OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Exercise 16 (B)

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(B)

Question 1. State the equation of the line which has the y-intercept
(i) 2 and slope 7;
(ii) -3 and slope -4;
(iii) -1 and is parallel to y = 5x – 7;
(iv) 2 and is inclined at 45° to the x-axis ;
(v) -5 and is equally inclined to the axes.
Answer:
(i) We are given the y-intercept \( b = 2 \) and the slope of the line \( m = 7 \). We use the slope-intercept form of a line, \( y = mx + b \).
Substituting the given values, the equation of the line is \( y = 7x + 2 \). This form directly shows where the line crosses the y-axis and how steep it is.
(ii) Here, the y-intercept \( b = -3 \) and the slope \( m = -4 \).
Using the slope-intercept form \( y = mx + b \), the equation of the line is \( y = -4x - 3 \). This line slopes downwards and crosses the y-axis below the origin.
(iii) The y-intercept is given as \( -1 \), so \( b = -1 \). The line is parallel to \( y = 5x - 7 \).
When comparing \( y = 5x - 7 \) with the standard form \( y = mx + b \), we find that the slope of the given line is \( m = 5 \). Since parallel lines have the same slope, the slope of our required line is also \( 5 \).
Using the slope-intercept form \( y = mx + b \), the equation of the line is \( y = 5x - 1 \). This means both lines have the same steepness but cross the y-axis at different points.
(iv) The y-intercept is given as \( 2 \), so \( b = 2 \). The required line makes an angle of \( 45^\circ \) with the positive x-axis.
The slope \( m \) of a line is given by \( \tan \theta \), where \( \theta \) is the angle it makes with the x-axis.
Here, \( \theta = 45^\circ \), so \( m = \tan 45^\circ = 1 \).
Using the slope-intercept form \( y = mx + b \), the equation of the line is \( y = x + 2 \). A slope of 1 means the line rises at a 45-degree angle.
(v) The y-intercept is \( -5 \). The required line is equally inclined to the axes.
This means the angle it makes with the x-axis can be \( 45^\circ \) or \( 135^\circ \). Therefore, the slope \( m \) can be \( \tan 45^\circ = 1 \) or \( \tan 135^\circ = -1 \). So, \( m = \pm 1 \).
Using the slope-intercept form \( y = mx + b \), the equation of the line is \( y = \pm x - 5 \). This means there are two possible lines that fit this description, one with a positive slope and one with a negative slope, both crossing the y-axis at the same point.
In simple words: We used the formula \( y = mx + b \) to find the line's equation. \( b \) is where the line crosses the y-axis, and \( m \) is its steepness. For parallel lines, the steepness is the same. For lines making an angle, we use \( m = \tan(\text{angle}) \). If a line is equally inclined to axes, its slope can be 1 or -1.

🎯 Exam Tip: Remember the standard forms of a straight line equation: slope-intercept \( y=mx+b \), point-slope \( y-y_1 = m(x-x_1) \). Choose the correct form based on the given information. For parallel lines, slopes are equal; for perpendicular lines, slopes are negative reciprocals.

Question 2. What will be the value of m and c if the straight line y = mx + c passes through the points (3, -4) and (-1, 2)?
Answer: The given equation of the straight line is \( y = mx + c \) ...(1).
Since this line passes through the point \( (3, -4) \), we can substitute these values into the equation:
\( -4 = 3m + c \) ...(2)
The line also passes through the point \( (-1, 2) \), so substituting these values gives:
\( 2 = -m + c \) ...(3)
To find \( m \) and \( c \), we can solve these two equations simultaneously. Subtract equation (3) from equation (2):
\( (-4) - (2) = (3m + c) - (-m + c) \)
\( -6 = 4m \)
\( \implies \) \( m = \frac{-6}{4} = -\frac{3}{2} \)
Now, substitute the value of \( m \) into equation (3) to find \( c \):
\( 2 = -(-\frac{3}{2}) + c \)
\( 2 = \frac{3}{2} + c \)
\( \implies \) \( c = 2 - \frac{3}{2} = \frac{4-3}{2} = \frac{1}{2} \)
So, the values are \( m = -\frac{3}{2} \) and \( c = \frac{1}{2} \). The line crosses the y-axis at 0.5 and has a downward slope.
In simple words: We have a line's equation with unknown \( m \) (slope) and \( c \) (y-intercept). Since the line goes through two specific points, we put those points' x and y values into the equation. This gave us two simple equations. We solved these two equations together to find \( m \) and \( c \).

🎯 Exam Tip: When a line passes through multiple points, each point provides an equation. Solving the system of equations will give the values of the unknown parameters of the line. Double-check your arithmetic, especially with negative signs.

Question 3. Find the equation of the straight line through the given point P and having the given slope m if
(i) P(-4, 7), m = \( -\sqrt{3} \);
(ii) P(-1, -5), m = \( \frac{-6}{11} \).
Answer:
(i) We use the one-point form of a line equation, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the given point and \( m \) is the slope.
Given point \( P(-4, 7) \) so \( x_1 = -4, y_1 = 7 \), and slope \( m = -\sqrt{3} \).
Substitute these values into the formula:
\( y - 7 = -\sqrt{3}(x - (-4)) \)
\( y - 7 = -\sqrt{3}(x + 4) \)
\( y - 7 = -\sqrt{3}x - 4\sqrt{3} \)
\( \implies \) \( \sqrt{3}x + y = 7 - 4\sqrt{3} \). This line has a negative slope, meaning it goes down from left to right.
(ii) Again, using the one-point form \( y - y_1 = m(x - x_1) \).
Given point \( P(-1, -5) \) so \( x_1 = -1, y_1 = -5 \), and slope \( m = \frac{-6}{11} \).
Substitute these values into the formula:
\( y - (-5) = \frac{-6}{11}(x - (-1)) \)
\( y + 5 = \frac{-6}{11}(x + 1) \)
Now, multiply both sides by 11 to clear the fraction:
\( 11(y + 5) = -6(x + 1) \)
\( 11y + 55 = -6x - 6 \)
\( \implies \) \( 6x + 11y + 61 = 0 \). This is the general form of the line's equation.
In simple words: We used a special formula for a line when we know one point it goes through and its slope. We just put the x and y of the point and the slope number into the formula to get the line's equation.

🎯 Exam Tip: The point-slope form \( y - y_1 = m(x - x_1) \) is incredibly useful when you have a point and a slope. Remember to handle negative signs carefully when substituting coordinates. Always simplify the equation to its standard form like \( Ax + By + C = 0 \).

Question 4. Find the equation of the line through the point (1, -2) making an angle of 135° with the x-axis.
Answer: The line passes through the point \( (1, -2) \). It makes an angle of \( 135^\circ \) with the x-axis.
The slope \( m \) of a line is found using the formula \( m = \tan \theta \), where \( \theta \) is the angle with the positive x-axis.
Here, \( \theta = 135^\circ \).
So, \( m = \tan 135^\circ \). We know \( \tan 135^\circ = \tan (90^\circ + 45^\circ) = -\cot 45^\circ = -1 \).
Therefore, the slope of the line is \( m = -1 \).
Now we have a point \( (1, -2) \) and a slope \( m = -1 \). We can use the point-slope form \( y - y_1 = m(x - x_1) \).
Substitute the values:
\( y - (-2) = -1(x - 1) \)
\( y + 2 = -x + 1 \)
Rearrange the equation to the general form:
\( \implies \) \( x + y + 1 = 0 \). This line has a negative slope and passes through the point (1, -2).
In simple words: First, we found how steep the line is by using the angle it makes with the x-axis. The slope is \( \tan(135^\circ) \), which is -1. Then, we used the point-slope formula with this steepness and the given point to find the line's equation.

🎯 Exam Tip: When given an angle, calculate the slope using \( m = \tan \theta \). Be careful with angles in different quadrants, as the tangent function can be negative. Always remember the relationships between angles and trigonometric values.

Question 5. Find the equation to the straight line passes through
(i) the origin and perpendicular to x + 2y = 4;
(ii) the point (4, 3) and parallel to 3x + 4y = 12;
(iii) the point (4, 5) and (a) parallel to, (b) perpendicular to 3x – 2y + 5 = 0.
Answer:
(i) The given line is \( x + 2y = 4 \) ...(1).
To find its slope, we rewrite it in slope-intercept form \( y = mx + b \):
\( 2y = -x + 4 \)
\( y = -\frac{1}{2}x + 2 \)
The slope of the given line (1) is \( m_1 = -\frac{1}{2} \).
The required line is perpendicular to line (1). If two lines are perpendicular, the product of their slopes is -1. So, the slope of the required line \( m_2 \) is:
\( m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{1}{2}} = 2 \).
The required line passes through the origin, which is \( (0, 0) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 0 = 2(x - 0) \)
\( \implies \) \( y = 2x \). This line goes through the center of the coordinate system.
(ii) The given line is \( 3x + 4y = 12 \) ...(1).
To find its slope, rewrite it in slope-intercept form \( y = mx + b \):
\( 4y = -3x + 12 \)
\( y = -\frac{3}{4}x + 3 \)
The slope of the given line (1) is \( m_1 = -\frac{3}{4} \).
The required line is parallel to line (1). Parallel lines have the same slope. So, the slope of the required line is \( m_2 = -\frac{3}{4} \).
The required line passes through the point \( (4, 3) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 3 = -\frac{3}{4}(x - 4) \)
Multiply both sides by 4 to clear the fraction:
\( 4(y - 3) = -3(x - 4) \)
\( 4y - 12 = -3x + 12 \)
Rearrange to the general form:
\( \implies \) \( 3x + 4y - 24 = 0 \). This line is equally steep as the first one but shifted to pass through (4,3).
(iii) The given line is \( 3x - 2y + 5 = 0 \) ...(1).
To find its slope, rewrite it in slope-intercept form \( y = mx + b \):
\( -2y = -3x - 5 \)
\( y = \frac{3}{2}x + \frac{5}{2} \)
The slope of the given line (1) is \( m_1 = \frac{3}{2} \).
(a) The required line is parallel to line (1). So, its slope is \( m_2 = \frac{3}{2} \).
It passes through the point \( (4, 5) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 5 = \frac{3}{2}(x - 4) \)
Multiply both sides by 2:
\( 2(y - 5) = 3(x - 4) \)
\( 2y - 10 = 3x - 12 \)
Rearrange to the general form:
\( \implies \) \( 3x - 2y - 2 = 0 \). This line maintains the same steepness.
(b) The required line is perpendicular to line (1). So, its slope \( m_2 \) is:
\( m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \).
It passes through the point \( (4, 5) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 5 = -\frac{2}{3}(x - 4) \)
Multiply both sides by 3:
\( 3(y - 5) = -2(x - 4) \)
\( 3y - 15 = -2x + 8 \)
Rearrange to the general form:
\( \implies \) \( 2x + 3y - 23 = 0 \). This line cuts the first line at a 90-degree angle.
In simple words: For each part, we first found the slope of the given line. If the new line needed to be parallel, it kept the same slope. If it needed to be perpendicular, its slope was the "negative flip" of the first line's slope. Then, we used the point-slope formula with the new slope and the given point to write the equation of the required line.

🎯 Exam Tip: Always convert the given line equation to slope-intercept form \( y=mx+b \) to easily identify its slope. Remember that parallel lines have equal slopes (\( m_1=m_2 \)), and perpendicular lines have slopes that are negative reciprocals (\( m_1 \cdot m_2 = -1 \)).

Question 6. Find the equation to the line which is perpendicular to the line \( \frac{x}{a} - \frac{y}{b} = 1 \) at the point where it meets the x-axis.
Answer: The given equation of the line is \( \frac{x}{a} - \frac{y}{b} = 1 \) ...(1).
First, find the point where this line meets the x-axis. A line meets the x-axis when \( y = 0 \).
Substitute \( y = 0 \) into equation (1):
\( \frac{x}{a} - \frac{0}{b} = 1 \)
\( \frac{x}{a} = 1 \)
\( \implies \) \( x = a \).
So, the line meets the x-axis at the point \( (a, 0) \). The required line must pass through this point.
Next, find the slope of the given line (1). Rewrite the equation in slope-intercept form \( y = mx + c \):
\( -\frac{y}{b} = 1 - \frac{x}{a} \)
\( \frac{y}{b} = \frac{x}{a} - 1 \)
\( y = \frac{b}{a}x - b \)
The slope of the given line (1) is \( m_1 = \frac{b}{a} \).
The required line is perpendicular to line (1). So, the slope of the required line \( m_2 \) is:
\( m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{b}{a}} = -\frac{a}{b} \).
Now, we have the point \( (a, 0) \) and the slope \( m_2 = -\frac{a}{b} \). Use the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 0 = -\frac{a}{b}(x - a) \)
Multiply both sides by \( b \):
\( by = -a(x - a) \)
\( by = -ax + a^2 \)
Rearrange to the general form:
\( \implies \) \( ax + by = a^2 \). This equation represents the line that is perpendicular to the given line and passes through its x-intercept.
In simple words: We first found the spot where the given line crosses the x-axis by setting \( y \) to zero. This gave us a point \( (a, 0) \). Then, we found the steepness (slope) of the given line. Because our new line is perpendicular, its slope is the negative inverse of the first line's slope. Finally, using this new slope and the point \( (a, 0) \), we wrote down the equation for the required line.

🎯 Exam Tip: To find the x-intercept, set \( y=0 \). To find the y-intercept, set \( x=0 \). Perpendicular lines have slopes \( m_1 \) and \( m_2 \) such that \( m_1 m_2 = -1 \). Always convert to slope-intercept form to easily find the slope.

Question 7. Find the equation to the two lines through the point (4, 5) which make an acute angle of 45° with the line 2x – y + 7 = 0.
Answer: The given line is \( 2x - y + 7 = 0 \).
First, find the slope of this given line. Rewrite it in slope-intercept form \( y = mx + b \):
\( y = 2x + 7 \).
So, the slope of the given line is \( m_1 = 2 \).
Let the slope of the required lines be \( m \).
The required lines make an angle of \( 45^\circ \) with the given line. The formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m \) is:
\( \tan \theta = \left| \frac{m_1 - m}{1 + m_1 m} \right| \)
We are given \( \theta = 45^\circ \) and \( m_1 = 2 \).
\( \tan 45^\circ = 1 \).
So, \( 1 = \left| \frac{2 - m}{1 + 2m} \right| \)
This means \( \frac{2 - m}{1 + 2m} = 1 \) or \( \frac{2 - m}{1 + 2m} = -1 \).

**Case 1: Taking the positive sign**
\( \frac{2 - m}{1 + 2m} = 1 \)
\( 2 - m = 1 + 2m \)
\( 1 = 3m \)
\( \implies \) \( m = \frac{1}{3} \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (4, 5) \) and slope \( m = \frac{1}{3} \):
\( y - 5 = \frac{1}{3}(x - 4) \)
\( 3(y - 5) = x - 4 \)
\( 3y - 15 = x - 4 \)
\( \implies \) \( x - 3y + 11 = 0 \). This is the equation of the first line.

**Case 2: Taking the negative sign**
\( \frac{2 - m}{1 + 2m} = -1 \)
\( 2 - m = -(1 + 2m) \)
\( 2 - m = -1 - 2m \)
\( 3 = -m \)
\( \implies \) \( m = -3 \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (4, 5) \) and slope \( m = -3 \):
\( y - 5 = -3(x - 4) \)
\( y - 5 = -3x + 12 \)
Rearrange to the general form:
\( \implies \) \( 3x + y - 17 = 0 \). This is the equation of the second line.
So, there are two possible lines, one with a positive slope and one with a negative slope, that pass through (4,5) and make a 45-degree angle with the given line.
In simple words: First, we found the slope of the given line. Then, we used a special formula that relates the slopes of two lines to the angle between them. Since we knew the angle (45 degrees) and one slope, we could find two possible slopes for our new lines. Finally, we used these two slopes and the given point (4, 5) to write the equations for the two lines.

🎯 Exam Tip: When dealing with angles between lines, remember the formula \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \). The absolute value indicates that you should consider both positive and negative cases for \( \frac{m_1 - m_2}{1 + m_1 m_2} \) to find all possible lines, as an angle can be formed in two directions.

Question 8. The line through A(4, 7) with gradient m meets the x-axis at P and the y-axis at R. The line through B(8, 3) with gradient \( \frac{-1}{m} \) meets the x-axis at Q and the y-axis at S. Find in terms of m, the coordinates of P, Q, R and S. Obtain expressions for OP.OQ and OR.OS, where O is the point (0, 0).
Answer:
**1. Equation of the line through A(4, 7) with gradient \( m \):**
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 7 = m(x - 4) \) ...(1)

**Coordinates of P (x-intercept of line (1)):**
Line (1) meets the x-axis at \( y = 0 \).
\( 0 - 7 = m(x - 4) \)
\( -7 = mx - 4m \)
\( 4m - 7 = mx \)
\( x = \frac{4m - 7}{m} \).
So, \( P = \left(\frac{4m - 7}{m}, 0\right) \).

**Coordinates of R (y-intercept of line (1)):**
Line (1) meets the y-axis at \( x = 0 \).
\( y - 7 = m(0 - 4) \)
\( y - 7 = -4m \)
\( y = 7 - 4m \).
So, \( R = (0, 7 - 4m) \).

**2. Equation of the line through B(8, 3) with gradient \( -\frac{1}{m} \):**
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 3 = -\frac{1}{m}(x - 8) \) ...(2)

**Coordinates of Q (x-intercept of line (2)):**
Line (2) meets the x-axis at \( y = 0 \).
\( 0 - 3 = -\frac{1}{m}(x - 8) \)
\( -3m = -(x - 8) \)
\( 3m = x - 8 \)
\( x = 8 + 3m \).
So, \( Q = (8 + 3m, 0) \).

**Coordinates of S (y-intercept of line (2)):**
Line (2) meets the y-axis at \( x = 0 \).
\( y - 3 = -\frac{1}{m}(0 - 8) \)
\( y - 3 = \frac{8}{m} \)
\( y = 3 + \frac{8}{m} = \frac{3m + 8}{m} \).
So, \( S = \left(0, \frac{3m + 8}{m}\right) \).

**3. Expressions for OP.OQ and OR.OS:**
The origin O is \( (0, 0) \).
\( OP \) is the distance from \( O(0, 0) \) to \( P\left(\frac{4m - 7}{m}, 0\right) \). Since P is on the x-axis, \( OP = \left| \frac{4m - 7}{m} \right| \).
\( OQ \) is the distance from \( O(0, 0) \) to \( Q(8 + 3m, 0) \). Since Q is on the x-axis, \( OQ = \left| 8 + 3m \right| \).
Therefore, \( OP \cdot OQ = \left| \frac{4m - 7}{m} \right| \cdot \left| 8 + 3m \right| = \left| \frac{(4m - 7)(8 + 3m)}{m} \right| \).

\( OR \) is the distance from \( O(0, 0) \) to \( R(0, 7 - 4m) \). Since R is on the y-axis, \( OR = \left| 7 - 4m \right| \).
\( OS \) is the distance from \( O(0, 0) \) to \( S\left(0, \frac{3m + 8}{m}\right) \). Since S is on the y-axis, \( OS = \left| \frac{3m + 8}{m} \right| \).
Therefore, \( OR \cdot OS = \left| 7 - 4m \right| \cdot \left| \frac{3m + 8}{m} \right| = \left| \frac{(7 - 4m)(3m + 8)}{m} \right| \).
This problem helps to understand how the slope influences the intercepts of a line.
In simple words: We found where the first line crosses the x-axis (point P) and the y-axis (point R). We did the same for the second line to find points Q and S. Then, we calculated the distances from the origin (O) to P, Q, R, and S. Finally, we multiplied the x-distances (OP and OQ) and the y-distances (OR and OS) together. All answers are shown using \( m \), which is the slope of the first line.

🎯 Exam Tip: To find x-intercepts, set \( y=0 \). To find y-intercepts, set \( x=0 \). Remember that distance from the origin is the absolute value of the coordinate. Pay close attention to algebraic simplification and the use of the absolute value when dealing with distances.

Question 9. Write down the slopes of the lines
(i) joining P(1, 1) and Q(2, 3);
(ii) joining L(-p, q) and M(r, s);
(iii) parallel to the line joining A(-1, 5) and B(-6, -7);
(iv) perpendicular to the line joining B(2, -3) and S(-4, 1).
Answer:
The slope of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\( m = \frac{y_2 - y_1}{x_2 - x_1} \).

(i) For points P(1, 1) and Q(2, 3):
\( x_1 = 1, y_1 = 1 \) and \( x_2 = 2, y_2 = 3 \).
Slope \( m = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \). This line rises at a steep angle.

(ii) For points L(-p, q) and M(r, s):
\( x_1 = -p, y_1 = q \) and \( x_2 = r, y_2 = s \).
Slope \( m = \frac{s - q}{r - (-p)} = \frac{s - q}{r + p} \). This slope is expressed in terms of the given variables.

(iii) First, find the slope of the line joining A(-1, 5) and B(-6, -7):
\( x_1 = -1, y_1 = 5 \) and \( x_2 = -6, y_2 = -7 \).
Slope \( m_{AB} = \frac{-7 - 5}{-6 - (-1)} = \frac{-12}{-6 + 1} = \frac{-12}{-5} = \frac{12}{5} \).
Since the required line is parallel to line AB, its slope is the same:
Slope \( m_{parallel} = \frac{12}{5} \). Parallel lines always have the same steepness.

(iv) First, find the slope of the line joining B(2, -3) and S(-4, 1):
\( x_1 = 2, y_1 = -3 \) and \( x_2 = -4, y_2 = 1 \).
Slope \( m_{BS} = \frac{1 - (-3)}{-4 - 2} = \frac{1 + 3}{-6} = \frac{4}{-6} = -\frac{2}{3} \).
Since the required line is perpendicular to line BS, its slope is the negative reciprocal:
Slope \( m_{perpendicular} = -\frac{1}{m_{BS}} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \). A perpendicular line will cross the first line at a 90-degree angle.
In simple words: To find the slope (steepness) of a line between two points, we use a formula: difference in y-coordinates divided by difference in x-coordinates. For parallel lines, their slopes are exactly the same. For lines that are perpendicular (at right angles), the slope of one is the negative flip of the other.

🎯 Exam Tip: Master the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Be careful with negative signs when substituting coordinates. Remember the conditions for parallel (\( m_1=m_2 \)) and perpendicular (\( m_1 m_2 = -1 \)) lines to find unknown slopes.

Question 10. Find the equations of the lines joining the points
(i) A(1, 1) and B(2, 3);
(ii) L(a, b) and M(b, a);
(iii) P(3, 3) and Q(7, 6).
Answer:
To find the equation of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \), we use the two-point form:
\( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \).

(i) For points A(1, 1) and B(2, 3):
\( x_1 = 1, y_1 = 1 \) and \( x_2 = 2, y_2 = 3 \).
First, find the slope: \( m = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \).
Now, use the point-slope form with point A(1, 1) and slope \( m = 2 \):
\( y - 1 = 2(x - 1) \)
\( y - 1 = 2x - 2 \)
Rearrange to the general form:
\( \implies \) \( 2x - y - 1 = 0 \). This line passes through both points A and B.

(ii) For points L(a, b) and M(b, a):
\( x_1 = a, y_1 = b \) and \( x_2 = b, y_2 = a \).
First, find the slope: \( m = \frac{a - b}{b - a} \).
Since \( a - b = -(b - a) \), if \( a \neq b \), then \( m = \frac{-(b - a)}{b - a} = -1 \).
If \( a = b \), the points are the same, and it's not a unique line, but if they are distinct, \( m = -1 \).
Now, use the point-slope form with point L(a, b) and slope \( m = -1 \):
\( y - b = -1(x - a) \)
\( y - b = -x + a \)
Rearrange to the general form:
\( \implies \) \( x + y - a - b = 0 \) or \( x + y = a + b \). This line always has a slope of -1.

(iii) For points P(3, 3) and Q(7, 6):
\( x_1 = 3, y_1 = 3 \) and \( x_2 = 7, y_2 = 6 \).
First, find the slope: \( m = \frac{6 - 3}{7 - 3} = \frac{3}{4} \).
Now, use the point-slope form with point P(3, 3) and slope \( m = \frac{3}{4} \):
\( y - 3 = \frac{3}{4}(x - 3) \)
Multiply both sides by 4 to clear the fraction:
\( 4(y - 3) = 3(x - 3) \)
\( 4y - 12 = 3x - 9 \)
Rearrange to the general form:
\( \implies \) \( 3x - 4y + 3 = 0 \). This line goes through both point P and Q.
In simple words: To find the equation of a line that passes through two given points, we first calculate its steepness (slope). We do this by dividing the difference in the y-coordinates by the difference in the x-coordinates. Then, using this slope and one of the points, we apply the point-slope formula to get the final equation of the line.

🎯 Exam Tip: The two-point form for the equation of a line is fundamental. Make sure you can derive the slope accurately and then apply it with either of the two points to obtain the line's equation. Always check your work by substituting the second point into your final equation.

Question 11. Given A(10, 4), B(-4, 9) and C(-2, -1) of △ABC, find
(i) the equation of the side AB;
(ii) the equation of the median through A;
(iii) the equation of the altitude through B;
(iv) the equation of the perpendicular bisector of the side AB.
Answer:
(i) **Equation of side AB:**
Use the two-point form with A(10, 4) and B(-4, 9):
\( x_1 = 10, y_1 = 4 \) and \( x_2 = -4, y_2 = 9 \).
\( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \)
\( y - 4 = \frac{9 - 4}{-4 - 10}(x - 10) \)
\( y - 4 = \frac{5}{-14}(x - 10) \)
Multiply both sides by -14:
\( -14(y - 4) = 5(x - 10) \)
\( -14y + 56 = 5x - 50 \)
Rearrange to the general form:
\( \implies \) \( 5x + 14y - 106 = 0 \). This line forms one side of the triangle.

(ii) **Equation of the median through A:**
A median from A connects A to the midpoint of the opposite side BC.
Let D be the midpoint of BC. Using the midpoint formula:
\( D = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right) = \left(\frac{-4 + (-2)}{2}, \frac{9 + (-1)}{2}\right) = \left(\frac{-6}{2}, \frac{8}{2}\right) = (-3, 4) \).
Now find the equation of the line joining A(10, 4) and D(-3, 4).
Notice that both points A and D have the same y-coordinate (4). This means the line is a horizontal line.
Its equation is simply \( y = 4 \). A horizontal line has a slope of zero.
\( \implies \) \( y = 4 \). This median is a flat line across the triangle.

(iii) **Equation of the altitude through B:**
An altitude from B is perpendicular to the opposite side AC.
First, find the slope of side AC using A(10, 4) and C(-2, -1):
\( x_A = 10, y_A = 4 \) and \( x_C = -2, y_C = -1 \).
Slope of AC \( m_{AC} = \frac{-1 - 4}{-2 - 10} = \frac{-5}{-12} = \frac{5}{12} \).
Since the altitude BE is perpendicular to AC, its slope \( m_{BE} \) is the negative reciprocal of \( m_{AC} \):
\( m_{BE} = -\frac{1}{m_{AC}} = -\frac{1}{\frac{5}{12}} = -\frac{12}{5} \).
Now, use the point-slope form with point B(-4, 9) and slope \( m_{BE} = -\frac{12}{5} \):
\( y - 9 = -\frac{12}{5}(x - (-4)) \)
\( y - 9 = -\frac{12}{5}(x + 4) \)
Multiply both sides by 5:
\( 5(y - 9) = -12(x + 4) \)
\( 5y - 45 = -12x - 48 \)
Rearrange to the general form:
\( \implies \) \( 12x + 5y + 3 = 0 \). This line drops from B at a right angle to side AC.

(iv) **Equation of the perpendicular bisector of side AB:**
A perpendicular bisector of AB passes through the midpoint of AB and is perpendicular to AB.
First, find the midpoint of AB, let's call it F, using A(10, 4) and B(-4, 9):
\( F = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{10 + (-4)}{2}, \frac{4 + 9}{2}\right) = \left(\frac{6}{2}, \frac{13}{2}\right) = \left(3, \frac{13}{2}\right) \).
Next, find the slope of side AB. From part (i), \( m_{AB} = \frac{5}{-14} = -\frac{5}{14} \).
Since the perpendicular bisector is perpendicular to AB, its slope \( m_{perp} \) is the negative reciprocal of \( m_{AB} \):
\( m_{perp} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{5}{14}} = \frac{14}{5} \).
Now, use the point-slope form with midpoint \( F\left(3, \frac{13}{2}\right) \) and slope \( m_{perp} = \frac{14}{5} \):
\( y - \frac{13}{2} = \frac{14}{5}(x - 3) \)
To clear fractions, multiply by 10 (LCM of 2 and 5):
\( 10y - 10\left(\frac{13}{2}\right) = 10\left(\frac{14}{5}\right)(x - 3) \)
\( 10y - 65 = 28(x - 3) \)
\( 10y - 65 = 28x - 84 \)
Rearrange to the general form:
\( \implies \) \( 28x - 10y - 19 = 0 \). This line cuts side AB exactly in half and at a right angle.
In simple words: We found different lines related to the triangle. For side AB, we used two points. For the median from A, we found the middle of side BC first, then used A and that midpoint. For the altitude from B, we found the slope of side AC, then found the perpendicular slope, and used point B. For the perpendicular bisector of AB, we found the middle of AB and the perpendicular slope to AB, then used those to get the line's equation.

🎯 Exam Tip: Break down complex geometry problems into smaller, manageable steps. Remember the definitions: a median connects a vertex to the midpoint of the opposite side; an altitude is perpendicular from a vertex to the opposite side; a perpendicular bisector passes through the midpoint and is perpendicular to the side. Apply the midpoint, slope, and point-slope formulas correctly.

Question 12. The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. AB produced cuts the y-axis at P and CB produced cuts the x-axis at Q. Find the co-ordinates of the points P and Q. Find the equation of the straight line joining the mid-points of AC and OB (where O is the origin), and verify that this line passes through the midpoint of PQ.
Answer:
**1. Coordinates of P:**
P is the point where the line AB cuts the y-axis. First, find the equation of line AB using points A(4, 0) and B(2, 2):
\( y - y_A = \frac{y_B - y_A}{x_B - x_A}(x - x_A) \)
\( y - 0 = \frac{2 - 0}{2 - 4}(x - 4) \)
\( y = \frac{2}{-2}(x - 4) \)
\( y = -1(x - 4) \)
\( y = -x + 4 \)
\( \implies \) \( x + y = 4 \) ...(1)
To find P (y-intercept), set \( x = 0 \) in equation (1):
\( 0 + y = 4 \)
\( y = 4 \).
So, the coordinates of P are \( (0, 4) \).

**2. Coordinates of Q:**
Q is the point where the line CB cuts the x-axis. First, find the equation of line CB using points C(0, 6) and B(2, 2):
\( y - y_C = \frac{y_B - y_C}{x_B - x_C}(x - x_C) \)
\( y - 6 = \frac{2 - 6}{2 - 0}(x - 0) \)
\( y - 6 = \frac{-4}{2}x \)
\( y - 6 = -2x \)
\( \implies \) \( 2x + y = 6 \) ...(2)
To find Q (x-intercept), set \( y = 0 \) in equation (2):
\( 2x + 0 = 6 \)
\( 2x = 6 \)
\( x = 3 \).
So, the coordinates of Q are \( (3, 0) \).

**3. Equation of the line joining midpoints of AC and OB:**
Let R be the midpoint of AC. Using A(4, 0) and C(0, 6):
\( R = \left(\frac{4 + 0}{2}, \frac{0 + 6}{2}\right) = \left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3) \).
Let S be the midpoint of OB (where O is the origin (0, 0)). Using O(0, 0) and B(2, 2):
\( S = \left(\frac{0 + 2}{2}, \frac{0 + 2}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1) \).
Now, find the equation of the line RS using R(2, 3) and S(1, 1):
\( y - y_R = \frac{y_S - y_R}{x_S - x_R}(x - x_R) \)
\( y - 3 = \frac{1 - 3}{1 - 2}(x - 2) \)
\( y - 3 = \frac{-2}{-1}(x - 2) \)
\( y - 3 = 2(x - 2) \)
\( y - 3 = 2x - 4 \)
\( \implies \) \( 2x - y - 1 = 0 \) ...(3). This line connects the centers of two segments.

**4. Verify that line (3) passes through the midpoint of PQ:**
Let T be the midpoint of PQ. Using P(0, 4) and Q(3, 0):
\( T = \left(\frac{0 + 3}{2}, \frac{4 + 0}{2}\right) = \left(\frac{3}{2}, 2\right) \).
Now, substitute the coordinates of T into equation (3) to see if it holds true:
\( 2x - y - 1 = 0 \)
\( 2\left(\frac{3}{2}\right) - 2 - 1 = 0 \)
\( 3 - 2 - 1 = 0 \)
\( 0 = 0 \).
Since \( 0 = 0 \), the line joining the midpoints of AC and OB indeed passes through the midpoint of PQ. This demonstrates a geometric property of these points.
In simple words: We found where the line AB crosses the y-axis (P) and where line CB crosses the x-axis (Q). Then, we found the middle points of AC and OB. We made a new line using these two midpoints. Finally, we checked if this new line also passed through the middle point of PQ, and it did!

🎯 Exam Tip: This problem combines multiple concepts: finding line equations, intercepts, and midpoints. Work systematically through each part. Remember that a line passes through a point if substituting the point's coordinates into the line's equation results in a true statement (e.g., \( 0=0 \)).

Question 13. A line through the point (3, 0) meets the variable line y = tx at right angles at the point P. Find, in terms of t, the coordinates of P. Find the value of k for which P lies on the curve \( x^2 + y^2 = kx \).
Answer:
**1. Coordinates of P:**
Let the given line be \( l_1: y = tx \) ...(1). Its slope is \( m_1 = t \).
A second line, \( l_2 \), passes through \( (3, 0) \) and meets \( l_1 \) at right angles at point P.
Since \( l_2 \) is perpendicular to \( l_1 \), its slope \( m_2 \) is the negative reciprocal of \( m_1 \):
\( m_2 = -\frac{1}{t} \).
Now, find the equation of line \( l_2 \) using the point-slope form with point \( (3, 0) \) and slope \( m_2 = -\frac{1}{t} \):
\( y - 0 = -\frac{1}{t}(x - 3) \)
Multiply both sides by \( t \):
\( ty = -1(x - 3) \)
\( ty = -x + 3 \)
\( \implies \) \( x + ty = 3 \) ...(2).
Point P is the intersection of \( l_1 \) and \( l_2 \). We solve equations (1) and (2) simultaneously.
Substitute \( y = tx \) from (1) into (2):
\( x + t(tx) = 3 \)
\( x + t^2x = 3 \)
\( x(1 + t^2) = 3 \)
\( \implies \) \( x = \frac{3}{1 + t^2} \).
Now substitute the value of \( x \) back into (1) to find \( y \):
\( y = t\left(\frac{3}{1 + t^2}\right) = \frac{3t}{1 + t^2} \).
So, the coordinates of P are \( \left(\frac{3}{1 + t^2}, \frac{3t}{1 + t^2}\right) \). These coordinates show how the position of P changes with the parameter t.

**2. Value of k for which P lies on \( x^2 + y^2 = kx \):**
Substitute the coordinates of P into the equation of the curve \( x^2 + y^2 = kx \):
\( \left(\frac{3}{1 + t^2}\right)^2 + \left(\frac{3t}{1 + t^2}\right)^2 = k\left(\frac{3}{1 + t^2}\right) \)
\( \frac{9}{(1 + t^2)^2} + \frac{9t^2}{(1 + t^2)^2} = k\left(\frac{3}{1 + t^2}\right) \)
Combine the terms on the left side:
\( \frac{9 + 9t^2}{(1 + t^2)^2} = k\left(\frac{3}{1 + t^2}\right) \)
Factor out 9 from the numerator on the left:
\( \frac{9(1 + t^2)}{(1 + t^2)^2} = k\left(\frac{3}{1 + t^2}\right) \)
Simplify the left side:
\( \frac{9}{1 + t^2} = k\left(\frac{3}{1 + t^2}\right) \)
Multiply both sides by \( (1 + t^2) \) (assuming \( 1 + t^2 \neq 0 \), which is always true for real \( t \)):
\( 9 = 3k \)
\( \implies \) \( k = 3 \).
So, the value of \( k \) is 3. This means that for any value of t, the point P always lies on the specific circle \( x^2 + y^2 = 3x \).
In simple words: First, we found the meeting point (P) of two lines. One line was \( y = tx \), and the other passed through \( (3, 0) \) and was perpendicular to the first. We used their equations to find P's x and y values in terms of \( t \). Then, we put P's coordinates into the curve equation \( x^2 + y^2 = kx \) to find what \( k \) had to be for P to always sit on that curve. We found \( k = 3 \).

🎯 Exam Tip: This question tests your ability to work with variable slopes and coordinates. Remember that the product of slopes of perpendicular lines is -1. When finding the intersection point, solve the two line equations simultaneously. When a point lies on a curve, its coordinates must satisfy the curve's equation.

Question 14. The point P is the foot of the perpendicular from A(0, t) to the line whose equation is y = tx. Determine
(i) the equation of the line AP,
(ii) the co-ordinates of P,
(iii) the area of △OAP, where O is the origin.
Answer:
**1. Equation of the line AP:**
The given line is \( l_1: y = tx \) ...(1). Its slope is \( m_1 = t \).
The line AP is perpendicular to \( l_1 \) and passes through A(0, t).
Since AP is perpendicular to \( l_1 \), its slope \( m_{AP} \) is the negative reciprocal of \( m_1 \):
\( m_{AP} = -\frac{1}{t} \).
Now, use the point-slope form for line AP with point A(0, t) and slope \( m_{AP} = -\frac{1}{t} \):
\( y - t = -\frac{1}{t}(x - 0) \)
\( y - t = -\frac{x}{t} \)
Multiply both sides by \( t \):
\( ty - t^2 = -x \)
Rearrange to the general form:
\( \implies \) \( x + ty = t^2 \) ...(2). This line connects point A to the perpendicular foot P.

**2. Coordinates of P:**
P is the foot of the perpendicular, meaning it is the intersection point of line \( l_1 \) (given line) and line AP. Solve equations (1) and (2) simultaneously.
Substitute \( y = tx \) from (1) into (2):
\( x + t(tx) = t^2 \)
\( x + t^2x = t^2 \)
\( x(1 + t^2) = t^2 \)
\( \implies \) \( x = \frac{t^2}{1 + t^2} \).
Now substitute the value of \( x \) back into (1) to find \( y \):
\( y = t\left(\frac{t^2}{1 + t^2}\right) = \frac{t^3}{1 + t^2} \).
So, the coordinates of P are \( \left(\frac{t^2}{1 + t^2}, \frac{t^3}{1 + t^2}\right) \). These coordinates change as t changes.

**3. Area of △OAP:**
O is the origin \( (0, 0) \). A is \( (0, t) \). P is \( \left(\frac{t^2}{1 + t^2}, \frac{t^3}{1 + t^2}\right) \).
We can use the determinant formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Let \( (x_1, y_1) = (0, 0) \) (O)
Let \( (x_2, y_2) = (0, t) \) (A)
Let \( (x_3, y_3) = \left(\frac{t^2}{1 + t^2}, \frac{t^3}{1 + t^2}\right) \) (P)
Area \( = \frac{1}{2} \left| 0(t - \frac{t^3}{1 + t^2}) + 0(\frac{t^3}{1 + t^2} - 0) + \frac{t^2}{1 + t^2}(0 - t) \right| \)
Area \( = \frac{1}{2} \left| 0 + 0 + \frac{t^2}{1 + t^2}(-t) \right| \)
Area \( = \frac{1}{2} \left| -\frac{t^3}{1 + t^2} \right| \)
Since \( t^2 \ge 0 \), \( 1 + t^2 > 0 \). The absolute value \( |-\frac{t^3}{1 + t^2}| = \frac{|t^3|}{1 + t^2} \).
Area \( = \frac{1}{2} \frac{|t^3|}{1 + t^2} \) square units.
Alternatively, since OAP forms a right-angled triangle at P (AP is perpendicular to OP since O is on the line y=tx), we can use base and height. The line \( y=tx \) passes through the origin. AP is perpendicular to \( y=tx \). So, \( \triangle OPA \) is a right-angled triangle with the right angle at P.
The base OP can be calculated as the distance from O to P. The height AP can be calculated as the distance from A to P.
However, a simpler method involves using O as a vertex at the origin. The base can be taken as OA on the y-axis, and the height would be the x-coordinate of P (since P is the foot of perpendicular to the y-axis).
Base \( OA = |t| \). Height \( = |x_P| = \left|\frac{t^2}{1 + t^2}\right| = \frac{t^2}{1 + t^2} \).
Area \( = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |t| \times \frac{t^2}{1 + t^2} = \frac{1}{2} \frac{|t^3|}{1 + t^2} \) square units. This area changes based on the value of t.
In simple words: First, we found the equation of line AP. This line starts at A and goes to P, touching the line \( y=tx \) at a right angle. To do this, we used the perpendicular slope and point A. Second, we found the exact spot of P by solving the equations of line AP and the given line. Third, we calculated the size of the triangle OAP, where O is the origin. We used a formula for triangle area with O, A, and P as its corners.

🎯 Exam Tip: For problems involving the "foot of the perpendicular," remember that this point is the intersection of the given line and the line perpendicular to it from the external point. The area of a triangle with one vertex at the origin can be efficiently calculated using the formula \( \frac{1}{2} |x_1 y_2 - x_2 y_1| \) if two other vertices are \( (x_1, y_1) \) and \( (x_2, y_2) \). For a general triangle, the determinant method works well.

Question 15. Find the equation of the line joining the origin to the point of intersection of 4x + 3y = 8 and x + y = 1.
Answer: First, we need to find where the two lines \( 4x + 3y = 8 \) and \( x + y = 1 \) cross each other.
Let's call the first equation (1) and the second equation (2).
Multiply equation (2) by 3: \( 3(x + y) = 3(1) \implies 3x + 3y = 3 \). Let's call this (3).
Now, subtract equation (3) from equation (1):
\( (4x + 3y) - (3x + 3y) = 8 - 3 \)
\( \implies 4x + 3y - 3x - 3y = 5 \)
\( \implies x = 5 \)
Substitute \( x = 5 \) back into equation (2):
\( 5 + y = 1 \)
\( \implies y = 1 - 5 \)
\( \implies y = -4 \)
So, the point where the two lines intersect is \( (5, -4) \).
Next, we need to find the equation of the line that connects the origin \( (0, 0) \) to this intersection point \( (5, -4) \).
We use the two-point form for the equation of a line, \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \).
Using \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (5, -4) \):
\( y - 0 = \frac{-4 - 0}{5 - 0} (x - 0) \)
\( \implies y = \frac{-4}{5} x \)
To remove the fraction, multiply both sides by 5:
\( 5y = -4x \)
\( \implies 4x + 5y = 0 \)
This is the required equation of the line. This equation represents a direct relationship between x and y, passing through the central point of the graph.
In simple words: First, find the point where the two given lines meet by solving their equations. Then, find the equation of a new line that goes through the starting point (origin) and that meeting point.

🎯 Exam Tip: Always double-check your calculations for the intersection point. A small error there will lead to an incorrect final line equation. Remember the origin is always (0,0).

Question 16. Find the equation to the straight line which passes through the point of intersection of the lines 3x + 4y – 1 = 0 and 5x + 8y – 3 = 0 and is perpendicular to the line 4x – 2y + 3 = 0.
Answer: We need to find the equation of a line that meets two conditions: it passes through a specific point and has a certain slope.
First, let's find the point where the lines \( 3x + 4y - 1 = 0 \) (1) and \( 5x + 8y - 3 = 0 \) (2) intersect.
To solve these equations simultaneously, we can multiply equation (1) by 2:
\( 2(3x + 4y - 1) = 2(0) \implies 6x + 8y - 2 = 0 \) (3)
Now, subtract equation (3) from equation (2):
\( (5x + 8y - 3) - (6x + 8y - 2) = 0 - 0 \)
\( \implies 5x + 8y - 3 - 6x - 8y + 2 = 0 \)
\( \implies -x - 1 = 0 \)
\( \implies -x = 1 \)
\( \implies x = -1 \)
Substitute \( x = -1 \) into equation (1):
\( 3(-1) + 4y - 1 = 0 \)
\( \implies -3 + 4y - 1 = 0 \)
\( \implies 4y - 4 = 0 \)
\( \implies 4y = 4 \)
\( \implies y = 1 \)
So, the point of intersection is \( P(-1, 1) \).
Next, let's find the slope of the line \( 4x - 2y + 3 = 0 \).
Rearrange it into the slope-intercept form \( y = mx + c \):
\( -2y = -4x - 3 \)
\( \implies y = \frac{-4x - 3}{-2} \)
\( \implies y = 2x + \frac{3}{2} \)
The slope of this line is \( m_1 = 2 \).
The required line is perpendicular to this line. If two lines are perpendicular, the product of their slopes is -1.
So, the slope of the required line \( m_2 \) is:
\( m_2 = -\frac{1}{m_1} = -\frac{1}{2} \)
Finally, we use the point-slope form \( y - y_1 = m(x - x_1) \) with the intersection point \( (-1, 1) \) and the required slope \( -\frac{1}{2} \).
\( y - 1 = -\frac{1}{2}(x - (-1)) \)
\( \implies y - 1 = -\frac{1}{2}(x + 1) \)
Multiply both sides by 2:
\( 2(y - 1) = -(x + 1) \)
\( \implies 2y - 2 = -x - 1 \)
Move all terms to one side:
\( x + 2y - 2 + 1 = 0 \)
\( \implies x + 2y - 1 = 0 \)
This is the equation of the line that satisfies both conditions. The steps systematically break down a complex problem into simpler, manageable parts.
In simple words: Find where the first two lines cross. Then, find the slope of the third line and figure out the slope of a line that is perfectly straight across from it. Use these two pieces of information to write the equation of the new line.

🎯 Exam Tip: Pay close attention to negative signs when solving simultaneous equations and when calculating perpendicular slopes. A small arithmetic error can change the entire equation.