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

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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### 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°

(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) $\frac { 1 }{ \sqrt { 3 } }$

(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.

(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.

(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).

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

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

#### 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

We know that in the equation
y = mx + c, m is the slope and c is the y-intercept.
Now using this, we find,

#### 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.

Equation of line PQ is 3y – 3x + 7 = 0
Writing in form of y = mx + c

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.

Slope of the line y = x + 1 after comparing
it with y = mx + c, m = 1

#### Question- 10

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

Equation of line is $\frac { y }{ 2 } =x-p$
It passes through the points (-4, 4)
and It will satisfy the equation

#### Question -11

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

∵ Point (a, 2a) lies on the line
$\frac { y }{ 2 } =3x-6$
∴this point will satisfy the equation

#### 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.

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)
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.

∴ 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) = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

#### Question- 14

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

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).

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)

m = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$
Given
(i) (0, -2), (3, 4)
(ii) (3, -7), (-1, 8)

#### Question- 17

The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :
(ii) The equation of EF
(iii) The coordinates of the point where the line EF intersects the x-axis.

Co-ordinates of points E (0, 4) and F (3, 7) are given, then

#### Question- 18

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

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.

The two given points are P (5, 1), Q(1, -1).
∴ Slope of the line (m)

#### 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)

Given That
A(a, 3), B (2, 1) and C (5, a) are collinear.
Slope of AB = Slope of BC

#### 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)

Points (h, 4) and (½, k) lie on the line passing
through A(-1, -1) and B(2, 5)

#### 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)

Given that
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)

#### Question- 23

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

⇒ D is mid point of BC
∴ D is $\left( \frac { 7+1 }{ 2 } ,\frac { 8-10 }{ 2 } \right)$
i.e (4, -1)

#### Question -24

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

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)

#### Question -25

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

x-intercept = 6
∴ The line will pass through the point (6, 0)
y -intercept = -4 ⇒ c = -4

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)

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

#### 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.

line x – 2y – 11 = 0 passes through y-axis
x = 0,
Now substituting the value of x in the equation x – 2y – 11 = 0

#### Question -28

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

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

#### 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)

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

#### 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

(i) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB. (2010)

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

#### 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

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

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

#### 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?

(i) Co-ordinates of point P are

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

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.

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

5x + 7y = 3 …(i)
2x – 3 y = 7 …(ii)
Multiply (i) by 3 and (ii) by 7,
15x + 21y = 9
14x – 21y = 49

### 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.

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.

In equation
6x + 5 y – 7 = 0
⇒ 5y = -6x + 7

#### Question -3

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

In equation 2x – by + 5 = 0
⇒ – by = – 2x – 5
⇒ y = $\\ \frac { 2 }{ b }$ + $\\ \frac { 5 }{ b }$

#### 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)

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

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.

Given
In the equation 3x + by + 5 = 0

#### 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).

Slope of the line passing through the points
(-2, 3) and (4, 1) = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } }$

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)

Gradient (m1) of the line passing through the
points A (-2, 3) and B (4, b)

#### 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.

In the line 3x + y = 4 …(i)
⇒ y = – 3x + 4
Slope (m1) = – 3
In the line x – ay + 7 = 0…..(ii)

⇒ -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.

In the given line 2x – 3y – 7 = 0
⇒ 3y = 2x – 7
⇒ $y=\frac { 2 }{ 3 } x-\frac { 7 }{ 3 }$

#### Question -10

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

In the line 2x + 5y + 7 = 0
⇒ 5y = – 2x – 7
⇒ $y=\frac { -2 }{ 5 } x-\frac { 7 }{ 5 }$

#### 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.

In the given line 3x – 4y + 12 = 0
⇒ 4y = 3x + 12
⇒ y = $\\ \frac { 3 }{ 4 } x+3$

#### Question-12

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

In the given equation 3x + 5y + 15 = 0
⇒ 5y = – 3x – 15
⇒ y = $\\ \frac { -3 }{ 5 } x-3$

#### Question -13

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

In the given line 3x + 8y = 12
⇒ 8y = -3x + 12
⇒ $y=\frac { -3 }{ 8 } x+\frac { 12 }{ 8 }$

#### 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.

(i) In the line 4x – 3y + 12 = 0 …(i)
3y = 4x + 12
⇒ y = $\\ \frac { 4 }{ 3 } x+4$

#### 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).

The given line 2x + 5y – 7 = 0
5y = -2x + 7
⇒ $y=\frac { -2 }{ 5 } x+\frac { 7 }{ 5 }$

#### 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).

In the given line 3x + 2y – 8=0
⇒ 2y = – 3x + 8
so ⇒ y = $\\ \frac { -3 }{ 2 } x+4$

#### 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.

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)

#### 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)

⇒ 4y = 7 – 3x

#### 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.

In the equation 4x – 3y – 5 = 0,
⇒ 3y = 4x – 5
⇒ $y=\frac { 4 }{ 3 } x-\frac { 5 }{ 3 }$

#### Question -20

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

Given that
Slope of the line through (0, 0) and (2, 3)

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).

Given that
Slope of the line through (-2, 6) and (4, 8)

#### 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.

Slope (m1) of line by joining the points
A(1, 3), B (3, -1) = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } }$

#### 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).

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) = $\\ \frac { 7 }{ 4 }$

#### 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

Given, A (-1, 3), B (4, 2), C (3, -2)
(i) Coordinates of centroid G

#### Question-25

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

The slope (m1) of the line joining the points (-3, 2) and (9, 1)

#### 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)

Vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, -6)

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.)

#### 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

(i) D is the mid-point of BC
Co-ordinates of D will

#### Question-29

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

Slope of the line joining the points (1, 2) and (5, -6)
${ m }_{ 1 }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -6-2 }{ 5-1 } =\frac { -8 }{ 4 } =-2$

#### 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)

Coordinates of A are (7, -3), of B = (1, 9)
(i) ∴ slope (m)

#### 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.

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

#### 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)

Co-ordinates of A (3, 6), C (-1, 2)
Slope of AC (m1) = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$
$\frac { 2-6 }{ -1-3 } =\frac { -4 }{ -4 } =1$

#### Question -33

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

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

#### 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.

Let the co-ordinates of Q be (α, β) and let the line x – 4y – 6 = 0 is the

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

#### 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.

OA = $\sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } }$
$\sqrt { { (3-0) }^{ 2 }+{ (0) }^{ 2 } }$

### 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

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) $- \frac { 1 }{ \sqrt { 3 } }$

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

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

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

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°

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

Slope of the line passing through the points (2, 5) and (k, 3) is 2, then

#### 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

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

Slope of the line joining the points (2, 5), (-3, 6)

#### 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) $\\ \frac { 3 }{ 2 }$

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

#### 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) $- \frac { 4 }{ 3 }$

slope of a line perpendicular to the line 3x = 4y + 11 is

#### 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) $- \frac { 1 }{ 4 }$

lines 2x + 3y = 5 and kx – 6y = 7 are parallel
Slope of 2x + 3y = 5 = Slope of kx – 6y = 7
⇒ 3y – 2x + 5

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) $- \frac { 2 }{ 3 }$

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

are perpendicular to each other

### 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.

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.

Slope of the line 3y + 2x = 12
⇒ 3y = 12 – 2x
⇒ 3y = -2x + 12

#### Question -3

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

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.

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,

#### Question -5

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

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

#### Question- 6

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

Slope of the line joining the points P (4, 0) and Q (7, 3)

#### 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 $- \frac { 3 }{ 7 }$

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

#### 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.

Given Equation of lines are

#### Question -9

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

The equation of the line is x – 2y + 8 = 0
⇒ 2y = x + 8

#### Question-10

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

Equations of the line is
3y = 5 – x ⇒ 3y = -x + 5

#### 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.

Let slope of the line joining the points A (1, 2) and B (6, 7) be m1

#### 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.

Slope of line AC (m1)

#### Question-13

A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:

(i) the co-ordinates of A and B
(ii) the equation of the line AB

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.

#### 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.

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

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.

Co-ordinates of A are (3, -2).

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

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)