formula and rationalize fractions properly.

Question 24. Write down the equation of the line whose gradient is \( \frac{3}{2} \) and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2: 3.
Answer: Given, P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3.
Co-ordinates of point P are
\( \left( \frac{2 \times 3 + 3 \times (-2)}{2 + 3}, \frac{2 \times (-4) + 3 \times 6}{2 + 3} \right) \)
\( = \left( \frac{6 - 6}{5}, \frac{-8 + 18}{5} \right) \)
\( = \left( 0, 2 \right) = (x_1, y_1) \)

Slope of the required line = m = \( -\frac{3}{2} \)

The required equation of the line is given by
y - y1 = m(x - x1)
y - 2 = \( \frac{3}{2} \)(x - 0)
2y - 4 = 3x
2y = 3x + 4
In simple words: First we found the point P using the section formula. Then we used the point-slope form to write the equation of the line.

📝 Teacher's Note: Section formula helps find a point that divides a line in a given ratio. Show students with a rope and two marks - the dividing point is between them.

🎯 Exam Tip: Always write the section formula first, then substitute values carefully. Check your arithmetic twice - small mistakes lose marks.

Question 25. The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Answer: Let A = (6, 4) and B = (7, -5)

Slope of the line AB = \( \frac{-5 - 4}{7 - 6} = -9 \)

(x₁, y₁) = (6, 4)

The equation of the line AB is given by
y - y₁ = m(x - x₁)
y - 4 = -9(x - 6)
y - 4 = -9x + 54
9x + y = 58 ...(1)

Now, given that the ordinate of the required point is -23.
Putting y = -23 in (1), we get,
9x - 23 = 58
9x = 81
x = 9

Thus, the co-ordinates of the required point is (9, -23).
In simple words: We found the equation of the line first. Then we put y = -23 (the given ordinate) to find x.

📝 Teacher's Note: Ordinate means y-coordinate. Abscissa means x-coordinate. Remind students of these terms regularly.

🎯 Exam Tip: Write "ordinate = y-coordinate" at the start. This shows you understand the question. Then substitute carefully.

Question 26. Points A and B have coordinates (7, -3) and (1, 9) respectively. Find: (i) the slope of AB. (ii) the equation of the perpendicular bisector of the line segment AB. (iii) the value of 'p' if (-2, p) lies on it.
Answer: Given points are A(7, -3) and B(1, 9).

(i) Slope of AB = \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - (-3)}{1 - 7} = \frac{12}{-6} = -2 \)

(ii) Slope of perpendicular bisector = \( \frac{-1}{-2} = \frac{1}{2} \)

Mid-point of AB = \( \left( \frac{7 + 1}{2}, \frac{-3 + 9}{2} \right) = (4, 3) \)

∴ Equation of perpendicular bisector is:
y - 3 = \( \frac{1}{2} \)(x - 4)
2y - 6 = x - 4
x - 2y + 2 = 0

(iii) Point (-2, p) lies on x - 2y + 2 = 0.
∴ -2 - 2p + 2 = 0
⇒ 2p = 0
⇒ p = 0
In simple words: Perpendicular bisector is a line that cuts another line at 90 degrees and passes through its middle point. Its slope is negative reciprocal of the original line's slope.

📝 Teacher's Note: Use a ruler and draw two perpendicular lines. Show that their slopes multiply to give -1. This makes the concept clear.

🎯 Exam Tip: For perpendicular lines, multiply their slopes - the answer should be -1. This is a quick check for your work.

Question 27. A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid-point of AB. Find the (i) coordinates of A and B (ii) slope of line AB (iii) equation of line AB.
Answer:

[Diagram: This diagram shows a coordinate system with point P(2, -3) marked, and points A and B on the x-axis and y-axis respectively, forming a line through P.]

(i) Let the co-ordinates be A(x, 0) and B(0, y).

Mid-point of A and B is given by \( \left( \frac{x + 0}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right) \)

⇒ \( (2, -3) = \left( \frac{x}{2}, \frac{y}{2} \right) \)

⇒ \( \frac{x}{2} = 2 \) and \( \frac{y}{2} = -3 \)

⇒ x = 4 and y = -6
∴ A = (4, 0) and B = (0, -6)

(ii) Slope of line AB, m = \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 0}{0 - 4} = \frac{-6}{-4} = \frac{3}{2} = 1\frac{1}{2} \)

(iii) Equation of line AB, using A(4, 0)
y - 0 = \( \frac{3}{2} \)(x - 4)
2y = 3x - 12
In simple words: Since A is on x-axis, its y-coordinate is 0. Since B is on y-axis, its x-coordinate is 0. We used midpoint formula backwards to find the coordinates.

📝 Teacher's Note: Draw the axes and mark the points clearly. Show that points on x-axis have y = 0, and points on y-axis have x = 0.

🎯 Exam Tip: When a point is on an axis, one coordinate is always zero. Remember: x-axis means y = 0, y-axis means x = 0.

Question 28. The equation of a line 3x + 4y - 7 = 0. Find: (i) the slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x - y + 2 = 0 and 3x + y - 10 = 0.
Answer: 3x + 4y - 7 = 0 ...(1)
4y = -3x + 7
y = \( \frac{-3}{4} \)x + \( \frac{7}{4} \)

(i) Slope of the line = m = \( -\frac{3}{4} \)

(ii) Slope of the line perpendicular to the given line = \( \frac{-1}{-\frac{3}{4}} = \frac{4}{3} \)

Solving the equations x - y + 2 = 0 and 3x + y - 10 = 0, we get x = 2 and y = 4.
So, the point of intersection of the two given lines is (2, 4).

Given that a line with slope \( \frac{4}{3} \) passes through point (2, 4).

Thus, the required equation of the line is
y - 4 = \( \frac{4}{3} \)(x - 2)
3y - 12 = 4x - 8
4x - 3y + 4 = 0
In simple words: We found the slope from the equation by writing it in y = mx + c form. Then we found where two given lines meet, and wrote the perpendicular line through that point.

📝 Teacher's Note: To find slope from ax + by + c = 0, rearrange to y = mx + c form. The coefficient of x gives the slope.

🎯 Exam Tip: Always check that perpendicular slopes multiply to give -1. This confirms your answer is correct.

Question 29. ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find: (i) Co-ordinates of A (ii) Equation of diagonal BD
Answer:

[Diagram: This diagram shows parallelogram ABCD with vertices labeled, and point O at the intersection of diagonals AC and BD.]

In parallelogram ABCD, A(x, y), B(5, 8), C(4, 7) and D(2, -4).
The diagonals of the parallelogram bisect each other.
O is the point of intersection of AC and BD
Since O is the midpoint of BD, its coordinates will be
\( \left( \frac{2 + 5}{2}, \frac{-4 + 8}{2} \right) \) or \( \left( \frac{7}{2}, \frac{4}{2} \right) \) or \( \left( \frac{7}{2}, 2 \right) \)

(i)
Since O is the midpoint of AC also,
\( \frac{x + 4}{2} = \frac{7}{2} ⇒ x + 4 = 7 ⇒ x = 7 - 4 = 3 \)

and \( \frac{y + 7}{2} = 2 ⇒ y + 7 = 14 ⇒ y = 14 - 7 = 7 \)

Thus, coordinates of A are (3, 7)

(ii)
Equation of BD will be
\( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \)

⇒ y - y₁ = \( \frac{y_2 - y_1}{x_2 - x_1} \) × (x - x₁)

⇒ y + 4 = \( \frac{8 + 4}{5 - 2} \) × (x - 2)

⇒ y + 4 = \( \frac{12}{3} \) × (x - 2)

⇒ y + 4 = 4(x - 2)
⇒ y + 4 = 4x - 8
⇒ 4x - y = 12
In simple words: In a parallelogram, the diagonals cut each other into two equal parts. We used this property to find the missing vertex A.

📝 Teacher's Note: Draw a parallelogram with diagonals. Show that they cross at the middle point of each diagonal. This property is very useful.

🎯 Exam Tip: In parallelogram problems, always remember that diagonals bisect each other. Use midpoint formula to set up equations.

Question 30. Given equation of the line L₁ is y = 4.
(i)Write the slope of the line L₂ if L₂ is the bisector of angle O
(ii)Write the coordinates of point P
(iii)Find the equation of L₂

[Diagram: Coordinate system showing horizontal line L₁, point P on the line, and line L₂ passing through origin O and point P, with L₂ appearing to bisect the angle between positive x-axis and positive y-axis at 45°]

Answer:
(i)
Equation of line L₁ is y = 4
∴ L₂ is the bisector of ∠O
∴ ∠POX = 45°
Slope = tan 45° = 1
Let coordinates of P be (x, y)
∴ P lies on L₁

(ii)
∴ Slope of L₂ = \( \frac{y_2 - y_1}{x_2 - x_1} \)
1 = \( \frac{4 - 0}{x - 0} \) ⇒ 1 = \( \frac{4}{x} \)
⇒ x = 4
∴ coordinates of P are (4, 4)

(iii)
Equation of L₂ is
y - y₁ = m{x - x₁}
⇒ y - 4 = 1(x - 4)
⇒ y - 4 = x - 4
⇒ x = y

In simple words: L₁ is a horizontal line at y = 4. L₂ goes through the origin and makes a 45° angle, so its slope is 1. Since P is on both lines, we find where they meet.

📝 Teacher's Note: Draw the angle bisector carefully. Show that tan 45° = 1 means the line goes up 1 unit for every 1 unit to the right. This makes a perfect diagonal.

🎯 Exam Tip: Write "slope = tan 45° = 1" clearly. Always show the substitution step when finding coordinates. The final equation can be written as x = y or y = x.

Question 31.
(i) equation of AB
(ii) equation of CD

[Diagram: Coordinate plane showing line segment AB from A(-5, 4) to B(3, 3), and line CD that appears perpendicular to AB, with point D at (-3, 0)]

Answer:
(i) Slope of AB = \( \frac{3 - 4}{3 - (-5)} = \frac{-1}{8} \)
∴ Equation of AB is given by
y - 4 = -\( \frac{1}{8} \)(x - (-5))
8y - 32 = -(x + 5)
8y - 32 = -x - 5
x + 8y = 27

(ii) AB and CD are perpendicular to each other.
Thus, product of their slopes = -1
Slope of AB × Slope of CD = -1
⇒ \( \frac{-1}{8} \) × Slope of CD = -1
⇒ Slope of CD = 8
Now, from graph we have coordinates of D = (-3, 0)
∴ Equation of line CD is given by
y - 0 = 8(x + 3)
y = 8x + 24

In simple words: First we find the slope of line AB using the two points. Then since CD is perpendicular, its slope is the negative reciprocal. We use point D to find CD's equation.

📝 Teacher's Note: Remind students that perpendicular lines have slopes that multiply to give -1. If one slope is 1/8, the perpendicular slope is -8, not 8.

🎯 Exam Tip: Always write "product of slopes = -1" for perpendicular lines. Show the calculation clearly. Use point-slope form: y - y₁ = m(x - x₁).

Question 32. Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.

Answer:
Slope of 3x + 5y = 1 is given by -\( \frac{3}{5} \) =
⇒ Slope of line perpendicular to 3x + 5y = 1 = -\( \frac{1}{\text{Slope of 3x + 5y = 1}} \) = -\( \frac{1}{-\frac{3}{5}} \) = \( \frac{5}{3} \)

Now, x-intercept = -3
y = mx + c
⇒ 0 = \( \frac{5}{3} \) × (-3) + c
⇒ c = 5

∴ Equation of required line is given by y = \( \frac{5}{3} \)x + 5
i.e. 3y = 5x + 15
i.e. 5x - 3y + 15 = 0

In simple words: The given line has slope -3/5. A perpendicular line has slope 5/3. Since it crosses the x-axis at -3, we find the y-intercept and write the equation.

📝 Teacher's Note: Show students how to get slope from ax + by = c form: slope = -a/b. For perpendicular lines, flip and change sign of the slope fraction.

🎯 Exam Tip: Write "perpendicular slope = negative reciprocal" clearly. Show that x-intercept means y = 0 at that point. Final answer can be in any standard form.

Question 33. A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M

Answer:
(i) The equation of the line passing through the points P(-1, 4) and Q(5, -2) is
y - 4 = \( \frac{-2 - 4}{5 - (-1)} \)[x - (-1)]
i.e. y - 4 = \( \frac{-6}{6} \)(x + 1)
i.e. y - 4 = -1(x + 1)
i.e. y - 4 = -x - 1
i.e. x + y = 3

(ii) The line x + y = 3 cuts x-axis at point A. Hence, its y-coordinate is 0.
And, x-coordinate is given by
x + 0 = 3 ⇒ x = 3
So, the coordinates of A are (3, 0).
The line x + y = 3 cuts y-axis at point B. Hence, its x-coordinate is 0.
And, y-coordinate is given by
0 + y = 3 ⇒ y = 3
So, the coordinates of B are (0, 3).

(iii) Since M is the mid-point of line segment AB,
So, coordinates of M = \( \left(\frac{3 + 0}{2}, \frac{0 + 3}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}\right) \)

In simple words: We find the line equation using two points. Then we find where this line crosses the x-axis (point A) and y-axis (point B). The midpoint M is halfway between A and B.

📝 Teacher's Note: Emphasize that x-intercept means y = 0, and y-intercept means x = 0. Use the midpoint formula: M = ((x₁+x₂)/2, (y₁+y₂)/2).

🎯 Exam Tip: Always substitute y = 0 for x-intercept and x = 0 for y-intercept. Show the midpoint formula calculation step by step with correct brackets.

Question 34. In the given figure, line AB meets y-axis at point A. Line through C(2, 10) and D intersects line AB at right angle at point R Find:
(i) equation of line AB
(ii) equation of line CD
(iii) co-ordinates of points E and D

[Diagram: Coordinate plane showing line AB meeting y-axis at A, with points B(6, 8), C(2, 10), and other intersection points marked]

Answer:
(i) Slope of line AB = m = \( \frac{8 - 6}{6 - 0} = \frac{2}{-6} = -\frac{1}{3} \)
The y-intercept of the line AB is 6.
Thus, the equation of the given line is given by the slope-intercept form
y = mx + c
i.e. y = -\( \frac{1}{3} \)x + 6
i.e. 3y = -x + 18
i.e. x + 3y = 18, which is the required equation.

In simple words: We find the slope using two points on line AB. Since we know where it crosses the y-axis (at 6), we can write the equation directly.

📝 Teacher's Note: Point out that when a line meets the y-axis, that point gives us the y-intercept directly. Use slope formula carefully with correct signs.

🎯 Exam Tip: Write slope = (y₂ - y₁)/(x₂ - x₁) clearly. If you know the y-intercept, use y = mx + c form first, then convert to standard form.

Question 35. A line through point P(4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Answer:
Since a line through point P meets x-axis at point A and y-axis at point B, coordinates of A are (x, 0) and coordinates of B are (0, y).
Now, BP = 2PA
\( \Rightarrow \frac{BP}{PA} = \frac{2}{1} \)
\( \Rightarrow \) P divides AB in the ratio 2 : 1.

So, the coordinates of P are \( \left(\frac{2 \times x + 1 \times 0}{2 + 1}, \frac{2 \times 0 + 1 \times y}{2 + 1}\right) = \left(\frac{2x}{3}, \frac{y}{3}\right) \)

But, coordinates of P are (4, 3).
\( \Rightarrow \frac{2x}{3} = 4 \Rightarrow 2x = 12 \Rightarrow x = 6 \) and \( \frac{y}{3} = 3 \Rightarrow y = 9 \)
\( \Rightarrow \) Coordinates of A are (6, 0) and coordinates of B are (0, 9).

Slope of line AB = \( \frac{9 - 0}{0 - 6} = \frac{9}{-6} = -\frac{3}{2} \)

Thus, the equation of line AB is given by
\( y - 0 = -\frac{3}{2}(x - 6) \)
i.e. \( 2y = -3x + 18 \)
i.e. \( 3x + 2y = 18 \)

In simple words: We found where the line crosses the x and y axes. Then we used the fact that P divides the line in a 2:1 ratio to find the coordinates. Finally, we found the slope and wrote the equation.

📝 Teacher's Note: Show students how the section formula works. When a point divides a line in ratio m:n, the coordinates are calculated using the section formula. This is very common in coordinate geometry.

🎯 Exam Tip: Always write the section formula clearly. Write coordinates of A as (x, 0) and B as (0, y) since they lie on axes. Show all steps of calculation for full marks.

Question 36. Find the equation of line through the intersection of lines 2x - y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Answer:
Since the line passing through the x-axis makes an angle of 30° with the positive direction of the x-axis,
the slope of the line is given by \( \tan 30° = \frac{1}{\sqrt{3}} \).

The intersection of the lines 2x - y = 1 and 3x + 2y = -9 is given by solving the equations simultaneously.
So, multiplying equation 2x - y = 1 by 2, we get,
4x - 2y = 2
Now add this resultant to the second equation 3x + 2y = -9.
\( \Rightarrow 7x = -7 \Rightarrow x = -1 \)
Substituting the value of x in 2x - y = 1, we get y = -3.
Thus, the intersection of the lines is (-1, -3).

To find the equation of the required line, we use the slope-point form, so we get
\( y - (-3) = \frac{1}{\sqrt{3}}[x - (-1)] \)
i.e. \( y + 3 = \frac{1}{\sqrt{3}}(x + 1) \)
i.e. \( y = \frac{x}{\sqrt{3}} + \frac{1}{\sqrt{3}} - 3 \)

In simple words: First we found where the two given lines meet. Then we used the angle given to find the slope. Finally, we wrote the equation using the point and slope.

📝 Teacher's Note: Teach students that tan θ gives the slope when a line makes angle θ with x-axis. Always solve the system of equations step by step to find intersection point.

🎯 Exam Tip: Write tan 30° = 1/√3 clearly. Show the solving of simultaneous equations step by step. Use slope-point form correctly with proper substitution.

Question 37. Find the equation of the line through the Points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear.
Answer:
Slope of line AB = \( m = \frac{2 - 3}{0 - (-1)} = \frac{-1}{1} = -1 \)

Using the slope-point form, the equation of line AB is given by
y - y₁ = m(x - x₁)
i.e. y - 3 = -1[x - (-1)]
i.e. y - 3 = -1(x + 1)
i.e. y - 3 = -x - 1
i.e. x + y = 2

Now, slope of line BC = \( \frac{1 - 2}{1 - 0} = \frac{-1}{1} = -1 \)

Since, Slope of line AB = Slope of line BC, points A, B and C are collinear.

In simple words: We found the slope between A and B, then wrote the line equation. We then checked if C also lies on the same line by comparing slopes. Same slope means they are on the same straight line.

📝 Teacher's Note: Explain that collinear means "on the same line". When slopes between different pairs of points are equal, all points lie on the same straight line.

🎯 Exam Tip: Always calculate slope first. Use slope-point form correctly. To prove collinear, show that slopes of AB and BC are equal. Write "points are collinear" as conclusion.

Question 38. Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2), find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.
Answer:

(i) Let (x, y) be the coordinates of D.
We know that the diagonals of a parallelogram bisect each other.
∴ Mid-point of diagonal AC = Mid-point of diagonal BD
\( \Rightarrow \left(\frac{3 + 3}{2}, \frac{6 + 2}{2}\right) = \left(\frac{5 + x}{2}, \frac{10 + y}{2}\right) \)
\( \Rightarrow (3, 4) = \left(\frac{5 + x}{2}, \frac{10 + y}{2}\right) \)
\( \Rightarrow \frac{5 + x}{2} = 3 \Rightarrow 5 + x = 6 \Rightarrow x = 1 \) and \( \frac{10 + y}{2} = 4 \Rightarrow 10 + y = 8 \Rightarrow y = -2 \)
∴ Coordinates of D are (1, -2).

(ii) Length of diagonal BD = \( \sqrt{(1 - 5)^2 + (-2 - 10)^2} \)
= \( \sqrt{(-4)^2 + (-12)^2} \)
= \( \sqrt{16 + 144} \)
= \( \sqrt{160} \)
= \( 4\sqrt{10} \) units

(iii) Slope of side AB = \( m = \frac{10 - 6}{5 - 3} = \frac{4}{2} = 2 \)
Thus, the equation of side AB is given by
y - 6 = 2(x - 3)
i.e. y - 6 = 2x - 6
i.e. 2x - y = 0
i.e. y = 2x

In simple words: In a parallelogram, the diagonals cut each other in half. We used this property to find the fourth vertex. Then we used the distance formula and slope-point form.

📝 Teacher's Note: Draw a parallelogram and show that diagonals bisect each other. This property is the key to finding the fourth vertex. Make sure students understand midpoint formula.

🎯 Exam Tip: Write "diagonals of parallelogram bisect each other" as the key property. Use midpoint formula correctly. For distance, always write the formula first, then substitute values.

Question 39. In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.
(i) Write the coordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC.
Answer:

(i) The line intersects the x-axis where y = 0.
Hence, the coordinates of A are (4, 0).

(ii) Length of AB = \( \sqrt{[4 - (-2)]^2 + (0 - 3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \) units

Length of AC = \( \sqrt{[4 - (-2)]^2 + (0 + 4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \) units

(iii) Let k be the required ratio which divides the line segment joining the coordinates A(4, 0) and Q(-2, -4).
Let the coordinates of Q be x and y.
∴ x = \( \frac{k(-2) + 1(4)}{k + 1} \) and y = \( \frac{k(-4) + 0}{k + 1} \)
Q lies on the y-axis where x = 0.
\( \Rightarrow \frac{-2k + 4}{k + 1} = 0 \)
\( \Rightarrow -2k + 4 = 0 \)
\( \Rightarrow 2k = 4 \)
\( \Rightarrow k = \frac{4}{2} = \frac{2}{1} \)
Thus, the required ratio is 2 : 1.

(iv) Slope of line AC = \( m = \frac{-4 - 0}{-2 - 4} = \frac{-4}{-6} = \frac{2}{3} \)
Thus, the equation of the line AC is given by
\( y - 0 = \frac{2}{3}(x - 4) \)
i.e. \( 3y = 2x - 8 \)
i.e. \( 2x - 3y = 8 \)

In simple words: We found where line AC crosses the x-axis (that's point A). Then we calculated distances using distance formula. We found the ratio using section formula. Finally, we used slope and point to write the line equation.

📝 Teacher's Note: Use the coordinate plane diagram to help students visualize the problem. Show them how to read coordinates from graphs. The section formula is important for finding ratios.

🎯 Exam Tip: Read coordinates carefully from the graph. Use distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). For ratio problems, use section formula and the fact that one coordinate is known (like x = 0 on y-axis).

Question 40.
Answer:
(i) Slope of PQ = \( \frac{3 - k}{1 - 3k - 6} \)
\( \Rightarrow \frac{1}{2} = \frac{3 - k}{-3k - 5} \)
\( \Rightarrow -3k - 5 = 2(3 - k) \)
\( \Rightarrow -3k - 5 = 6 - 2k \)
\( \Rightarrow k = -11 \)

(ii) Substituting k in P and Q, we get
P(6, k) = P(6, -11) and Q(1 - 3k, 3) = Q(34, 3)
\( \therefore \) Mid-point of PQ = \( \left(\frac{6 + 34}{2}, \frac{-11 + 3}{2}\right) = \left(\frac{40}{2}, \frac{-8}{2}\right) = (20, -4) \)
In simple words: We found the value of k by using the given slope condition. Then we used that k value to find the exact coordinates and calculated the midpoint.

📝 Teacher's Note: Show students how to use the slope formula step by step. Many students forget to cross multiply properly when solving for k.

🎯 Exam Tip: Always substitute the found value back to get exact coordinates. Write the midpoint formula clearly and show each calculation step.

Question 41.
Answer:
i. Since A lies on the X-axis, let the co-ordinates of A be (x, 0).
Since B lies on the Y-axis, let the co-ordinates of B be (0, y).
Let m = 1 and n = 2
Using Section formula,
Coordinates of P = \( \left(\frac{1(0) + 2(x)}{1 + 2}, \frac{1(y) + 2(0)}{1 + 2}\right) \)
\( \Rightarrow (4, -1) = \left(\frac{2x}{3}, \frac{y}{3}\right) \)
\( \Rightarrow \frac{2x}{3} = 4 \) and \( \frac{y}{3} = -1 \)
\( \Rightarrow x = 6 \) and \( y = -3 \)
So, the co-ordinates of A are (6, 0) and that of B are (0, -3).

ii. Slope of AB = \( \frac{-3 - 0}{0 - 6} = \frac{-3}{-6} = \frac{1}{2} \)
\( \Rightarrow \) Slope of line perpendicular to AB = m = -2
P = (4, -1)
Thus, the required equation is
y – y₁ = m(x – x₁)
\( \Rightarrow y - (-1) = -2(x - 4) \)
\( \Rightarrow y + 1 = -2x + 8 \)
\( \Rightarrow 2x + y = 7 \)
In simple words: First we found where A and B are located using the section formula. Then we found the slope of line AB and used it to get the perpendicular line equation.

📝 Teacher's Note: Remind students that perpendicular slopes multiply to give -1. If one slope is 1/2, the perpendicular slope is -2.

🎯 Exam Tip: Always write "coordinates of A are..." and "coordinates of B are..." clearly. Use the point-slope form y - y₁ = m(x - x₁) for the equation.