OP Malhotra Class 10 Maths Solutions Chapter 11 Coordinate Geometry Exercise 11 (C)

 

Question 1. State the slope of the line
(i) \( 3x + 4y = 9 \)
(ii) \( 3x + 2y - 4 \)
Answer:
(i) The given equation is \( 3x + 4y = 9 \).
To find the slope, we convert this into the slope-intercept form, \( y = mx + c \).
First, rearrange the equation to isolate \( 4y \):
\( 4y = -3x + 9 \)
Now, divide both sides by 4:
\( y = \frac{-3}{4}x + \frac{9}{4} \)
Comparing this with \( y = mx + c \), we see that the slope \( m \) is \( \frac{-3}{4} \). The y-intercept is \( \frac{9}{4} \).
(ii) The given equation is \( 3x + 2y - 4 = 0 \).
We convert this to the slope-intercept form, \( y = mx + c \).
First, isolate \( 2y \):
\( 2y = -3x + 4 \)
Now, divide both sides by 2:
\( y = \frac{-3}{2}x + \frac{4}{2} \)
\( y = \frac{-3}{2}x + 2 \)
Comparing this with \( y = mx + c \), the slope \( m \) is \( \frac{-3}{2} \). The y-intercept is 2.
In simple words: To find the slope, change the equation to the form where 'y' is by itself. The number next to 'x' is the slope.

🎯 Exam Tip: Always convert the linear equation to the form \( y = mx + c \) to easily identify the slope (\( m \)) and y-intercept (\( c \)). Remember to handle signs carefully when moving terms across the equals sign.

 

Question 2. Given \( 3x + 2y + 4 = 0 \)
(i) Express the equation in the form \( y = mx + c \).
(ii) Find the slope and the y-intercept of the line \( 3x + 2y + 4 = 0 \).
(iii) Use your answer to (ii) above and on a graph paper draw the graph of the straight line \( 3x + 2y + 4 = 0 \).
Answer:
(i) The given equation is \( 3x + 2y + 4 = 0 \).
To express it in the form \( y = mx + c \), we need to isolate \( y \).
First, move the \( 3x \) and \( 4 \) terms to the right side:
\( 2y = -3x - 4 \)
Next, divide the entire equation by 2:
\( y = \frac{-3}{2}x - \frac{4}{2} \)
\( y = \frac{-3}{2}x - 2 \)
This is the equation in the form \( y = mx + c \).
(ii) From the equation \( y = \frac{-3}{2}x - 2 \), by comparing with \( y = mx + c \):
The slope \( (m) \) is \( \frac{-3}{2} \).
The y-intercept \( (c) \) is \( -2 \).
(iii) To draw the graph, we use the y-intercept and the slope. The y-intercept is -2, meaning the line crosses the y-axis at the point \( (0, -2) \). The slope is \( \frac{-3}{2} \), which means for every 2 units moved to the right on the x-axis, the line goes down 3 units on the y-axis. Another point can be found by starting from \( (0, -2) \) and moving 2 units right to \( (2, -2) \), then 3 units down to \( (2, -5) \). Connecting \( (0, -2) \) and \( (2, -5) \) will draw the line. Alternatively, we can find the x-intercept by setting \( y = 0 \): \( 3x + 2(0) + 4 = 0 \implies 3x + 4 = 0 \implies 3x = -4 \implies x = -\frac{4}{3} \approx -1.33 \). The graph shows the line passing through \( (0, -2) \) and \( (- \frac{4}{3}, 0) \).
In simple words: First, rewrite the equation so that 'y' is alone on one side. This shows you the slope (number with 'x') and where the line crosses the y-axis (the constant number). Then, use these two points to draw the line on a graph.

🎯 Exam Tip: When drawing a graph, plotting the y-intercept (where \( x = 0 \)) and the x-intercept (where \( y = 0 \)) often provides the easiest two points to accurately draw the line. Always label your axes.

 

Question 3. State the equation of the line which has the y-intercept
(a) 2 and a slope 7 ;
(b) - 3 and a slope – 4 ;
(c) – 1 and is parallel to \( y = 5x – 7 \) ;
(d) 2 and is inclined at \( 45^\circ \) to the x-axis ;
(e) – 5 and is equally inclined to the axes.
Answer:
We know that the equation of a line is generally given by \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept.
(a) Given y-intercept \( (c) = 2 \) and slope \( (m) = 7 \).
Substitute these values into the equation:
\( y = 7x + 2 \)
Rearranging this to the general form:
\( 7x - y + 2 = 0 \)
(b) Given y-intercept \( (c) = -3 \) and slope \( (m) = -4 \).
Substitute these values into the equation:
\( y = -4x - 3 \)
Rearranging this to the general form:
\( 4x + y + 3 = 0 \)
(c) Given y-intercept \( (c) = -1 \). The line is parallel to \( y = 5x - 7 \).
Parallel lines have the same slope. The slope of \( y = 5x - 7 \) is \( 5 \).
So, the slope of our required line \( (m) = 5 \).
Substitute \( m = 5 \) and \( c = -1 \) into \( y = mx + c \):
\( y = 5x - 1 \)
Rearranging this to the general form:
\( 5x - y - 1 = 0 \)
(d) Given y-intercept \( (c) = 2 \). The line is inclined at \( 45^\circ \) to the x-axis.
The slope \( (m) \) of a line inclined at an angle \( \theta \) to the x-axis is \( \tan \theta \).
So, \( m = \tan 45^\circ = 1 \).
Substitute \( m = 1 \) and \( c = 2 \) into \( y = mx + c \):
\( y = 1x + 2 \)
\( y = x + 2 \)
Rearranging this to the general form:
\( x - y + 2 = 0 \)
(e) Given y-intercept \( (c) = -5 \). The line is equally inclined to the axes.
A line equally inclined to the axes forms an angle of \( 45^\circ \) or \( 135^\circ \) with the x-axis. Thus, its slope can be \( \tan 45^\circ = 1 \) or \( \tan 135^\circ = -1 \).
Case 1: Slope \( (m) = 1 \).
Substitute \( m = 1 \) and \( c = -5 \) into \( y = mx + c \):
\( y = 1x - 5 \)
\( y = x - 5 \)
Rearranging this to the general form:
\( x - y - 5 = 0 \)
Case 2: Slope \( (m) = -1 \).
Substitute \( m = -1 \) and \( c = -5 \) into \( y = mx + c \):
\( y = -1x - 5 \)
\( y = -x - 5 \)
Rearranging this to the general form:
\( x + y + 5 = 0 \)
In simple words: The basic rule for a line is \( y = mx + c \). 'm' is how steep the line is, and 'c' is where it crosses the 'y' line. Just put in the given numbers for 'm' and 'c' to get your equation. If the line is parallel, it has the same steepness. If it's equally tilted, it's either going up or down at a \( 45^\circ \) angle.

🎯 Exam Tip: Remember that parallel lines have equal slopes, and the slope of a line inclined at \( \theta \) to the x-axis is \( \tan \theta \). A line "equally inclined to the axes" means its slope is \( \pm 1 \).

 

Question 4. 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 equation of the line is \( y = mx + c \). Let's call this Equation (i).
The line passes through two points: \( (3, -4) \) and \( (-1, 2) \).
We can substitute the x and y values of each point into Equation (i) to get two new equations.
Using point \( (3, -4) \):
\( -4 = m(3) + c \)
\( -4 = 3m + c \) (Equation (ii))
Using point \( (-1, 2) \):
\( 2 = m(-1) + c \)
\( 2 = -m + c \) (Equation (iii))
Now we have a system of two linear equations with two variables, \( m \) and \( c \).
From Equation (iii), we can express \( c \) in terms of \( m \):
\( c = 2 + m \)
Substitute this expression for \( c \) into Equation (ii):
\( -4 = 3m + (2 + m) \)
\( -4 = 4m + 2 \)
Subtract 2 from both sides:
\( -4 - 2 = 4m \)
\( -6 = 4m \)
Divide by 4 to find \( m \):
\( m = \frac{-6}{4} \)
\( m = \frac{-3}{2} \)
Now, substitute the value of \( m \) back into the expression for \( c \):
\( c = 2 + m \)
\( c = 2 + \left(\frac{-3}{2}\right) \)
\( c = \frac{4}{2} - \frac{3}{2} \)
\( c = \frac{1}{2} \)
So, the values are \( m = \frac{-3}{2} \) and \( c = \frac{1}{2} \).
In simple words: When a line passes through two points, you can use each point's coordinates in the line's formula to make two separate equations. Then, solve these two equations to find the values of 'm' (slope) and 'c' (y-intercept). This tells you the specific equation for that line.

🎯 Exam Tip: For problems involving two points and the general line equation \( y = mx + c \), always set up two simultaneous equations by substituting each point. Solving these equations is a standard method to find \( m \) and \( c \).

 

Question 5. The graph of the equation \( y = mx + c \) passes through the points \( (1, 4) \) and \( (-2, -5) \). Determine the values of m and c.
Answer:
The equation of the line is given as \( y = mx + c \).
The line passes through point \( (1, 4) \). Substituting these coordinates:
\( 4 = m(1) + c \)
\( 4 = m + c \) (Equation i)
The line also passes through point \( (-2, -5) \). Substituting these coordinates:
\( -5 = m(-2) + c \)
\( -5 = -2m + c \) (Equation ii)
Now we have a system of two linear equations:
1. \( m + c = 4 \)
2. \( -2m + c = -5 \)
To solve for \( m \) and \( c \), subtract Equation (ii) from Equation (i):
\( (m + c) - (-2m + c) = 4 - (-5) \)
\( m + c + 2m - c = 4 + 5 \)
\( 3m = 9 \)
Divide by 3:
\( m = \frac{9}{3} \)
\( m = 3 \)
Now substitute \( m = 3 \) back into Equation (i):
\( 3 + c = 4 \)
Subtract 3 from both sides:
\( c = 4 - 3 \)
\( c = 1 \)
So, the values are \( m = 3 \) and \( c = 1 \). These values help form the specific line equation \( y = 3x + 1 \).
In simple words: When you know two points on a line, you can set up two math puzzles using the line's formula. Solve those puzzles to find the line's slope ('m') and where it crosses the 'y' axis ('c').

🎯 Exam Tip: When using two points to find \( m \) and \( c \), carefully set up the simultaneous equations. Subtraction is often effective for eliminating \( c \) if its coefficients are the same, simplifying the calculation for \( m \).

 

Question 6. Find the equation of the straight line through the given point P and having the given slope m
(a) P \( (-4, 7) \); m = \( -\sqrt{3} \);
(b) P \( (-1, -5) \), m = \( \frac{-6}{11} \).
Answer:
We use the point-slope form of a linear equation, which is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the given point and \( m \) is the slope.
(a) Given point P \( (-4, 7) \) and slope \( m = -\sqrt{3} \).
Here, \( x_1 = -4 \) and \( y_1 = 7 \).
Substitute these values into the point-slope form:
\( y - 7 = -\sqrt{3}(x - (-4)) \)
\( y - 7 = -\sqrt{3}(x + 4) \)
Now, distribute \( -\sqrt{3} \) on the right side:
\( y - 7 = -\sqrt{3}x - 4\sqrt{3} \)
To get the equation in the general form \( Ax + By + C = 0 \), move all terms to one side:
\( \sqrt{3}x + y - 7 + 4\sqrt{3} = 0 \)
(b) Given point P \( (-1, -5) \) and slope \( m = \frac{-6}{11} \).
Here, \( x_1 = -1 \) and \( y_1 = -5 \).
Substitute these values into the point-slope form:
\( y - (-5) = \frac{-6}{11}(x - (-1)) \)
\( y + 5 = \frac{-6}{11}(x + 1) \)
Multiply both sides by 11 to remove the fraction:
\( 11(y + 5) = -6(x + 1) \)
\( 11y + 55 = -6x - 6 \)
Move all terms to one side to get the general form \( Ax + By + C = 0 \):
\( 6x + 11y + 55 + 6 = 0 \)
\( 6x + 11y + 61 = 0 \)
In simple words: If you know one point on a line and how steep it is (its slope), you can use a special formula called the point-slope form. Just plug in the point's numbers and the slope's number, then tidy up the equation. This gives you the line's rule.

🎯 Exam Tip: The point-slope form \( y - y_1 = m(x - x_1) \) is incredibly useful when a point and slope are given. Remember to correctly handle negative signs for coordinates and slopes, especially when simplifying to the general form.

 

Question 7. Find the equation to the straight line passing through :
(a) the origin and perpendicular to \( x + 2y = 4 \);
(b) the point \( (4, 3) \) and parallel to \( 3x + 4y = 12 \);
(c) the point \( (4, 5) \) and (i) parallel to, (ii) perpendicular to \( 3x – 2y + 5 = 0 \).
Answer:
(a) The line passes through the origin \( (0, 0) \). It is perpendicular to \( x + 2y = 4 \).
First, find the slope of the given line \( x + 2y = 4 \). Convert it to \( y = mx + c \) form:
\( 2y = -x + 4 \)
\( y = \frac{-1}{2}x + 2 \)
The slope of this line \( (m_1) \) is \( \frac{-1}{2} \).
For a line perpendicular to this, its slope \( (m_2) \) will be the negative reciprocal of \( m_1 \):
\( m_2 = -\frac{1}{m_1} = -\frac{1}{(\frac{-1}{2})} = 2 \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (0, 0) \) and slope \( m_2 = 2 \):
\( y - 0 = 2(x - 0) \)
\( y = 2x \)
Rearranging: \( 2x - y = 0 \)
(b) The line passes through point \( (4, 3) \) and is parallel to \( 3x + 4y = 12 \).
First, find the slope of the given line \( 3x + 4y = 12 \). Convert it to \( y = mx + c \) form:
\( 4y = -3x + 12 \)
\( y = \frac{-3}{4}x + 3 \)
The slope of this line \( (m_1) \) is \( \frac{-3}{4} \).
Since our required line is parallel, its slope \( (m_2) \) is the same: \( m_2 = \frac{-3}{4} \).
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (4, 3) \) and slope \( m_2 = \frac{-3}{4} \):
\( y - 3 = \frac{-3}{4}(x - 4) \)
Multiply both sides by 4:
\( 4(y - 3) = -3(x - 4) \)
\( 4y - 12 = -3x + 12 \)
Move all terms to one side:
\( 3x + 4y - 12 - 12 = 0 \)
\( 3x + 4y - 24 = 0 \)
(c) The line passes through point \( (4, 5) \). We need to find the equation when it's (i) parallel and (ii) perpendicular to \( 3x – 2y + 5 = 0 \).
First, find the slope of the given line \( 3x – 2y + 5 = 0 \):
\( -2y = -3x - 5 \)
\( 2y = 3x + 5 \)
\( y = \frac{3}{2}x + \frac{5}{2} \)
The slope of this line \( (m_1) \) is \( \frac{3}{2} \).
(i) For the line parallel to \( 3x – 2y + 5 = 0 \):
The slope \( (m_2) \) will be the same: \( m_2 = \frac{3}{2} \).
Using point \( (4, 5) \) and slope \( m_2 = \frac{3}{2} \) in point-slope form:
\( y - 5 = \frac{3}{2}(x - 4) \)
Multiply by 2:
\( 2(y - 5) = 3(x - 4) \)
\( 2y - 10 = 3x - 12 \)
Move all terms to one side:
\( 3x - 2y - 12 + 10 = 0 \)
\( 3x - 2y - 2 = 0 \)
(ii) For the line perpendicular to \( 3x – 2y + 5 = 0 \):
The slope \( (m_3) \) will be the negative reciprocal of \( m_1 \):
\( m_3 = -\frac{1}{m_1} = -\frac{1}{(\frac{3}{2})} = -\frac{2}{3} \)
Using point \( (4, 5) \) and slope \( m_3 = -\frac{2}{3} \) in point-slope form:
\( y - 5 = -\frac{2}{3}(x - 4) \)
Multiply by 3:
\( 3(y - 5) = -2(x - 4) \)
\( 3y - 15 = -2x + 8 \)
Move all terms to one side:
\( 2x + 3y - 15 - 8 = 0 \)
\( 2x + 3y - 23 = 0 \)
In simple words: To find a line's equation, you need its steepness (slope) and a point it goes through. If a line is parallel to another, they have the same slope. If it's perpendicular, its slope is flipped and has the opposite sign. Once you have the slope and a point, use the point-slope formula to write the equation.

🎯 Exam Tip: Carefully determine the slope of the reference line first. Remember that for perpendicular lines, \( m_1 m_2 = -1 \), and for parallel lines, \( m_1 = m_2 \). Always double-check your sign changes and fraction inversions.

 

Question 8. Find the equations of the line joining the points
(a) A \( (1, 1) \) and B \( (2, 3) \);
(b) P \( (3, 3) \) and Q \( (7, 6) \);
(c) L \( (a, b) \) and M \( (b, a) \).
Answer:
To find the equation of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \), we first find the slope \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Then, we use the point-slope form \( y - y_1 = m(x - x_1) \).
(a) Given points A \( (1, 1) \) and B \( (2, 3) \).
First, calculate the slope \( m \):
\( m = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \)
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 \( Ax + By + C = 0 \):
\( 2x - y - 2 + 1 = 0 \)
\( 2x - y - 1 = 0 \)
(b) Given points P \( (3, 3) \) and Q \( (7, 6) \).
First, calculate the slope \( m \):
\( m = \frac{6 - 3}{7 - 3} = \frac{3}{4} \)
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 \( Ax + By + C = 0 \):
\( 3x - 4y - 9 + 12 = 0 \)
\( 3x - 4y + 3 = 0 \)
(c) Given points L \( (a, b) \) and M \( (b, a) \).
First, calculate the slope \( m \):
\( m = \frac{a - b}{b - a} \)
Since \( (a - b) = -(b - a) \), we can write:
\( m = \frac{-(b - a)}{b - a} = -1 \) (assuming \( b \neq a \))
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 \( Ax + By + C = 0 \):
\( x + y - b - a = 0 \)
\( x + y = a + b \)
In simple words: To find the line's rule when you have two points, first figure out its steepness (slope) by using the change in y divided by the change in x. Then, pick one of the points and use this slope in the point-slope formula. Finally, rearrange the equation to its simplest form.

🎯 Exam Tip: When using the two-point formula, ensure consistent subtraction of coordinates (e.g., \( y_2 - y_1 \) and \( x_2 - x_1 \)). For algebraic points, simplify the slope expression first before applying the point-slope form.

 

Question 9. The lines represented by \( 3x + 4y = 8 \) and \( px + 2y = 7 \) are parallel. Find the value of p.
Answer:
For two lines to be parallel, their slopes must be equal.
First, find the slope of the first line: \( 3x + 4y = 8 \).
Convert to \( y = mx + c \) form:
\( 4y = -3x + 8 \)
\( y = \frac{-3}{4}x + \frac{8}{4} \)
\( y = \frac{-3}{4}x + 2 \)
So, the slope of the first line \( (m_1) \) is \( \frac{-3}{4} \). This line tells us about the steepness and y-intercept for the first relationship.
Next, find the slope of the second line: \( px + 2y = 7 \).
Convert to \( y = mx + c \) form:
\( 2y = -px + 7 \)
\( y = \frac{-p}{2}x + \frac{7}{2} \)
So, the slope of the second line \( (m_2) \) is \( \frac{-p}{2} \).
Since the lines are parallel, their slopes must be equal:
\( m_1 = m_2 \)
\( \frac{-3}{4} = \frac{-p}{2} \)
To solve for \( p \), multiply both sides by 4 (or cross-multiply):
\( -3 \times 2 = -p \times 4 \)
\( -6 = -4p \)
Divide both sides by -4:
\( p = \frac{-6}{-4} \)
\( p = \frac{3}{2} \)
In simple words: Parallel lines always have the same steepness. Find the 'steepness number' (slope) for both lines. Since they are parallel, these numbers must be equal. Use this fact to figure out the missing value 'p'.

🎯 Exam Tip: The key to solving problems with parallel lines is to equate their slopes. Always convert both equations to the \( y = mx + c \) form to correctly extract their slopes before comparing them.

 

Question 10. The coordinates of two points E and F are \( (0, 4) \) and \( (3, 7) \) respectively. Find :
(i) the gradient of EF ;
(ii) the equation of EF ;
(iii) the coordinates of the point where the line EF intersects the x-axis.
Answer:
Given coordinates of two points E \( (0, 4) \) and F \( (3, 7) \).
(i) The gradient (slope) \( m \) of a line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Here, \( x_1 = 0, y_1 = 4 \) and \( x_2 = 3, y_2 = 7 \).
\( m = \frac{7 - 4}{3 - 0} = \frac{3}{3} = 1 \)
So, the gradient of EF is 1.
(ii) To find the equation of line EF, we can use the point-slope form \( y - y_1 = m(x - x_1) \).
Using point E \( (0, 4) \) and slope \( m = 1 \):
\( y - 4 = 1(x - 0) \)
\( y - 4 = x \)
Rearranging to the general form \( Ax + By + C = 0 \):
\( x - y + 4 = 0 \)
(iii) The line EF intersects the x-axis at a point where the y-coordinate is 0. Let this point be P \( (x, 0) \).
Substitute \( y = 0 \) into the equation of line EF: \( x - y + 4 = 0 \).
\( x - 0 + 4 = 0 \)
\( x + 4 = 0 \)
\( x = -4 \)
So, the coordinates of the point where line EF intersects the x-axis are \( (-4, 0) \). This point represents where the line crosses the horizontal axis.
In simple words: For a line between two points, first find its steepness (gradient) by dividing the change in 'y' by the change in 'x'. Then, use this steepness and one point to write the line's rule. To find where it crosses the 'x' line, just set 'y' to zero in the line's rule and solve for 'x'.

🎯 Exam Tip: When finding the point where a line intersects the x-axis, always remember that the y-coordinate at this point is zero. Similarly, for the y-axis intersection, the x-coordinate is zero. This simplifies the equation significantly.

 

Question 11. P, Q, R have coordinates \( (-2, 1) \), \( (2, 2) \) and \( (6, -2) \) respectively. Write down
(i) the gradient of QR ;
(ii) the equation of the line through P perpendicular to QR.
Answer:
Given coordinates of points: P \( (-2, 1) \), Q \( (2, 2) \) and R \( (6, -2) \).
(i) To find the gradient of QR, we use points Q \( (2, 2) \) and R \( (6, -2) \).
The gradient \( m \) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Let \( x_1 = 2, y_1 = 2 \) and \( x_2 = 6, y_2 = -2 \).
\( m_{\text{QR}} = \frac{-2 - 2}{6 - 2} = \frac{-4}{4} = -1 \)
So, the gradient of QR is \( -1 \).
(ii) We need the equation of the line through P \( (-2, 1) \) that is perpendicular to QR.
The slope of QR is \( m_{\text{QR}} = -1 \).
For a line perpendicular to QR, its slope \( m_{\text{perp}} \) is the negative reciprocal of \( m_{\text{QR}} \):
\( m_{\text{perp}} = -\frac{1}{m_{\text{QR}}} = -\frac{1}{(-1)} = 1 \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point P \( (-2, 1) \) and slope \( m_{\text{perp}} = 1 \):
\( y - 1 = 1(x - (-2)) \)
\( y - 1 = 1(x + 2) \)
\( y - 1 = x + 2 \)
Rearrange to the general form \( Ax + By + C = 0 \):
\( x - y + 2 + 1 = 0 \)
\( x - y + 3 = 0 \)
In simple words: First, find how steep the line QR is. Then, for a line that cuts QR at a right angle, its steepness will be the opposite and flipped version of QR's steepness. Finally, use this new steepness and point P to write down the equation for the new line.

🎯 Exam Tip: When dealing with perpendicular lines, remember that their slopes multiply to \( -1 \). This means if one slope is \( m \), the perpendicular slope is \( -\frac{1}{m} \). Be careful with signs in the calculation.

 

Question 12. A line \( 3x - 4y + 12 = 0 \) meets the x-axis at the point P. Find the equation of the line through P, perpendicular to the line \( 3x + 5y - 15 = 0 \).
Answer:
First, find the coordinates of point P where the line \( 3x - 4y + 12 = 0 \) meets the x-axis.
When a line meets the x-axis, the y-coordinate is 0.
Substitute \( y = 0 \) into the equation \( 3x - 4y + 12 = 0 \):
\( 3x - 4(0) + 12 = 0 \)
\( 3x + 12 = 0 \)
\( 3x = -12 \)
\( x = \frac{-12}{3} \)
\( x = -4 \)
So, the coordinates of point P are \( (-4, 0) \).
Next, find the slope of the line \( 3x + 5y - 15 = 0 \). Convert it to \( y = mx + c \) form:
\( 5y = -3x + 15 \)
\( y = \frac{-3}{5}x + \frac{15}{5} \)
\( y = \frac{-3}{5}x + 3 \)
The slope of this line \( (m_1) \) is \( \frac{-3}{5} \).
The required line is perpendicular to this line. So, its slope \( (m_2) \) will be the negative reciprocal of \( m_1 \):
\( m_2 = -\frac{1}{m_1} = -\frac{1}{(\frac{-3}{5})} = \frac{5}{3} \)
Now, we have point P \( (-4, 0) \) and the slope \( m_2 = \frac{5}{3} \). Use the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 0 = \frac{5}{3}(x - (-4)) \)
\( y = \frac{5}{3}(x + 4) \)
Multiply both sides by 3 to remove the fraction:
\( 3y = 5(x + 4) \)
\( 3y = 5x + 20 \)
Rearrange to the general form \( Ax + By + C = 0 \):
\( 5x - 3y + 20 = 0 \)
In simple words: First, find the exact spot (point P) where the first line crosses the 'x' line. Next, figure out how steep the second line is. Because your new line must cut the second line at a right angle, its steepness will be the opposite of the second line's steepness, and its fraction will be upside down. Finally, use this new steepness and point P to write the rule for your new line.

🎯 Exam Tip: Remember that a line meeting the x-axis means \( y=0 \), and meeting the y-axis means \( x=0 \). Perpendicular slopes are negative reciprocals, so be precise with the calculation \( -\frac{1}{m} \).

 

Question 13. The line segment joining P \( (5, -2) \) and Q \( (9, 6) \) is divided in the ratio 3 : 1 by a point A on it. Find the equation of a line through the point A perpendicular to the line \( x - 3y + 4 = 0 \).
Answer:
First, find the coordinates of point A that divides the line segment joining P \( (5, -2) \) and Q \( (9, 6) \) in the ratio 3 : 1.
Using the section formula, if a point \( (x, y) \) divides the segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m_1 : m_2 \):
\( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \) and \( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \)
Here, \( (x_1, y_1) = (5, -2) \), \( (x_2, y_2) = (9, 6) \), \( m_1 = 3 \), \( m_2 = 1 \).
\( x_A = \frac{3(9) + 1(5)}{3 + 1} = \frac{27 + 5}{4} = \frac{32}{4} = 8 \)
\( y_A = \frac{3(6) + 1(-2)}{3 + 1} = \frac{18 - 2}{4} = \frac{16}{4} = 4 \)
So, the coordinates of point A are \( (8, 4) \).
Next, find the slope of the line \( x - 3y + 4 = 0 \). Convert it to \( y = mx + c \) form:
\( -3y = -x - 4 \)
\( 3y = x + 4 \)
\( y = \frac{1}{3}x + \frac{4}{3} \)
The slope of this line \( (m_1) \) is \( \frac{1}{3} \).
The required line through point A is perpendicular to this line. So, its slope \( (m_2) \) will be the negative reciprocal of \( m_1 \):
\( m_2 = -\frac{1}{m_1} = -\frac{1}{(\frac{1}{3})} = -3 \)
Now, we have point A \( (8, 4) \) and the slope \( m_2 = -3 \). Use the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 4 = -3(x - 8) \)
\( y - 4 = -3x + 24 \)
Move all terms to one side to get the general form \( Ax + By + C = 0 \):
\( 3x + y - 4 - 24 = 0 \)
\( 3x + y - 28 = 0 \)
In simple words: First, find the exact location of point A using the section formula, since it divides the line in a specific ratio. Next, determine the steepness of the given line. Because your new line must cut the given line at a right angle, its steepness will be the opposite and upside-down version of the given line's steepness. Finally, use this new steepness and point A to write the equation for the new line.

🎯 Exam Tip: The section formula is crucial for finding the coordinates of a point that divides a line segment in a given ratio. Be careful with calculations involving negative signs and fractions, especially when determining the perpendicular slope.

 

Question 14. Write down the equation of the line parallel to \( x - 2y + 8 = 0 \) passing through the point \( (1, 2) \).
Answer:
First, find the slope of the given line: \( x - 2y + 8 = 0 \).
Convert the equation to the slope-intercept form \( y = mx + c \):
\( -2y = -x - 8 \)
\( 2y = x + 8 \)
\( y = \frac{1}{2}x + \frac{8}{2} \)
\( y = \frac{1}{2}x + 4 \)
The slope of this line \( (m_1) \) is \( \frac{1}{2} \).
Since the required line is parallel to the given line, its slope \( (m_2) \) will be the same:
\( m_2 = \frac{1}{2} \)
The required line passes through the point \( (1, 2) \).
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (1, 2) \) and slope \( m_2 = \frac{1}{2} \):
\( y - 2 = \frac{1}{2}(x - 1) \)
Multiply both sides by 2 to eliminate the fraction:
\( 2(y - 2) = 1(x - 1) \)
\( 2y - 4 = x - 1 \)
Rearrange the terms to the general form \( Ax + By + C = 0 \):
\( x - 2y - 1 + 4 = 0 \)
\( x - 2y + 3 = 0 \)
In simple words: To find the equation of a new line that is parallel to an existing one and passes through a certain point, first find the steepness (slope) of the existing line. Since parallel lines have the same steepness, use this slope with the given point in the point-slope formula to write the equation of your new line.

🎯 Exam Tip: When a line is parallel to another, their slopes are identical. Ensure you correctly extract the slope from the given equation by rewriting it in the \( y = mx + c \) form before applying it to the new line's equation.

 

Question 15. Write down the gradient and intercept on the y-axis of the line \( \frac{x}{3} + \frac{y}{4} = 1 \).
Answer:
The given equation of the line is \( \frac{x}{3} + \frac{y}{4} = 1 \).
To find the gradient (slope) and y-intercept, we need to convert this equation to the slope-intercept form \( y = mx + c \).
First, isolate the term with \( y \):
\( \frac{y}{4} = 1 - \frac{x}{3} \)
\( \frac{y}{4} = -\frac{x}{3} + 1 \)
Now, multiply both sides by 4:
\( y = 4 \left(-\frac{x}{3} + 1\right) \)
\( y = -\frac{4}{3}x + 4 \)
Comparing this with \( y = mx + c \):
The gradient \( (m) \) is \( -\frac{4}{3} \).
The y-intercept \( (c) \) is \( 4 \).
This means the line descends as it moves from left to right and crosses the y-axis at the point \( (0, 4) \).
In simple words: To find the steepness (gradient) and where a line crosses the 'y' axis (y-intercept) from its equation, just rearrange the equation so 'y' is by itself. The number with 'x' is the gradient, and the number without 'x' is the y-intercept.

🎯 Exam Tip: Equations in the form \( \frac{x}{a} + \frac{y}{b} = 1 \) are in intercept form, where \( a \) is the x-intercept and \( b \) is the y-intercept. While this immediately gives you the y-intercept, converting it to \( y = mx + c \) is necessary to find the slope correctly.

 

Question 16.
(i) Lines \( 2x - by + 5 = 0 \) and \( ax + 3y = 2 \) are parallel. Find the relation connecting a and b.
(ii) Find the equation of the line through \( (1, 3) \) making an intercept of 5 on the y-axis.
Answer:
(i) For two lines to be parallel, their slopes must be equal.
First line: \( 2x - by + 5 = 0 \)
Convert to \( y = mx + c \) form:
\( -by = -2x - 5 \)
\( by = 2x + 5 \)
\( y = \frac{2}{b}x + \frac{5}{b} \)
The slope of the first line \( (m_1) \) is \( \frac{2}{b} \).
Second line: \( ax + 3y = 2 \)
Convert to \( y = mx + c \) form:
\( 3y = -ax + 2 \)
\( y = \frac{-a}{3}x + \frac{2}{3} \)
The slope of the second line \( (m_2) \) is \( \frac{-a}{3} \).
Since the lines are parallel, \( m_1 = m_2 \):
\( \frac{2}{b} = \frac{-a}{3} \)
Cross-multiply:
\( 2 \times 3 = -a \times b \)
\( 6 = -ab \)
So, the relation connecting a and b is \( ab = -6 \).
(ii) The line passes through \( (1, 3) \) and makes an intercept of 5 on the y-axis.
An intercept of 5 on the y-axis means the line passes through the point \( (0, 5) \). This is the point where the line crosses the y-axis.
Now we have two points: \( (1, 3) \) and \( (0, 5) \).
First, find the slope \( m \) using these two points:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{0 - 1} = \frac{2}{-1} = -2 \)
Since the line passes through \( (0, 5) \), its y-intercept \( (c) \) is 5.
Now use the slope-intercept form \( y = mx + c \):
\( y = -2x + 5 \)
Rearrange to the general form \( Ax + By + C = 0 \):
\( 2x + y - 5 = 0 \)
In simple words: (i) For parallel lines, their steepness (slope) must be the same. Find the slope for each line, set them equal, and that will show you how 'a' and 'b' are connected. (ii) If a line crosses the 'y' axis at 5, it means it passes through the point (0, 5). Now, use this point and the other given point to find the line's steepness, then write its full equation.

🎯 Exam Tip: Remember that "making an intercept of \( k \) on the y-axis" means the line passes through the point \( (0, k) \). This gives you two points if another point is provided, which is enough to find both the slope and the equation of the line.

 

Question 17.
(i) Find the value of p, given that the line \( \frac{y}{2} = x - p \) passes through the point \( (-4, 4) \).
(ii) Given that the line \( \frac{y}{2} = x - p \) and the line \( ax + 5 = 3y \) are parallel, find the value of a.
Answer:
(i) The equation of the line is \( \frac{y}{2} = x - p \).
The line passes through the point \( (-4, 4) \). This means when \( x = -4 \), \( y = 4 \).
Substitute these values into the equation:
\( \frac{4}{2} = -4 - p \)
\( 2 = -4 - p \)
To solve for \( p \), add 4 to both sides:
\( 2 + 4 = -p \)
\( 6 = -p \)
\( p = -6 \)
So, the value of p is \( -6 \).
(ii) We are given two parallel lines: \( \frac{y}{2} = x - p \) and \( ax + 5 = 3y \). We need to find the value of a.
First, find the slope of the first line, \( \frac{y}{2} = x - p \). Convert it to \( y = mx + c \) form:
\( y = 2(x - p) \)
\( y = 2x - 2p \)
The slope of the first line \( (m_1) \) is \( 2 \).
Next, find the slope of the second line, \( ax + 5 = 3y \). Convert it to \( y = mx + c \) form:
\( 3y = ax + 5 \)
\( y = \frac{a}{3}x + \frac{5}{3} \)
The slope of the second line \( (m_2) \) is \( \frac{a}{3} \).
Since the lines are parallel, their slopes must be equal:
\( m_1 = m_2 \)
\( 2 = \frac{a}{3} \)
Multiply both sides by 3:
\( a = 2 \times 3 \)
\( a = 6 \)
So, the value of a is 6.
In simple words: (i) If a line goes through a point, you can put that point's numbers into the line's equation to find any missing value, like 'p'. (ii) If two lines are parallel, they have the same steepness (slope). Find the steepness for both lines, set them equal, and then solve for the unknown 'a'.

🎯 Exam Tip: Remember that for a line to pass through a point, the coordinates of that point must satisfy the line's equation. For parallel lines, always equate their slopes after converting them to the \( y = mx + c \) form.

 

Question 18. A line intersects x-axis at \( (-2, 0) \) and cuts off an intercept of 3 from the positive side of y-axis. Write the equation of the line.
Answer:
The line intersects the x-axis at \( (-2, 0) \). This is one point on the line.
The line cuts off an intercept of 3 from the positive side of the y-axis. This means it passes through the point \( (0, 3) \). This is another point on the line.
Now we have two points: \( (x_1, y_1) = (-2, 0) \) and \( (x_2, y_2) = (0, 3) \).
First, find the slope \( m \) of the line using these two points:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - (-2)} = \frac{3}{0 + 2} = \frac{3}{2} \)
Now, use the slope-intercept form \( y = mx + c \). We already know the y-intercept \( (c) \) is 3.
Substitute \( m = \frac{3}{2} \) and \( c = 3 \):
\( y = \frac{3}{2}x + 3 \)
To remove the fraction, multiply the entire equation by 2:
\( 2y = 3x + 6 \)
Rearrange to the general form \( Ax + By + C = 0 \):
\( 3x - 2y + 6 = 0 \)
In simple words: You're given two points: one where the line crosses the 'x' axis and one where it crosses the 'y' axis. Use these two points to find the steepness (slope) of the line. Since you already know where it crosses the 'y' axis (the y-intercept), you can easily write the full equation of the line.

🎯 Exam Tip: Remember that an "x-intercept of \( a \)" means the point \( (a, 0) \), and a "y-intercept of \( b \)" means the point \( (0, b) \). Having two points allows you to calculate the slope and then use either the point-slope or slope-intercept form.

 

Question 19.
(a) In the adjoining figure, write down (i) the coordinates of the points A, B and C; (ii) the equation of the line through A, parallel to BC.
(b) In what ratio is the join of A \( (0,3) \) and B \( (4, 1) \) divided by the x-axis? (AB produced if necessary). Write the co-ordinates of the point where AB intersects the x-axis.
Answer:
(a) (i) From the figure, we can read the coordinates of the points:
Point A is at \( (2, 3) \).
Point B is at \( (-1, 2) \).
Point C is at \( (3, 0) \).
(ii) We need the equation of the line through A \( (2, 3) \) parallel to BC.
First, find the slope of line BC using points B \( (-1, 2) \) and C \( (3, 0) \):
\( m_{\text{BC}} = \frac{0 - 2}{3 - (-1)} = \frac{-2}{3 + 1} = \frac{-2}{4} = \frac{-1}{2} \)
Since the required line is parallel to BC, its slope \( (m) \) will be the same: \( m = \frac{-1}{2} \).
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point A \( (2, 3) \) and slope \( m = \frac{-1}{2} \):
\( y - 3 = \frac{-1}{2}(x - 2) \)
Multiply both sides by 2:
\( 2(y - 3) = -1(x - 2) \)
\( 2y - 6 = -x + 2 \)
Rearrange to the general form \( Ax + By + C = 0 \):
\( x + 2y - 6 - 2 = 0 \)
\( x + 2y - 8 = 0 \)
(b) Given points A \( (0, 3) \) and B \( (4, 1) \). Let the x-axis divide the line segment AB at point P \( (x, 0) \) in the ratio \( m_1 : m_2 \).
Since P lies on the x-axis, its y-coordinate is 0. Using the section formula for the y-coordinate:
\( y_P = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \)
\( 0 = \frac{m_1(1) + m_2(3)}{m_1 + m_2} \)
For this fraction to be 0, the numerator must be 0:
\( m_1 + 3m_2 = 0 \)
\( m_1 = -3m_2 \)
\( \frac{m_1}{m_2} = -3 \)
The ratio is \( 3:1 \), but the negative sign indicates that the division is external, meaning P lies outside the segment AB on the line produced. The ratio of the lengths PR:RQ is \( 3:1 \).
Now, find the x-coordinate of P using the section formula:
\( x_P = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \)
Using \( m_1 = -3 \) and \( m_2 = 1 \) (for external division, one ratio is negative):
\( x_P = \frac{(-3)(4) + (1)(0)}{-3 + 1} = \frac{-12 + 0}{-2} = \frac{-12}{-2} = 6 \)
So, the coordinates of the point P where AB intersects the x-axis are \( (6, 0) \).

In simple words: (a) For the figure, simply read the 'x' and 'y' numbers for points A, B, and C. Then, for the line through A that's parallel to BC, find how steep BC is. Use that same steepness with point A to write the line's rule. (b) To find where a line crosses the 'x' axis and the ratio, use a special formula for dividing a line segment. The 'y' value will be zero at the 'x' axis.

🎯 Exam Tip: When reading coordinates from a graph, ensure you align with the correct scale on both axes. For section formula with external division, one of the ratio components is treated as negative, which is crucial for correct calculation.

 

Question 20. Write down the equation of the line AB, through (3, 2), perpendicular to the line \( 2y = 3x + 5 \). AB meets the x-axis at A and y-axis at B. Write down the coordinates of A and B. Calculate the area of \( \Delta OAB \) where O is the origin.
Answer: First, we find the slope of the given line. The equation \( 2y = 3x + 5 \) can be written as \( y = \frac{3}{2}x + \frac{5}{2} \). So, its slope \( m_1 = \frac{3}{2} \).
The line AB is perpendicular to this given line. When two lines are perpendicular, the product of their slopes is -1. So, the slope of line AB, \( m_2 = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3} \).
Line AB passes through the point (3, 2). Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 2 = -\frac{2}{3}(x - 3) \)
Now, we multiply by 3 on both sides to clear the fraction:
\( 3(y - 2) = -2(x - 3) \)
\( 3y - 6 = -2x + 6 \)
Rearranging the terms to get the standard form of the equation:
\( 2x + 3y - 6 - 6 = 0 \)
\( 2x + 3y - 12 = 0 \)
So, the equation of line AB is \( 2x + 3y = 12 \).
To find where line AB meets the x-axis (point A), we set \( y = 0 \):
\( 2x + 3(0) = 12 \)
\( 2x = 12 \)
\( x = 6 \)
So, the coordinates of A are (6, 0).
To find where line AB meets the y-axis (point B), we set \( x = 0 \):
\( 2(0) + 3y = 12 \)
\( 3y = 12 \)
\( y = 4 \)
So, the coordinates of B are (0, 4).
The origin O is at (0, 0). The area of a triangle with vertices at the origin and on the axes can be found using the formula \( \frac{1}{2} \times \text{base} \times \text{height} \).
Here, OA is the base (length 6 units) and OB is the height (length 4 units).
Area of \( \Delta OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 6 \times 4 = 12 \) square units. When finding the area of a triangle, remember to always use positive lengths for the base and height.
In simple words: We first found the slope of the original line. Then, we used it to find the slope of the new line (because they are perpendicular) and its equation. After that, we found where this new line crosses the X and Y axes to get points A and B. Finally, we calculated the area of the triangle formed by A, B, and the origin.

🎯 Exam Tip: Always remember that the product of slopes of two perpendicular lines is -1. For area calculations involving axes, use the absolute lengths of the intercepts as base and height.

 

Question 21.
(i) Write down the coordinates of the point P that divides the line joining A (- 4, 1) and B (17, 10) in the ratio 1: 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB?
Answer:
(i) To find the coordinates of point P that divides the line segment joining A(-4, 1) and B(17, 10) in the ratio \( m_1 : m_2 = 1 : 2 \), we use the section formula:
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \) and \( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
For the x-coordinate of P:
\( x = \frac{(1 \times 17) + (2 \times -4)}{1 + 2} = \frac{17 - 8}{3} = \frac{9}{3} = 3 \)
For the y-coordinate of P:
\( y = \frac{(1 \times 10) + (2 \times 1)}{1 + 2} = \frac{10 + 2}{3} = \frac{12}{3} = 4 \)
So, the coordinates of point P are (3, 4). This point is an internal division of the line segment.

(ii) To calculate the distance OP, where O is the origin (0, 0) and P is (3, 4), we use the distance formula:
\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( OP = \sqrt{(3 - 0)^2 + (4 - 0)^2} \)
\( OP = \sqrt{3^2 + 4^2} \)
\( OP = \sqrt{9 + 16} \)
\( OP = \sqrt{25} \)
\( OP = 5 \) units.

(iii) To find the ratio in which the y-axis divides the line segment AB, let R be the point of division on the y-axis. The x-coordinate of any point on the y-axis is 0.
Let the ratio be \( m_1 : m_2 \). We use the x-coordinate from the section formula:
\( x_R = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \)
We know \( x_R = 0 \), A(-4, 1) and B(17, 10).
\( 0 = \frac{(m_1 \times 17) + (m_2 \times -4)}{m_1 + m_2} \)
Multiplying both sides by \( (m_1 + m_2) \):
\( 0 = 17m_1 - 4m_2 \)
\( 4m_2 = 17m_1 \)
\( \frac{m_1}{m_2} = \frac{4}{17} \)
So, the y-axis divides the line segment AB in the ratio 4:17. This means the y-axis is much closer to point A in terms of its dividing proportion.
In simple words: For part (i), we used a special formula to find the exact spot P when a line is cut into a certain ratio. For part (ii), we used another formula to measure the straight distance from the start point (origin) to point P. For part (iii), we figured out where the y-axis cuts the line AB and what fraction each part is.

🎯 Exam Tip: Remember the section formula for internal division. For a point on the x-axis, the y-coordinate is 0; for a point on the y-axis, the x-coordinate is 0. Use this fact in the section formula to find unknown ratios or coordinates.

 

Question 1.
(a) The mid-point of the line segment AB shown in the diagram is (4, -3). Write down the coordinates of A, B.

(b) Match the equations A, B, C, D with the lines L1, L2, L3, L4, whose graphs are roughly drawn in figure below.
A = \( y = 2x \)
B = \( y - 2x + 2 = 0 \)
C = \( 3x + 2y - 6 = 0 \)
D = \( y = 2 \)

(c) Write down the equation of the line whose gradient is \( \frac { 3 }{ 2 } \) and it passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3.
Answer:
(a) The midpoint of the line segment AB is given as (4, -3). The diagram shows point A lies on the x-axis (so its y-coordinate is 0) and point B lies on the y-axis (so its x-coordinate is 0).
Let the coordinates of A be (x, 0) and B be (0, y).
Using the midpoint formula \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \):
For the x-coordinate: \( 4 = \frac{x + 0}{2} \Rightarrow 8 = x \). So A is (8, 0).
For the y-coordinate: \( -3 = \frac{0 + y}{2} \Rightarrow -6 = y \). So B is (0, -6).
Thus, the coordinates of A are (8, 0) and B are (0, -6). These points show the line crossing the axes.

(b) We need to match each equation to the correct line based on its slope and y-intercept:
(i) Equation A: \( y = 2x \)
This line has a slope of 2 and passes through the origin (0, 0) because its y-intercept is 0. From the given graph, line L3 passes through the origin.
So, A \( \rightarrow \) L3.

(ii) Equation B: \( y - 2x + 2 = 0 \Rightarrow y = 2x - 2 \)
This line has a slope of 2 and a y-intercept of -2. From the graph, line L4 has a positive slope and crosses the y-axis below the origin.
So, B \( \rightarrow \) L4.

(iii) Equation C: \( 3x + 2y - 6 = 0 \Rightarrow 2y = -3x + 6 \Rightarrow y = -\frac{3}{2}x + 3 \)
This line has a slope of \( -\frac{3}{2} \) (negative) and a y-intercept of 3 (positive). From the graph, line L2 has a negative slope and crosses the y-axis above the origin.
So, C \( \rightarrow \) L2.

(iv) Equation D: \( y = 2 \)
This line is a horizontal line with a slope of 0, meaning it is parallel to the x-axis and passes through y = 2. From the graph, line L1 is a horizontal line at y=2.
So, D \( \rightarrow \) L1. A constant y-value always makes a horizontal line.

(c) The gradient (slope) of the line is given as \( \frac{3}{2} \).
The line passes through point P, which divides the line segment joining A(-2, 6) and B(3, -4) in the ratio 2:3.
First, find the coordinates of P using the section formula:
\( x_P = \frac{(2 \times 3) + (3 \times -2)}{2 + 3} = \frac{6 - 6}{5} = 0 \)
\( y_P = \frac{(2 \times -4) + (3 \times 6)}{2 + 3} = \frac{-8 + 18}{5} = \frac{10}{5} = 2 \)
So, the coordinates of point P are (0, 2).
Now, we have the slope \( m = \frac{3}{2} \) and a point P(0, 2). We can use the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 2 = \frac{3}{2}(x - 0) \)
\( 2(y - 2) = 3x \)
\( 2y - 4 = 3x \)
Rearranging the terms to standard form:
\( 3x - 2y + 4 = 0 \).
In simple words: For part (a), we used the midpoint rule to find the coordinates of points A and B on the axes. For part (b), we matched each line equation to its graph by checking its slope and where it crosses the y-axis. For part (c), we found the exact spot of point P by using the section formula, and then used that point and the given slope to write the equation of the line.

🎯 Exam Tip: When matching lines to equations, always check both the slope (for direction and steepness) and the y-intercept (where it crosses the Y-axis) to confirm your answer. For section formula, remember to handle negative coordinates carefully.

 

Question 2.
(a) Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
(b) (i) The line \( 4x - 3y + 12 = 0 \), meets x-axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line perpendicular to \( 4x - 3y + 12 = 0 \).
Answer:
(a) Point A is (7, 3). Point B is on the x-axis, and its x-coordinate (abscissa) is 11. So, the y-coordinate of B is 0, making B(11, 0).
To find the distance between A(7, 3) and B(11, 0), we use the distance formula:
\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( AB = \sqrt{(11 - 7)^2 + (0 - 3)^2} \)
\( AB = \sqrt{4^2 + (-3)^2} \)
\( AB = \sqrt{16 + 9} \)
\( AB = \sqrt{25} \)
\( AB = 5 \) units.
This distance represents the shortest path between the two points.

(b) (i) The line is \( 4x - 3y + 12 = 0 \). It meets the x-axis at point A. When a line meets the x-axis, the y-coordinate is always 0.
Substitute \( y = 0 \) into the equation:
\( 4x - 3(0) + 12 = 0 \)
\( 4x + 12 = 0 \)
\( 4x = -12 \)
\( x = -3 \)
So, the coordinates of A are (-3, 0).

(ii) We need the equation of a line perpendicular to \( 4x - 3y + 12 = 0 \). First, find the slope of the given line. Rewrite the equation in slope-intercept form \( y = mx + c \):
\( 3y = 4x + 12 \)
\( y = \frac{4}{3}x + 4 \)
The slope of this line, \( m_1 = \frac{4}{3} \).
The slope of a line perpendicular to it, \( m_2 \), will be the negative reciprocal of \( m_1 \):
\( m_2 = -\frac{1}{m_1} = -\frac{1}{4/3} = -\frac{3}{4} \).
The required line also passes through point A (-3, 0), which we found in part (b)(i).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 0 = -\frac{3}{4}(x - (-3)) \)
\( y = -\frac{3}{4}(x + 3) \)
Multiplying by 4 on both sides:
\( 4y = -3(x + 3) \)
\( 4y = -3x - 9 \)
Rearranging into standard form:
\( 3x + 4y + 9 = 0 \).
In simple words: In part (a), we found the simple straight-line distance between two points. In part (b)(i), we found where a line crosses the 'x' line by setting 'y' to zero. In part (b)(ii), we found the slope of the first line, then used it to get the slope of a line that cuts it at a right angle, and finally wrote the equation for this new line using the point found in (i).

🎯 Exam Tip: When finding where a line intersects the x-axis, always set y=0. For finding the equation of a perpendicular line, remember the relationship between their slopes: \( m_1 m_2 = -1 \).

 

Question 3.
(a) The centre O, of a circle has the coordinates (4, 5) and one point on the circumference is (8, 10). Find the coordinates of the other end of the diameter of the circle through this point.
(b) Find the equation of a line, which has the y-intercept 4 and is parallel to the line \( 2x - 3y = 1 \). Find the coordinates of the point, where it cuts the x-axis.
Answer:
(a) The centre of the circle, O, is (4, 5). One end of the diameter on the circumference is B(8, 10). Let the other end of the diameter be A(x, y).
Since AB is the diameter, the centre O is the midpoint of AB. We can use the midpoint formula to find the coordinates of A.
Midpoint formula: \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \)
For the x-coordinate of O:
\( 4 = \frac{x + 8}{2} \)
\( 8 = x + 8 \)
\( x = 0 \)
For the y-coordinate of O:
\( 5 = \frac{y + 10}{2} \)
\( 10 = y + 10 \)
\( y = 0 \)
So, the coordinates of the other end of the diameter, A, are (0, 0). This means the diameter passes through the origin.

(b) The line has a y-intercept of 4, meaning it passes through the point (0, 4). So, \( c = 4 \).
The line is parallel to \( 2x - 3y = 1 \). First, find the slope of this given line. Rewrite it in slope-intercept form \( y = mx + c \):
\( 3y = 2x - 1 \)
\( y = \frac{2}{3}x - \frac{1}{3} \)
The slope of this line, \( m_1 = \frac{2}{3} \).
Since the required line is parallel to it, its slope \( m_2 \) will be the same:
\( m_2 = \frac{2}{3} \).
Now, using the slope \( m = \frac{2}{3} \) and y-intercept \( c = 4 \), the equation of the line is \( y = \frac{2}{3}x + 4 \).
To find where this line cuts the x-axis, we set \( y = 0 \):
\( 0 = \frac{2}{3}x + 4 \)
\( -4 = \frac{2}{3}x \)
Multiply by 3:
\( -12 = 2x \)
\( x = -6 \)
So, the coordinates of the point where the line cuts the x-axis are (-6, 0). This is the x-intercept of the line.
In simple words: For part (a), we used the idea that the center of a circle is exactly halfway between the ends of its diameter. For part (b), we used the given y-intercept and the slope of a parallel line to write its equation. Then, we found where this new line crosses the x-axis by setting 'y' to zero.

🎯 Exam Tip: Remember that parallel lines have the same slope. When finding coordinates, use the appropriate formulas (midpoint, distance, section) and clearly show your substitution steps.

 

Question 4.
(a) Find the equation of the line passing through (0, 4) and parallel to the line \( 3x + 5y + 15 = 0 \).
(b) In the given figure line APB meets the x-axis at A, y-axis at B. P is the point (-4, 2) and AP : PB = 1 : 2. Write down the coordinates of A and B.

(c) The centre of a circle of radius 13 units is the point (3, 6). P (7, 9) is a point inside the circle. APB is a chord of the circle such that AP = PB. Calculate the length of AB.
Answer:
(a) The line passes through (0, 4), which means its y-intercept \( c = 4 \).
The line is parallel to \( 3x + 5y + 15 = 0 \). To find its slope, we rearrange the equation into \( y = mx + c \) form:
\( 5y = -3x - 15 \)
\( y = -\frac{3}{5}x - 3 \)
The slope of this given line, \( m_1 = -\frac{3}{5} \).
Since the required line is parallel, it will have the same slope: \( m_2 = -\frac{3}{5} \).
Using the slope \( m = -\frac{3}{5} \) and y-intercept \( c = 4 \), the equation of the line is:
\( y = -\frac{3}{5}x + 4 \)
Multiplying by 5 to clear the fraction:
\( 5y = -3x + 20 \)
Rearranging into standard form:
\( 3x + 5y - 20 = 0 \). This shows the relationship between x and y coordinates on the line.

(b) Point P(-4, 2) divides the line segment AB in the ratio AP : PB = 1 : 2. Point A is on the x-axis (so \( y_A = 0 \)), and point B is on the y-axis (so \( x_B = 0 \)).
Let A be \( (x_A, 0) \) and B be \( (0, y_B) \). We use the section formula for the coordinates of P(x, y):
\( x_P = \frac{m_1 x_B + m_2 x_A}{m_1 + m_2} \) and \( y_P = \frac{m_1 y_B + m_2 y_A}{m_1 + m_2} \)
For the x-coordinate of P:
\( -4 = \frac{(1 \times 0) + (2 \times x_A)}{1 + 2} \)
\( -4 = \frac{2x_A}{3} \)
\( -12 = 2x_A \)
\( x_A = -6 \)
So, the coordinates of A are (-6, 0).
For the y-coordinate of P:
\( 2 = \frac{(1 \times y_B) + (2 \times 0)}{1 + 2} \)
\( 2 = \frac{y_B}{3} \)
\( y_B = 6 \)
So, the coordinates of B are (0, 6).

(c) The circle has its centre O at (3, 6) and a radius of 13 units. P(7, 9) is a point inside the circle. APB is a chord, and AP = PB, which means P is the midpoint of the chord AB.
A key property of circles is that the line from the center to the midpoint of a chord is perpendicular to the chord. Therefore, OP \( \perp \) AB.
First, calculate the length of OP using the distance formula between O(3, 6) and P(7, 9):
\( OP = \sqrt{(7 - 3)^2 + (9 - 6)^2} \)
\( OP = \sqrt{4^2 + 3^2} \)
\( OP = \sqrt{16 + 9} \)
\( OP = \sqrt{25} = 5 \) units.
Now consider the right-angled triangle formed by the center O, the midpoint of the chord P, and one end of the chord A (or B). OA is the radius of the circle.
By Pythagoras theorem, \( OA^2 = OP^2 + AP^2 \).
We know \( OA = \text{radius} = 13 \) and \( OP = 5 \).
\( 13^2 = 5^2 + AP^2 \)
\( 169 = 25 + AP^2 \)
\( AP^2 = 169 - 25 \)
\( AP^2 = 144 \)
\( AP = \sqrt{144} = 12 \) units.
Since P is the midpoint of the chord AB, \( AB = 2 \times AP \).
\( AB = 2 \times 12 = 24 \) units. This length tells us how long the chord is within the circle.
In simple words: For part (a), we used the rules for parallel lines (same slope) and y-intercept to find the line's equation. For part (b), we used the section formula with the given ratio to find the coordinates of points A and B on the axes. For part (c), we found the distance from the circle's center to the midpoint of the chord. Then, using Pythagoras' theorem, we calculated half the chord's length and doubled it to find the full length.

🎯 Exam Tip: For problems involving chords and the center of a circle, always remember that the line segment from the center to the midpoint of a chord is perpendicular to the chord, forming a right-angled triangle that allows you to use the Pythagorean theorem.

 

Question 5.
(a) Calculate the ratio in which the line joining A (6, 5) and B (4, - 3) is divided by the line \( y = 2 \).
(b) In this figure, AB and CD are the lines \( 2x - y + 6 = 0 \) and \( x - 2y = 4 \) respectively.

(i) Write down the coordinates of A, B, C, D;
(ii) Prove that the triangles OAB and ODC are similar;
(iii) Is figure ABCD cyclic? Give reasons for your answer.
(c) ABCD is a rhombus. The coordinates of A and C are (3, 6) and (- 1, 2) respectively. Write down the equation of BD.
Answer:
(a) Let the line \( y = 2 \) divide the line segment joining A(6, 5) and B(4, -3) in the ratio \( m_1 : m_2 \). The y-coordinate of the division point is 2.
Using the section formula for the y-coordinate:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( 2 = \frac{(m_1 \times -3) + (m_2 \times 5)}{m_1 + m_2} \)
Multiplying both sides by \( (m_1 + m_2) \):
\( 2(m_1 + m_2) = -3m_1 + 5m_2 \)
\( 2m_1 + 2m_2 = -3m_1 + 5m_2 \)
Move all \( m_1 \) terms to one side and \( m_2 \) terms to the other:
\( 2m_1 + 3m_1 = 5m_2 - 2m_2 \)
\( 5m_1 = 3m_2 \)
\( \frac{m_1}{m_2} = \frac{3}{5} \)
So, the line \( y = 2 \) divides the line segment AB in the ratio 3:5. This means the point is closer to A than to B proportionally.

(b) (i) First, let's find the coordinates of A, B, C, D from their respective line equations and axis intersections.
For line AB: \( 2x - y + 6 = 0 \)
Point A is on the y-axis (where \( x=0 \)): \( 2(0) - y + 6 = 0 \Rightarrow -y + 6 = 0 \Rightarrow y = 6 \). So, A = (0, 6).
Point B is on the x-axis (where \( y=0 \)): \( 2x - 0 + 6 = 0 \Rightarrow 2x = -6 \Rightarrow x = -3 \). So, B = (-3, 0).
For line CD: \( x - 2y = 4 \)
Point C is on the y-axis (where \( x=0 \)): \( 0 - 2y = 4 \Rightarrow y = -2 \). So, C = (0, -2).
Point D is on the x-axis (where \( y=0 \)): \( x - 2(0) = 4 \Rightarrow x = 4 \). So, D = (4, 0).
The coordinates are: A(0, 6), B(-3, 0), C(0, -2), D(4, 0).

(ii) We need to prove that \( \triangle OAB \) and \( \triangle ODC \) are similar. O is the origin (0, 0).
First, find the lengths of the sides from the origin to the points on the axes:
\( OA = \text{length from (0,0) to (0,6)} = 6 \) units.
\( OB = \text{length from (0,0) to (-3,0)} = 3 \) units (we use absolute value for length).
\( OC = \text{length from (0,0) to (0,-2)} = 2 \) units.
\( OD = \text{length from (0,0) to (4,0)} = 4 \) units.
Now, let's look at the ratios of corresponding sides:
\( \frac{OA}{OD} = \frac{6}{4} = \frac{3}{2} \)
\( \frac{OB}{OC} = \frac{3}{2} \)
So, \( \frac{OA}{OD} = \frac{OB}{OC} \). This shows that two pairs of sides are in proportion.
Also, \( \angle AOB \) and \( \angle DOC \) are both vertically opposite angles and are both \( 90^\circ \) (since A and C are on the y-axis, and B and D are on the x-axis).
Since two pairs of sides are proportional and the included angles are equal, by the SAS (Side-Angle-Side) similarity criterion, \( \triangle OAB \sim \triangle ODC \). This shows that the two triangles have the same shape.

(iii) If \( \triangle OAB \sim \triangle ODC \), then their corresponding angles are equal. Specifically, \( \angle OAB = \angle ODC \) and \( \angle OBA = \angle OCD \).
These equal angles mean that the points A, B, C, D lie on a circle, making ABCD a cyclic quadrilateral. A key property for similarity leading to concyclic points is that angles subtended by a chord from different points on the circumference are equal. In this case, \( \angle OAB \) and \( \angle ODC \) can be seen as angles subtended by the imaginary chord BC and AD respectively, leading to concyclic points. When two similar triangles are formed with a common vertex (O) and the other vertices (A,B and D,C) are on intersecting lines, the quadrilateral formed by these four non-origin points (A,B,C,D) is cyclic. Thus, ABCD is a cyclic quadrilateral because opposite angles sum to 180 degrees (implicit from angle equalities derived from similarity) and angles subtended by the same chord are equal.

(c) ABCD is a rhombus. The coordinates of A are (3, 6) and C are (-1, 2).
In a rhombus, the diagonals bisect each other at right angles. This means that diagonal BD is perpendicular to diagonal AC and passes through the midpoint of AC.
First, find the midpoint of AC, let's call it M:
\( M = (\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}) = (\frac{3 + (-1)}{2}, \frac{6 + 2}{2}) = (\frac{2}{2}, \frac{8}{2}) = (1, 4) \).
Next, find the slope of AC:
\( m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{2 - 6}{-1 - 3} = \frac{-4}{-4} = 1 \).
Since BD is perpendicular to AC, the slope of BD, \( m_{BD} \), will be the negative reciprocal of \( m_{AC} \):
\( m_{BD} = -\frac{1}{1} = -1 \).
Now, we have the slope \( m_{BD} = -1 \) and a point M(1, 4) that BD passes through. Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 4 = -1(x - 1) \)
\( y - 4 = -x + 1 \)
Rearranging into standard form:
\( x + y - 4 - 1 = 0 \)
\( x + y - 5 = 0 \). The diagonals of a rhombus are important for its unique properties.
In simple words: For part (a), we used the y-coordinate of the dividing line to find the ratio. For part (b)(i), we found where each line crosses the x and y axes. For (b)(ii), we showed that the triangles are similar by comparing their side ratios and the angle between them. For (b)(iii), we used the similarity to explain why the points make a circle (are cyclic). For part (c), because it's a rhombus, we knew its diagonals cut each other in half at right angles. So, we found the middle point of one diagonal, got its slope, and used that to write the equation of the other diagonal.

🎯 Exam Tip: Remember the properties of similar triangles (SAS criterion) and cyclic quadrilaterals (angles subtended by the same arc are equal). For rhombuses, diagonals bisect each other perpendicularly, which is key for finding the equation of the other diagonal.

 

Question 6.
(a) A (10, 5), B (6, - 3) and C (2, 1) are the vertices of a triangle ABC. L is the mid-point of AB, and M is the mid-point of AC. Write down the coordinates of L and M. Show that \( LM = \frac { 1 }{ 2 } BC \).
(b) Write down the equation of the line whose gradient is \( \frac { 3 }{ 2 } \) and which passes through point P that divides the line segment joining A (- 2, 6) and B (3, - 4) in the ratio 2 : 3.
Answer:
(a) Given vertices of \( \triangle ABC \) are A(10, 5), B(6, -3), and C(2, 1).
L is the mid-point of AB. Using the midpoint formula \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \):
\( L = (\frac{10 + 6}{2}, \frac{5 + (-3)}{2}) = (\frac{16}{2}, \frac{2}{2}) = (8, 1) \).
M is the mid-point of AC. Using the midpoint formula:
\( M = (\frac{10 + 2}{2}, \frac{5 + 1}{2}) = (\frac{12}{2}, \frac{6}{2}) = (6, 3) \).
The coordinates of L are (8, 1) and M are (6, 3).
Now, let's calculate the lengths of LM and BC using the distance formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
Length of BC (between B(6, -3) and C(2, 1)):
\( BC = \sqrt{(2 - 6)^2 + (1 - (-3))^2} = \sqrt{(-4)^2 + (1 + 3)^2} = \sqrt{16 + 4^2} = \sqrt{16 + 16} = \sqrt{32} \).
\( \sqrt{32} \) can be simplified as \( \sqrt{16 \times 2} = 4\sqrt{2} \) units.
Length of LM (between L(8, 1) and M(6, 3)):
\( LM = \sqrt{(6 - 8)^2 + (3 - 1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \).
\( \sqrt{8} \) can be simplified as \( \sqrt{4 \times 2} = 2\sqrt{2} \) units.
Comparing the lengths, we see that \( LM = 2\sqrt{2} \) and \( BC = 4\sqrt{2} \).
Therefore, \( LM = \frac{1}{2} BC \). This illustrates the midpoint theorem in coordinate geometry.

(b) The gradient (slope) of the line is given as \( \frac{3}{2} \).
The line passes through point P, which divides the line segment joining A(-2, 6) and B(3, -4) in the ratio \( m_1 : m_2 = 2 : 3 \).
First, find the coordinates of P using the section formula:
\( x_P = \frac{(2 \times 3) + (3 \times -2)}{2 + 3} = \frac{6 - 6}{5} = 0 \)
\( y_P = \frac{(2 \times -4) + (3 \times 6)}{2 + 3} = \frac{-8 + 18}{5} = \frac{10}{5} = 2 \)
So, the coordinates of point P are (0, 2).
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point P(0, 2) and slope \( m = \frac{3}{2} \):
\( y - 2 = \frac{3}{2}(x - 0) \)
\( 2(y - 2) = 3x \)
\( 2y - 4 = 3x \)
Rearranging into standard form:
\( 3x - 2y + 4 = 0 \).
In simple words: For part (a), we found the middle points L and M of two sides of the triangle. Then we measured the length of the line connecting L and M, and the length of the third side BC, to show that LM is half of BC. For part (b), we used a formula to find the exact point P where a line segment is divided in a given ratio. Then, using that point and the given slope, we wrote the equation of the line.

🎯 Exam Tip: The Midpoint Theorem states that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side. Always show your distance calculations clearly for full marks.

 

Question 7.
(a) Find the equation of a line passing through the point (- 2, 3) and having the x-intercept of 4 units.
(b) A (1, 4), B (3, 2), and C (7, 5) are the vertices of a triangle ABC. Find
(i) The coordinates of the centroid G of \( \triangle ABC \).
(ii) The equation of a line through G and parallel to AB.
Answer:
(a) The line passes through (-2, 3). An x-intercept of 4 units means the line also passes through the point (4, 0).
First, calculate the slope (m) of the line using the two points (-2, 3) and (4, 0):
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 3}{4 - (-2)} = \frac{-3}{4 + 2} = \frac{-3}{6} = -\frac{1}{2} \).
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with the point (-2, 3) and slope \( m = -\frac{1}{2} \):
\( y - 3 = -\frac{1}{2}(x - (-2)) \)
\( y - 3 = -\frac{1}{2}(x + 2) \)
Multiply both sides by 2 to clear the fraction:
\( 2(y - 3) = -1(x + 2) \)
\( 2y - 6 = -x - 2 \)
Rearranging into standard form:
\( x + 2y - 6 + 2 = 0 \)
\( x + 2y - 4 = 0 \). This equation describes all points lying on this straight line.

(b) Given vertices: A(1, 4), B(3, 2), and C(7, 5).
(i) To find the coordinates of the centroid G of \( \triangle ABC \), we use the centroid formula:
\( G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}) \)
\( G = (\frac{1 + 3 + 7}{3}, \frac{4 + 2 + 5}{3}) \)
\( G = (\frac{11}{3}, \frac{11}{3}) \).
So, the coordinates of the centroid G are \( (\frac{11}{3}, \frac{11}{3}) \). The centroid is the balancing point of the triangle.

(ii) We need the equation of a line through G and parallel to AB. First, find the slope of line AB:
\( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{3 - 1} = \frac{-2}{2} = -1 \).
Since the required line is parallel to AB, its slope will be the same: \( m = -1 \).
The line passes through the centroid G \( (\frac{11}{3}, \frac{11}{3}) \). Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - \frac{11}{3} = -1(x - \frac{11}{3}) \)
To simplify, multiply by 3:
\( 3(y - \frac{11}{3}) = -3(x - \frac{11}{3}) \)
\( 3y - 11 = -3x + 11 \)
Rearranging into standard form:
\( 3x + 3y - 11 - 11 = 0 \)
\( 3x + 3y - 22 = 0 \).
In simple words: For part (a), we found two points on the line, then used them to calculate its slope, and finally wrote the line's equation. For part (b)(i), we used a special formula to find the centroid (the average position) of the triangle's corners. For part (b)(ii), we found the slope of side AB, and since our new line is parallel, it has the same slope. Then, we used the centroid's coordinates and this slope to write the equation of the new line.

🎯 Exam Tip: Remember the centroid formula for a triangle, as it's a direct application. When finding the equation of a parallel line, make sure to use the exact same slope as the reference line. Simplify fractions and clear denominators for final equations.

 

Question 8.
(a) A straight line passes through the points P (- 1, 4) and Q (5, - 2). It intersects the coordinate axes at points A and B. M is the mid-point of the segment AB. Find
(i) The equation of the line.
(ii) The coordinates of A and B.
(iii) The coordinates of M.

(b) Find the value of k for which the lines \( kx - 5y + 4 = 0 \) and \( 4x - 2y + 5 = 0 \) are perpendicular to each other.
Answer:
(a) (i) The line passes through P(-1, 4) and Q(5, -2).
First, calculate the slope (m) of the line:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 4}{5 - (-1)} = \frac{-6}{5 + 1} = \frac{-6}{6} = -1 \).
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point P(-1, 4) and slope \( m = -1 \):
\( y - 4 = -1(x - (-1)) \)
\( y - 4 = -1(x + 1) \)
\( y - 4 = -x - 1 \)
Rearranging into standard form:
\( x + y - 4 + 1 = 0 \)
\( x + y - 3 = 0 \). This is the equation of the line.

(ii) To find the coordinates of A and B, we find where the line \( x + y - 3 = 0 \) intersects the x-axis and y-axis.
Point A is on the x-axis (where \( y=0 \)):
\( x + 0 - 3 = 0 \)
\( x = 3 \). So, A = (3, 0).
Point B is on the y-axis (where \( x=0 \)):
\( 0 + y - 3 = 0 \)
\( y = 3 \). So, B = (0, 3).
The coordinates of A are (3, 0) and B are (0, 3). These are the points where the line crosses the coordinate axes.

(iii) M is the mid-point of the segment AB. Using the midpoint formula with A(3, 0) and B(0, 3):
\( M = (\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}) = (\frac{3 + 0}{2}, \frac{0 + 3}{2}) = (\frac{3}{2}, \frac{3}{2}) \).
So, the coordinates of M are \( (\frac{3}{2}, \frac{3}{2}) \) or (1.5, 1.5). This is the exact center of the line segment AB.

(b) Given lines: \( kx - 5y + 4 = 0 \) and \( 4x - 2y + 5 = 0 \).
For the first line, \( kx - 5y + 4 = 0 \):
\( 5y = kx + 4 \)
\( y = \frac{k}{5}x + \frac{4}{5} \)
The slope \( m_1 = \frac{k}{5} \).
For the second line, \( 4x - 2y + 5 = 0 \):
\( 2y = 4x + 5 \)
\( y = \frac{4}{2}x + \frac{5}{2} = 2x + \frac{5}{2} \)
The slope \( m_2 = 2 \).
Since the lines are perpendicular to each other, the product of their slopes must be -1:
\( m_1 \times m_2 = -1 \)
\( (\frac{k}{5}) \times 2 = -1 \)
\( \frac{2k}{5} = -1 \)
\( 2k = -5 \)
\( k = -\frac{5}{2} \). The value of k determines how steeply the first line rises or falls.
In simple words: For part (a), we found the slope of the line using the two given points, then wrote its equation. We then found where this line crosses the 'x' and 'y' axes to get points A and B. Finally, we found the exact middle point M of the line segment AB. For part (b), we found the slopes of both lines. Since they are at right angles to each other, we multiplied their slopes and set the result to -1 to find the value of 'k'.

🎯 Exam Tip: Always convert equations to \( y = mx + c \) form to easily identify the slope. For perpendicular lines, the product of slopes is -1. For coordinates on axes, remember that one coordinate will always be zero.

 

Question 9.
(a) KM is a straight line of 13 units. If K has coordinates (2, 5) and M has co-ordinates (x, -7), find the possible values of x.
(b) The line joining P (- 4, 5) and Q (3, 2), intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find :
(i) the ratio PR : RQ.
(ii) the co-ordinates of R.
(iii) the area of quadrilateral PMNQ.
(c) 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:
(a) Given coordinates of K are (2, 5) and M are (x, -7). The length KM is 13 units.
We use the distance formula: \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
So, \( 13 = \sqrt{(x - 2)^2 + (-7 - 5)^2} \)
\( 13 = \sqrt{(x - 2)^2 + (-12)^2} \)
Squaring both sides:
\( 13^2 = (x - 2)^2 + 144 \)
\( 169 = (x - 2)^2 + 144 \)
\( (x - 2)^2 = 169 - 144 \)
\( (x - 2)^2 = 25 \)
Taking the square root on both sides, we get:
\( x - 2 = \pm 5 \)
This gives two possibilities:
Case 1: \( x - 2 = 5 \implies x = 7 \)
Case 2: \( x - 2 = -5 \implies x = -3 \)
Therefore, the possible values for x are 7 or -3. The distance formula is very useful for finding lengths in coordinate geometry.
(b) Given points P(-4, 5) and Q(3, 2). The line PQ intersects the y-axis at R. Since R is on the y-axis, its x-coordinate is 0. Let R be (0, y).
(i) To find the ratio PR : RQ, let R divide PQ in the ratio \( m_1 : m_2 \). Using the section formula for the x-coordinate:
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \)
\( 0 = \frac{m_1(3) + m_2(-4)}{m_1 + m_2} \)
\( 0 = 3m_1 - 4m_2 \)
\( 3m_1 = 4m_2 \)
\( \frac{m_1}{m_2} = \frac{4}{3} \)
So, the y-axis divides the line segment AB in the ratio 4:3.
(ii) Now we use this ratio (4:3) with the section formula for the y-coordinate to find R's y-coordinate:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( y = \frac{4(2) + 3(5)}{4 + 3} \)
\( y = \frac{8 + 15}{7} \)
\( y = \frac{23}{7} \)
So, the coordinates of R are \( (0, \frac{23}{7}) \).
(iii) PM and QN are perpendiculars from P and Q to the x-axis. This means M and N are points on the x-axis.
P = (-4, 5), so M = (-4, 0) (projecting P onto the x-axis)
Q = (3, 2), so N = (3, 0) (projecting Q onto the x-axis)
The quadrilateral PMNQ is a trapezium because PM and QN are both vertical lines, making them parallel. The lengths of the parallel sides are PM = 5 units and QN = 2 units. The distance between them along the x-axis (the height of the trapezium) is MN = \( |3 - (-4)| = |3 + 4| = 7 \) units.
The area of a trapezium is given by: \( \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \)
\( \text{Area of PMNQ} = \frac{1}{2} \times (PM + QN) \times MN \)
\( \text{Area} = \frac{1}{2} \times (5 + 2) \times 7 \)
\( \text{Area} = \frac{1}{2} \times 7 \times 7 \)
\( \text{Area} = \frac{49}{2} = 24.5 \) square units.
(c) Given vertices of triangle PQR are P(3, 4), Q(7, -2), R(-2, -1). We need to find the equation of the median through R. A median connects a vertex to the midpoint of the opposite side. The side opposite to R is PQ.
First, find the midpoint of PQ. Let's call it L.
Coordinates of L = \( (\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}) = (\frac{3 + 7}{2}, \frac{4 + (-2)}{2}) = (\frac{10}{2}, \frac{2}{2}) = (5, 1) \)
Now, find the equation of the line passing through R(-2, -1) and L(5, 1). This is our median.
Slope of median RL (m) = \( \frac{y_L - y_R}{x_L - x_R} = \frac{1 - (-1)}{5 - (-2)} = \frac{1 + 1}{5 + 2} = \frac{2}{7} \)
Using the point-slope form \( y - y_1 = m(x - x_1) \) with R(-2, -1) and slope \( m = \frac{2}{7} \):
\( y - (-1) = \frac{2}{7}(x - (-2)) \)
\( y + 1 = \frac{2}{7}(x + 2) \)
Multiply both sides by 7 to remove the fraction:
\( 7(y + 1) = 2(x + 2) \)
\( 7y + 7 = 2x + 4 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 2x - 7y + 4 - 7 = 0 \)
\( 2x - 7y - 3 = 0 \)
In simple words: For part (a), we used the distance formula to find the possible x-values. For part (b), we used the section formula to find how the y-axis divides the line, then found the exact point. We recognized that the shape formed was a trapezium and used its area formula. For part (c), we found the middle point of the opposite side and then the equation of the line connecting it to the vertex.

🎯 Exam Tip: When dealing with coordinate geometry problems, always clearly label your points and write down the formulas before substituting values. For perpendiculars to an axis, remember the coordinates of the projected points. For ratio problems, drawing a quick sketch can help visualize if the point is internal or external to the segment, though it's typically internal for these questions.

 

Question 10.
The graph of a linear equation in x and y, passes through A (- 1, - 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and \( (\frac{1}{2}, k) \).
(b) In the given figure, write,
(i) The coordinates of A, B and C.
(ii) The equation of the line through A and parallel to BC.
Answer:
(a) The line passes through points A(-1, -1) and B(2, 5).
First, we find the slope (m) of this line:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-1)}{2 - (-1)} = \frac{5 + 1}{2 + 1} = \frac{6}{3} = 2 \)
Now, we find the equation of the line using the point-slope form \( y - y_1 = m(x - x_1) \). Using point A(-1, -1):
\( y - (-1) = 2(x - (-1)) \)
\( y + 1 = 2(x + 1) \)
\( y + 1 = 2x + 2 \)
\( y = 2x + 1 \)
The line passes through point (h, 4). Substitute y = 4 into the equation:
\( 4 = 2h + 1 \)
\( 3 = 2h \)
\( h = \frac{3}{2} = 1.5 \)
The line also passes through point \( (\frac{1}{2}, k) \). Substitute \( x = \frac{1}{2} \) into the equation:
\( k = 2(\frac{1}{2}) + 1 \)
\( k = 1 + 1 \)
\( k = 2 \)
So, the values are h = 1.5 and k = 2. These points (h,4) and (1/2, k) must lie on the same straight line.
(b) (i) Based on the provided figure from the source, the coordinates of the points are:
A = (2, 3)
B = (-1, 2)
C = (3, 0)
(ii) We need to find the equation of a line that passes through A(2, 3) and is parallel to BC. First, calculate the slope of line segment BC.
Using points B(-1, 2) and C(3, 0):
Slope of BC (m_BC) = \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2}{3 - (-1)} = \frac{-2}{3 + 1} = \frac{-2}{4} = -\frac{1}{2} \)
Since the required line is parallel to BC, its slope (m_required) will be the same as m_BC:
\( m_{\text{required}} = -\frac{1}{2} \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with point A(2, 3) and slope \( m_{\text{required}} = -\frac{1}{2} \):
\( y - 3 = -\frac{1}{2}(x - 2) \)
Multiply both sides by 2 to clear the fraction:
\( 2(y - 3) = -1(x - 2) \)
\( 2y - 6 = -x + 2 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x + 2y - 6 - 2 = 0 \)
\( x + 2y - 8 = 0 \)
In simple words: For the first part, we found the line equation from two points, then used it to find missing coordinates. For the second part, we looked at the picture to get point values. Then, we found the slope of BC and used that same slope with point A to make the equation for the parallel line.

🎯 Exam Tip: Remember that parallel lines have the same slope. When reading coordinates from a graph, always double-check the values to avoid errors. The point-slope form is often the quickest way to write a line's equation if you have a point and a slope.

 

Question 11.
(a) The line segment joining A (2, 3) and B (6, - 5) is intercepted by the x-axis at the point K. Write the ordinate of the point K. Hence find the ratio in which K divides AB.
(b) If the lines \( y = 3x + 7 \) and \( 2y + px = 3 \) are perpendicular to each other, find the value of p.
(c) Find the coordinates of the centroid of a triangle whose vertices are : A (-1, 3), B (1, -1) and C (5, 1).
Answer:
(a) The line segment joins A(2, 3) and B(6, -5). It is intercepted by the x-axis at point K. Any point on the x-axis has a y-coordinate (ordinate) of 0. So, the ordinate of point K is 0. Let K be (x, 0).
To find the ratio in which K divides AB, let K divide AB in the ratio \( m_1 : m_2 \). We use the section formula for the y-coordinate, since we know its value (0):
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( 0 = \frac{m_1(-5) + m_2(3)}{m_1 + m_2} \)
\( 0 = -5m_1 + 3m_2 \)
\( 5m_1 = 3m_2 \)
\( \frac{m_1}{m_2} = \frac{3}{5} \)
Thus, the x-axis divides the line segment AB in the ratio 3:5. This means K is closer to B than to A.
(b) We are given two lines: \( y = 3x + 7 \) and \( 2y + px = 3 \).
For the first line, \( y = 3x + 7 \), its slope \( m_1 = 3 \) (by comparing with \( y = mx + c \)).
For the second line, \( 2y + px = 3 \), we rearrange it into the slope-intercept form:
\( 2y = -px + 3 \)
\( y = -\frac{p}{2}x + \frac{3}{2} \)
So, the slope of the second line \( m_2 = -\frac{p}{2} \).
Since the lines are perpendicular to each other, the product of their slopes must be -1:
\( m_1 \times m_2 = -1 \)
\( 3 \times (-\frac{p}{2}) = -1 \)
\( -\frac{3p}{2} = -1 \)
\( 3p = 2 \)
\( p = \frac{2}{3} \)
(c) The vertices of the triangle are A(-1, 3), B(1, -1), and C(5, 1).
To find the coordinates of the centroid G, we use the centroid formula:
\( G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}) \)
Substitute the coordinates of A, B, and C:
\( G = (\frac{-1 + 1 + 5}{3}, \frac{3 + (-1) + 1}{3}) \)
\( G = (\frac{5}{3}, \frac{3}{3}) \)
\( G = (\frac{5}{3}, 1) \)
In simple words: For (a), we used the y-coordinate to find how the x-axis splits the line. For (b), we found the slopes of both lines and used the rule that perpendicular lines have slopes that multiply to -1 to find 'p'. For (c), we added up all the x-coordinates and divided by 3, and did the same for the y-coordinates, to find the centroid.

🎯 Exam Tip: When a line intersects an axis, remember that one coordinate will be zero (x-coordinate is 0 for y-axis, y-coordinate is 0 for x-axis). For perpendicular lines, be careful with the negative reciprocal of the slope. Centroid calculation is a simple average of coordinates.

 

Question 12.
(a) The mid-point of the line segment joining (2a, 4) and (- 2, 2b) is (1, 2a + 1). Find the values of a and b.
(b) Find the equation of the line parallel to the line \( 3x + 2y = 8 \) and passing through the point (0, 1).
(c) If the line joining the points A (4, - 5) and B (4, 5) is divided by the point P such that \( \frac{\text{AP}}{\text{AB}} = \frac{2}{5} \), find the co-ordinates of P.
Answer:
(a) Let the two points be \( (x_1, y_1) = (2a, 4) \) and \( (x_2, y_2) = (-2, 2b) \). The midpoint is \( (x_m, y_m) = (1, 2a + 1) \).
Using the midpoint formula, \( x_m = \frac{x_1 + x_2}{2} \) and \( y_m = \frac{y_1 + y_2}{2} \).
For the x-coordinate:
\( 1 = \frac{2a + (-2)}{2} \)
\( 1 = \frac{2a - 2}{2} \)
\( 2 = 2a - 2 \)
\( 2a = 4 \)
\( a = 2 \)
For the y-coordinate:
\( 2a + 1 = \frac{4 + 2b}{2} \)
Substitute \( a = 2 \) into this equation:
\( 2(2) + 1 = \frac{4 + 2b}{2} \)
\( 4 + 1 = \frac{4 + 2b}{2} \)
\( 5 = \frac{4 + 2b}{2} \)
\( 10 = 4 + 2b \)
\( 2b = 6 \)
\( b = 3 \)
So, the values of a and b are 2 and 3 respectively. Checking the result can be done by plugging 'a' and 'b' back into the midpoint definition.
(b) The given line is \( 3x + 2y = 8 \). To find its slope, rearrange it into the slope-intercept form \( y = mx + c \):
\( 2y = -3x + 8 \)
\( y = -\frac{3}{2}x + 4 \)
The slope of this line is \( m_1 = -\frac{3}{2} \).
The required line is parallel to this given line, so its slope \( m_2 \) will be the same:
\( m_2 = -\frac{3}{2} \)
The required line passes through the point (0, 1). Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 1 = -\frac{3}{2}(x - 0) \)
\( y - 1 = -\frac{3}{2}x \)
Multiply both sides by 2 to remove the fraction:
\( 2(y - 1) = -3x \)
\( 2y - 2 = -3x \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 3x + 2y - 2 = 0 \)
(c) The points are A(4, -5) and B(4, 5). Point P divides the line segment AB such that \( \frac{\text{AP}}{\text{AB}} = \frac{2}{5} \). This means that the length AP is 2 parts out of a total of 5 parts for AB. Therefore, the length PB will be \( 5 - 2 = 3 \) parts. So, point P divides AB in the ratio AP : PB = 2 : 3. Let P have coordinates (x, y).
Using the section formula with \( m_1 = 2 \) and \( m_2 = 3 \):
For the x-coordinate of P:
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2(4) + 3(4)}{2 + 3} = \frac{8 + 12}{5} = \frac{20}{5} = 4 \)
For the y-coordinate of P:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{2(5) + 3(-5)}{2 + 3} = \frac{10 - 15}{5} = \frac{-5}{5} = -1 \)
The coordinates of P are (4, -1). This means P is located directly below the other points, as expected when the x-coordinates are the same.
In simple words: For (a), we used the midpoint formula to set up two equations and solved for 'a' and 'b'. For (b), we found the slope of the given line, then used that same slope with the given point to write the parallel line's equation. For (c), we first figured out the ratio in which P divides the line segment, then used the section formula to find P's coordinates.

🎯 Exam Tip: When using the midpoint formula, ensure you equate the correct coordinates (x with x, y with y). For parallel lines, remember their slopes are identical. In ratio division problems, clarify the ratio \( m_1:m_2 \) carefully from the given information (like AP/AB) before applying the section formula.

 

Question 13.
(a) If A = (- 4, 3) and B (8, - 6),
(i) Find the length of AB.
(ii) In what ratio is the line joining AB, divided by the x-axis?
(b) Co-ordinates of two points are A (7, - 3) and B (1, 9)
(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:
(a) (i) The points are A(-4, 3) and B(8, -6). To find the length of AB, we use the distance formula:
\( \text{AB} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \text{AB} = \sqrt{(8 - (-4))^2 + (-6 - 3)^2} \)
\( \text{AB} = \sqrt{(8 + 4)^2 + (-9)^2} \)
\( \text{AB} = \sqrt{(12)^2 + (-9)^2} \)
\( \text{AB} = \sqrt{144 + 81} \)
\( \text{AB} = \sqrt{225} \)
\( \text{AB} = 15 \) units. This distance represents the straight-line length between A and B.
(ii) The line joining A(-4, 3) and B(8, -6) is divided by the x-axis. Any point on the x-axis has a y-coordinate of 0. Let the point of division be K(x, 0). Let K divide AB in the ratio \( m_1 : m_2 \). We use the section formula for the y-coordinate:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( 0 = \frac{m_1(-6) + m_2(3)}{m_1 + m_2} \)
\( 0 = -6m_1 + 3m_2 \)
\( 6m_1 = 3m_2 \)
\( \frac{m_1}{m_2} = \frac{3}{6} = \frac{1}{2} \)
So, the x-axis divides the line segment AB in the ratio 1:2. The line crosses the x-axis one-third of the way from A to B.
(b) (i) The coordinates of two points are A(7, -3) and B(1, 9). To find the slope of AB, we use the slope formula:
\( \text{Slope of AB (m)} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - (-3)}{1 - 7} = \frac{9 + 3}{-6} = \frac{12}{-6} = -2 \)
(ii) To find the equation of the perpendicular bisector of the line segment AB, we need two things: its midpoint and its slope.
First, find the midpoint M of AB:
\( M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) = (\frac{7 + 1}{2}, \frac{-3 + 9}{2}) = (\frac{8}{2}, \frac{6}{2}) = (4, 3) \)
Next, find the slope of the perpendicular bisector. The slope of AB is \( m_{\text{AB}} = -2 \). The slope of a line perpendicular to AB (\( m_{\perp} \)) is the negative reciprocal of \( m_{\text{AB}} \):
\( m_{\perp} = -\frac{1}{m_{\text{AB}}} = -\frac{1}{-2} = \frac{1}{2} \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with the midpoint M(4, 3) and slope \( m_{\perp} = \frac{1}{2} \):
\( y - 3 = \frac{1}{2}(x - 4) \)
Multiply both sides by 2:
\( 2(y - 3) = 1(x - 4) \)
\( 2y - 6 = x - 4 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x - 2y - 4 + 6 = 0 \)
\( x - 2y + 2 = 0 \)
(iii) If the point (-2, p) lies on the perpendicular bisector \( x - 2y + 2 = 0 \), it must satisfy the equation. Substitute \( x = -2 \) and \( y = p \) into the equation:
\( (-2) - 2(p) + 2 = 0 \)
\( -2 - 2p + 2 = 0 \)
\( -2p = 0 \)
\( p = 0 \)
In simple words: For part (a), we calculated the distance between two points to find the length of AB. Then, we used the fact that the x-axis means y=0 to find how the line is split. For part (b), we found the slope of the line AB, then found the middle point. Using the opposite of the slope and the middle point, we wrote the equation for the perpendicular line. Finally, we put the given point's x-value into this equation to find 'p'.

🎯 Exam Tip: A perpendicular bisector passes through the midpoint of a line segment and is perpendicular to it. Make sure to use both these properties. Always simplify fractions for slopes and ratios. For finding p, ensure the point is substituted correctly into the equation.

 

Question 14.
Find the equation of a line with x-intercept = 5 and passing through the point (4, - 7).
Answer:
An x-intercept of 5 means the line passes through the point (5, 0).
The line also passes through the point (4, -7).
First, find the slope (m) of the line using these two points:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 0}{4 - 5} = \frac{-7}{-1} = 7 \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \). We can use either point, let's use (5, 0):
\( y - 0 = 7(x - 5) \)
\( y = 7x - 35 \)
Rearrange the equation into the standard form \( Ax + By + C = 0 \):
\( 7x - y - 35 = 0 \)
This equation describes all points on the line. The x-intercept tells us one point directly.
In simple words: We used the x-intercept to get a point (5,0). With this and the other point (4,-7), we found the line's steepness (slope). Then, using one point and the slope, we wrote the equation for the line.

🎯 Exam Tip: An x-intercept is where the line crosses the x-axis (y=0). A y-intercept is where it crosses the y-axis (x=0). Use these to form points when they are given, and then apply the standard line equation formulas.

 

Question 15.
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:
(i) Point A is on the x-axis, so its y-coordinate is 0. Let A = (x, 0).
Point B is on the y-axis, so its x-coordinate is 0. Let B = (0, y).
P(2, -3) is the midpoint of AB. We use the midpoint formula: \( x_m = \frac{x_1 + x_2}{2} \) and \( y_m = \frac{y_1 + y_2}{2} \).
For the x-coordinate of P:
\( 2 = \frac{x + 0}{2} \)
\( 4 = x \)
So, the coordinates of A are (4, 0).
For the y-coordinate of P:
\( -3 = \frac{0 + y}{2} \)
\( -6 = y \)
So, the coordinates of B are (0, -6).
The coordinates of A are (4, 0) and B are (0, -6).
(ii) Now we find the slope of line AB using A(4, 0) and B(0, -6):
\( \text{Slope of AB (m)} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 0}{0 - 4} = \frac{-6}{-4} = \frac{3}{2} \)
The slope tells us how steep the line is. For every 2 units moved horizontally, the line moves up 3 units.
(iii) To find the equation of line AB, we can use the point-slope form \( y - y_1 = m(x - x_1) \). Using point A(4, 0) and slope \( m = \frac{3}{2} \):
\( y - 0 = \frac{3}{2}(x - 4) \)
\( y = \frac{3}{2}(x - 4) \)
Multiply both sides by 2 to clear the fraction:
\( 2y = 3(x - 4) \)
\( 2y = 3x - 12 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 3x - 2y - 12 = 0 \)
In simple words: First, we used the midpoint formula to work backward from the midpoint to find the x- and y-intercepts (points A and B). Then, with those two points, we calculated the slope of the line. Finally, we used one of the points and the slope to write down the equation of the line.

🎯 Exam Tip: When a point is on the x-axis, its y-coordinate is zero. When on the y-axis, its x-coordinate is zero. This is crucial for problems involving intercepts or points on axes. The midpoint formula links the coordinates of two endpoints and their midpoint.

 

Question 16.
The equation of a line is \( 3x + 4y - 7 = 0 \). 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:
(i) The given line is \( 3x + 4y - 7 = 0 \). To find its slope, rearrange the equation into the slope-intercept form \( y = mx + c \):
\( 4y = -3x + 7 \)
\( y = -\frac{3}{4}x + \frac{7}{4} \)
The slope of this line is \( m = -\frac{3}{4} \).
(ii) First, we need to find the point of intersection of the lines \( x - y + 2 = 0 \) and \( 3x + y - 10 = 0 \). We can solve these two linear equations simultaneously. Add the two equations together:
\( (x - y + 2) + (3x + y - 10) = 0 \)
\( x - y + 2 + 3x + y - 10 = 0 \)
\( 4x - 8 = 0 \)
\( 4x = 8 \)
\( x = 2 \)
Now, substitute \( x = 2 \) into the first equation \( x - y + 2 = 0 \):
\( 2 - y + 2 = 0 \)
\( 4 - y = 0 \)
\( y = 4 \)
So, the point of intersection is (2, 4). This is the common point for both lines.
Next, we need the slope of a line perpendicular to \( 3x + 4y - 7 = 0 \). From part (i), the slope of this line is \( m_1 = -\frac{3}{4} \). The slope of a perpendicular line (\( m_{\perp} \)) is the negative reciprocal:
\( m_{\perp} = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3} \)
Now, we have a point (2, 4) and a slope \( m_{\perp} = \frac{4}{3} \). We can find the equation of the line using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 4 = \frac{4}{3}(x - 2) \)
Multiply both sides by 3:
\( 3(y - 4) = 4(x - 2) \)
\( 3y - 12 = 4x - 8 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 4x - 3y - 8 + 12 = 0 \)
\( 4x - 3y + 4 = 0 \)
In simple words: First, we found the steepness of the given line. Then, we solved two equations to find the exact point where they cross. Since the new line needs to be at a right angle, we took the opposite flip of the first line's steepness. Finally, we used this new steepness and the crossing point to write the equation for the new line.

🎯 Exam Tip: Solving simultaneous equations is key to finding the intersection point of two lines. Remember that the slopes of perpendicular lines multiply to -1. Always convert equations to slope-intercept form (y=mx+c) to easily identify the slope.

 

Question 17.
ABC is a triangle and G (4, 3) is the centroid of the triangle. If A = (1, 3), B = (4, b) and C = (a, 1), find 'a' and 'b'. Find the length of side BC.
Answer:
The vertices of the triangle are A(1, 3), B(4, b), and C(a, 1). The centroid G is given as (4, 3).
The formula for the centroid \( (x_g, y_g) \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
\( x_g = \frac{x_1 + x_2 + x_3}{3} \)
\( y_g = \frac{y_1 + y_2 + y_3}{3} \)
Substitute the known x-coordinates and \( x_g \):
\( 4 = \frac{1 + 4 + a}{3} \)
\( 4 = \frac{5 + a}{3} \)
Multiply by 3:
\( 12 = 5 + a \)
\( a = 12 - 5 \)
\( a = 7 \)
Substitute the known y-coordinates and \( y_g \):
\( 3 = \frac{3 + b + 1}{3} \)
\( 3 = \frac{4 + b}{3} \)
Multiply by 3:
\( 9 = 4 + b \)
\( b = 9 - 4 \)
\( b = 5 \)
So, the values are a = 7 and b = 5. This tells us the exact coordinates of points B and C are B(4, 5) and C(7, 1).
Now, we need to find the length of side BC using the coordinates B(4, 5) and C(7, 1). We use the distance formula:
\( \text{BC} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \text{BC} = \sqrt{(7 - 4)^2 + (1 - 5)^2} \)
\( \text{BC} = \sqrt{(3)^2 + (-4)^2} \)
\( \text{BC} = \sqrt{9 + 16} \)
\( \text{BC} = \sqrt{25} \)
\( \text{BC} = 5 \) units.
In simple words: We used the centroid formula, which averages all x-coordinates and all y-coordinates, to find the unknown 'a' and 'b'. Once we had all the point values, we used the distance formula to find the length of the side BC.

🎯 Exam Tip: The centroid is the balancing point of a triangle. The centroid formula is a straightforward way to find missing coordinates or verify the centroid's position. Always double-check calculations for signs and arithmetic, especially when substituting values.

 

Question 18.
ABCD is a parallelogram when A (x, y), B (5, 8) and C (4, 7) and D (2, -4). Find :
(i) Coordinates of A
(ii) Equation of diagonal BD.
Answer:
(i) In a parallelogram, the diagonals bisect each other. This means that the midpoint of diagonal AC is the same as the midpoint of diagonal BD. Let this common midpoint be M.
Coordinates of A are (x, y) and C are (4, 7).
Midpoint of AC: \( M_{\text{AC}} = (\frac{x + 4}{2}, \frac{y + 7}{2}) \)
Coordinates of B are (5, 8) and D are (2, -4).
Midpoint of BD: \( M_{\text{BD}} = (\frac{5 + 2}{2}, \frac{8 + (-4)}{2}) = (\frac{7}{2}, \frac{4}{2}) = (3.5, 2) \)
Since \( M_{\text{AC}} = M_{\text{BD}} \), we can equate their coordinates:
Equating x-coordinates:
\( \frac{x + 4}{2} = 3.5 \)
\( x + 4 = 7 \)
\( x = 7 - 4 \)
\( x = 3 \)
Equating y-coordinates:
\( \frac{y + 7}{2} = 2 \)
\( y + 7 = 4 \)
\( y = 4 - 7 \)
\( y = -3 \)
Therefore, the coordinates of A are (3, -3). The property of diagonals bisecting each other is a key characteristic of parallelograms.
(ii) To find the equation of diagonal BD, we use the coordinates B(5, 8) and D(2, -4).
First, find the slope (m) of BD:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 8}{2 - 5} = \frac{-12}{-3} = 4 \)
Now, use the point-slope form \( y - y_1 = m(x - x_1) \). Using point D(2, -4) and slope m=4:
\( y - (-4) = 4(x - 2) \)
\( y + 4 = 4x - 8 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 4x - y - 8 - 4 = 0 \)
\( 4x - y - 12 = 0 \)
In simple words: For part (i), we used the rule that the middle point of the two diagonals in a parallelogram is the same. We found the midpoint of BD, then used it to figure out the missing coordinates of A. For part (ii), with the coordinates of B and D, we first found the slope (steepness) of the diagonal BD. Then, using one of the points and the slope, we wrote the equation for that line.

🎯 Exam Tip: Remember that for a parallelogram, the midpoints of both diagonals coincide. This is a common property used to find unknown coordinates. When finding the equation of a line, always clearly state the two points or one point and the slope you are using.

 

Question 19.
Given equation of line L1 is \( y = 4 \).
(i) Write the slope of line L2 if L2 is the bisector of angle O.
(ii) Write the coordinates of point P.
(iii) Find the equation of L2.
Answer:
(i) From the figure (referring to the image in the source content), L2 is shown passing through the origin (O) and bisecting the angle between the positive x-axis and the positive y-axis. The angle between these axes is \( 90^\circ \). Therefore, L2 makes an angle of \( \frac{90^\circ}{2} = 45^\circ \) with the positive x-axis.
The slope (m) of a line is given by \( m = \tan \theta \).
So, the slope of L2 is \( m = \tan 45^\circ = 1 \).
(ii) Line L1 is given by the equation \( y = 4 \). Line L2 passes through the origin (0, 0) and has a slope of 1, so its equation is \( y = x \). Point P is the intersection of L1 and L2.
To find the coordinates of P, substitute the value of y from L1 into the equation of L2:
Since \( y = 4 \) and \( y = x \), then \( x = 4 \).
So, the coordinates of point P are (4, 4). This point is equidistant from both axes.
(iii) Line L2 passes through the origin (0, 0) and has a slope of 1. Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 0 = 1(x - 0) \)
\( y = x \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x - y = 0 \)
In simple words: For (i), we understood that "bisector of angle O" means it cuts the 90-degree angle between the axes in half, making a 45-degree angle. The slope of a 45-degree line is 1. For (ii), we found where the horizontal line y=4 crosses the line L2 (which is y=x), giving us the point P. For (iii), we used the origin (0,0) and the slope of 1 to write the equation of L2.

🎯 Exam Tip: The slope of a line that makes a \( 45^\circ \) angle with the x-axis is 1. The equation of such a line passing through the origin is \( y = x \). Understanding the relationship between angle, slope, and standard line forms is important.

 

Question 20.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Answer:
The given points are A(-4, 6) and B(8, -3).
(i) The line segment AB is divided by the y-axis. Any point on the y-axis has an x-coordinate of 0. Let the point of intersection be R(0, y). Let R divide AB in the ratio \( m_1 : m_2 \). We use the section formula for the x-coordinate, as we know its value (0):
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \)
\( 0 = \frac{m_1(8) + m_2(-4)}{m_1 + m_2} \)
\( 0 = 8m_1 - 4m_2 \)
\( 8m_1 = 4m_2 \)
\( \frac{m_1}{m_2} = \frac{4}{8} = \frac{1}{2} \)
So, the y-axis divides the line segment AB in the ratio 1:2. This means the intersection point is one-third of the way from A to B.
(ii) Now that we have the ratio \( m_1 : m_2 = 1:2 \), we can find the y-coordinate of the point of intersection R(0, y) using the section formula for y-coordinates:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( y = \frac{1(-3) + 2(6)}{1 + 2} \)
\( y = \frac{-3 + 12}{3} \)
\( y = \frac{9}{3} \)
\( y = 3 \)
The coordinates of the point of intersection R are (0, 3).
(iii) To find the length of AB, we use the distance formula with points A(-4, 6) and B(8, -3):
\( \text{AB} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \text{AB} = \sqrt{(8 - (-4))^2 + (-3 - 6)^2} \)
\( \text{AB} = \sqrt{(8 + 4)^2 + (-9)^2} \)
\( \text{AB} = \sqrt{(12)^2 + (-9)^2} \)
\( \text{AB} = \sqrt{144 + 81} \)
\( \text{AB} = \sqrt{225} \)
\( \text{AB} = 15 \) units. This is the exact length of the line segment AB.
In simple words: For (i), we used the fact that any point on the y-axis has an x-coordinate of zero. We then applied the section formula to find the ratio in which the y-axis cuts the line. For (ii), we used this ratio with the section formula again, this time for the y-coordinates, to find the exact point where the line crosses the y-axis. For (iii), we used the distance formula to calculate the straight-line length between the two given points.

🎯 Exam Tip: Remember that "dividing by the y-axis" implies the x-coordinate of the intersection point is zero. The distance formula is essential for finding lengths of segments, and you should be comfortable with squaring negative numbers (they always become positive). The section formula is versatile for finding points that divide segments in a given ratio.

 

Question 21. 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-ordinate of Q.
Answer:
(i) Here, the angle \( \theta \) is \( 45^\circ \). The slope of the line (m) is found using \( \tan \theta \).
Slope \( (m) = \tan 45^\circ = 1 \). This means the line rises at a 45-degree angle from the x-axis.
(ii) We use the point-slope form \( y - y_1 = m(x - x_1) \). Given point P(5, 3) and slope \( m = 1 \).
\( y - 3 = 1(x - 5) \)
\( y - 3 = x - 5 \)
\( y = x - 5 + 3 \)
\( y = x - 2 \) or \( x - y - 2 = 0 \). This is the equation of the line.
(iii) When the line intersects the y-axis, the x-coordinate of that point (Q) is 0.
Substitute \( x = 0 \) into the equation \( y = x - 2 \).
\( y = 0 - 2 \)
\( y = -2 \)
So, the coordinates of Q are \( (0, -2) \). This point is where the line crosses the y-axis.
In simple words: The line goes up at a 45-degree slant, so its slope is 1. Using this, we find the line's rule (equation). To find where it crosses the y-axis, we just set the x-value to zero in our rule.

🎯 Exam Tip: Remember that the slope of a line inclined at \( \theta \) to the x-axis is \( \tan \theta \). For intersections with the y-axis, always set \( x = 0 \), and for the x-axis, set \( y = 0 \).

 

Question 22.
(a) AB is a diameter of a circle with centre C = (- 2, 5). If A = (3, - 7), find the coordinates of B.
(b) In \( \triangle ABC \), A (3, 5), B (7, 8) and C (1, - 10). Find the equation of the median through A.
Answer:
(a) For a circle, the center is always the midpoint of its diameter. Here, C is the midpoint of AB.
Let the coordinates of B be \( (x, y) \).
Using the midpoint formula: \( C_x = \frac{A_x + B_x}{2} \) and \( C_y = \frac{A_y + B_y}{2} \).
For the x-coordinate: \( -2 = \frac{3 + x}{2} \)
\( -4 = 3 + x \)
\( x = -4 - 3 \)
\( x = -7 \).
For the y-coordinate: \( 5 = \frac{-7 + y}{2} \)
\( 10 = -7 + y \)
\( y = 10 + 7 \)
\( y = 17 \).
So, the coordinates of B are \( (-7, 17) \). This point is on the opposite side of the circle from A, through the center.
(b) A median of a triangle connects a vertex to the midpoint of the opposite side. The median through A will connect A to the midpoint of BC. Let's call this midpoint D.
Coordinates of B are (7, 8) and C are (1, -10).
Midpoint D of BC: \( D_x = \frac{7+1}{2} = \frac{8}{2} = 4 \)
\( D_y = \frac{8+(-10)}{2} = \frac{-2}{2} = -1 \).
So, the coordinates of D are \( (4, -1) \).
Now, we need the equation of the line passing through A(3, 5) and D(4, -1).
First, calculate the slope of AD: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m = \frac{-1 - 5}{4 - 3} = \frac{-6}{1} = -6 \).
Now use the point-slope form with A(3, 5) and slope \( m = -6 \):
\( y - y_1 = m(x - x_1) \)
\( y - 5 = -6(x - 3) \)
\( y - 5 = -6x + 18 \)
\( y + 6x = 18 + 5 \)
\( 6x + y = 23 \) or \( 6x + y - 23 = 0 \). This is the equation for the median AD.
In simple words: For part (a), the center of a circle is exactly in the middle of its diameter. We use this "midpoint rule" to find the coordinates of the other end of the diameter. For part (b), a median line goes from one corner of a triangle to the middle of the side across from it. We first find the middle point of side BC, then find the rule for the line that goes through corner A and this middle point.

🎯 Exam Tip: For problems involving diameters and centers, the center is always the midpoint. For medians, always calculate the midpoint of the opposite side first, then find the equation of the line joining the vertex to that midpoint.

 

Question 23. In the figure given below, the line segment AB meets X-axis at A and Y-axis at B. The point P (- 3, 4) on AB divides it in the ratio 2 : 3. Find the coordinates of A and B.
Answer:
Let the coordinates of A be \( (x, 0) \) since it lies on the x-axis.
Let the coordinates of B be \( (0, y) \) since it lies on the y-axis.
Point P(-3, 4) divides the line segment AB in the ratio 2:3. We can use the section formula:
\( P_x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \) and \( P_y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \).
Here, \( (x_1, y_1) = (x, 0) \), \( (x_2, y_2) = (0, y) \), \( (P_x, P_y) = (-3, 4) \), and \( m_1 = 2, m_2 = 3 \).
For the x-coordinate:
\( -3 = \frac{2 \times 0 + 3 \times x}{2 + 3} \)
\( -3 = \frac{3x}{5} \)
\( 3x = -3 \times 5 \)
\( 3x = -15 \)
\( x = -5 \).
So, the coordinates of A are \( (-5, 0) \). This point is on the negative side of the x-axis.
For the y-coordinate:
\( 4 = \frac{2 \times y + 3 \times 0}{2 + 3} \)
\( 4 = \frac{2y}{5} \)
\( 2y = 4 \times 5 \)
\( 2y = 20 \)
\( y = 10 \).
So, the coordinates of B are \( (0, 10) \). This point is on the positive side of the y-axis.
In simple words: We know point P splits line AB into a 2:3 ratio. Point A is on the x-axis (y is zero), and B is on the y-axis (x is zero). We use a special formula called the section formula to work backward from P's coordinates and the ratio to find the exact locations of A and B.

🎯 Exam Tip: When a point divides a line segment in a given ratio, the section formula is key. Remember that points on the x-axis have a y-coordinate of 0, and points on the y-axis have an x-coordinate of 0.

 

Question 24.
(a) Calculate the ratio in which the line joining A (-4, 2) and B (3, 6) is divided by point P (x, 3). Also find (i) x (ii) Length of AP.
(b) Find the value of 'a' for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line.
Answer:
(a) Let point P(x, 3) divide the line segment joining A(-4, 2) and B(3, 6) in the ratio \( k:1 \).
Using the section formula for the y-coordinate (since we know the y-coordinate of P is 3):
\( P_y = \frac{k y_2 + 1 y_1}{k + 1} \)
\( 3 = \frac{k(6) + 1(2)}{k + 1} \)
\( 3(k + 1) = 6k + 2 \)
\( 3k + 3 = 6k + 2 \)
\( 3 - 2 = 6k - 3k \)
\( 1 = 3k \)
\( k = \frac{1}{3} \).
So, the ratio is \( 1:3 \).
(i) Now use the section formula for the x-coordinate with \( k = \frac{1}{3} \) (ratio \( 1:3 \)):
\( x = \frac{1 \times 3 + 3 \times (-4)}{1 + 3} \)
\( x = \frac{3 - 12}{4} \)
\( x = \frac{-9}{4} \).
So, the x-coordinate of P is \( -\frac{9}{4} \). The point P is \( (-\frac{9}{4}, 3) \).
(ii) To find the length of AP, use the distance formula between A(-4, 2) and P\( (-\frac{9}{4}, 3) \).
Length \( AP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( AP = \sqrt{(-\frac{9}{4} - (-4))^2 + (3 - 2)^2} \)
\( AP = \sqrt{(-\frac{9}{4} + \frac{16}{4})^2 + (1)^2} \)
\( AP = \sqrt{(\frac{7}{4})^2 + 1^2} \)
\( AP = \sqrt{\frac{49}{16} + 1} = \sqrt{\frac{49}{16} + \frac{16}{16}} \)
\( AP = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4} \). The distance is a positive length.
(b) For points A(a, 3), B(2, 1), and C(5, a) to be collinear (lie on the same straight line), the slope of AB must be equal to the slope of BC.
Slope of AB \( = \frac{1 - 3}{2 - a} = \frac{-2}{2 - a} \).
Slope of BC \( = \frac{a - 1}{5 - 2} = \frac{a - 1}{3} \).
Equating the slopes:
\( \frac{-2}{2 - a} = \frac{a - 1}{3} \)
Cross-multiply:
\( -2 \times 3 = (a - 1)(2 - a) \)
\( -6 = 2a - a^2 - 2 + a \)
\( -6 = -a^2 + 3a - 2 \)
Rearrange into a quadratic equation:
\( a^2 - 3a - 6 + 2 = 0 \)
\( a^2 - 3a - 4 = 0 \)
Factor the quadratic equation:
\( a^2 - 4a + a - 4 = 0 \)
\( a(a - 4) + 1(a - 4) = 0 \)
\( (a + 1)(a - 4) = 0 \)
So, \( a = -1 \) or \( a = 4 \).
If \( a = -1 \), the points are A(-1, 3), B(2, 1), C(5, -1).
If \( a = 4 \), the points are A(4, 3), B(2, 1), C(5, 4).
Let's find the equation of the line using \( a = 4 \). Points are A(4, 3), B(2, 1), C(5, 4).
Using points A(4, 3) and B(2, 1).
Slope \( m = \frac{1 - 3}{2 - 4} = \frac{-2}{-2} = 1 \).
Using the point-slope form with A(4, 3) and \( m = 1 \):
\( y - y_1 = m(x - x_1) \)
\( y - 3 = 1(x - 4) \)
\( y - 3 = x - 4 \)
\( x - y - 1 = 0 \). This is the equation of the line when \( a=4 \).
(The problem statement does not specify which value of 'a' to use, but usually a unique value is implied. If a value does not satisfy the condition, it is discarded, but here both seem valid for collinearity).
In simple words: In part (a), we find how point P cuts the line between A and B using the y-coordinate. Then we use this ratio to find the x-coordinate of P and the length from A to P. In part (b), if three points are on the same straight line, the slope between any two pairs of points must be the same. We use this idea to find the missing value 'a' and then write down the rule for that line.

🎯 Exam Tip: For collinear points, equating slopes is a standard method. Be careful with algebraic manipulations in solving quadratic equations. When calculating length, use the distance formula accurately, especially with fractions.

 

Question 25. Three vertices of a parallelogram ABCD taken in order ae 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) In a parallelogram, the diagonals bisect each other, meaning they share the same midpoint.
Let the coordinates of the fourth vertex D be \( (x, y) \).
The midpoint of diagonal AC is: \( (\frac{3+3}{2}, \frac{6+2}{2}) = (\frac{6}{2}, \frac{8}{2}) = (3, 4) \).
The midpoint of diagonal BD is: \( (\frac{5+x}{2}, \frac{10+y}{2}) \).
Since these midpoints are the same:
\( \frac{5+x}{2} = 3 \implies 5+x = 6 \implies x = 1 \).
\( \frac{10+y}{2} = 4 \implies 10+y = 8 \implies y = -2 \).
So, the coordinates of the fourth vertex D are \( (1, -2) \). This point completes the parallelogram.
(ii) To find the length of diagonal BD, we use the distance formula between B(5, 10) and D(1, -2).
Length \( BD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( BD = \sqrt{(1 - 5)^2 + (-2 - 10)^2} \)
\( BD = \sqrt{(-4)^2 + (-12)^2} \)
\( BD = \sqrt{16 + 144} = \sqrt{160} \).
\( \sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10} \approx 12.65 \) units.
(iii) To find the equation of side AB, we use points A(3, 6) and B(5, 10).
First, calculate the slope of AB: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m = \frac{10 - 6}{5 - 3} = \frac{4}{2} = 2 \).
Now use the point-slope form with A(3, 6) and slope \( m = 2 \):
\( y - y_1 = m(x - x_1) \)
\( y - 6 = 2(x - 3) \)
\( y - 6 = 2x - 6 \)
\( 2x - y = 0 \). This equation describes the line segment AB.
In simple words: For a parallelogram, the middle point of one diagonal is the same as the middle point of the other. We use this fact to find the missing corner D. Then, we use the distance rule to measure how long the diagonal BD is. Finally, we find the rule (equation) for the line forming side AB using its two corner points.

🎯 Exam Tip: The property that diagonals of a parallelogram bisect each other is crucial for finding missing vertices. Remember that two points are needed to find the equation of a line, either directly or by first calculating the slope.

 

Question 26. In the given figure ABC is a triangle and BC is parallel to the y-axis. AB and AC intersects the y-axis at P and Q respectively.
(i) Write the coordinates of A.
(ii) Find the length of AB and AC.
Answer:
(i) From the figure (assuming it's similar to typical grid diagrams where points are marked), the coordinates of A are (4, 0). Point A lies on the x-axis, so its y-coordinate is 0.
(ii) Let's assume the coordinates from the provided image on page 50, where B is (-2, 3) and C is (-2, -4). A is (4, 0).
Length of AB: Use the distance formula between A(4, 0) and B(-2, 3).
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( AB = \sqrt{(-2 - 4)^2 + (3 - 0)^2} \)
\( AB = \sqrt{(-6)^2 + (3)^2} \)
\( AB = \sqrt{36 + 9} = \sqrt{45} \approx 6.71 \) units. This is the length of one side of the triangle.
Length of AC: Use the distance formula between A(4, 0) and C(-2, -4).
\( AC = \sqrt{(-2 - 4)^2 + (-4 - 0)^2} \)
\( AC = \sqrt{(-6)^2 + (-4)^2} \)
\( AC = \sqrt{36 + 16} = \sqrt{52} \approx 7.21 \) units. This is the length of another side.
(iii) To find the ratio in which Q divides AC, we use the fact that Q lies on the y-axis, so its x-coordinate is 0. Let Q divide AC in the ratio \( k:1 \). A(4, 0) and C(-2, -4).
Using the section formula for the x-coordinate of Q:
\( Q_x = \frac{k x_2 + 1 x_1}{k + 1} \)
\( 0 = \frac{k(-2) + 1(4)}{k + 1} \)
\( 0 = -2k + 4 \)
\( 2k = 4 \)
\( k = 2 \).
So, the ratio in which Q divides AC is \( 2:1 \). This means Q is closer to C than to A.
(iv) To find the equation of the line AC, use points A(4, 0) and C(-2, -4).
Slope of AC: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m = \frac{-4 - 0}{-2 - 4} = \frac{-4}{-6} = \frac{2}{3} \).
Using the point-slope form with A(4, 0) and slope \( m = \frac{2}{3} \):
\( y - y_1 = m(x - x_1) \)
\( y - 0 = \frac{2}{3}(x - 4) \)
\( 3y = 2(x - 4) \)
\( 3y = 2x - 8 \)
\( 2x - 3y - 8 = 0 \). This is the equation for line AC.
In simple words: We first look at the picture to find the location of point A. Then, we use the distance rule to calculate the lengths of sides AB and AC of the triangle. Next, we figure out how point Q splits the line AC, knowing that Q is on the y-axis. Finally, we find the rule (equation) for the straight line that connects A and C.

🎯 Exam Tip: When points are given in a figure, always explicitly state the coordinates you are using from the figure. Remember the distance formula and the section formula are essential tools for coordinate geometry problems.

 

Question 27.
(a) The slope of a line joining P (6, k) and Q (1 - 3k, 3) is \( \frac { 1 }{ 2 } \). Find (i) k (ii) Midpoint of PQ, using the value of 'k' found in (i).
(b) A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1: 2. (i) Find the coordinates of A and B. (ii) Find the equation of the line through P and perpendicular to AB.
Answer:
(a) (i) Given points P(6, k) and Q(1 - 3k, 3). The slope is \( \frac{1}{2} \).
The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
\( \frac{1}{2} = \frac{3 - k}{(1 - 3k) - 6} \)
\( \frac{1}{2} = \frac{3 - k}{-3k - 5} \)
Cross-multiply:
\( 1(-3k - 5) = 2(3 - k) \)
\( -3k - 5 = 6 - 2k \)
\( -5 - 6 = -2k + 3k \)
\( -11 = k \).
So, the value of k is -11.
(ii) Now, substitute \( k = -11 \) back into the coordinates of P and Q.
P becomes \( (6, -11) \).
Q becomes \( (1 - 3(-11), 3) = (1 + 33, 3) = (34, 3) \).
Midpoint of PQ is \( (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) \).
Midpoint \( = (\frac{6 + 34}{2}, \frac{-11 + 3}{2}) \)
Midpoint \( = (\frac{40}{2}, \frac{-8}{2}) \)
Midpoint \( = (20, -4) \). This is the exact center point between P and Q.
(b) (i) Let A be \( (x, 0) \) on the x-axis and B be \( (0, y) \) on the y-axis.
Point P(4, -1) divides AB in the ratio 1:2. So \( m_1 = 1, m_2 = 2 \).
Using the section formula:
For the x-coordinate:
\( 4 = \frac{1 \times 0 + 2 \times x}{1 + 2} \)
\( 4 = \frac{2x}{3} \)
\( 12 = 2x \)
\( x = 6 \).
So, the coordinates of A are \( (6, 0) \).
For the y-coordinate:
\( -1 = \frac{1 \times y + 2 \times 0}{1 + 2} \)
\( -1 = \frac{y}{3} \)
\( y = -3 \).
So, the coordinates of B are \( (0, -3) \).
(ii) First, find the slope of line AB. A(6, 0), B(0, -3).
Slope of AB \( = \frac{-3 - 0}{0 - 6} = \frac{-3}{-6} = \frac{1}{2} \).
The line required passes through P(4, -1) and is perpendicular to AB.
If two lines are perpendicular, the product of their slopes is -1.
Slope of perpendicular line \( = \frac{-1}{\text{slope of AB}} = \frac{-1}{1/2} = -2 \).
Now use the point-slope form for the required line with P(4, -1) and slope \( m = -2 \):
\( y - y_1 = m(x - x_1) \)
\( y - (-1) = -2(x - 4) \)
\( y + 1 = -2x + 8 \)
\( 2x + y + 1 - 8 = 0 \)
\( 2x + y - 7 = 0 \). This is the equation of the line perpendicular to AB and passing through P.
In simple words: For part (a), we are given two points and the slope of the line that connects them. We use the slope formula to find the missing number 'k', then use 'k' to find the exact middle point of the line. For part (b), we know point P divides line AB, and A is on the x-axis while B is on the y-axis. We use a division rule to find A and B's locations. Then, we find the rule for a new line that passes through P and is at a right angle (perpendicular) to the line AB.

🎯 Exam Tip: Clearly distinguish between the ratio formula for a point on a line segment and the slope formula. When dealing with perpendicular lines, remember that their slopes multiply to -1. Be careful with negative signs in calculations.