OP Malhotra Class 10 Maths Solutions Chapter 11 Coordinate Geometry Ex 11(c)

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S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(c)

Question 1.
State the slope of the line
(i) 3x + 4y = 9
(ii) 3x + 2y – 4
Solution:
(i) 3x + 4y = 9 ⇒ 4y = – 3x + 9
⇒ y = \(\frac { -3 }{ 4 }\)x + \(\frac { 9 }{ 4 }\)
Which is in the form of y = mx + c
∴ Slope (m) = \(\frac { -3 }{ 4 }\)

(ii) 3x + 2y = 4 ⇒ 2y = – 3x + 4
⇒ y = \(\frac { -3 }{ 2 }\)x + \(\frac { 4 }{ 2 }\) ⇒ y = \(\frac { -3 }{ 2 }\)x + 2
∴ Slope (m) = \(\frac { -3 }{ 2 }\)

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.
Solution:
(i) 3x + 2y + 4 = 0 ⇒ 2y – -3x – 4
⇒ y = \(\frac { -3 }{ 2 }\)x – 2 which is in the form of y – mx + c

(ii) Slope (m) = \(\frac { -3 }{ 2 }\) and y-intercept (c) = – 2

(iii) The line has been drawn on the graph paper as given here.

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° to the x-axis ;
(e) – 5 and is equally inclined to the axes.
Solution:
We know that the equation of a line is y = mx + c
Where m is the slope and c is the y-intercept.
Therefore
(a) y-intercept (c) = 2
slope (m) = 7
∴ Equation of the line will be y = 7x + 2⇒7x-y-t-2 = 0

(b) Slope (m) = – 4
and y-intercept (c) = – 3
∴ Equation of the line will be y = – 4x – 3 or 4x + y + 3 = 0

(c) y-intercept (c) = – 1
∵ The line is parallel to the line y = 5x – 7
Slope (m) = 5
∴ The required line will be
y = 5x – 1
(∵ parallel line have equal slopes)

(d) y-intercept (c) = 2
∵ The equation inclined at 45° to the x-axis
∴ Slope (m) = tan θ = tan 45° = 1
∴ Equation of the required line will be
y = 1 x + 2 ⇒ y = x + 2
x – y + 2 = 0

(e) y-intercept (c) = – 5
∵ The line is equally inclined to the axes
∴ Slope (m) = tan θ = + 1
∴ Equation of the line will be
y – 1x – 5 ⇒ y = x – 5
x – y = 5
or y = – 1x – 5 ⇒ y = – x – 5 or x + y + 5 = 0

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) ?
Solution:
The equation of line is y = mx + c … (i)
∵ It passes through the points (3, – 4) and (-1, 2)
Substituting the values of x and y in (i)
– 4 = 3 m + c … (ii)
and 2 = – 1m + c ⇒ 2 = – m + c … (iii)
Subtracting we get,
– 6 = 4 m ⇒ m = \(\frac { -6 }{ 4 }\) = \(\frac { -3 }{ 2 }\)
From (ii)
– 4 = 3 × (\(\frac { -3 }{ 2 }\)) + c ⇒ – 4\(\frac { -9 }{ 2 }\) + c
⇒ c = – 4 + \(\frac { -9 }{ 2 }\) = \(\frac { -8+9 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
∴ m = \(\frac { -3 }{ 2 }\) and c = \(\frac { 1 }{ 2 }\)

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.
Solution:
∵ The line y = mx + c passes through the points (1, 4) and (-2, -5)
∴ Substituting the values of x and y, we get
4 = 1m + c ⇒ 4 = m + c
– 5 = – 2m + c ⇒ – 5 = – 2m + c … (ii)
Subtracting, we get
9 = 3m ⇒ m = \(\frac { 9 }{ 3 }\) = 3
From (i)
4 = 3 + c ⇒ c = 4 – 3 = 1
Hence m = 3 and c = 1

Question 6.
Find the equation of the straight line through the given point P and having the given slope m if
(a) P (- 4, 7); m = – \(\sqrt{3}\);
(b) P (- 1, – 5), m = \(\frac { -6 }{ 11 }\).
Solution:
(a) The equation of the line will be y – y₁ = m (x – x₁)
⇒ Here it passes through P (- 4, 7) and m = – \(\sqrt{3}\)
∴ y – 7 = \(\sqrt{3}\) [x – (- 4)]
⇒ y – 7 = – \(\sqrt{3}\) x – 4\(\sqrt{3}\)
⇒ \(\sqrt{3}\)x + y = 7 – 4\(\sqrt{3}\)

(b) ∵ The line passes through the point P (- 1, – 5)
and has slope = \(\frac { -6 }{ 11 }\)
∴ Equation of the line will be
y – y₁ = m (x – x₁).
⇒ y + 5 = \(\frac { -6 }{ 11 }\) (x + 1)
⇒ 11 y + 55 = – 6x – 6
⇒ 6x + 11y + 55 + 6 = 0
⇒ 6x + 11y + 61 = 0

Question 7.
Find the equation to the straight line passing through :
(a) the origin and perpendicular to x + 2v = 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.
Solution:
(a) Slope of the line x + 2y = 4 ⇒ 2y = – x + 4
⇒ y = \(\frac { -1 }{ 2 }\)x + 2
m₁ = \(\frac { -1 }{ 2 }\) and slope of the line perpendicular
∵ The required equation passes through the origin (0, 0)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 0 = 2(x – 0) ⇒ y – 2x
⇒ 2x – y = 0

(b) Slope of the given line 3x + 4y = 12
⇒ 4y = – 3x + 12 ⇒ y = \(\frac { -3 }{ 4 }\)x + 3
m = \(\frac { -3 }{ 4 }\)
∴ Slope of the line parallel to the given line = \(\frac { -3 }{ 4 }\)
∵ It passes through the point (4, 3)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { -3 }{ 4 }\)(x – 4)
⇒ 4y – 12 = – 3x + 12
⇒ 3x + 4y = 12 + 12
⇒ 3x + 4y = 24
⇒ 3x + 4y – 24 = 0

(c) Slope of the given line 3x – 2y + 5 = 0
⇒ 2y = 3x + 5 ⇒ y = \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 2 }\)
m = \(\frac { 3 }{ 2 }\)
(i) Slope of the required line which is parallel to the given line = m = \(\frac { 3 }{ 2 }\)
∵ It passes through the point (4, 5)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 5 = \(\frac { 3 }{ 2 }\) (x – 4) ⇒ 2y – 10 = 3x – 12
⇒ 3x – 2y + 10 – 12 = 0
⇒ 3x – 2y – 2 = 0

(ii) Slope of the line perpendicular to the given line = \(\frac { -1 }{ m }\)
m₁ = \(\frac { -2 }{ 3 }\)
∵ It passes through the point (4, 5)
∴ Equation of the line will be
y – y₁ = m (x – x₁) ⇒ y – 5 = \(\frac { -2 }{ 3 }\) (x – 4)
⇒ 3y – 15 = – 2x + 8 ⇒ 2x + 3y – 15 – 8 = 0
⇒ 2x + 3y – 23 = 0 ⇒ 2x + 3y = 23

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).
Solution:
The equation of a line which passes through two points is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
(a) ∵ The required line passes through the points
A (1, 1) and B (2, 3)
∴ Equation of the line will be
y – 1 = \(\frac{3-1}{2-1}\) (x – 1) ⇒ y – 1 = \(\frac { 2 }{ 1 }\) (x – 1)
⇒ y – 1 = 2x – 2 ⇒ 2x – y – 2 + 1 = 0
⇒ 2x – y – 1 = 0

(b) ∵ The required line passes through the points P (3, 3) and Q (7, 6)
∴ Equation of the line will be
⇒ y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 3 = \(\frac{6-3}{7-3}\) (x – 3)
⇒ y – 3 = \(\frac { 3 }{ 4 }\) (x – 3) ⇒ 4y – 12 = 3x – 9
3x – 4y + 12 – 9 = 0 ⇒ 3x – 4y + 3 = 0

(c) ∵ The required line passes through L (a, b) and M (b, a)
∴ Equation of the line will be
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 3 = \(\frac{a-b}{b-a}\) (x – a)
⇒ y – b = \(\frac{(a-b)}{-(a-b)}\) (x – a)
⇒ y – b = – 1 (x – a) ⇒ y – b = – x + a
⇒ x + y = a + b

Question 9.
The lines represented by 3x + 4y = 8 and px + 2y = 7 are parallel. Find the value of p.
Solution:
Slope of line 3x + 4y = 8 ⇒ 4y = – 3x + 8 – 3
⇒ y = \(\frac { -3 }{ 4 }\)x + 2
∴ (m₁) = \(\frac { -3 }{ 4 }\)
and slope of line px + 2y = 7 ⇒ 2y = – px + 7
⇒ y = \(\frac{-p}{2} x+\frac{7}{2}\)
m₂ = \(\frac { -p }{ 2 }\)
∵ The lines are parallel
∴ Their slopes are equal
∴ m₁ = m₂ ⇒ y = \(\frac{-3}{4}=\frac{-p}{2}\)
⇒ p = \(\frac{3 \times 2}{4}=\frac{3}{2}\)

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.
Solution:
Co-ordinates of two points are E (0, 4), F (3, 7)
(i) ∴ Slope (W) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{7-4}{3-0}=\frac{3}{3}\) = 1
∴ Gradient of EF = 1

(ii) Now equation of EF will be
⇒ y – y₁ = m(x – x₁)
⇒ y – 7 = 1 (x – 3) ⇒ y – 7 = x – 3
⇒ x – y + 7 – 3 = 0 ⇒ x – y + 4 = 0

(iii) Let the line EF intersect the x-axis at P then
y-co-ordinates of P will be 0
Let the co-ordinates of P be (x₁, 0)
∵ P lies on EF
∴ It will satsify the equation of EF
x – 0 + 4 = 0 ⇒ x + 4 = 0 ⇒ x = – 4
∴ Co-ordinates of P will be (- 4, 0)

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.
Solution:
Co-ordinates of points are P (- 2, 1), Q (2, 2) and R (6, – 2)
(i) Gradient of QR (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-2-2}{6-2}\) = \(\frac { -4 }{ 4 }\) = 1

(ii) Slope of line perpendicular to QR = \(\frac { -1 }{ m }\)
= -(-1) = 1
∵ It is passes through P
∴ Equation of the line will be
⇒ y – y₁ = m(x – x₁)
⇒ y – 1 = 1(x + 2) ⇒ y – 1 = x + 2
x – y + 2 + 1 = 0 ⇒ x – y + 3 = 0

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
Solution:
The line 3x – 4y + 12 = 0 meets x-axis at P
∴ y-coordinate of P will be 0
Let the co-ordinates of P be (x, 0)
∵ It lies on the line 3x – 4y+ 12 = 0
∴ It will satisfy the equation
∴ 3x – 4 x 0 + 12 = 0 ⇒ 3x + 12 = 0
⇒ 3x = – 12 ⇒ x = \(\frac { -12 }{ 3 }\) = – 4
∴ Co-ordinates of P will be (- 4, 0)
Now slope of the line 3x + 5y – 15 = 0 ⇒ 5y = – 3x + 15
⇒ y = \(\frac { -3 }{ 5 }\)x + 3
(m) = \(\frac { -3 }{ 5 }\)
∴ Slope of the line perpendicular to this line
= \(\frac{-1}{m}=-\left(\frac{-5}{3}\right)=\frac{5}{3}\)
∴ Equation of the perpendicular line through P will be
y – y₁ = m (x – x₁) ⇒ y – 0 = \(\frac { 5 }{ 3 }\)(x + 4)
⇒ y = \(\frac { 5 }{ 3 }\) (x + 4) ⇒ 3y = 5x + 20
⇒ 5x – 3y + 20 = 0

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.
Solution:
Point A divides the line joining the point P (5, -2) and Q (9, 6) in the ratio 3 : 1
∴ m₁ = m₂ = 3 : 1
Now co-ordinates of A will be
\(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)
= \(\left(\frac{3 \times 9+1 \times 5}{3+1}, \frac{3 \times 6+1 \times(-2)}{3+1}\right)\)
= \(\left(\frac{27+5}{4}, \frac{18-2}{4}\right) \text { or }\left(\frac{32}{4}, \frac{16}{4}\right)\) or (8, 4)
Slope of the line x – 3y + 4 = 0 ⇒ 3y = x + 4
⇒ y = \(\frac { 1 }{ 3 }\)x + \(\frac { 4 }{ 3 }\) will be (m) = \(\frac { 1 }{ 3 }\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\)
= – (\(\frac { 3 }{ 1 }\)) = – 3
Equation of the line through A will be
y – y₁ = m(x – x₁)
⇒ y – 4 = – 3(x – 8) ⇒ y – 4 = – 3x + 24
⇒ 3x + y = 24 + 4
⇒ 3x + y = 28
⇒ y + 3x – 28 = 0

Question 14.
Write down the equation of the line parallel to x – 2y + 8 = 0 passing through the point (1, 2).
Solution:
The slope of line x – 2y + 8 = 0
⇒ 2y = x + 8
⇒ y = \(\frac { 1 }{ 2 }\)x + 4
(m) = \(\frac { 1 }{ 2 }\)
∴ Slope of the line parallel to it = \(\frac { 1 }{ 2 }\)
∵ It passes through the point (1, 2)
∴ Equation of the line will be
y – y₁ = m(x – x₁) ⇒ y – 2 = \(\frac { 1 }{ 2 }\) (x – 1)
⇒ 2y – 4 = x – 1 ⇒ x – 2y + 4 – 1 = 0
⇒ x – 2y + 3 = 0

Question 15.
Write down the gradient and intercept on the v-axis of the line \(\frac { x }{ 3 }\) + \(\frac { y }{ 4 }\) = 1.
Solution:
The equation of the line is given
\(\frac { x }{ 3 }\) + \(\frac { y }{ 4 }\) = 1
⇒ 4x + 3y = 12 ⇒ 3y = – 4x + 12 – 4
⇒ y = \(\frac { -4 }{ 3 }\)x + 4
∴ Gradient (m) = \(\frac { -4 }{ 3 }\)
and y-intercept (c) = 4

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.
Solution:
(i) 2x – by + 5 = 0 ⇒ by = 2x + 5
⇒ y = \(\frac { 2 }{ b }\)x + \(\frac { 5 }{ b }\)
∴ Slope (m₁) = \(\frac { 2 }{ b }\)
ax + 3y = 2 ⇒ 3y = – ax + 2
⇒ y = \(\frac{-a}{3}+\frac{2}{3}\)
∴ Slope (m₂) = \(\frac { -a }{ 3 }\)
∵ These lines are parallel
∴ m₁ = m₂
⇒ \(\frac { 2 }{ b }\) = \(\frac { -a }{ 3 }\)
⇒ – ab = 6
⇒ ab = – 6

(ii) We know that equation of a line is y = mx + c where m is slope and c is y-intercept
It passes through (1,3) and has ^-intercept = 5
∵ It will also pass through (0, 5)
∴ Slop(m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{5-3}{0-1}=\frac{2}{-1}\) = – 2
∴ Equation of the line y = mx + c ⇒ y = – 2x + 5
⇒ 2x + y = 5

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.
Solution:
(i) Equation of the line is given \(\frac { y }{ 2 }\) = x – p
∵ It passes through the point (- 4, 4)
∴ \(\frac { 4 }{ 2 }\) = – 4 – P ⇒ 2 = – 4 – p
⇒ p = – 4 – 2 ⇒ p = – 6
∴ p = – 6

(ii) In equation
\(\frac { y }{ 2 }\) = x – p ⇒ y = 2x – 2p
Slope (m₁) = 2
and in equation ax + 5 = 3y
⇒ y = \(\frac { a }{ 3 }\)x + \(\frac { 5 }{ 3 }\)
Slope (m₂) = \(\frac { a }{ 3 }\)
∵ These two lines are parallel
∴ m₁ = m₂
⇒ 2 = \(\frac { a }{ 3 }\) ⇒ 6 = a
∴ a = 6

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.
Solution:
The line passes through a point (- 2, 0) and has its y-intercept = 3
∴ It will pass through (0, 3)
∴ Slope of the line (m) = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{3-0}{0-(-2)}=\frac{3}{2}\)
∴ Equation of the line will be y = mx + c
y = \(\frac { 3 }{ 2 }\)x + 3 ⇒ 2y = 3x + 6

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.
Solution:
(a) In the figure,
(i) Co-ordinates of A are (2, 3), of B are (- 1,2) and of C are (3, 0)
(ii) Equation of a line through A and parallel to BC
Slope of BC (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-2}{3+1}=\frac{-2}{4}\) = \(\frac { -1 }{ 2 }\)
∴ Slope of the required line = \(\frac { 1 }{ 2 }\)
∴ y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { -1 }{ 2 }\) (x – 2)
⇒ 2y – 6 = – x + 2
⇒ x + 2y = 2 + 6 ⇒ x + 2y = 8
⇒ x + 2y – 8 = 0

(b) Let ratio be m₁ : m₂ which divides the join of
A (0, 3) and B (4, 1) by the x-axis at P
∵ P lies on x-axis
∴ Co-ordinate of P be (x, 0)

∴ Co-ordinates of P will be (6, 0)

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 ∆OAB where O is the origin.
Solution:
In the given line 2y = 3x + 5 ⇒ y = \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 2 }\)
Slope (m₁) = \(\frac { 3 }{ 2 }\)
∴ Slope of the line AB perpendicular to the
given line (m₂) = \(\frac{-1}{m_1}=\frac{-2}{3}\)
∵ It passes through (3, 2)
∴ Equation of the line will be y – y₁ = m(x – x₁)
⇒ y – 2 = \(\frac { -2 }{ 3 }\) (x – 3) ⇒ 3y – 6 = – 2x + 6
⇒ 2x + 3y = 6 + 6 ⇒ 2x + 3y = 12
⇒ 2x + 3y – 12 = 0
∵ This line meets x-axis at A and y-axis at B
∴ Substituting the values of y = 0 and x = 0
If y = 0, then
2x + 0 x y = 12
⇒ 2x = 12
⇒ x = \(\frac { 12 }{ 2 }\) = 6
If x = 0, then
2 x 0 + 3y = 12 ⇒ 0 + 3y = 12
⇒ 3y = 12 ⇒ y = \(\frac { 12 }{ 3 }\) = 4
∴ Co-ordinates of A and B will be A (6,0) and B (0, 4)
∵ O is the origin
∴ Area of AOAB = \(\frac { 1 }{ 2 }\) x OA x OB
= \(\frac { 1 }{ 2 }\) x 6 x 4 = 12 sq. units

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?
Solution:
(i) Let co-ordinates of point P be (x, y)
∵ P divides the line joining A (- 4, 1) and B(17, 10) in the ratio 1 : 2 i.e., m₁ : m₂ = 1 : 2
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{1 \times 17+2 \times(-4)}{1+2}\)
= \(\frac{17-8}{3}=\frac{9}{3}\) = 3
and y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{1 \times 10+2 \times 1}{1+2}\)
= \(\frac{10+2}{3}=\frac{12}{3}\) = 4
∴ Co-ordinates of P are (3, 4)

(ii) ∵ O is the origin (0, 0)
∴ OP = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(3-0)^2+(4-0)^2}=\sqrt{(3)^2+(4)^2}\)
= \(\sqrt{9+16}=\sqrt{25}\) = 5

(iii) Let the ratio be m₁ : m₂, in which y-axis divides the line segment AB. Let R be the point on y-axis which divides AB in the ratio m₁ : m₂
Co-ordinates of R be (0, y)
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\)
0 = \(\frac{m_1 \times 17+m_2 \times(-4)}{m_1+m_2}\)
⇒ 17m₁ – 4m₂ = 0 ⇒ 17m₁ = 4m₂
⇒ \(\frac{m_1}{m_2}=\frac{4}{17}\)
∴ Ratio = 4 : 17

Self Evaluation And Revision
(LATEST ICSE QUESTIONS)

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, 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.
Solution:
(a) The mid-point of the line segment AB as shown in the figure is (4, -3). Point A lies on x-axis and point B lies on y-axis

Let the co-ordinates of A be (x, 0) and of B be (0, y)
∴ 4 = \(\frac { x+0 }{ 2 }\) ⇒ x + 0 = 8 ⇒ x = 8
and – 3 = \(\frac { 0+y }{ 2 }\) ⇒ – 6 = y
∴ Co-ordinates of A are (8, 0) and of B are (0, – 6)

(b) We are given four equations of 4 lines
i.e., Equation A = y = 2x
Equation B ≡ y – 2x + 2 = 0
Equation C ≡ 3x + 2y – 6
and Equation D ≡ y – 2
In equation A, x = 0, then y = 0
or it passes through the origin 0 (0, 0)
But line L₃ in the graph, is passing through the origin O
∴ A → L₃
In equation B, y – 2x + 2 = 0 ⇒ y = 2x – 2
Slope (m) = 2 and C = – 2
We see that line L₄ has negative y-intercept and positive slope,
∴ B → L₄
In equation C, 3x + 2y = 6
⇒ 2y = – 3x + 6 ⇒ y = \(\frac { -3 }{ 2 }\)x + 3
Slope (m) = \(\frac { -3 }{ 2 }\) and y-intercept (C) = 3
We see that line L₂ has negative slope and positive y-intercept
∴ C → L₂
In equation D, y = 2
Which is parallel to x-axis at a distance of 2 unit on the positive side of y-axis and we see line L, is parallel to x-axis
∴ D → L₁

(c) The gradient of a line = \(\frac { 3 }{ 2 }\)
and it passes through a point P
∵ P divides the line segment joining A (-2, 6)
and B (3, -4) in the ratio 2 : 3
Let co-ordinates of P be (x, y), then
x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times(3)+3 \times(-2)}{2+3}\)
= \(\frac { 6-6 }{ 5 }\) = 0
and y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times(-4)+3 \times 6}{2+3}\)
= \(\frac{-8+18}{5}=\frac{10}{5}\) = 2
∴ Co-ordinates of P are (0, 2)
Now equation of line through P will be
y – y₁ = m(x – x₂) ⇒ y – 2 = \(\frac { 3 }{ 2 }\) (x – 0)
⇒ 2y – 4 = 3x ⇒ 3x – 2y + 4 = 0

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 the x- axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y +12 = 0.
Solution:
(a) ∵ Point B is on x-axis is and its abscissa is 11
∴ Co-ordiantes of B will be (11, 0)
∴ Distance between A (7, 3) and B (11, 0)
= \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(11-7)^2+(0-3)^2}=\sqrt{(4)^2+(-3)^2}\)
= \(\sqrt{16+9}=\sqrt{25}\)
= 5 units

(b) (i) The line 4x – 3y + 12 = 0, meets x-axis at A
Let co-ordinates of A be (x, 0), then
4 × x – 3 x 0 + 12 = 0 ⇒ 4x + 12 = 0
⇒ 4x = – 12
⇒ x = \(\frac { -12 }{ 4 }\) = – 3
∴ Co-ordinates of A are (- 3, 0)

(ii) In equation 4x – 3y + 12 = 0
⇒ 3y = 4x + 12 ⇒ y = \(\frac { 4 }{ 3 }\)x + 4
Slope (m₁) = \(\frac { 4 }{ 3 }\)
∴ Slope of the line perpendicular to it m₂ = \(\frac{-1}{m_1}\)
= \(\frac { -3 }{ 4 }\)
∵ This line passes through A (- 3, 0)
∴ Equation of the line will be
y – y₁ = m(x – x₁) ⇒ y – 0 = \(\frac { -3 }{ 4 }\) (x + 3)
⇒ 4y = – 3x – 9 ⇒ 3x + 4y + 9 = 0

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.
Solution:
(a) Co-ordinates O, the centre of a circle are (4, 5)
One point on the circle is B (8, 10)

BO is joined and produced to A such that AB
is the diameter of the circle
Let co-ordinates of A be (x, b)
∴ O is the mid-point of AB,
∴ 4 = \(\frac { 8+x }{ 2 }\) ⇒ 8 = 8 + x ⇒ x = 8 – 8 = 0
and 5 = \(\frac { 10+y }{ 2 }\) ⇒ 10 = 10 + y
⇒ y = 10 – 10 = 0
∴ Co-ordinates of A will be (0, 0)

(b) y-intercept of a line (C) = 4
and parallel to the line whose equation is 2x – 3y = 7 ⇒ 3y = 2x + 7
y = \(\frac { 2 }{ 3 }\)x + \(\frac { 7 }{ 3 }\)
∴ Slope (m₁) = \(\frac { 2 }{ 3 }\)
Now the slope of the line which is parallel to
it = m₂ = m₁ = \(\frac { 2 }{ 2 }\)
∴ Equation of the line will be y = mx + c
⇒ y = \(\frac { 2 }{ 3 }\)x + 4 ⇒ 3y = 2x + 12
Let this line cut the x-axis at A
∴ Ordinate of A will be O
Let abscissa of A be x
But it will satisfy the equation
3 x 0 = 2x + 12 ⇒ 0 = 2x + 12
⇒ 2x = – 12 ⇒ x = \(\frac { -12 }{ 2 }\) = – 6
∴ Co-ordinates of A are (- 6, 0)

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.
Solution:
(a) In the given equation of a line is + 5y + 15 = 0 ⇒ 5y = – 3x – 15
⇒ y = \(\frac { -3 }{ 5 }\)x – 3
∴ Slope (m₁) = \(\frac { -3 }{ 5 }\)
Now slope of the required line which is parallel to the given line = m₂ = \(\frac { -3 }{ 5 }\) (∵ m₁ = m₂)
∵ This line passes through the point (0, 4)
∴ Equation of the line will be
y – y₁ = m(x – x₁) ⇒ y – 4 = \(\frac { -3 }{ 5 }\) (x – 0)
⇒ 5y – 20 = – 3x + 0 ⇒ 3x + 5y = 0 + 20
⇒ 3x + 5y = 20

(b) P (- 4,2) is a point on the line AB such that P divides it in the ratio 1 : 2 i.e., AP = PB = 1 : 2 and the line intersects x-axis at A and y-axis at B
Let co-ordinates of A be (x, 0) and of B be (0, y)
∴ – 4 = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{1 \times 0+2 \times x}{1+2}=\frac{2 x}{3}\)
∴ x = \(\frac{-4 \times 3}{2}\)
and 2 = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{1 \times y+2 \times 0}{1+2}=\frac{y}{3}\)
∴ y = 2 x 3 = 6
∴ Co-ordinates of A are (- 6, 0) and of B are (0, 6)

(c) Radius of a circle = 13 units
Centre of the circle is O (3, 6)
P (7, 9) is a point inside the circle and APB is a chord of the circle Joint OP and OA

∵ P is the midpoint of chord AB and OP is joined
∴ OP ⊥ AB
Now length of OP = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(7-3)^2+(9-6)^2}\)
= \(\sqrt{(4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}\)
= 5 units
Now in right angled AOAP,
OA² = OP² + AP² (Pythagoras theorem)
⇒ (13)² = (5)² + AP²
⇒ 169 = 25 + AP² ⇒ AP² = 169 – 25 = 144 = (12)²
∴ AP = 12 units
∵ P is mid-point of AB
∴ AB = 2AP = 2 x 12 = 24 units

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.
Solution:
(a) Let points A (6, 5) and B (4, – 3) are joined and a line y = 2 divides it in the ratio m₁ : m₂
∵ The point which divides the line segment AB lies on y = 2
∴ 2 = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{m_1(-3)+m_2 \times 5}{m_1+m_2}\)
⇒ 2 = \(\frac{-3 m_1+5 m_2}{m_1+m_2}\)
⇒ 2m₁ + 2m₂ = – 3m₁ + 5m₂
⇒ 2m₁ + 3m₁ = 5m₂ – 2m₂ ⇒ 5m₁ = 3m₂
⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
Ratio = 3 : 5

(b) (i) Equation of line AB is 2x – y + 6 = 0 and of CD is x – 2y = 4
Which intersect x-axis at B and D and y-axis at A and C respectively

∵ A lies on y-axis
∴ x-coordinates of A = 0
∴ From the equation 2x – y + 6 = 0
⇒ 2 x 0 – y + 6 = 0
⇒ – y + 6 = 0 ⇒ y = 6
∴ Co-ordinates of A are (0, 6)
∵ B lies on x-axis
∴ Its y-coordinates = 0
From the equation 2x – y + 6 = 0
2x – 0 + 6 = 0 ⇒ 2x = – 6
⇒ x = \(\frac { -6 }{ 2 }\) = – 3
∴ Co-ordinates of B are (-3, 0)
∵ C lies on y-axis
∴ Its x-coordinates = 0
From the equation x – 2y = 4
⇒ 0 – 2y = 4 ⇒ y = \(\frac { 4 }{ -2 }\) = -2
∴ Co-ordinates of C are (0, – 2)
∵ D lies on x-axis
∴ Its y-coordinates = 0
From the equation x – 2y = 4
⇒ x – 2 x 0 = 4 ⇒ x – 0 = 4 ⇒ x = 4
Co-ordinates of D are (4, 0)

(ii) In ∆OAB and ∆ODC,
∠AOB = ∠COD
(Vertically opposite angles or each = 90°)
∵ \(\frac{\mathrm{OA}}{\mathrm{OB}}=\frac{6}{3}=\frac{2}{1}=\frac{\mathrm{OD}}{\mathrm{OC}} \frac{4}{2}=\frac{2}{1}\)
∴ \(\frac{\mathrm{OA}}{\mathrm{OB}}=\frac{\mathrm{OD}}{\mathrm{OC}}\)
∆OAB ~ ∆ODC (SAS axiom)

(iii) ∴ ∠OAB = ∠ODC
and ∠OBA = ∠OCB
But these are angles in the same segment
∴ A, B, C and D are concyclic
Hence ABCD is a cyclic quadrilateral

(c) In a rhombus ABCD,
Co-ordinates of A are (3, 6) and of C are (- 1, 2)
∴ Slop of AC (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-6}{-1-3}\)
= \(\frac { -4 }{ -4 }\) = 1
∵ AC and BD bisect each other at right angles
∴ Slope of BD = \(\frac { -1 }{ m }\) = – 1
and mid-point of AC = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
or \(\left(\frac{3-1}{2}, \frac{6+2}{2}\right) \text { or }\left(\frac{2}{2}, \frac{8}{2}\right)\) or (1, 4)
∴ Equation of the line BD will be,
y – y₁ = m(x – x₁)
y – 4 = – 1 (x – 1)
⇒ y – 4 = – x + 1
⇒ x + y = 1 + 4 ⇒ x + y = 5
⇒ x + y – 5 = 0

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.
Solution:
(a) Vertices of a A ABC are A (10, 5), B (6. -3) and C (2, 1)
L is the mid point of AB and M is the mid-point of AC
∴ Co-ordinates of L will be

We see that LM = \(\frac { 1 }{ 2 }\) BC

(b) ∵ P divides the line segment joining the points A (-2, 6) and B (3, -4) in the ratio 2 : 3
∵ Co-ordinates of P will be
\(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\) or
\(\left(\frac{2 \times 3+3 \times(-2)}{2+3}, \frac{2 \times(-4)+3 \times 6}{2+3}\right)\) or
\(\left(\frac{6-6}{5}, \frac{-8+18}{5}\right) \text { or }\left(0, \frac{10}{5}\right)\) or (0, 2)
∴ Gradient (slope) of the line = \(\frac { 3 }{ 2 }\)
∴ Equation of the line will be
(y – y₁) = m (x – x₁) ⇒ y – 2 = \(\frac { 3 }{ 2 }\) (x – 0)
⇒ 2y – 4 = 3x ⇒ 3x – 2y + 4 = 0

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 ∆ABC.
(ii) The equation of a line through G and parallel to AB.
Solution:
∵ x-intercept = 4
∴ Coordinates of the point on x-axis = (4, 0)
Now slope of the line joining the points (-2, 3) and (4, 0)
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-3}{4+2}=\frac{-3}{6}=\frac{-1}{2}\)
∴ Equation of the line will be
y – y₁ = m (x – x₁)
⇒ y – 3 = \(\frac { -1 }{ 2 }\) (x + 2)
⇒ 2y – 6 = – x – 2
⇒ x + 2y – 6 + 2 = 0
⇒ x + 2y – 4 = 0

(b) Vertices of a ∆ABC are A (1,4), B (3,2) and C (7, 5)
(i) ∴ Coordinates of the centroid G of ∆ABC
= \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
or \(\left(\frac{1+3+7}{3}, \frac{4+2+5}{3}\right)\) or \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

(ii) Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{3-1}=\frac{-2}{2}\) = – 1
Slope of the line parallel to AB from G = – 1
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – \(\frac { 11 }{ 3 }\) = – 1(x – \(\frac { 11 }{ 3 }\)
\(\frac{3 y-11}{3}\) = – \(\frac{(3 x-11)}{3}\)
⇒ 3y – 11 = – 1 (3x -11)
⇒ 3y – 11 = – 3x + 11
⇒ 3x + 3y – 11 – 11 = 0
⇒ 3x + 3y – 22 = 0

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 – 5j + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other.
Solution:
(a) A line passes through two points P (- 1, 4) and Q (5, -2) which intersects the axes at A and B. M is the mid-point of AB
Now slope of the line (m) = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-2-4}{5+1}=\frac{-6}{6}\) = – 1

(i) ∴ Equation of the line will be
y – y₁ = m (x – x₁) ⇒ y – 4 = – 1 (x + 1)
⇒ y – 4 = – x – 1 ⇒ x + y – 4 + 1 = 0
⇒ x + y – 3 = 0

(ii) ∵ A lies on x-axis and B lies on y-axis
If A lies on x-axis then y-coordinates = 0
∴ x + 0 – 3 = 0 ⇒ x = 3
∴ Coordinates of A will be (3, 0)
∵ B lies on y-axis, then x-coordinate = 0
∴ 0 + y – 3 = 0 ⇒ y = 3
∴ Co-ordinates of B will be (0, 3)

(iii) ∵ M is the mid-point of AB
Co-ordinates of M will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
or \(\left(\frac{0+3}{2}, \frac{3+0}{2}\right)\) or \(\left(\frac{3}{2}, \frac{3}{2}\right)\) or (1.5, 1.5)

(b) In line kx – 5y + 4 = 0 ⇒5y = kx + 4
⇒ y = \(\frac{k x}{5}+\frac{4}{5}\)
Slope (m₁) = \(\frac { k }{ 5 }\)
and in line 4x – 2y + 5 = 0 ⇒ 2y = 4x + 5
⇒ y = \(\frac { 4 }{ 2 }\)x + \(\frac { 5 }{ 2 }\) ⇒ y = 2x + \(\frac { 5 }{ 2 }\)
∴ Slope (m₂) = 2
∵ The lines are perpendicular to each other
∴ m₁ x m₂ = – 1 ⇒ \(\frac { k }{ 5 }\) x 2 = – 1
⇒ k = \(\frac{-1 \times 5}{2}=\frac{-5}{2}\)

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.
Solution:
(a) KM is a straight line and length of KM = 13 units coordinates of K are (2, 5) and of M are (x, – 7)
∴ Length of KM = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
13 = \(\sqrt{(x-2)^2+(-7-5)^2}\)
⇒ 13 = \(\sqrt{(x-2)^2+(-12)^2}\)
Squaring both sides
169 = (x – 2)² + (- 12)²
169 = x² – 4x + 4 + 144
⇒ x² – 4x + 148 – 169 = 0
⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
⇒ x (x – 7) + 3 (x – 7) = 0
⇒ (x – 7) (x + 3) = 0
∴ Either x – 7 = 0, then x = 7
or x + 3 = 0, then x = – 3
∴ Value of x are 7 or – 3

(b) The line joining the points P (- 4, 5) and Q (3, 2) intersects x-axis at R.
PM and QN are the perpendiculars drawn from P and Q on x-axis
Let R divides PQ in the ratio m₁ : m₂
∵ R lies on y-axis
(i) ∴ Its x-coordinates will be 0
∴ 0 = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \Rightarrow 0=\frac{m_1 \times 3+m_2 \times(-4)}{m_1+m_2}\)
⇒ 3m₁ – 4m₂ = 0 ⇒ 3m₁ = 4m₂
⇒ \(\frac{m_1}{m_2}=\frac{4}{3}\) ⇒ m₁ : m₂ = 4 : 3
∴ PR : RQ = 4 : 3

(ii) y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{4 \times 2+3 \times 5}{4+3}\)
= \(\frac{8+15}{7}=\frac{23}{7}\)
∴ Co-ordinates of R will be \(\left(0, \frac{23}{7}\right)\)

(iii) Now area of quadrilateral PMNQ or trapezium PMNQ
= \(\frac { 1 }{ 2 }\) (PM + QN) x MN
= \(\frac { 1 }{ 2 }\) (5 + 2) x 7 = \(\frac { 1 }{ 2 }\) x 7 x 7 = \(\frac { 49 }{ 2 }\)
= 24.5 sq. units

(c) The vertices of APQR are P (3, 4), Q (7, – 2) and R (- 2, – 1)
Let L is the mid point of PQ
∴ Co-ordinates of L are \(\left(\frac{3+7}{2}, \frac{4-2}{2}\right)\) or \(\left(\frac{10}{2}, \frac{2}{2}\right)\) or (5, 1)
∴ Slope of RQ = \(\frac{y_2-y_1}{x_2-x_1}=\frac{1+1}{5+2}=\frac{2}{7}\)
∴ Equation of median RL will be
y – y₁ = m (x – x₁)
⇒ y + 1 = \(\frac { 2 }{ 7 }\) (x + 2) ⇒ 7y + 7 = 2x + 4
⇒ 2x – 7y – 7 + 4 = 0
⇒ 2x – 7 – 3 = 0

Question 10.
(a) Use a graph for this question:
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 || to BC.

Solution:
(a) Plot the points A (- 1, – 1) and B (2, 5) and join them. On this line two more points are taken P (h, 4) and Q (\(\frac { 1 }{ 2 }\), k)
We see that in P
If y = 4, then x = 1.5
∴ h = 1.5
and in Q, if x = \(\frac { 1 }{ 2 }\)
Then y = 2
∴ k = 2

(b) (i) From the figure, we find that
Co-ordinates of A are (2, 3) of B are (- 1, 2) and of C are (3, 0)

(ii) Now slope of BC
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-2}{3+1}=\frac{-2}{4}=\frac{-1}{2}\)
∴ Slope of line drawn from Aparallel to BC = \(\frac { -1 }{ 2 }\)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { -1 }{ 2 }\) (x – 2)
⇒ 2y – 6 = – x + 2 ⇒ x + 2y = 2 + 6 ⇒ x + 2y = 8

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).
Solution:
The slope of the line joining the points A (2, 3) and B (6, – 5) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-5-3}{6-2}\)
= \(\frac { -8 }{ 4 }\) = – 2
∵ This line is intercepted by x-axis at K
∴ Its ordinate = 0
Let K divides the line in the ratio m₁ : m₂
∴ x = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\)
⇒ 0 = \(\frac{m_1 \times-5+m_2 \times 3}{m_1+m_2} \Rightarrow \frac{-5 m_1+3 m_2}{m_1+m_2}\) = 0
⇒ \(-5 m_1+3 m_2=0 \Rightarrow 5 m_1=3 m_2\)
⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
Ratio = 3 : 5

(b) In the line y = 3x + 7
Slope (m₁) = 3
and in the line 2y + px = 3 ⇒ 2y = – px + 3
⇒ y = \(\frac{-p}{2} x+\frac{3}{2}\)
∴ Slope (m₂) = \(\frac { – p }{ 2 }\)
∵ Lines are perpendicular to each other
∴ m₁m₂ = – 1 ⇒ 3 x \(\frac { – p }{ 2 }\) = – 1
⇒ P = \(\frac{-1 \times 2}{-3}=\frac{2}{3}\)
Hence p = \(\frac { 2 }{ 3 }\)

(c) Vertices of a AABC are A (- 1, 3), B (1, – 1), C (5, 1)
Let G be the centroid of the triangle ABC
∴ Co-ordinates of G will be
\(\left(\frac{x_1+x_2+x_2}{3}, \frac{y_1+y_2+y_3}{3}\right)\) or
\(\left(\frac{-1+1+5}{3}, \frac{3-1+1}{3}\right) \text { or }\left(\frac{5}{3}, \frac{3}{3}\right) \text { or }\left(\frac{5}{3}, 1\right)\)

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 \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{2}{5}\), find the co-ordinates of P.
Solution:
(a) ∵ (1, 2a + 1) is the mid-point of the line joining the points (2a, 4) and (- 2, 2b)
∵ 1 = \(\frac { 2a – 2 }{ 2 }\) ⇒ 2a – 2 = 2
⇒ 2a = 2 + 2 = 4 ⇒ a = \(\frac { 4 }{ 2 }\) = 2
and 2a + 1 = \(\frac { 4+2b }{ 2 }\) ⇒ 4a + 2 = 4 + 2b
⇒ 4 x 2 + 2 – 4 = 2b ⇒ 10 – 4 = 2b
⇒ 2b = 6 ⇒ b = \(\frac { 6 }{ 2 }\) = 3
Hence a = 2, b = 3

(b) In line 3x + 2y = 8 ⇒ 2y = – 3x + 8
⇒ y = \(\frac { -3 }{ 2 }\)x + 4
Slope (m₁) = \(\frac { 3 }{ 2 }\)
∴ Slope of the line which is parallel to the given line
= m₂ = \(\frac { -3 }{ 2 }\) (∵ m₁ = m₂)
∵ It passes through the point (0, 1)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
y – 1 = \(\frac { -3 }{ 2 }\) (x – 0) ⇒ 2y – 2 = – 3x
⇒ 3x + 2y – 2 = 0

(c) Two points are given A (4, – 5), B (4, 5)
P is a point on the line AB such that
\(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{2}{5}\)
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{AB}-\mathrm{AP}}=\frac{2}{5-2}=\frac{2}{3}\)
or m₁ : m₂ = 2 : 3
Let coordinates of P be (x, y)
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times 4+3 \times 4}{2+3}\)
= \(\frac{8+12}{5}=\frac{20}{5}\) = 4
y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times 5+3 \times(-5)}{2+3}\)
= \(\frac{10-15}{5}=\frac{-5}{5}\) = – 1
∴ Coordinates of P are (4, – 1)

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) Points A and B have coordinates (7, – 3) and (1, 9) respectively. Find :
(i) the slope of AB;
(ii) the equation of the perpendicular bisector of the line segment AB;
(iii) the value of ‘p’ if (- 2, p) lies on it.
Solution:
(a) Co-ordinates of A are (- 4, 3) and of B are (8, – 6)
(i) ∴ Length of AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{[8-(-4)]^2+(-6-3)^2}\)
= \(\sqrt{(8+4)^2+(-6-3)^2}=\sqrt{(12)^2+(-9)^2}\)
= \(\sqrt{144+81}=\sqrt{225}\)
∵ AB is intersected by x-axis
∴ The y-coordinates of the point of intersection = 0
Let this point divides AB in the ratio m₁ : m₂
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\) ⇒ 0 = \(\frac{m_1 \times 8+m_2 \times(-4)}{m_1+m_2}\)
⇒ \(\frac{8 m_1-4 m_2}{m_1+m_2}=0 \Rightarrow 8 m_1-4 m_2=0\)
⇒ 8m₁ = 4m₂ ⇒ \(\frac{m_1}{m_2}=\frac{4}{8}=\frac{1}{2}\)
Ratio = 1 : 2

(b) Co-ordinates of two points are A (7, -3) and B (1, 9)
(i) Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}=\frac{9-(-3)}{1-7}\)
= \(\frac{9+3}{1-7}=\frac{12}{-6}\)

(ii) ∵ Slope of perpendicular to AB
\(\frac{-1}{m}=-\left(\frac{-1}{2}\right)=\frac{1}{2}\)
Co-ordinates of mid-point of AB
= \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \text { or }\left(\frac{7+1}{2}, \frac{-3+9}{2}\right)\) or (\(\frac { 8 }{ 2 }\), \(\frac { 6 }{ 2 }\)) or (4, 3)
Equation of the perpendicular line will be
y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { 1 }{ 2 }\) (x – 4) ⇒ 2y – 6 = x – 4
⇒ x – 2y + 6 – 4 = 0 ⇒ x – 2y + 2 = 0

(iii) ∵ Point (- 2, p) lies on the line x – 2y + 2 = 0
∴ It will satisfy it
∴ – 2 – 2p + 2 = 0 ⇒ – 2p = 0 ⇒ p = 0
Hence p = 0

Question 14.
Find the equation of a line with x intercept = 5 and passing through the point (4, – 7).
Solution:
Given : x intercept = 5
∴ Line passes through (5, 0)
∴ Equation is
y = \(\frac { – 7 }{ – 1 }\)(x – 5)
y = 7x – 35 ⇒ 7x – y – 35 = 0

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.

Solution:
As P (2, – 3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)
∴ By mid point formula \(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) is the mid-point of AB. Comparing coordinates of mid-point x
∴ \(\frac { x }{ 2 }\) = 2 ⇒ x = 4
\(\frac { y }{ 2 }\) = – 3 ⇒ y = – 6

(i) ∴ Coordinates of A is (4, 0)
Coordinates of B is (0, -6)

(ii) Slope of line AB
= \(\frac{\text { Difference of } y \text { coordinates }}{\text { Difference of } x \text { coordinates }}\)
= \(\frac{0-(-6)}{4-0}\)
∴ Slope of line AB = \(\frac { 6 }{ 4 }\) = \(\frac { 3 }{ 2 }\)

(iii) Equation of line AB is
y – (- 6) = (Slope of AB) (x – 0)
y + 6 =\(\frac { 3 }{ 2 }\) (x)
2y+ 12 = 3x
⇒ 3x – 2y – 12 = 0

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.
Solution:
Given line 3x + 4y – 1 = 0
(i) Slope of line = – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } y}\)
Slope of line = \(\frac { -3 }{ 4 }\)

(ii) Point of intersection of the lines
x – y + 2 = 0
and 3x + y – 10 = 0,
Adding both we get,
4x – 8 = 0
4x = 8
⇒ x = 2
and putting the value of x in above equation 2 – y + 2 = 0 and y = 4
∴ Point of intersection is (2, 4)
Slope of any line perpendicular to 3x + 4y – 7 = 0 = \(\left(\frac{-4}{3}\right)=\frac{4}{3}\)
∴ Equation of line whose slope is \(\frac { 4 }{ 3 }\) and passing through (2, 4) is given by
y – 4 = \(\frac { 4 }{ 3 }\) (x – 2), [y – y₁ = m (x – x₁)]
⇒ 3 (y – 4) = 4 (x – 2)
⇒ 3y – 12 = 4x – 8
⇒ 4x – 3y – 8 + 12 = 0
⇒ 4x – 3y + 4 = 0

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.
Solution:

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.
Solution:

(i) Coordinates of O = \(\left(\frac{5+2}{2}, \frac{8-4}{2}\right)\) = (3.5, 2)
For the line AC
3.5 = \(\frac { x+4 }{ 2 }\)
⇒ x + 4 = 7
⇒ x = 7 – 4 = 3
x = 3, y = – 3

2 = \(\frac { y+7 }{ 2 }\)
⇒ y + 7 = 4
⇒ y = 4 – 7 = – 3
Thus, the coordinates of A are (3, -3)

(ii) Equation of diagonal BD is given by
y – 8 = \(\frac{-4-8}{2-5}\) (x – 5)
y – 8 = \(\frac { -12 }{ -3 }\) (x – 5)
⇒ y – 8 = 4x – 20
⇒ 4x – y – 12 = 0

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.

Solution:
Line L₁ is y = 4 (given)
Coordinates of O is (0, 0)
(i) L₂ is the bisector of angle O
∴ θ = \(\frac { 90° }{ 2 }\) = 45°
m = tan θ = tan 45° = 1

(ii) OAPB is square
[∵ OB = BP = 4 as slope of line L₂ is 1]
∴ Coordinates of P is (4, 4)

(iii) Equation of L₂ :
y – o = \(\frac{4-0}{4-0}\) (x – 0)
⇒ y = \(\frac { 4 }{ 4 }\) (x)
⇒ y = x
⇒ x – y = 0

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.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1
So, 0 = \(\frac{m \times 8-4 \times 1}{m+1}\) ⇒ 8m – 4 = 0
⇒ m = \(\frac { 4 }{ 8 }\) ⇒ m = \(\frac { 1 }{ 2 }\)
So, required ratio = \(\frac { 1 }{ 2 }\) : 1 or 1 : 2

(ii) Also, y = \(\frac{1 \times(-3)+2 \times 6}{1+2}=\frac{9}{3}\) = 3
So, coordinates of the point of intersection are (0, 3)

(iii) AB = \(\sqrt{(8+4)^2+(-3-6)^2}\)
= \(\sqrt{144+81}\) = 15 units

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.
Solution:
(i) Here θ = 45°
So, slope of the line = tan θ = tan 45° = 1

(ii) Equation of the line using point slope is
y – y₁ = m(x – x₁)
⇒ y – 3 = 1(x – 5) ⇒ y – 3 = x – 5
⇒ y = x – 2 or x – y – 2 = 0

(iii) Let the coordinates of Q be (0, y)
Then m = \(\frac{y_2-y_1}{x_2-x_1}\)
⇒ 1 = \(\frac{3-y}{5-0}\)
⇒ 5 = 3 – y ⇒ y = – 2
So, coordinates of Q are (0, – 2)

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 ∆ABC, A (3, 5), B (7, 8) and C (1, – 10).
Find the equation of the median through A.
Solution:
(a) AC = \(\sqrt{(3+2)^2+(-7-5)^2}\)
= \(\sqrt{5^2+12^2}=\sqrt{25+144}\) = \(\sqrt{169}\) = 13 units

∵ AB is diameter and C is mid point of AB
Let co-ordinate of B are (x, y)
∴ \(\frac { 3+x }{ 2 }\) = – 2 and \(\frac { y – 7 }{ 2 }\) = 5
3 + x = – 4 and y – 7 = 10
x = – 4 – 3 and y = 10 + 7
x = – 7 and y = 17
∴ B is (- 7, 17)

(b) AD is median
⇒ D is mid point of BC

∴ D is \(\left(\frac{7+1}{2}, \frac{8-10}{2}\right)\)
i.e. (4 , – 1)
Slope of AD(m) = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{5+1}{3-4}=\frac{6}{-1}\) = – 6
∴ Equation of AD
y – y₁ = m (x – x₁)
y + 1 = – 6(x – 4)
y + 1 = – 6x + 24
y + 6x = – 1 + 24
6x + y = 23
6x + y – 23 = 0

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.

Solution:
Let A (x, 0) and B (0, y)

3x = – 15 and 2y = 20
x = – 5 and y = 10
∴ Co-ordinates of A are (- 5, 0) and B are (0, 10)

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.
Solution:
(a) Let ratio = k : 1

(b) A (a, 3), B(2, 1) and C(5, a) are collinear.
Slope of AB = Slope of BC
\(\frac{1-3}{2-a}=\frac{a-1}{5-2}\)
= \(\frac{-2}{2-a}=\frac{a-1}{3}\)
– 6 = (a – 1) (2 – a) (Cross-multipication)
– 6 = 2o – a² – 2 + a
– 6 = 3a – a² – 2
a² – 3a+ 2 – 6 = 0
a² – 3a – 4 = 0
a² – 4a + a – 4 = 0
a(a – 4) + (a – 4) = 0
(a + 1) (a – 4) = 0
a = – 1, or a = 4
a = – 1 (∵ does not satisfy the equation)
∴ a = 4
Slope of BC = \(\frac{a-1}{5-2}=\frac{4-1}{3}=\frac{3}{3}\) = 1 = m
Equation of BC ; (y – 1) = 1(x – 2)
y – 1 = x – 2 ⇒ x – y = – 1 + 2
x – y = 1
x – y – 1 = 0

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.
Solution:
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2)
(i) We need to find the co-ordinates of D We know that the diagonals of a parallelogram bisect each other
Let (x, y) be the co-ordinates of D
∴ Mid-point of diagonal AC = \(\left(\frac{3+3}{2}, \frac{6+2}{2}\right)\) = (3, 4)
And, mid-point of diagonal BD
= \(\left(\frac{5+x}{2}, \frac{10+y}{2}\right)\)

Thus, we have
\(\frac{5+x}{2}=3 \text { and } \frac{10+y}{2}=4\)
⇒ 5 + x = 6 and 10 + y = 8
⇒ x = 1 and y = – 2
∴ Coordinate of D = (1, – 2)

(ii) Length of diagonal BD
= \(\sqrt{(1-5)^2+(-2-10)^2}\)
= \(\sqrt{(4)^2+(-12)^2}\)
= \(\sqrt{16+144}=\sqrt{160}\) units
= 12.65 units

(iii) Equation of the side joining A (3, 6) and D (1, -2) is given by
\(\frac{x-3}{3-1}=\frac{y-6}{6+2}\)
⇒ \(\frac{x-3}{2}=\frac{y-6}{8}\)
⇒ 4(x – 3) ⇒ y – 6 ⇒ 4x – 12 = y – 6
⇒ 4x – y = 6
Thus, the equation of the side joining A (3, 6) and D (1, -2) is 4x – y = 6

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.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC.

Solution:
(i) The line intersects the x-axis where, y = 0
The co-ordinates of A are (4, 0)

(ii) Length of AB = \(\sqrt{[4-(-2)]^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}\) = 6.71 units
Length of AC = \(\sqrt{[4-(-2)]^2+(0+4)^2}\)
= \(\sqrt{36+16}=\sqrt{52}\)
= 7.21 units

(iii) Let k be the required ratio which divides the line segment joining the co-ordinates A (4, 0) and C (- 2, – 4)
Let the co-ordinates of Q be x and y
∴ x = \(\frac{k(-2)+1(4)}{k+1}\) and y = \(\frac{k(-4)+0}{k+1}\)
Q lies on the y-axis where x = 0
⇒ \(\frac{-2 k+4}{k+1}\) ⇒ – 2k + 4 = 0
⇒ 2k = 4
The ratio is 2 : 1

(iv) The equation of line AC is
\(\frac{x-4}{4+2}=\frac{y-0}{0+4}\)
⇒ \(\frac{x-4}{6}=\frac{y}{4}\)
⇒ \(\frac{x-4}{3}=\frac{y}{2}\)
⇒ 2(x – 4) = 3y
⇒ 2x – 8 = 3y
⇒ 2x – 3y = 8
The equation of the line AC is 2x – 3y = 8

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 ‘A’ 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.

Solution:
(a) (i) Slope of PQ = \(\frac{3-k}{1-3 k-6}\)
\(\frac{1}{2}=\frac{3-k}{-3 k-5}\)
⇒ -3k – 5 = 2(3 – k)
⇒ -3k – 5 = 6 – 2k
⇒ k = 1

(ii) Substitute k in P and Q.
∴ P (6, k) = P (6, 1)
and Q (1 – 3k, 3) = Q (- 2, 3)
Midpoint of PQ = \(\left(\frac{6-2}{2}, \frac{1+3}{2}\right)\)
= \(\left(\frac{4}{2}, \frac{4}{2}\right)\)
= (2, 2)

(b) (i) Since, A lies on the x-axis, let the coordinates of A be (x, 0).
Since B lies on the y-axis, let the coordinates of B be (0, y).
Let m = 1 and n = 2.
Using section formula,
Coordinates of P = \(\left[\frac{1(0)+2(x)}{1+2}, \frac{1 y+2(0)}{1+2}\right]\)
∴ (4, – 1) = \(\left(\frac{2 x}{3}, \frac{y}{3}\right)\)
⇒ \(\frac { 2x }{ 3 }\) = 4 and \(\frac { y }{ 3 }\) = – 1
⇒ x = 6 and y = – 3
So, the coordinates of A are (6, 0) and that of B are (0, -3).

(ii) Slope of AB = \(\frac{0-(-3)}{6-0}\)
= \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(Slope of line through P)
(Slope of AB) = – 1
⇒ Slope of line through P\(\frac { 1 }{ 2 }\) = – 1
(Since the lines are perpendicular)
⇒ Slope of line through P = – 2
Let the equation of the line through P be y = mx + c
⇒ y = – 2x + c
c is the x-intercept
Put y = 0 and x = c in (i) to find c
y = – 2x + c ⇒ 0 = – 2c + c ⇒ c = 0
So, the equation of the line is y = – 2x