Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 13 Equation Of A Straight Line

Exercise 13.1

Answer 2.
Answer: The line \( 3y = 5x - 7 \) passes through \( (p,6) \)
\( \therefore (p,6) \) will satisfy the equation of line
\( 3(6) = 5(p) - 7 \)
\( \Rightarrow 25 = 5p \)
\( \Rightarrow p = 5 \)
In simple words: We put the point (p,6) into the line equation. This gives us 18 = 5p - 7, which solves to p = 5.

📝 Teacher's Note: When a point lies on a line, it must satisfy the line equation. Substitute the point coordinates and solve for the unknown value.

🎯 Exam Tip: Always write "will satisfy the equation" first. Then substitute carefully and solve step by step.

Answer 3.
Answer: The line \( 3y = 5x - 7 \) passes through \( (p,6) \)
\( \therefore (p,6) \) will satisfy the equation of line
\( 3(6) = 5(p) - 7 \)
\( \Rightarrow 25 = 5p \)
\( \Rightarrow p = 5 \)
In simple words: This is the same as Answer 2. We substitute the y-value 6 into the equation and find p = 5.

📝 Teacher's Note: This appears to be a duplicate of the previous question. Make sure students understand the substitution process clearly.

🎯 Exam Tip: Double-check your substitution. Put y = 6 into the equation and solve for p step by step.

Answer 4.
Answer: The line \( 4x = 11 - 3y \) passes through the point \( (a,5) \)
\( \therefore (a,5) \) will satisfy the equation of line
\( 4(a) = 11 - 3(5) \)
\( \Rightarrow 4a = -4 \)
\( \Rightarrow a = -1 \)
In simple words: We put the point (a,5) into the line equation. This gives us 4a = 11 - 15 = -4, so a = -1.

📝 Teacher's Note: Be careful with negative numbers. When y = 5, we get -3(5) = -15, so 11 - 15 = -4.

🎯 Exam Tip: Watch for negative signs. Calculate 11 - 3(5) = 11 - 15 = -4 carefully.

Answer 5.
Answer: The line \( y = 6 - \frac{3x}{2} \) passes through the point \( (r,3) \)
\( \therefore (r,3) \) will satisfy the equation of line
\( 3 = 6 - \frac{3r}{2} \)
\( \Rightarrow -3 = -\frac{3r}{2} \)
\( \Rightarrow r = 2 \)
In simple words: We substitute y = 3 into the equation. After solving, we get r = 2.

📝 Teacher's Note: When dealing with fractions, multiply both sides by 2 to clear the fraction: -6 = -3r, so r = 2.

🎯 Exam Tip: Move all terms to one side first: 3 - 6 = -3r/2, then solve. Don't forget to handle fractions properly.

Answer 6.
Answer: The line \( \frac{3 + 5y}{2} = \frac{4x - 7}{2} \) passes through the point \( (1,k) \)
\( \therefore (1,k) \) will satisfy the equation of line
\( \frac{3 + 5k}{2} = \frac{4(1) - 7}{3} \)
\( \Rightarrow 9 + 15k = 6 \)
\( \Rightarrow 15k = -15 \)
\( \Rightarrow k = -1 \)
In simple words: We substitute x = 1 into the equation and solve for k. We get k = -1.

📝 Teacher's Note: Cross multiply to clear fractions. The equation becomes 3(3 + 5k) = 2(4 - 7) = 2(-3) = -6.

🎯 Exam Tip: Be careful with fraction equations. Cross multiply properly and solve step by step.

Answer 7.
Answer: Let the point of intersection of AB and line \( 4x + 4y = 11 \) be the point P(a,b)
Also given \( 4x + 3y = 11 \) bisects line segment AB
\( \therefore AP : PB = 1 : 1 \)
Coordinates of P are:
\( P(a,b) = P\left(\frac{6 + 4}{2}, \frac{m - 9}{2}\right) = P\left(5, \frac{m - 9}{2}\right) \)
Since P(a,b) lies on the line \( 4x + 3y = 11 \), P will satisfy the equation of line
\( 4(5) + 3\left(\frac{m - 9}{2}\right) = 11 \)
\( \frac{3m - 27}{2} = 11 - 20 \)
\( \Rightarrow 3m - 27 = -18 \)
\( \Rightarrow 3m = 9 \)
\( \Rightarrow m = 3 \)
In simple words: The line cuts AB into two equal parts. We find the middle point and use it in the line equation to find m = 3.

📝 Teacher's Note: When a line bisects a segment, it passes through the midpoint. Use the midpoint formula to find coordinates.

🎯 Exam Tip: Write "bisects means AP:PB = 1:1" clearly. Then find midpoint coordinates and substitute into the line equation.

Answer 8.
Answer: Let the point of intersection of AB and the line \( 2x - 5y + 31 = 0 \) be the point R(a,b)
Also, given the line \( 2x - 5y + 31 = 0 \) bisects the line segment AB
\( \therefore AR : RB = 1 : 1 \)
Coordinates of R are:
\( R(a,b) = R\left(\frac{-4 + p}{2}, \frac{5 + 9}{2}\right) = R\left(\frac{-4 + p}{2}, 7\right) \)
\( \because R(a,b) \) lies on the line \( 2x - 5y + 31 = 0 \)
\( \therefore \) R will satisfy the equation of the line
\( 2\left(\frac{-4 + p}{2}\right) - 5(7) + 31 = 0 \)
\( \Rightarrow (-4 + p) - 4 = 0 \)
\( \Rightarrow p = 8 \)
In simple words: The line cuts AB in half. We find the middle point and put it in the line equation to get p = 8.

📝 Teacher's Note: Remember that when simplifying, -35 + 31 = -4. So we get (-4 + p) - 4 = 0.

🎯 Exam Tip: Calculate the midpoint carefully. Then substitute both coordinates into the line equation and solve for the unknown.

Answer 9.
Answer: Let the point of intersection of AB and the line \( 3x + 4y = 18 \) be the point P(a,b)
Also, given the line \( 3x + 4y = 18 \) bisects the line segment AB
\( \therefore AP : PB = 1 : 1 \)
Coordinates of P are:
\( P(a,b) = P\left(\frac{3 - 7}{2}, \frac{7 + z}{2}\right) = P\left(-2, \frac{7 + z}{2}\right) \)
\( \because P(a,b) \) lies on the line \( 3x + 4y = 18 \)
\( \therefore \) P will satisfy the equation of the line
\( 3(-2) + 4\left(\frac{7 + z}{2}\right) = 18 \)
\( \Rightarrow 14 + 2z = 24 \)
\( \Rightarrow 2z = 10 \)
\( \Rightarrow z = 5 \)
In simple words: The line bisects AB, so it passes through the middle point. We use this to find z = 5.

📝 Teacher's Note: After substituting, we get -6 + 2(7 + z) = 18, which simplifies to -6 + 14 + 2z = 18.

🎯 Exam Tip: Find the midpoint first, then substitute into the line equation. Simplify step by step to avoid mistakes.

Answer 10.
Answer: Let the point of intersection of PQ and \( 5x - 3y + 1 = 0 \) be the point R(a,b)
Also given the line \( 5x - 3y + 1 = 0 \) divides the line segment PQ in the ratio 2 : 3
i.e. PR : RQ = 2 : 5
Coordinates of R are:
\( R(a,b) = R\left(\frac{14 + 6}{5}, \frac{18 + 3m}{5}\right) = R\left(4, \frac{18 + 3m}{5}\right) \)
\( \because R(a,b) \) lies on the line \( 5x - 3y + 1 = 0 \)
\( \therefore \) R will satisfy the equation of the line
\( 5(4) - 3\left(\frac{18 + 3m}{5}\right) + 1 = 0 \)
\( \Rightarrow -3\left(\frac{18 + 3m}{5}\right) = -21 \)
\( \Rightarrow 18 + 3m = 35 \)
\( \Rightarrow 3m = 17 \)
\( \Rightarrow m = \frac{17}{3} \)
In simple words: The line divides PQ in the ratio 2:3. We find the dividing point and use it in the line equation to get m = 17/3.

📝 Teacher's Note: Use the section formula when a line divides a segment in a given ratio. The coordinates depend on the ratio.

🎯 Exam Tip: Apply the section formula correctly for the given ratio. Then substitute the coordinates into the line equation.

Answer 11.
Answer: Let the point of intersection of AB and the line \( 7x - 8y = 4 \) be the point P(a,b)
Also, given the line \( 7x - 8y = 4 \) divides the line segment AB in the ratio 2 : 5
i.e. AP : PB = 2 : 5
Coordinates of P are:
\( P(a,b) = P\left(\frac{12 - 40}{7}, \frac{2k - 20}{7}\right) = P\left(-4, \frac{2k - 20}{7}\right) \)
\( \because P(a,b) \) lies on the line \( 7x - 8y = 4 \)
\( \therefore \) P will satisfy the equation of the line
\( 7(-4) - 8\left(\frac{2k - 20}{7}\right) = 4 \)
\( -8\left(\frac{2k - 20}{7}\right) = 32 \)
\( 2k - 20 = -28 \)
\( 2k = -8 \)
\( k = -4 \)
In simple words: The line divides AB in ratio 2:5. We find this dividing point and use it to get k = -4.

📝 Teacher's Note: When using the section formula, be careful with signs. The ratio 2:5 means we divide internally.

🎯 Exam Tip: Use the section formula for ratio 2:5. Substitute the point into the line equation and solve carefully for k.

Answer 12.
Answer: Let the point of intersection of PQ and the line \( 5x + 3y = 25 \) be the point R(x,y)
Also, given the line \( 5x + 3y = 25 \) divides the line segment PQ in the ratio 1 : 3
i.e. PR : RQ = 1 : 3
Coordinates of R are:
\( R(x,y) = R\left(\frac{5 + 3b}{4}, \frac{8 + 12}{4}\right) = R\left(\frac{5 + 3b}{4}, 5\right) \)
\( \because R(x,y) \) lies on the line \( 5x + 3y = 25 \)
\( \therefore \) R will satisfy the equation of the line
\( 5\left(\frac{5 + 3b}{4}\right) + 3(5) = 25 \)
\( \Rightarrow 5\left(\frac{5 + 3b}{4}\right) = 10 \)
\( \Rightarrow 5 + 3b = 8 \)
\( \Rightarrow 3b = 3 \)
\( \Rightarrow b = 1 \)
In simple words: The line divides PQ in ratio 1:3. We find the dividing point and solve to get b = 1.

📝 Teacher's Note: Apply the section formula for internal division in ratio 1:3. Then substitute and solve the linear equation.

🎯 Exam Tip: Use the section formula correctly. Remember to simplify 25 - 15 = 10 on the right side of the equation.

Answer 13.
Answer: Let the point P on the line segment AB be P(a,b)
Also, given that P(a,b) divides the line segment AB in the ratio 2 : 3
i.e. AP : PB = 2 : 3
Coordinates of P are:
\( P(a,b) = P\left(\frac{16 - 6}{5}, \frac{16 + 9}{5}\right) = P(2,5) \)
If P(a,b) lies on the line \( 7x - 2y = 4 \), then will satisfy the equation of the line
LHS \( 7(2) - 2(5) = 14 - 10 = 4 = \) RHS
Yes, the point P(2,5) lies on the line \( 7x - 2y = 4 \)
In simple words: We find point P using the ratio. Then we check if P(2,5) satisfies the line equation. It does, so the answer is yes.

📝 Teacher's Note: Use the section formula to find coordinates, then verify by substituting into the line equation.

🎯 Exam Tip: First find the point coordinates using the section formula. Then check if LHS = RHS when substituting into the line equation.

Ex 13.2

Answer 1.
Answer:
(a) Slope of line = m = tan θ
= tan 60°
= \( \sqrt{3} \)
=1.73

(b) Slope of line = m = tan θ
= tan50°
=1.19

(c) Slope of line = m = tan θ
=tan 45° = 1

(d) Slope of line = m = tan θ
=tan75°

tan(75°) = tan(45° + 30°) = \( \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°} \)

= \( \frac{1 + \frac{1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{1 - \frac{1}{\sqrt{3}}} \)

= \( \frac{2.73}{0.73} = \frac{273}{73} = 3.73 \)

(e) Slope of line = m = tan θ
= tan30° = \( \frac{1}{\sqrt{3}} \)

In simple words: The slope of a line tells us how steep it is. We find it by using the angle the line makes with the x-axis and calculating tan of that angle.

📝 Teacher's Note: Show students a ladder against a wall. The steeper the ladder, the bigger the angle and the bigger the slope. This helps them see the connection between angle and slope.

🎯 Exam Tip: Remember that slope = tan θ. Always convert the angle to the correct trigonometric ratio. Show your work step by step for compound angles like 75°.

Answer 2.
Answer:
(a) tan θ = 0.4663
∴ θ = 25°

(b) tan θ = 1.4281
∴ θ = 55°

(c) tan θ = 3.0777
∴ θ = 72°

(d) tan θ = 5.6713
∴ θ = 80°

(e) tan θ = 0.5317
∴ θ = 28°

In simple words: When we know the slope (tan value), we use inverse tan (tan⁻¹) to find the angle. This is the reverse process of finding slope from angle.

📝 Teacher's Note: Students often forget to use tan⁻¹ on their calculators. Show them how to press the right buttons. Practice this step many times.

🎯 Exam Tip: Use tan⁻¹ function on calculator to find the angle. Always check if your calculator is in degree mode, not radian mode. Round to the nearest degree.

Answer 3.
Answer:
(a) Slope of line = \( \frac{y_2 - y_1}{x_2 - x_1} \)
= \( \frac{8 - 5}{-1 - 2} \)
= -1

(b) Slope of line = \( \frac{y_2 - y_1}{x_2 - x_1} \)
= \( \frac{13 - 7}{5 - 3} \)
= 3

(c) Slope of line = \( \frac{y_2 - y_1}{x_2 - x_1} \)
= \( \frac{-7 + 1}{-9 - 5} = \frac{3}{2} \)
= 1.5

(d) Slope of line = \( \frac{y_2 - y_1}{x_2 - x_1} \)

(e) Slope of line = \( \frac{y_2 - y_1}{x_2 - x_1} \)
= \( \frac{0 - 5}{5 - 0} \)
= -1

In simple words: To find slope between two points, we subtract the y-values and divide by the difference of x-values. This tells us how much the line goes up or down.

📝 Teacher's Note: Teach students to be careful with signs. When the slope is negative, the line goes down from left to right. When positive, it goes up.

🎯 Exam Tip: Always write the slope formula first. Be very careful with negative signs when subtracting coordinates. Show each step clearly.

Answer 4.
Answer:
(a) A(x₁, y₁) = A(a²m², 2am)
B(x₂, y₂) = B(p²m², 2pm)

Slope of line AB = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{2pm - 2am}{p²m² - a²m²} \)

= \( \frac{2m(p - a)}{m²(p² - a²)} \)

= \( \frac{2}{m} \times \frac{(p - a)}{(p + a)(p - a)} \)

= \( \frac{2}{m(p + a)} \)

= \( \frac{2}{pm + am} \)

(b) A(x₁, y₁) = A(5pq, p²q)
B(x₂, y₂) = B(5qr, qr²)

Slope of line AB = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{qr² - p²q}{5qr - 5pq} \)

= \( \frac{q(r² - p²)}{5q(r - p)} \)

= \( \frac{1(r - p)(r + p)}{5(r - p)} \)

= \( \frac{r + p}{5} \)

In simple words: These are coordinate geometry problems where we find slope using the formula. We simplify the algebraic expressions step by step.

📝 Teacher's Note: Students struggle with algebraic simplification. Show them how to factor expressions like (r² - p²) = (r - p)(r + p). Practice factoring before doing these problems.

🎯 Exam Tip: Write the slope formula first. Then substitute the coordinates carefully. Factor algebraic expressions to simplify. Cancel common terms only when they are not zero.

Answer 5.
Answer:
(a) 3x - 2y = 5
3x - 5 = 2y
\( \frac{3}{2}x - \frac{5}{2} = y \)
y = mx + c
Slope = \( \frac{3}{2} \)

(b) x + 3y = 7
3y = -x + 7
y = \( -\frac{1}{3}x + \frac{7}{3} \)
Slope = \( -\frac{1}{3} \)

(c) 5x - y = 10
5x - 10 = y
Slope = 5

(d) 4x - 2y = 3
-2y = -4x + 3
y = \( 2x - \frac{3}{2} \)
Slope = 2

(e) 5x + 2y = 11
2y = -5x + 11
y = \( -\frac{5}{2}x + \frac{11}{2} \)
Slope = \( -\frac{5}{2} \)

In simple words: To find the slope from an equation, we change it to y = mx + c form. The coefficient of x is the slope (m).

📝 Teacher's Note: Show students that the slope is always the number multiplied with x when the equation is in y = mx + c form. Practice converting different forms.

🎯 Exam Tip: Always convert to y = mx + c form first. The coefficient of x is your slope. Be careful with signs when rearranging the equation.

Answer 6.
Answer:
When the lines are perpendicular to the product of their slopes is -1.
i.e. m₁ × m₂ = -1

(a) 2x - 3y = 4
3y = 2x - 4
y = \( \frac{2}{3}x - \frac{4}{3} \)

Slope m₁ = \( \frac{2}{3} \)

Required slope of line (m₂)
m₁ · m₂ = -1
⇒ m₂ = \( \frac{-1}{m_1} \)

In simple words: When two lines are perpendicular (meet at 90°), their slopes multiply to give -1. If one slope is known, we can find the other.

📝 Teacher's Note: Draw two perpendicular lines on the board and show their slopes. One will be positive, the other negative. Their product is always -1.

🎯 Exam Tip: Remember the rule: m₁ × m₂ = -1 for perpendicular lines. Find the slope of the given line first, then use this formula to find the required slope.

Answer 14.
Answer:
Let L(a,b) be the point on line segment PQ dividing it in the ratio 1 : 3
i.e. PL : LQ = 1 : 3
Coordinates of L are,

L(a,b) = L\( \left(\frac{11 + 9}{4}, \frac{-5 + 21}{4}\right) \)
= L(5, 4)

If L(a,b) lies on the line 2x+5y=20, then it will satisfy the equation of the line
LHS =2(5) + 5(4) = 10 + 20 = 30 ≠ RHS
No, L(a,b) does not lie on the line 2x + 5y = 20

In simple words: We found the coordinates of point L using the section formula. Then we checked if this point lies on the given line by putting the values in the equation.

📝 Teacher's Note: Teach students the section formula step by step. Show them how to substitute coordinates into an equation to check if a point lies on a line.

🎯 Exam Tip: Use section formula to find the dividing point. Then substitute these coordinates into the line equation. If LHS = RHS, the point lies on the line.

Answer 15.
Answer:
Let the point on x-axis be P(x,y) which divides the line segment AB in the ratio 1 : 2,
i.e. AP : PB = 1 : 2

Coordinates of P are,

P(x,y) = P\( \left(\frac{5 + 4}{3}, \frac{6 + 6}{3}\right) \)

x = 3, y=4

If P(x,y) lies on the line 3x - 4y + 5 =0, then it will satisfy the equation of the line.

LHS =3(3) - 4(4) + 5 = 9-9= 0= RHS

Yes, the point P lies on the line 3x-4y+5 = 0.

In simple words: We used the section formula to find point P that divides the line segment. Then we checked if this point satisfies the given line equation. Since LHS = RHS, it does lie on the line.

📝 Teacher's Note: Emphasize that when a point lies on a line, its coordinates must satisfy the line equation. This is a key concept for checking solutions.

🎯 Exam Tip: Apply section formula correctly with the given ratio. Always verify your answer by substituting back into the equation. Write "LHS = RHS" to show the point lies on the line.

Answer 7.
Answer: Slope of line AB = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{5 + 1}{-7 - 3} = -\frac{3}{5} \)

Slope of line parallel to AB
= Slope of AB
= \( -\frac{3}{5} \)
In simple words: Lines that are parallel have the same slope. So any line parallel to AB will also have slope \( -\frac{3}{5} \).

📝 Teacher's Note: Remind students that parallel lines have the same slope. Use railway tracks as an example - they never meet and have the same slope.

🎯 Exam Tip: Always write "parallel lines have equal slopes" in your answer. This statement gets marks even if your calculation has small errors.

Answer 8.
Answer: Slope of line MN = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{3 - 9}{-2 - 4} = \frac{-6}{-6} \)

= -1

Slope of line parallel to MN = Slope of MN = 1
In simple words: We found the slope of line MN first. Then any line parallel to it will have the same slope, which is 1.

📝 Teacher's Note: Show students that when we divide negative by negative, we get positive. This is why \( \frac{-6}{-6} = 1 \), not -1.

🎯 Exam Tip: Be careful with signs when calculating slope. Double-check your subtraction and division signs to avoid silly mistakes.

Answer 9.
Answer: Slope of line PQ = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{13 + 3}{7 - 11} = \frac{16}{-4} \)

= -4

Slope of line parallel to PQ
= Slope of PQ
= -4
In simple words: We calculated the slope of PQ by using the slope formula. Any line parallel to PQ will have the same slope value of -4.

📝 Teacher's Note: Help students remember that \( y_2 - y_1 \) means "final y minus starting y". Use a simple example on the board to show this.

🎯 Exam Tip: Write the slope formula first, then substitute the coordinates clearly. This shows the examiner you know the method.

Answer 10.
Answer: Slope of line AB

\( \Rightarrow \frac{-1}{3} = \frac{6 - 9}{12 - x} \)

\( \Rightarrow x - 12 = -9 \)

\( \Rightarrow x = 3 \)
In simple words: We used the given slope and two points to find the missing x-coordinate. Cross multiplication helped us solve for x.

📝 Teacher's Note: Show students how to cross multiply step by step. This is a common mistake area - they often forget to flip signs correctly.

🎯 Exam Tip: Always check your answer by substituting back into the slope formula. This catches calculation errors quickly.

Answer 11.
Answer: Slope of line PQ = \( \frac{y_2 - y_1}{x_2 - x_1} \)

\( \Rightarrow \frac{1}{3} = \frac{m - 5}{2 + 7} \)

\( \Rightarrow 3 = m - 5 \)
\( \Rightarrow m = 8 \)
In simple words: We knew the slope and both x-coordinates. So we could find the missing y-coordinate m by solving the equation.

📝 Teacher's Note: Emphasize that students should simplify the denominator first (2 + 7 = 9) before cross multiplying. This reduces errors.

🎯 Exam Tip: Write \( \Rightarrow \) symbols clearly to show each step. Examiners like to see logical progression in your working.

Answer 12.
Answer: Slope of line AB = \( \frac{y_2 - y_1}{x_2 - x_1} \)

\( \Rightarrow 1 = \frac{2p + 1 - 5}{p + 2} \)

\( \Rightarrow p + 2 = 2p - 4 \)
\( \Rightarrow 6 = p \)
In simple words: We set up the slope formula with the given slope value 1. Then we solved the equation to find p = 6.

📝 Teacher's Note: Show students how to collect like terms when solving linear equations. Move all p terms to one side and numbers to the other side.

🎯 Exam Tip: Always write your final answer clearly. Write "p = 6" or "The value of p is 6" to make it obvious to the examiner.

Answer 13.
Answer: Slope of line PQ = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{5 - 1}{6 - 8} \)

= \( \frac{4}{-2} \)

Slope = -2
Also, Slope of line PQ = tan θ
∴ tan θ = -2
θ = tan⁻¹(-2)
Inclination = tan⁻¹(-2)
In simple words: The slope of a line equals the tangent of its angle with the x-axis. So we use inverse tangent to find the angle.

📝 Teacher's Note: Explain that inclination is just a fancy word for the angle the line makes with the horizontal. Use a book tilted on a desk as an example.

🎯 Exam Tip: Always write "tan θ = slope" in your answer. Then use tan⁻¹ to find the angle. Don't forget the inverse tangent symbol.

Answer 14.
Answer: Slope of line PQ = \( \frac{y_2 - y_1}{x_2 - x_1} \)

= \( \frac{-5 - 7}{7 + 5} = -1 \)
Also, Slope of line AB = tan θ
∴ tan θ = -1
tan θ = tan (90° + 45°)
θ = 135°
Inclination = 135°
In simple words: When slope is -1, the line makes a 135° angle with the x-axis. This is 45° more than a right angle.

📝 Teacher's Note: Draw a coordinate plane and show students what 135° looks like. It's in the second quadrant, sloping downward from left to right.

🎯 Exam Tip: Remember that tan(-1) = 135°, not -45°. The angle is always measured as a positive value from the x-axis.

Answer 15.
Answer: (a) \( x = \frac{y}{2} - 5 \)
y = 2x + 10
m₁ = 2
Slope of required line(m₂) = m₁ = 2

(b) \( x = \frac{3y}{2} + 2 \)
3y = 2x - 4
\( y = \frac{2}{3}x - \frac{4}{3} \)
m₁ = \( \frac{2}{3} \)
Slope of required line (m₂) = m₁ = \( \frac{2}{3} \)
In simple words: We changed each equation to y = mx + c form to find the slope. Parallel lines have the same slope.

📝 Teacher's Note: Teach students to always convert equations to y = mx + c form first. The coefficient of x is the slope.

🎯 Exam Tip: Write "parallel lines have equal slopes" in your answer. Then clearly state what the slope value is.

Answer 15 (continued).
Answer: (c) \( \frac{3x}{4} + \frac{5y}{2} = 7 \)
10y = -3x + 28
\( y = \frac{-3}{10}x + \frac{14}{5} \)
m₁ = \( \frac{-3}{10} \)
Slope of required line (m₂) = m₁ = \( \frac{-3}{10} \)

(d) \( \frac{x}{4} + \frac{y}{3} = 1 \)
3x + 4y = 12
\( y = \frac{-3}{4}x + 3 \)
m₁ = \( \frac{-3}{4} \)
Slope of required line(m₂) = m₁ = \( \frac{-3}{4} \)

(e) \( \frac{2x}{5} + \frac{y}{3} = 2 \)
6x + 5y = 30
\( y = \frac{-6}{5}x + 6 \)
m₁ = \( \frac{-6}{5} \)
Slope of required line (m₂) = m₁ = \( \frac{-6}{5} \)
In simple words: For each part, we converted the equation to slope-intercept form and found the slope. Parallel lines have identical slopes.

📝 Teacher's Note: Show students how to clear fractions by multiplying through by the LCD. This makes the algebra much easier.

🎯 Exam Tip: Always convert to y = mx + c form. The coefficient of x gives you the slope directly. Don't try to find slope from other forms.

Answer 16.
Answer: When two lines are perpendicular to each other the product of their slope is -1.
i.e. m₁×m₂ = -1

(a) \( \frac{x}{2} + \frac{y}{3} = \frac{4}{3} \)
3x + 2y = 8
\( y = -\frac{3}{2}x + 4 \)
m₁ = \( -\frac{3}{2} \)
In simple words: For perpendicular lines, when you multiply their slopes together, you always get -1. This is the key rule to remember.

📝 Teacher's Note: Draw two perpendicular lines on the board and show students that one goes up while the other goes down. Their slopes have opposite signs.

🎯 Exam Tip: Always write "m₁ × m₂ = -1 for perpendicular lines" at the start of your answer. This shows you know the key concept.

Ex 13.3

Answer 3.
Answer:
Given: \( m = \tan 30° \)

\( m = \frac{1}{\sqrt{3}} \)

Equation of line is given by:
\( \frac{y - y_1}{x - x_1} = m \)

\( \frac{y - 5}{x - 2} = \frac{1}{\sqrt{3}} \)

\( x - 2 = \sqrt{3}y - 5\sqrt{3} \)

\( x - \sqrt{3}y - 2 + 5\sqrt{3} = 0 \)

\( x - \sqrt{3}y + (5\sqrt{3} - 2) = 0 \)

In simple words: We used the slope from tan 30° and the point-slope formula. This gives us the equation of the line passing through the given point.

📝 Teacher's Note: Remember that tan 30° = 1/√3. Use this value in the point-slope formula. Students often forget to simplify the constant term properly.

🎯 Exam Tip: Always write the final equation in the form ax + by + c = 0. Show all steps clearly from slope to final equation.

Answer 4.
Answer:
Given: \( m = \tan(180° - 60°) \)

\( m = -\tan 60° \)

\( m = -\sqrt{3} \)

Equation of line is given by:
\( \frac{y - y_1}{x - x_1} = m \)

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

\( y - 7 = -\sqrt{3}(x - 3) \)

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

\( \sqrt{3}x + y - 7 - 3\sqrt{3} = 0 \)

In simple words: When the angle is 180° - 60°, we get negative slope. We used this negative slope in the point-slope formula to find the line equation.

📝 Teacher's Note: Teach students that tan(180° - θ) = -tan θ. This is a key trigonometric identity for obtuse angles.

🎯 Exam Tip: Write "tan(180° - 60°) = -tan 60°" as the first step. This shows you know the identity and gets you marks.

Answer 5.
Answer:
Given: \( m = \tan 45° = 1 \)

Equation of line:
\( \frac{y - y_1}{x - x_1} = m \)

\( \frac{y - 3}{x - 8} = 1 \)

\( y - 3 = x - 8 \)

\( x - y - 5 = 0 \)

In simple words: Since tan 45° = 1, the slope is 1. This means the line goes up at a 45° angle. We used this slope with the given point.

📝 Teacher's Note: tan 45° = 1 is the easiest slope to remember. Show students that this line makes equal steps in x and y directions.

🎯 Exam Tip: Always write "tan 45° = 1" first. Then substitute directly in the point-slope formula for quick solution.

Answer 6.
Answer:
Given equation: \( 3x + 4y = 11 \)

Converting to slope form:
\( 4y = -3x + 11 \)

\( y = \frac{-3}{4}x + \frac{11}{4} \)

So, \( m = \frac{-3}{4} \)

Equation of required line:
\( \frac{y - y_1}{x - x_1} = m \)

\( \frac{y - 9}{x - 2} = \frac{-3}{4} \)

\( 4(y - 9) = -3(x - 2) \)

\( 4y - 36 = -3x + 6 \)

\( 3x + 4y = 42 \)

In simple words: We found the slope of the given line first. Then we used this same slope with the new point to get the parallel line equation.

📝 Teacher's Note: Parallel lines have the same slope. First find slope from given line, then use it with new point. This is the standard method.

🎯 Exam Tip: Always write "parallel lines have same slope" first. Then convert given line to y = mx + c form to find slope clearly.

Answer 7.
Answer:
Given equation: \( 3x + y = 9 \)

Converting to slope form:
\( y = -3x + 9 \)

So, \( m_1 = -3 \)

For perpendicular line: \( m_2 = \frac{-1}{m_1} = \frac{-1}{-3} = \frac{1}{3} \)

Equation of required line:
\( \frac{y - y_1}{x - x_1} = m_2 \)

\( \frac{y + 1}{x + 5} = \frac{1}{3} \)

\( 3(y + 1) = x + 5 \)

\( 3y + 3 = x + 5 \)

\( x - 3y + 2 = 0 \)

In simple words: Perpendicular lines have slopes that multiply to give -1. We found the opposite reciprocal of the given slope and used it with the new point.

📝 Teacher's Note: For perpendicular lines, if one slope is m, the other is -1/m. Make students practice this formula with simple examples first.

🎯 Exam Tip: Write "for perpendicular lines: m₁ × m₂ = -1" first. Then find m₂ = -1/m₁ clearly. This gets you method marks.

Answer 17.
Answer:
Let A = (-2, 6), B = (-2, 8), C = (6, 0)

Slope of AB: \( m_1 = \frac{8 - 6}{-2 - (-2)} = \frac{2}{0} = \text{undefined} \)

Since slope is undefined, AB is vertical.

Slope of BC: \( m_2 = \frac{0 - 8}{6 - (-2)} = \frac{-8}{8} = -1 \)

Slope of AC: \( m_3 = \frac{0 - 6}{6 - (-2)} = \frac{-6}{8} = \frac{-3}{4} \)

Since \( m_2 \times m_3 = (-1) \times \frac{-3}{4} = \frac{3}{4} \neq -1 \)

BC and AC are not perpendicular.

Since AB is vertical and BC has slope -1, they are perpendicular (vertical line is perpendicular to any non-vertical line).

Therefore, ABC forms a right-angled triangle with right angle at B.

In simple words: We found slopes of all three sides. When one line is vertical and another has a slope, they are always perpendicular. So this makes a right triangle.

📝 Teacher's Note: Vertical lines have undefined slope. They are perpendicular to any line with a defined slope. This is an important special case.

🎯 Exam Tip: When you get undefined slope, write "line is vertical". Then check if any other line has defined slope for perpendicularity.

Answer 18.
Answer:
Let P = (-1, 5), Q = (-1, -1), R = (1, 2), S = (-2, 1)

Slope of PQ: \( = \frac{-1 - 5}{-1 - (-1)} = \frac{-6}{0} = \text{undefined} \)

Since slope is undefined, PQ is vertical.

Slope of RS: \( = \frac{1 - 2}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3} \)

Since PQ is vertical and RS has slope \( \frac{1}{3} \), they are perpendicular.

Therefore: PQ ∥ RS is false, but PQ ⊥ RS is true.

Slope of QR: \( = \frac{2 - (-1)}{1 - (-1)} = \frac{3}{2} \)

Slope of SP: \( = \frac{5 - 1}{-1 - (-2)} = \frac{4}{1} = 4 \)

Since \( \frac{3}{2} \times 4 = 6 \neq -1 \)

QR and SP are neither parallel nor perpendicular.

In simple words: We checked all pairs of opposite sides. PQ and RS are perpendicular (not parallel). QR and SP are neither parallel nor perpendicular. So PQRS is not a special quadrilateral.

📝 Teacher's Note: For a quadrilateral to be a rectangle or rhombus, opposite sides must be parallel. Check this first before checking other properties.

🎯 Exam Tip: Calculate slopes systematically: opposite sides first, then adjacent sides. Write "parallel" when slopes are equal, "perpendicular" when product is -1.

Answer 19.
Answer:
To prove ABCD is a rhombus, we need to show opposite sides are parallel.

Let A = (0, 5), B = (1, 4), C = (5, 8), D = (4, 9)

Slope of AC: \( = \frac{8 - 5}{5 - 0} = \frac{3}{5} \)

Slope of BD: \( = \frac{9 - 4}{4 - 1} = \frac{5}{3} \)

Since \( \frac{3}{5} \times \frac{5}{3} = 1 \neq -1 \)

The diagonals AC and BD are not perpendicular.

Let me check if opposite sides are parallel:

Slope of AB: \( = \frac{4 - 5}{1 - 0} = -1 \)

Slope of DC: \( = \frac{8 - 9}{5 - 4} = -1 \)

So AB ∥ DC

Slope of AD: \( = \frac{9 - 5}{4 - 0} = 1 \)

Slope of BC: \( = \frac{8 - 4}{5 - 1} = 1 \)

So AD ∥ BC

Since both pairs of opposite sides are parallel, ABCD is a parallelogram.

In simple words: We found that opposite sides have the same slopes. This means they are parallel. When both pairs of opposite sides are parallel, we get a parallelogram.

📝 Teacher's Note: To prove a rhombus, check if it's a parallelogram first (opposite sides parallel), then check if all sides are equal or diagonals are perpendicular.

🎯 Exam Tip: Calculate slopes of opposite sides first. If they are equal, write "opposite sides are parallel". This proves it's at least a parallelogram.

Answer 20.
Answer:
Let M = (1, 7), N = (4, 3), O = (-2, -1), P = (-1, 5)

Slope of OM: \( = \frac{7 - (-1)}{1 - (-2)} = \frac{8}{3} \)

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

Product of slopes: \( \frac{8}{3} \times \frac{-2}{5} = \frac{-16}{15} \neq -1 \)

So OM and PN are not perpendicular.

Let me check all sides to see if MNOP is a square:

For a square, opposite sides must be parallel and adjacent sides must be perpendicular.

Slope of MN: \( = \frac{3 - 7}{4 - 1} = \frac{-4}{3} \)

Slope of OP: \( = \frac{5 - (-1)}{-1 - (-2)} = \frac{6}{1} = 6 \)

Since slopes are different, MN and OP are not parallel.

Therefore, MNOP is not a square.

In simple words: For a square, opposite sides must be parallel and all sides must be perpendicular to adjacent sides. We found this is not true here.

📝 Teacher's Note: For a square, check three things: opposite sides parallel, adjacent sides perpendicular, all sides equal length. If any fails, it's not a square.

🎯 Exam Tip: Write "For a square: opposite sides parallel and adjacent sides perpendicular" first. Then check these conditions systematically.

Answer 8.
Answer: Let l be the perpendicular bisector of AB

Step 1: Find slope of AB.
Slope of AB = \( \frac{-6-6}{4+2} = \frac{-12}{6} = -2 \)

Step 2: Find slope of perpendicular line.
Slope of l = \( \frac{1}{2} \) (negative reciprocal of -2)

Step 3: Find coordinates of midpoint P.
AP : PB = 1 : 1
P = \( \left(\frac{-2+4}{2}, \frac{6-6}{2}\right) = P(1,0) \)

Step 4: Find equation of line l.
Using point-slope form: \( \frac{y-y_1}{x-x_1} = \text{slope} \)
\( \frac{y-0}{x-1} = \frac{1}{2} \)
\( x - 1 = 2y \)
\( x - 2y - 1 = 0 \)

Final Answer: x - 2y - 1 = 0
In simple words: We found the middle point of line AB. Then we made a line through that point that cuts AB at 90 degrees. This is the perpendicular bisector.

📝 Teacher's Note: Show students that perpendicular slopes multiply to give -1. If one slope is -2, the perpendicular slope is 1/2. Draw this on the board to make it clear.

🎯 Exam Tip: Always find the midpoint first, then the perpendicular slope. Write the final equation in standard form ax + by + c = 0.

Answer 9.
Answer: Let MN be perpendicular bisector of AB

Step 1: Find coordinates of midpoint P.
AP : PB = 1 : 1
P = \( \left(\frac{3-1}{2}, \frac{5+7}{2}\right) = P(1,6) \)

Step 2: Find slope of AB.
Slope of AB = \( \frac{7-5}{-1-3} = \frac{2}{-4} = \frac{-1}{2} \)

Step 3: Find slope of perpendicular line MN.
Slope of MN = 2 (negative reciprocal of -1/2)

Step 4: Find equation of line MN.
Using point-slope form: \( \frac{y-y_1}{x-x_1} = \text{slope} \)
\( \frac{y-6}{x-1} = 2 \)
\( 2x - 2 = y - 6 \)
\( 2x - y + 4 = 0 \)
\( y - 2x = 4 \)

Final Answer: y - 2x = 4
In simple words: We found the middle point of AB. Then we drew a line through that point that is perpendicular (90 degrees) to AB.

📝 Teacher's Note: Remind students that if a line has slope m, the perpendicular line has slope -1/m. Use a T-square or set square to show 90-degree angles.

🎯 Exam Tip: Check your answer by verifying that the slopes multiply to give -1. Also check that the midpoint satisfies your equation.

Answer 10.
Answer: Given equations:
x + 3y = 6 ... (1)
2x - 3y = 12 ... (2)

Step 1: Solve the system of equations.
Adding (1) and (2): 3x = 18
So x = 6
And y = 0

Step 2: Find point of intersection.
Point of intersection = (6, 0)

Step 3: Find slope of given line.
From 5x + 2y = 10: slope = \( \frac{-5}{2} \)

Step 4: Find slope of required line.
Slope of required line = \( \frac{-5}{2} \) (parallel lines have same slope)

Step 5: Find equation of required line.
Using point-slope form: \( \frac{y-y_1}{x-x_1} = \text{slope} \)
\( \frac{y-0}{x-6} = \frac{-5}{2} \)
\( -5x + 30 = 2y \)
\( 5x + 2y - 30 = 0 \)

Final Answer: 5x + 2y - 30 = 0
In simple words: We found where the two given lines meet. Then we drew a line through that point that has the same slope as the third line.

📝 Teacher's Note: Show students how to solve simultaneous equations by elimination. Emphasize that parallel lines never meet and have the same slope.

🎯 Exam Tip: Always solve for the intersection point first. Then use the fact that parallel lines have identical slopes. Show all working clearly.

Answer 11.
Answer: Given equations:
x + 2y + 1 = 0 ... (1)
2x - 3y = 12 ... (2)

Step 1: Rewrite equations in standard form.
(1) can be rewritten as: 2x + 4y = -2 ... (3)
(2) can be rewritten as: 2x - 3y = 12 ... (4)

Step 2: Solve the system.
Subtracting (4) from (3): 7y = -14
y = -2
x = 3
Point of intersection = (3, -2)

Step 3: Find slope of given line.
From 2x + 3y = 9: slope = \( \frac{-2}{3} \)

Step 4: Find slope of required line.
Slope of required line = \( \frac{3}{2} \) (negative reciprocal for perpendicular)

Step 5: Find equation of required line.
Using point-slope form: \( \frac{y-y_1}{x-x_1} = m \)
\( \frac{y+2}{x-3} = \frac{3}{2} \)
\( 3x - 9 = 2y + 4 \)
\( 3x - 2y = 13 \)

Final Answer: 3x - 2y = 13
In simple words: We found where the two given lines cross. Then we drew a line through that point that cuts the third line at 90 degrees.

📝 Teacher's Note: Emphasize that perpendicular lines have slopes that multiply to give -1. If one slope is -2/3, the perpendicular slope is 3/2.

🎯 Exam Tip: Double-check that your slopes multiply to -1. Also verify that the intersection point satisfies both original equations.

Answer 12.
Answer: Given equations:
\( \frac{x}{10} + \frac{y}{5} = 14 \Rightarrow x + 2y = 140 \) ... (1)
\( \frac{x}{8} + \frac{y}{6} = 15 \Rightarrow 3x + 4y = 360 \) ... (2)

Step 1: Rewrite in standard form.
(1) can be rewritten as: 2x + 4y = 280 ... (3)
(2) can be rewritten as: 3x + 4y = 360 ... (4)

Step 2: Solve the system.
Subtracting (3) from (4): x = 80
y = 30
Point of intersection = (80, 30)

Step 3: Find slope of given line.
From x - 2y = 5: slope = \( \frac{1}{2} \)

Step 4: Find slope of required line.
Slope of required line = -2 (negative reciprocal for perpendicular)

Step 5: Find equation of required line.
Using point-slope form: \( \frac{y-y_1}{x-x_1} = m \)
\( \frac{y-30}{x-80} = -2 \)
\( -2x + 160 = y - 30 \)
\( 2x + y = 190 \)

Final Answer: 2x + y = 190
In simple words: We first changed the fraction equations to simple form. Then we found where they meet and drew a perpendicular line.

📝 Teacher's Note: Show students how to clear fractions by finding the LCM. Convert \( \frac{x}{10} + \frac{y}{5} = 14 \) by multiplying through by 10.

🎯 Exam Tip: Always clear fractions first to make calculations easier. Check that perpendicular slopes multiply to -1.

Answer 13.
Answer: Given lines:
px + 5y + 7 = 0
2y = 5x - 6

Step 1: Find slopes of both lines.
Slope of px + 5y + 7 = 0 is \( \frac{-p}{5} \)
Slope of 2y = 5x - 6 is \( \frac{5}{2} \)

Step 2: Use perpendicular condition.
Since the lines are perpendicular, product of slopes = -1
\( \frac{-p}{5} \times \frac{5}{2} = -1 \)
\( \frac{-p}{2} = -1 \)
p = 2

Final Answer: p = 2
In simple words: When two lines cut each other at 90 degrees, their slopes multiply to give -1. We used this rule to find p.

📝 Teacher's Note: Draw two perpendicular lines on the board and show students how their slopes are negative reciprocals. This visual helps a lot.

🎯 Exam Tip: Remember the perpendicular condition: \( m_1 \times m_2 = -1 \). Always convert equations to slope form first.

Answer 14.
Answer: Given lines:
3x - 2y + 4 = 0
3x + my + 6 = 0

Step 1: Find slopes of both lines.
Slope of 3x - 2y + 4 = 0 is \( \frac{3}{2} \)
Slope of 3x + my + 6 = 0 is \( \frac{-3}{m} \)

Step 2: Use parallel condition.
Since the lines are parallel, their slopes must be equal
\( \frac{3}{2} = \frac{-3}{m} \)
\( 3m = -6 \)
m = -2

Final Answer: m = -2
In simple words: Parallel lines never meet and have the same slope. We made both slopes equal to find m.

📝 Teacher's Note: Show students railway tracks as an example of parallel lines. They have the same slope and never meet.

🎯 Exam Tip: For parallel lines, slopes are equal: \( m_1 = m_2 \). For perpendicular lines, \( m_1 \times m_2 = -1 \).

Answer 15.
Answer: Given lines:
py = 2x + 5
qx + 3y = 2

Step 1: Find slopes of both lines.
Slope of py = 2x + 5 is \( \frac{2}{p} \)
Slope of qx + 3y = 2 is \( \frac{-q}{3} \)

Step 2: Use parallel condition.
Since the lines are parallel, their slopes must be equal
\( \frac{2}{p} = \frac{-q}{3} \)
pq = -6

Final Answer: pq = -6
In simple words: We made the slopes of both lines equal because parallel lines have the same slope. This gave us the relationship between p and q.

📝 Teacher's Note: Remind students to be careful with signs when finding slopes from ax + by + c = 0 form. The slope is -a/b.

🎯 Exam Tip: Always write the slope formula clearly: for ax + by + c = 0, slope = -a/b. Check your signs carefully.

Answer 16.
Answer: Given lines:
ay = 2x + 4
4y + bx = 2

Step 1: Find slopes of both lines.
Slope of ay = 2x + 4 is \( \frac{2}{a} \)
Slope of 4y + bx = 2 is \( \frac{-b}{4} \)

Step 2: Use perpendicular condition.
Since the lines are perpendicular, product of slopes = -1
\( \frac{2}{a} \times \frac{-b}{4} = -1 \)
\( \frac{-2b}{4a} = -1 \)
\( \frac{b}{2a} = 1 \)
b = 2a

Final Answer: b = 2a
In simple words: When lines cut at 90 degrees, their slopes multiply to give -1. This rule gave us the relationship between a and b.

📝 Teacher's Note: Draw perpendicular lines and show that one goes up while the other goes down. Their slopes have opposite signs and are reciprocals.

🎯 Exam Tip: For perpendicular lines, always check that \( m_1 \times m_2 = -1 \). Substitute your answer back to verify.

Answer 17.
Answer: Let PS be the median of triangle PQR from P

Step 1: Find coordinates of midpoint S.
RS : SQ = 1 : 1
S = \( \left(\frac{8-4}{2}, \frac{3+7}{2}\right) = S(2,5) \)

Step 2: Find equation of median PS.
Using two-point form: \( \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \)
\( \frac{y-3}{x-5} = \frac{5-3}{2-5} \)
\( \frac{y-3}{x-5} = \frac{2}{-3} \)
\( 2x - 10 = -3y + 9 \)
\( 2x + 3y = 19 \)

Final Answer: 2x + 3y = 19
In simple words: A median goes from a corner of a triangle to the middle of the opposite side. We found the middle point of QR, then drew a line from P to that point.

📝 Teacher's Note: Draw a triangle and show students how to find the midpoint of a side. Explain that every triangle has three medians.

🎯 Exam Tip: Always find the midpoint first using the midpoint formula. Then use two-point form to get the equation.

Answer 18.
Answer: Let CE be the median of triangle ABC from C

Step 1: Find coordinates of midpoint E.
AE : EB = 1 : 1
By using mid-point formula:
E = \( \left(\frac{8-2}{2}, \frac{5+1}{2}\right) = E(3,3) \)

Step 2: Find equation of median CE.
Using two-point form: \( \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \)
\( \frac{y-3}{x-3} = \frac{4-3}{5-3} \)
\( \frac{y-3}{x-3} = \frac{1}{2} \)
x - 3 = 2y - 6
x - 2y + 3 = 0
2y = x + 3

Final Answer: 2y = x + 3
In simple words: We found the middle point E of side AB. Then we drew a line from vertex C to this middle point E. This line is the median.

📝 Teacher's Note: Show students that in any triangle, a median divides the opposite side into two equal parts. Use a ruler to measure and verify.

🎯 Exam Tip: Use the midpoint formula correctly: \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \). Then apply two-point form for the line equation.

Answer 22.

[Diagram: A line segment PQ with points P(0,4) and Q(3,7), showing the calculation of slope and equation of the line]

Answer:

(i) Finding the slope and gradient:
Slope of PQ = \( \frac{7-4}{3-0} = \frac{3}{3} = 1 \)
\( \tan \theta = 1 \)
∴ Gradient = 1

(ii) Finding the equation of PQ:
Using the slope formula: \( \frac{y-y_1}{x-x_1} = \text{slope} \)
\( \frac{y-7}{x-3} = 1 \)
\( \Rightarrow x-3 = y-7 \)
\( \Rightarrow y = x+4 \)

(iii) Finding point A where line meets x-axis:
Let A(x,0) divide PQ in the ratio k:1
Using section formula:
Coordinates of A(x,0) = \( \left(\frac{3k}{k+1}, \frac{7k+4}{k+1}\right) \)
Since A is on x-axis, y-coordinate = 0
\( \frac{7k+4}{k+1} = 0 \)
\( 7k+4 = 0 \)
\( k = -\frac{4}{7} \)

In simple words: We found how steep the line is (slope = 1), wrote its equation (y = x + 4), and found where it crosses the x-axis using the section formula.

📝 Teacher's Note: Show students that slope is rise over run. When slope is 1, the line goes up 1 unit for every 1 unit right. Use graph paper to make this clear.

🎯 Exam Tip: Always write the slope formula first. For line equation, use point-slope form. Show all steps clearly to get full marks.

Answer 25.

Answer:

The altitude through X is perpendicular to YZ.

Step 1: Find slope of YZ
Slope of YZ = \( \frac{-4-4}{7+5} = \frac{-8}{12} = -\frac{2}{3} \)
∴ m = Slope of YZ = \( -\frac{2}{3} \)

Step 2: Find slope of perpendicular line
Slope of line perpendicular to YZ will be \( -\frac{1}{m} = \frac{3}{2} \)

Step 3: Find equation of altitude through X(4,9)
This line passes through X(4,9)
Using point-slope formula:
\( y-y_1 = m(x-x_1) \)
\( y-9 = \frac{3}{2}(x-4) \)
\( \Rightarrow 2y-18 = 3x-12 \)
\( \Rightarrow 2y = 3x+6 \)

In simple words: An altitude is a line from one corner that meets the opposite side at a right angle. We found the slope of YZ, then used the rule that perpendicular lines have slopes that multiply to give -1.

📝 Teacher's Note: Remind students that if one line has slope m, a perpendicular line has slope -1/m. This is a key formula they must remember.

🎯 Exam Tip: Write "perpendicular slopes multiply to give -1" clearly. Then find the perpendicular slope and use point-slope form. Show each step for full marks.