Equation of Straight Line ML Aggarwal Solutions ICSE Class-10 Maths

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Equation of Straight Line ML Aggarwal Solutions ICSE Class-10 Maths Chapter-12. We Provide Step by Step Answer of Exercise-12.1 ,  Exercise-12.2 , Equation of Straight Line , with MCQs and Chapter-Test Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  .  Visit official Website CISCE  for detail information about ICSE Board Class-10.

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 10th
Chapter-12 Equation of Straight Line
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-12.1, Exe-12.2, MCQ and Chapter Test Questions
Academic Session 2021-2022

Equation of Straight Line ML Aggarwal Solutions ICSE Class-10 Maths Chapter-12

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Exe-12.1

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 MCQS ,

Chapter Test


How to Solve Equation of Straight Line Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-12 Equation of Straight Line of ML Aggarwal Solution. Read the Chapter Carefully and then solve all example given in  your text book.

For more practice on Equation of Straight Line related problems /Questions / Exercise try to solve Equation of Straight Line  exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE)  / Concise Selina Publications ICSE  Mathematics. Get the formula of Equation of Straight Line for ICSE Class 10 Maths  to understand the topic more clearly in effective way.


Exercise-12.1, Equation of Straight Line Solutions of ML Aggarwal for ICSE Class-10

page-237

Question -1

Find the slope of a line whose inclination is
(i) 45°
(ii) 30°

Answer- 1

(i) tan 45° = 1
(ii) tan 30° = \frac { 1 }{ \sqrt { 3 } }

Question- 2

Find the inclination of a line whose gradient is

(i) 1
(ii) √3
(iii) \frac { 1 }{ \sqrt { 3 } }

Answer- 2

(i) tan θ = 1 ⇒ θ = 45°
(ii) tan θ = √3 ⇒ θ = 60°
(iii) tan θ = \frac { 1 }{ \sqrt { 3 } }  ⇒ θ = 30°

Question -3

Find the equation of a straight line parallel 1 to x-axis which is at a distance
(i) 2 units above it
(ii) 3 units below it.

Answer -3

(i) A line which is parallel to x-axis is y = a
⇒ y = 2
⇒ y – 2 = 0
(ii) A line which is parallel to x-axis is y = a
⇒ y = -3
⇒ y + 3 = 0

Question -4

Find the equation of a straight line parallel to y-axis which is at a distance of:
(i) 3 units to the right
(ii) 2 units to the left.

Answer- 4

(i) The equation of line parallel to y-axis is at a distance of 3 units to the right is x = 3 ⇒ x – 3 = 0
(ii) The equation of line parallel to y-axis at a distance of 2 units to the left is x = -2 ⇒ x + 2 = 0

Question -5

Find the equation of a straight line parallel to y-axis and passing through the point ( – 3, 5).

Answer- 5

The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3
⇒ x + 3 = 0

Question -6

Find the equation of the a line whose
(i) slope = 3, y-intercept = – 5
(ii) slope = - \frac { 2 }{ 7 } , y-intercept = 3
(iii) gradient = √3, y-intercept = - \frac { 4 }{ 3 }
(iv) inclination = 30°, y-intercept = 2

Answer -6

Equation of a line whose slope and y-intercept is given is
y = mx + c
where m is the slope and c is the y-intercept
(i) y = mx + c
⇒ y = 3x + (-5)
⇒ y = 3x – 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 6.1

Question- 7

Find the slope and y-intercept of the following lines:
(i) x – 2y – 1 = 0
(ii) 4x – 5y – 9 = – 0
(iii) 3x +5y + 7 = 0
(iv) \frac { x }{ 3 } +\frac { y }{ 4 } =1
(v) y – 3 = 0
(vi) x – 3 = 0

Answer- 7

We know that in the equation
y = mx + c, m is the slope and c is the y-intercept.
Now using this, we find,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 7

Question -8

The equation of the line PQ is 3y – 3x + 7 = 0
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.

Answer- 8

Equation of line PQ is 3y – 3x + 7 = 0
Writing in form of y = mx + c
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 8


page-238

Question -9

The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 9

Answer -9

Slope of the line y = x + 1 after comparing
it with y = mx + c, m = 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 9.1

Question- 10

Find the value of p, given that the line \frac { y }{ 2 } =x-p passes through the point ( – 4, 4) 

Answer -10

Equation of line is \frac { y }{ 2 } =x-p
It passes through the points (-4, 4)
and It will satisfy the equation
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 10

Question -11

Given that (a, 2a) lies on the line \frac { y }{ 2 } =3x-6 find the value of a.

Answer- 11

∵ Point (a, 2a) lies on the line
\frac { y }{ 2 } =3x-6
∴this point will satisfy the equation
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 11

Question- 12

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- 12

Equation of the line is y = mx + c
∴ it passes through the points (1, 4)
∴ 4 = m x 1 + c
⇒ 4 = m + c
hence ⇒ m + c = 4 … (i)
Again it passes through the point (-2, -5)
∴ 5 = m (-2) + c
⇒ 5 = -2 m + c
so ⇒ 2m – c = 5 …(ii)
Adding (i) and (ii)
3m = 9
⇒ m = 3
Substituting the value of m in (i)
3 + c = 4
⇒ c = 4 – 3 = 1
Hence m = 3, c = 1

Question -13

Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y-axis.

Answer- 13

∴ The line intersects y-axis making an intercept of -3
∴ the co-ordinates of point of intersection will be (0, -3)
Now the slope of line (m) = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 13

Question- 14

Find the equation of a straight line passing through ( – 1, 2) and whose slope is \\ \frac { 2 }{ 5 } .

Answer -14

Equation of the line will be
y-{ y }_{ 1 }=m(x-{ x }_{ 1 })
y-2=\frac { 2 }{ 5 } (x+1)
⇒ 5y – 10 = 2x + 2
so ⇒ 2x – 5y + 2 + 10 = 0
hence ⇒ 2x – 5y + 12 = 0

Question -15

Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).

Answer- 15

The equation of line whose slope is wand passes through a given point is
y-{ y }_{ 1 }=m(x-{ x }_{ 1 })
Here m = tan 60° = √3 and point is (0, -3)
∴ y + 3 = √3 (x – 0)
⇒ y + 3 = √3x
⇒ √3x – y – 3 = 0

Question- 16

Find the gradient of a line passing through the following pairs of points.
(i) (0, – 2), (3, 4)
(ii) (3, – 7), ( – 1, 8)

Answer -16

m = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }
Given
(i) (0, -2), (3, 4)
(ii) (3, -7), (-1, 8)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 16

Question- 17

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- 17

Co-ordinates of points E (0, 4) and F (3, 7) are given, then
(i) The gradient of EF
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 17
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 17.1

Question- 18

Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.

Answer- 18

Putting y = 0, we will get the intercept made on x-axis,
2x – 3y + 12 = 0
⇒ 2x – 3 × 0 + 12 = 0, ⇒ 2x – 0 + 2 = 0,⇒ 2x = -12
and ⇒ x = -6
and putting x = 0, we get the intercepts made on y-axis,
2x – 3y + 12 = 0
⇒ 2 × 0 – 3y + 12 = 0
so ⇒ -3y = -12
therefore ⇒ y= \frac { -12 }{ -3 }  = 4

Question -19

Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence, show that the points P, Q and R (11, 4) are collinear.

Answer- 19

The two given points are P (5, 1), Q(1, -1).
∴ Slope of the line (m)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 19

Question- 20

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. (2014)

Answer -20

Given That
A(a, 3), B (2, 1) and C (5, a) are collinear.
Slope of AB = Slope of BC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 20

Question -21

Use a graph paper 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 (½, k). (2005)

Answer -21

Points (h, 4) and (½, k) lie on the line passing
through A(-1, -1) and B(2, 5)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 21

Question -22

ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, – 4). Find
(i) the coordinates of A
(ii) the equation of the diagonal BD.(2011)

Answer -22

Given that
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 22

Question- 23

In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.(2013)

Answer- 23

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 23
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 23.1

Question -24

Find the equation of a line passing through the point ( – 2, 3) and having x-intercept 4 units. (2002)

Answer -24

x-intercept = 4
∴ Co-ordinates of the point will be (4, 0)
Now slope of the line passing through the points (-2, 3) and (4, 0)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 24

Question -25

Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.

Answer -25

x-intercept = 6
∴ The line will pass through the point (6, 0)
y -intercept = -4 ⇒ c = -4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 25


page-239

Question -26

A(2, 5), B(-1, 2) and C(5, 8) are the vertices of a triangle ABC, ‘M’ is a point on AB such that AM : MB = 1:2. Find the co-ordinates of M’. Hence find the equation of the line passing through the points C and M. (2018)

Answer- 26

Coordinates of the vertices of a ∆ ABC are A(2, 5), B(- 1, 2) and C (5, 8). Since M is a point on AB such that AM : MB = 1 : 2

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Answer-26

Question -27

Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.

Answer- 27

line x – 2y – 11 = 0 passes through y-axis
x = 0,
Now substituting the value of x in the equation x – 2y – 11 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 27

Question -28

Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.

Answer -28

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 28
Let the line containing the point P (3, 2)
passes through x-axis at A (x, 0) and y-axis at B (0, y)
OA = OB given
∴ x = y
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 28.1

Question- 29

Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)

Answer-29

 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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 29
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 29.1

Question -30

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 30
(i) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB. (2010)

Answer -30

Points A and B are on x-axis and y-axis respectively
Let co-ordinates of A be (X, O) and of B be (O, Y)
P (2, -3) is the midpoint of AB
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 30.1

Question- 31

M and N are two points on the X axis and Y axis respectively.
P(3, 2) divides the line segment MN in the ratio 2 : 3.
Find :
(i) the coordinates of M and N
(ii) slope of the line MN

Answer-31

Let the coordinates of M and N be (x, 0) and (0, y)

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Answer-31.1

Thus, the coordinates of M and N are M(5,0) and N(0, 5).

Hence, the slope of the line MN is – 1.

Question-32

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

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Q-32

Answer-32

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Ans-32

Question-33

(i) write down the coordinate of point P  that divide the line joining A (-4, 1) an 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-33

(i) Co-ordinates of point P are

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Ans-33.1 and 33.2

(iii) Let AB be divided by the point P(0, y) lying on y-axis in the ratio

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Ans-33.3

Thus, the ratio in which the y-axis divide the line AB is 4: 17.

Question-34

Find the equations of the diagonals of a rectangle whose sides are x = – 1, x = 2 , y = – 2 and y = 6.

Answer-34

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.1 Ans-34.1

The equations of sides of a rectangle whose equations are
x1 = -1, x2 = 2, y1 = -2, y2 = 6.
These lines form a rectangle when they intersect at A, B, C, D respectively
Co-ordinates of A, B, C and D will be
(-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.
AC and BD are its diagonals
(i) Slope of the diagonal AC

Question -35

Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x +7y = 3 and 2x – 3y = 7

Answer -35

5x + 7y = 3 …(i)
2x – 3 y = 7 …(ii)
Multiply (i) by 3 and (ii) by 7,
15x + 21y = 9
14x – 21y = 49
Adding we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1 32


Exercise 12.2 , Equation of Straight Line ML Aggarwal ICSE Class-10 

page-245

Question- 1

State which one of the following is true : The straight lines y = 3x – 5 and 2y = 4x + 7 are
(i) parallel
(ii) perpendicular
(iii) neither parallel nor perpendicular.

Answer- 1

Slope of line y = 3x – 5 = 3
and slope of line 2y = 4x + 7
⇒ y = 2x + \\ \frac { 7 }{ 2 }  = 2.
∴ Slope of both the lines are neither equal nor their product is – 1.
∴ These line are neither parallel nor perpendicular.

Question- 2

If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.

Answer -2

In equation
6x + 5 y – 7 = 0
⇒ 5y = -6x + 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 2

Question -3

Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b. 

Answer- 3

In equation 2x – by + 5 = 0
⇒ – by = – 2x – 5
⇒ y = \\ \frac { 2 }{ b }  + \\ \frac { 5 }{ b }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 3

Question -4

If the straight lines 3x – 5y = 7 and 4x + ay + 9 = O are perpendicular to one another, find the value of a (2018)

Answer-4

Given lines are
3x – 5y = 1 ……….(i) and 4x + ay + 9 = 0  …………(ii)
Slope of line (i) (m1) =  (3/5)=3/5
Slope of line (ii) (m2) = (4/𝑎)

Also, given that two lines are perpendicular to one and another
∴ (m1) (m2) = – 1

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.2 Ans-4
Hence, the value of a = 12/5 .

Question- 5

If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

Answer -5

Given
In the equation 3x + by + 5 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 7

Question-6

Is the line through ( – 2, 3) and (4, 1) perpendicular to the line 3x = y + 1 ?
Does the line 3x = y + 1 bisect the join of ( – 2, 3) and (4, 1). 

Answer -6

Slope of the line passing through the points
(-2, 3) and (4, 1) = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 8
Hence the line 3x = y + 1 bisects the line joining the points (-2, 3), (4, 1).

Question -7

The line through A ( – 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b. (2012)

Answer -7

Gradient (m1) of the line passing through the
points A (-2, 3) and B (4, b)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 9

Question-8

If the lines 3x + y = 4, x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.

Answer -8

In the line 3x + y = 4 …(i)
⇒ y = – 3x + 4
Slope (m1) = – 3
In the line x – ay + 7 = 0…..(ii)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 10
⇒ -b = -6 ⇒ b = 6
Hence a = 3, b = 6

Question -9

Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis. 

Answer -9

In the given line 2x – 3y – 7 = 0
⇒ 3y = 2x – 7
⇒ y=\frac { 2 }{ 3 } x-\frac { 7 }{ 3 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 11

Question -10

Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 

Answer -10

In the line 2x + 5y + 7 = 0
⇒ 5y = – 2x – 7
⇒ y=\frac { -2 }{ 5 } x-\frac { 7 }{ 5 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 12.1

Question -11

Find the equation of a straight. line perpendicular to the line 3x – 4y + 12 = 0 and having same y-intercept as 2x – y + 5 = 0.

Answer -11

In the given line 3x – 4y + 12 = 0
⇒ 4y = 3x + 12
⇒ y = \\ \frac { 3 }{ 4 } x+3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 13

Question-12

Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0. 

Answer -12

In the given equation 3x + 5y + 15 = 0
⇒ 5y = – 3x – 15
⇒ y = \\ \frac { -3 }{ 5 } x-3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 15

Question -13

Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point ( – 1, – 2).

Answer -13

In the given line 3x + 8y = 12
⇒ 8y = -3x + 12
⇒ y=\frac { -3 }{ 8 } x+\frac { 12 }{ 8 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 17
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 17.1

Question-14

(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.

Answer-14

(i) In the line 4x – 3y + 12 = 0 …(i)
3y = 4x + 12
⇒ y = \\ \frac { 4 }{ 3 } x+4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 18

Question -15

Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and ( – 4, 1).

Answer -15

The given line 2x + 5y – 7 = 0
5y = -2x + 7
⇒ y=\frac { -2 }{ 5 } x+\frac { 7 }{ 5 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 19
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 19.1

Question -16

Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2), (2, 2).

Answer -16

In the given line 3x + 2y – 8=0
⇒ 2y = – 3x + 8
so ⇒ y = \\ \frac { -3 }{ 2 } x+4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 20

Question -17

Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.

Answer -17

Let the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)
Substituting the value of y in the equation
2x + 5 × 0 – 4 = 0
⇒ 2x – 4 = 0
and ⇒ 2x = 4
so ⇒ x = \\ \frac { 4 }{ 2 }  = 2
Coordinates of the points of intersection will be (2, 0)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 21

Question -18

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. (2010)

Answer -18

(i) Equation of the line is 3x + 4y – 1 = 0

⇒ 4y = 7 – 3x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 22
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 22.1

Question -19

Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.

Answer -19

In the equation 4x – 3y – 5 = 0,
⇒ 3y = 4x – 5
⇒ y=\frac { 4 }{ 3 } x-\frac { 5 }{ 3 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 23
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 23.1

Question -20

Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, – 2) and (6, 4).

Answer -20

Given that
Slope of the line through (0, 0) and (2, 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 24


page-246

Question -21

Prove that the line through,( – 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).

Answer -21

Given that
Slope of the line through (-2, 6) and (4, 8)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 25

Question-22

Show that the triangle formed by the points A (1, 3), B (3, – 1) and C ( – 5, – 5) is a right angled triangle by using slopes.

Answer -22

Slope (m1) of line by joining the points
A(1, 3), B (3, -1) = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 26
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 26.1

Question -23

Find the equation of the line through the point ( – 1, 3) and parallel to the line joining the points (0, – 2) and (4, 5).

Answer -23

Slope of the line joining the points (0, -2) and (4, 5) = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } }
\frac { 5+2 }{ 4-0 } =\frac { 7 }{ 4 }
Slope of the line parallel to it passing through (-1, 3) = \\ \frac { 7 }{ 4 }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 27

Question-24

A ( – 1, 3), B (4, 2), C (3, – 2) are the vertices of a triangle.
(i) Find the coordinates of the centroid G of the triangle.
(ii) Find the equation of the line through G and parallel to AC

Answer-24

Given, A (-1, 3), B (4, 2), C (3, -2)
(i) Coordinates of centroid G
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 28
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 28.1

Question-25

Find the equation of the line through (0, – 3) and perpendicular to the line joining the points  (– 3, 2) and (9, 1).

Answer -25

The slope (m1) of the line joining the points (-3, 2) and (9, 1)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 31

Question-26

The vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, -6). Find :
(i) Slope of BC.
(ii) Equation of a line perpendicular to BC and passing through A.(2019)

Answer-26

Vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, -6)
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.2 Ans-26.1
Slope of the line perpendicular to BC = 7/8
Now, equation of the line perpendicular to BC and passing through A is

8y – 64 = 7x – 21
7x – 8y + 43 = 0

Question-27

The vertices of a triangle are A (10, 4), B (4, – 9) and C ( – 2, – 1). Find the equation of the altitude through A. 

(The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.)

Answer-27

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.2 Ans-28

Question-28

A (2, – 4), B (3, 3) and C ( – 1, 5) are the vertices of triangle ABC. Find the equation of :
(i) the median of the triangle through A
(ii) the altitude of the triangle through B

Answer -28

(i) D is the mid-point of BC
Co-ordinates of D will
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 33
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 33.1

Question-29

Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, – 6).

Answer -29

Slope of the line joining the points (1, 2) and (5, -6)
{ m }_{ 1 }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -6-2 }{ 5-1 } =\frac { -8 }{ 4 } =-2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 34

Question -30

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.(2008)

Answer -30

Coordinates of A are (7, -3), of B = (1, 9)
(i) ∴ slope (m)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 35
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 35.1

Question-31

The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.

Answer -31

Slope of BD (m1) = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }
\frac { 8-3 }{ 6-1 } =\frac { 5 }{ 5 } =1
Diagonal AC is perpendicular bisector of diagonal BD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 36

Question-32

ABCD is a rhombus. The co-ordinates of A and C are (3, 6) and ( – 1, 2) respectively. Write down the equation of BD. (2000)

Answer-32

Co-ordinates of A (3, 6), C (-1, 2)
Slope of AC (m1) = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }
\frac { 2-6 }{ -1-3 } =\frac { -4 }{ -4 } =1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 37
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 37.1

Question -33

Find the image of the point (1, 2) in the line x – 2y – 7 = 0

Answer -33

Draw a perpendicular from the point P(1, 2) on the line, x – 2y – 7 = 0
Let P’ is the image of P and let its
co-ordinates sue (α, β) slope of line x – 2y – 7 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 40
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 40.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 40.2

Question -34

If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.

Answer -34

Let the co-ordinates of Q be (α, β) and let the line x – 4y – 6 = 0 is the
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 41
perpendicular bisector of PQ and it intersects the line at M.
M is the mid point of PQ Now slope of line x – 4y – 6 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 41.1

Question -35

OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.

Answer -35

OA = \sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } }
\sqrt { { (3-0) }^{ 2 }+{ (0) }^{ 2 } }
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.2 42


MCQ of Equation of Straight Line of ML Aggarwal Solutions for ICSE Class-10

Choose the correct answer from the given four options (1 to 13) :

page-246

Question- 1

The slope of a line parallel to y-axis is
(a) 0
(b) 1
(c) – 1
(d) not defined

Answer -1

Slope of a line parallel to y-axis is not defined. (d)


page-247

Question -2

The slope of a line which makes an angle of 30° with the positive direction of x-axis is
(a) 1
(b) \frac { 1 }{ \sqrt { 3 } }
(c) √3
(d) - \frac { 1 }{ \sqrt { 3 } }

Answer -2

Slope of a line which makes an angle of 30°
with positive direction of x-axis = tan 30°
\frac { 1 }{ \sqrt { 3 } }  (b)

Question -3

The slope of the line passing through the points (0, – 4) and ( – 6, 2) is
(a) 0
(b) 1
(c) – 1
(d) 6

Answer -3

Slope of the line passing through the points (0, -4) and (-6, 2)
\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2+4 }{ -6-0 } =\frac { 6 }{ -6 } =-1  (c)

Question- 4

The slope of the line passing through the points (3, – 2) and ( – 7, – 2) is
(a) 0
(b) 1
(c) - \frac { 1 }{ 10 }
(d) not defined

Answer- 4

Slope of the line passing through the points (3, -2) and (-7, -2)
\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -2+2 }{ -7-3 } =\frac { 0 }{ -10 } =0  (a)

Question- 5

The slope of the fine passing through the points (3, – 2) and (3, – 4) is
(a) – 2
(b) 0
(c) 1
(d) not defined

Answer- 5

The slope of the line passing through (3, -2) and (3, -4)
\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -4+2 }{ 3-3 } =\frac { -2 }{ 0 }  (d)

Question -6

The inclination of the line y = √3x – 5 is
(a) 30°
(b) 60°
(c) 45°
(d) 0°

Answer -6

The inclination of the line y = √3x – 5 is
√3 = tan 60° = 60° (b)

Question- 7

If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is
(a) -2
(b) -1
(c) 1
(d) 2

Answer- 7

Slope of the line passing through the points (2, 5) and (k, 3) is 2, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS 7

Question- 8

The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is
(a) - \frac { 1 }{ 5 }
(b) \\ \frac { 1 }{ 5 }
(c) -5
(d) 5

Answer -8

Slope of the line parallel to the line passing through (0, 6) and (7, 3)
Slope of the line = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 3-6 }{ 7-0 } =\frac { -3 }{ 7 }  (b)

Question -9

The slope of a line perpendicular to the line passing through the points (2, 5) and ( – 3, 6) is
(a) - \frac { 1 }{ 5 }
(b) \\ \frac { 1 }{ 5 }
(c) -5
(d) 5

Answer -9

Slope of the line joining the points (2, 5), (-3, 6)
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line MCQ Ans-9

Question- 10

The slope of a line parallel to the line 2x + 3y – 7 = 0 is

(a) - \frac { 2 }{ 3 }
(b) \\ \frac { 2 }{ 3 }
(c) - \frac { 3 }{ 2 }
(d) \\ \frac { 3 }{ 2 }

Answer -10

The slope of a line parallel to the line 2x + 3y – 7 = 0
slope of the line
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line MCQ Ans-10

Question -11

The slope of a line perpendicular to the line 3x = 4y + 11 is
(a) \\ \frac { 3 }{ 4 }
(b) - \frac { 3 }{ 4 }
(c) \\ \frac { 4 }{ 3 }
(d) - \frac { 4 }{ 3 }

Answer -11

slope of a line perpendicular to the line 3x = 4y + 11 is
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line MCQ Ans-11

Question -12

If the lines 2x + 3y = 5 and kx – 6y = 7 are parallel, then the value of k is
(a) 4
(b) – 4
(c) \\ \frac { 1 }{ 4 }
(d) - \frac { 1 }{ 4 }

Answer -12

lines 2x + 3y = 5 and kx – 6y = 7 are parallel
Slope of 2x + 3y = 5 = Slope of kx – 6y = 7
⇒ 3y – 2x + 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS 12
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line MCQ Ans-12

so option (b) is correct

Question -13

If the line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is
(a) \\ \frac { 3 }{ 2 }
(b) - \frac { 3 }{ 2 }
(c) \\ \frac { 2 }{ 3 }
(d) - \frac { 2 }{ 3 }

Answer-13

line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0

are perpendicular to each other
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line MCQS 13


Chapter – Test  Equation of Straight Line Solutions of ML Aggarwal

page-249

Question- 1

Find the equation of a line whose inclination is 60° and y-intercept is – 4.

Answer- 1

Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
⇒ y – √3x – 4

Question -2

Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.

Answer- 2

Slope of the line 3y + 2x = 12
⇒ 3y = 12 – 2x
⇒ 3y = -2x + 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 2

Question -3

If the equation of a line is y – √3x + 1, find its inclination.

Answer -3

In the line
y = √3 x + 1
Slope = √3
⇒ tan θ = √3
⇒ θ = 60° (∵ tan 60° = √3)

Question -4

If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.

Answer- 4

The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points -4 = 2m + c …(i)
and 1 = -3m + c …(ii)
Subtracting we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 4

Question -5

If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.

Answer- 5

Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)
divides AC in the ratio of m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 5

Question- 6

Find the inclination of the line joining the points P (4, 0) and Q (7, 3).

Answer -6

Slope of the line joining the points P (4, 0) and Q (7, 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 6

Question -7

Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to - \frac { 3 }{ 7 }

Answer -7

Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5
Adding we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 7.1

Question -8

If the lines \frac { x }{ 3 } +\frac { y }{ 4 } =7 and 3x + ky = 11 are perpendicular to each other, find the value of k.

Answer -8

Given Equation of lines are
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 9

Question -9

Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).

Answer -9

The equation of the line is x – 2y + 8 = 0
⇒ 2y = x + 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 10

Question-10

Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.

Answer-10

Equations of the line is
3y = 5 – x ⇒ 3y = -x + 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test 11

Question-11

Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.

Answer-11

Let slope of the line joining the points A (1, 2) and B (6, 7) be m1
Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.

Question-12

The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.

Answer-12

Slope of line AC (m1)
The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.

Question-13

A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line Chapter-Test Q-13
(i) the co-ordinates of A and B
(ii) the equation of the line AB

Answer-13

A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (2, 1) divides BA in the ratio 3 : 1.
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line Chapter-Test Ans-13

Question-14

A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.

Answer -14

Let the line make intercept a and b with the
x-axis and y-axis respectively then the line passes through

ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line Chapter-Test Ans-14.2

Hence, the equation of the line is x3+y4=1  or 4x + 3y = 12

Question-15

If the coordinates of the vertex A of a square ABCD are (3, – 2) and the equation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.

Answer -15

Co-ordinates of A are (3, -2).
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line Chapter-Test Q-15
Diagonals AC and BD of the square ABCD
bisect each other at right angle at O.
∴ O is the mid-point of AC and BD Equation
ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line Chapter-Test Q-15.1

The co-ordinate of intersection of diagonal -O

3x-7y= -6 ……….(i)

7x+3y=15 ……..(ii)

multiply (i) by 3 and .(ii) by 7

9x – 21y  =  -18    …….(iii)

49x + 21y = 105 ………(iv)

Adding 3rd and 4th equation

— : End of Equation of Straight Line Solutions :–


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