NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 10
Chapter Name : Straight Lines

Exercise 10.1

Q1 Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7),
(5, – 5) and (– 4, –2). Also, nd its area.

Answer. Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given
quadrilateral can be drawn as

To nd the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (ΔABC) + area (ΔACD)
We know that the area of a triangle whose vertices are (x , y ) , (x , y ) , and (x , y ) is
1 1 2 2 3 3

1
|x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 ) |
2

Therefore, area of ΔABC
1
2
= | − 4(7 + 5) + 0(−5 − 5) + 5(5 − 7)| unit
2
1
2
= | − 4(12) + 5(−2)| unit
2

1
2
= | − 48 − 10| unit
2

1
2
= | − 58| unit
2

1
2
= × 58 unit
2
2
= 29 unit

Area of ΔACD
1
2
= | − 4(−5 + 2) + 5(−2 − 5) + (−4)(5 + 5)| unit
2
1
2
= | − 4(−3) + 5(−7) − 4(10)| unit
2
1
2
= |12 − 35 − 40| unit
2
1 2
= | − 63| unit
2

63 2
= unit
2

Thus, area (ABCD)
63 2 58+63 2 121 2
= (29 + ) unit = unit = unit
2 2 2

Page : 211 , Block Name : Exercise 10.1

Q2 The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the
base is at the origin.
Find vertices of the triangle.

Answer. Let ABC be the given equilateral triangle with side 2a.Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, -a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side
is perpendicular.
Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ΔAOC, we obtain
2 2 2
(AC) = (OA) + (OC)

2 2 2
⇒ (2a) = (OA) + a

2 2 2
⇒ 4a − a = (OA)

2 2
⇒ (OA) = 3a

⇒ OA = √3a

∴ Coordinates of point A = ( ±√3a, 0)

Thus, the vertices of the given equilateral triangle are (0, a), (0, −a), and (√3a, 0) or

(0, a), (0, −a), and

(−√3a, 0) .

Page : 211 , Block Name : Exercise 10.1

Q3 Find the distance between P (x , y ) and Q (x , y ) when :
1 1 2 2

(i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Answer. The given points are P(x , y ) 1 1 and
Q (x2 , y2 ) .
(i) When PQ is parallel to the y-axis, x 1 = x2 .
In this case, distance between P and Q
2 2
= √(x2 − x1 ) + (y2 − y1 )

2
= √(y2 − y1 )

= |y2 − y1 |

(ii) When PQ is parallel to the x-axis, y 1
= y
2
.
In this case, distance between P and Q
2 2
= √(x2 − x1 ) + (y2 − y1 )

2
= √(x2 − x1 )

= |x2 − x1 |

page : 211 , Block Name : Exercise 10.1

Q4 Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer. Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).
2 2 2 2
Accordingly, √(7 − a) + (6 − 0) = √(3 − a) + (4 − 0)

2 2
⇒ √49 + a − 14a + 36 = √9 + a − 6a + 16

2 2
⇒ √a − 14a + 85 = √a − 6a + 25

On squaring both sides, we obtain
2 2
a − 14a + 85 = a − 6a + 25

⇒ –14a + 6a = 25 – 85
⇒ –8a = –60
60 15
⇒ a = =
8 2

Thus, the required point on the x-axis is ( .
15
, 0)
2

page : 211, Block Name : Exercise 10.1

Q5 Find the slope of a line, which passes through the origin, and the mid-point of the
line segment joining the points P (0, – 4) and B (8, 0)

Answer. The coordinates of the mid-point of the line segment joining the points
P (0, –4) and B (8, 0) are
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is
given by m = .
y2 −y1
, x2 ≠ x1
x2 −x1

Therefore, the slope of the line passing through (0, 0) and (4, –2) is

.
−2−0 −2 1
= = −
4−0 4 2

Hence, the required slope of the line is − . 1

2

Page : 211 , Block Name : Exercise 10.1

Q6 Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and
(–1, –1) are the vertices of a right angled triangle.

Answer. The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).
It is known that the slope (m) of a non-vertical line passing through the points (x , y ) and (x , y ) is
1 1 2 2

given by m = .
y2 −y1
, x2 ≠ x1
x2 −x1

∴Slope of AB (m1) =
5−4
= −1
3−4

Slope of BC (m2) =
−1−5 −6 3
= =
−1−3 −4 2

Slope of CA (m3) =
4+1 5
= = 1
4+1 5

It is observed that m1m3 = –1
This shows that line segments AB and CA are perpendicular to each other
i.e., the given triangle is right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Page : 212 , Block Name : Exercise 10.1

Q7 Find the slope of the line, which makes an angle of 30° with the positive direction
of y-axis measured anticlockwise.

Answer. If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise,
then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° +
30° = 120°.

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60° = −√3

Page : 212 , Block Name : Exercise 10.1

Q8 Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear.

Answer. If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then
Slope of AB = Slope of BC

1−(−1) 5−1
⇒ =
2−x 4−2

1+1 4
⇒ =
2−x 2

2
⇒ = 2
2−x

⇒ 2 = 4 − 2x

⇒ 2x = 2

⇒ x = 1

Thus, the required value of x is 1.

Page : 212 , Block Name : Exercise 10.1

Q9 Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2)
are the vertices of a parallelogram.

Answer. Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

Slope of AB =
0+1 1
=
4+2 6

Slope of CD = 2−3

−3−3
=
−1

−6
=
1

6

⇒ Slope of AB = Slope of CD
⇒ AB and CD are parallel to each other.
Now, slope of BC = =
3−0 3
= = −3
3−4 −1

Slope of AD = =
2+1 3
= = −3
−3+2 −1

⇒ Slope of BC = Slope of AD
⇒ BC and AD are parallel to each other.
Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a
parallelogram.
Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Page : 212 , Block Name : Exercise 10.1

Q10 Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).

−2−(−1)
Answer. The slope of the line joining the points (3, –1) and (4, –2) is m = 4−3
= −2 + 1 = −1 .
Now, the inclination (θ ) of the line joining the points (3, –1) and (4, – 2) is given by
tan θ= –1
⇒ θ = (90° + 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

Page : 212 , Block Name : Exercise 10.1

Q11 The slope of a line is double of the slope of another line. If tangent of the angle
between them is , nd the slopes of the lines.
1

3

Answer. Let m and m be the slopes of the two given lines such that.m = 2m.
1 1

We know that if θisthe angle between the lines l1 and l2 with slopes m1 and m2, then tan θ = ∣∣ .
m2 −m1
∣
1+m1 m2 ∣

It is given that the tangent of the angle between the two lines is 1

3
.
1 ∣ m−2m ∣
∴ =
3 ∣ 1+(2m)⋅m ∣

1 ∣ −m ∣
⇒ =
3 ∣ 1+2m2 ∣

1 −m 1 −m m
⇒ = 2
or = −( 2
) = 2
3 1+2m 3 1+2m 1+2m

Case I
1 −m
⇒ =
3 1+2m2

2
⇒ 1 + 2m = −3m

2
⇒ 2m + 3m + 1 = 0

2
⇒ 2m + 2m + m + 1 = 0

⇒ (m + 1)(2m + 1) = 0

1
⇒ m = −1 or m = −
2

If m = –1, then the slopes of the lines are –1 and –2.
If m = − , then the slopes of the lines are − and –1.
1 1

2 2

Case II
1 m
= 2
3 1+2m

2
⇒ 2m + 1 = 3m

2
⇒ 2m − 3m + 1 = 0

2
⇒ 2m − 2m − m + 1 = 0

⇒ 2m(m − 1) − 1(m − 1) = 0

⇒ (m − 1)(2m − 1) = 0

1
⇒ m = 1 or m =
2

If m = 1, then the slopes of the lines are 1 and 2.
If m = , then the slopes of the lines are and 1.
1 1

2 2

Hence, the slopes of the lines are –1 and –2 or − 1

2
and –1 or 1 and 2 or 1

2
and 1 .

Page : 212 , Block Name : Exercise 10.1

Q12 A line passes through (x , y ) and (h, k).If slope of the line is m, show that k − y
1 1 1 = m (h − x1 ) .

k−y1
Answer. The slope of the line passing through (x , y ) and (h, k) is 1 1 .
h−x1

It is given that the slope of the line is m.
k−y1
∴ = m
h−x1

⇒ k − y1 = m (h − x1 )

Hence, k − y1 = m (h − x1 )

Page : 212 , Block Name : Exercise 10.1

Q13 If three points (h, 0), (a, b) and (0, k) lie on a line, show that
a b
+ = 1
h k

Answer. If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then
Slope of AB = Slope of BC
b−0 k−b
=
a−h 0−a

b k−b
⇒ =
a−h −a

⇒ −ab = (k − b)(a − h)

⇒ −ab = ka − kh − ab + bh

⇒ ka + bh = kh

On dividing both sides by kh, we obtain
ka bh kh
+ =
kh kh kh

a b
⇒ + = 1
h k

a b
Hence, + = 1
h k

Page : 212 , Block Name : Exercise 10.1

Q14 Consider the following population and year graph (Fig 10.10), nd the slope of the
line AB and using it, nd what will be the population in the year 2010?

Answer. Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is \*\frac{97-92}
{1995-1985}=\frac{5}{10}=\frac{1}{2}\)
Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through
point C (2010, y).
∴Slope of AB = Slope of BC
1 y−97
⇒ =
2 2010−1995

1 y−97
⇒ =
2 15

15
⇒ = y − 97
2

⇒ y − 97 = 7.5

⇒ y = 7.5 + 97 = 104.5

Thus, the slope of line AB is 1

2
, while in the year 2010, the population will be 104.5 crores.

Page : 212 , Block Name : Exercise 10.1

Exercise 10.2

Q1 Write the equations for the x-and y-axes.

Answer. The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is x = 0.

Page : 219 , Block Name : Exercise 10.2

Q2 Passing through the point (– 4, 3) with slope 1

2

Answer.
We know that the equation of the line passing through

point (x0 , y0 ) , whose slope is m, is (y − y0 ) = m (x − x0 )

Thus, the equation of the line passing through point (−4,

1
3), whose slope is
2
1
(y − 3) = (x + 4)
2

2(y − 3) = x + 4

2y − 6 = x + 4

i.e., x − 2y + 10 = 0

Page : 219 , Block Name : Exercise 10.2

Q3 Passing through (0, 0) with slope m.

Answer.
We know that the equation of the line passing through

point (x0 , y0 ) , whose slope is m, is (y − y0 ) = m (x − x0 )

Thus, the equation of the line passing through point (0, 0), whose slope is m,is
(y – 0) = m(x – 0)
i.e., y = mx

Page : 219 , Block Name : Exercise 10.2

Q4 Passing through (2, 2√3) and inclined with the x-axis at an angle of 75 . ∘

Answer. The slope of the line that inclines with the x-axis at an angle of 75° is
m = tan 75°

1 √3+1

∘ ∘ 1+
∘ ∘ tan 45 +tan 30 √3 √3 √3+1
⇒ m = tan(45 + 30 ) = ∘ ∘
= = =
1−tan 45 ⋅tan 30 1 √3−1
1−1⋅ √3−1
√3
√3

We know that the equation of the line passing through

point (x0 , y0 ) , whose slope is m, is (y − y0 ) = m (x − x0 )

Thus, if a line passes though (2, 2√3) and inclines with

∘
the x -axis at an angle of 75 , then the equation of the

line is given as

√3+1
(y − 2√3) = (x − 2)
√3−1

(y − 2√3)(√3 − 1) = (√3 + 1)(x − 2)

y(√3 − 1) − 2√3(√3 − 1) = x(√3 + 1) − 2(√3 + 1)

(√3 + 1)x − (√3 − 1)y = 2√3 + 2 − 6 + 2√3

(√3 + 1)x − (√3 − 1)y = 4√3 − 4

i.e.. (√3 + 1)x − (√3 − 1)y = 4(√3 − 1)

Page : 219 , Block Name : Exercise 10.2

Q5 Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

Answer. It is known that if a line with slope m makes x-intercept d, then the equation of the line is given
as
y = m(x – d)
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.
The slope of the line is given as m = –2
Thus, the required equation of the given line is
y = –2 [x – (–3)]
y = –2x – 6
i.e., 2x + y + 6 = 0

Page : 219 , Block Name : Exercise 10.2

Q6 Intersecting the y-axis at a distance of 2 units above the origin and making an
angle of 30 with positive direction of the x-axis.
∘

Answer. It is known that if a line with slope m makes y-intercept c, then the equation of the line is given
as
y = mx + c
∘
Here, c = 2 and m = tan 30

Thus, the required equation of the given line is
1
y = x + 2
√3

x+2√3
y =
√3

√3y = x + 2√3

i.e., x − √3y + 2√3 = 0

Page : 219 , Block Name : Exercise 10.2

Q7 Passing through the points (–1, 1) and (2, – 4).

Answer. It is known that the equation of the line passing through

y2 −y1
points (x1 , y1 ) and (x2 , y2 ) is y − y1 = (x − x1 )
x2 −x1

Therefore, the equation of the line passing through the points (–1, 1) and(2, –4) is
−4−1
(y − 1) = (x + 1)
2+1

−5
(y − 1) = (x + 1)
3

3(y − 1) = −5(x + 1)

3y − 3 = −5x − 5

i.e., 5x + 3y + 2 = 0

Page : 219 , Block Name : Exercise 10.2

Q8 Perpendicular distance from the origin is 5 units and the angle made by the
perpendicular with the positive x-axis is 30 .
∘

Answer. If p is the length of the normal from the origin to a line and ω is the angle made by the normal
with the positive direction of the x-axis,
then the equation of the line is given by xcos ω + y sin ω = p.
Here, p = 5 units and ω = 30°
Thus, the required equation of the given line is
x cos 30° + y sin 30° = 5
√3 1
x + y ⋅ = 5
2 2

i.e., √3x + y = 10

Page : 220 , Block Name : Exercise 10.2

Q9 The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

Answer. It is given that the vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5).
Let RL be the median through vertex R. Accordingly, L is the mid-point of PQ.
By mid-point formula, the coordinates of point L are given by (
2−2 1+3
, ) = (0, 2)
2 2

It is known that the equation of the line passing through

y2 −y1
points (x1 , y1 ) and (x2 , y2 ) is y − y1 = (x − x1 )
x2 −x1

Therefore, the equation of RL can be determined by substituting (x1, y1) = (4, 5) and (x2, y2) = (0, 2).
2−5
Hence, y − 5 = (x − 4)
0−4

−3
⇒ y − 5 = (x − 4)
−4

⇒ 4(y − 5) = 3(x − 4)

⇒ 4y − 20 = 3x − 12

⇒ 3x − 4y + 8 = 0

Thus, the required equation of the median through vertex R is 3x − 4y + 8 = 0

Page : 220 , Block Name : Exercise 10.2

Q10 Find the equation of the line passing through (–3, 5) and perpendicular to the line
through the points (2, 5) and (–3, 6).

Answer. The slope of the line joining the points (2, 5) and (–3, 6) is m =
6−5 1
=
−3−2 −5

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are
negative reciprocals of each other.
Therefore, slope of the line perpendicular to the line = − = − = 5
m
1 1

−1
( )
5

through the points (2, 5) and (–3, 6)
Now, the equation of the line passing through point (–3, 5), whose slope is 5, is
(y − 5) = 5(x + 3)

y − 5 = 5x + 15

i.e., 5x − y + 20 = 0

Page : 220 , Block Name : Exercise 10.2

Q11 A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides
it in the ratio 1: n. Find the equation of the line.

Answer. According to the section formula,
the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio
1: n is given by
n(1)+1(2) n(0)+1(3) n+2 3
( , ) = ( , )
1+n 1+n n+1 n+1

The slope of the line joining the points (1, 0) and (2, 3) is
3−0
m = = 3
2−1

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are
negative reciprocals of each other.
Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3)
1 1
= − = −
m 3

Now, the equation of the line passing through
1
n+2 3 −
( , ) and whose slope is 3
is given by
n+1 n+1

3 −1 n+2
(y − ) = (x − )
n+1 3 n+1

⇒ 3[(n + 1)y − 3] = −[x(n + 1) − (n + 2)]

⇒ 3(n + 1)y − 9 = −(n + 1)x + n + 2

⇒ (1 + n)x + 3(1 + n)y = n + 11

Page : 220 , Block Name : Exercise 10.2

Q12 Find the equation of a line that cuts off equal intercepts on the coordinate axes
and passes through the point (2, 3).

Answer. The equation of a line in the intercept form is
x y
+ = 1 … (i)
a b

Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes. This means that a = b.
Accordingly, equation (i) reduces to
x y
+ = 1
a a

⇒ x + y = a . . . (ii)

Since the given line passes through point (2, 3), equation (ii) reduces to
2+3=a⇒a=5
On substituting the value of a in equation (ii), we obtain
x + y = 5, which is the required equation of the line

Page : 220 , Block Name : Exercise 10.2

Q13 Find equation of the line passing through the point (2, 2) and cutting off intercepts
on the axes whose sum is 9.

Answer. The equation of a line in the intercept form is
x y
+ = 1 (i)
a b

Here, a and b are the intercepts on x and y axes respectively.
It is given thata + b = 9 ⇒ b = 9 – a … (ii)
From equations (i) and (ii), we obtain
x y
+ = 1 … (iii)
a 9−a

It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to
2 2
+ = 1
a 9−a

1 1
⇒ 2( + ) = 1
a 9−a

9−a+a
⇒ 2( ) = 1
a(9−a)

18
⇒ = 1
2
9a−a

2
⇒ 18 = 9a − a

2
⇒ a − 9a + 18 = 0

2
⇒ a − 6a − 3a + 18 = 0

⇒ a(a − 6) − 3(a − 6) = 0

⇒ (a − 6)(a − 3) = 0

⇒ a = 6 or a = 3

If a = 6 and b = 9 – 6 = 3, then the equation of the line is
x y
+ = 1 ⇒ x + 2y − 6 = 0
6 3

If a = 3 and b = 9 – 3 = 6, then the equation of the line is
x y
+ = 1 ⇒ 2x + y − 6 = 0
3 6

Page : 220 , Block Name : Exercise 10.2

Q14 Find equation of the line through the point (0, 2) making an angle with the positive x-axis.
2π

3

Also, nd the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the
origin.

Answer.
2π

The slope of the line making an angle 3
with the positive

2π
m = tan( ) = − √3
3

Now, the equation of the line passing through point (0, 2)
and having a slope − √3 is (y − 2) = −√3(x − 0)

y − 2 = −√3x

i.e., √3x + y − 2 = 0

The slope of line parallel to line √3x + y − 2 = 0 is − √3

It is given that the line parallel to line √3x + y − 2 = 0

crosses the y-axis 2 units below the origin i.e., it passes through point (0, –2).
Hence, the equation of the line passing through point (0, –2) and having a slope −√3 is
y − (−2) = −√3(x − 0)

y + 2 = −√3x

√3x + y + 2 = 0

Page : 220 , Block Name : Exercise 10.2

Q15 The perpendicular from the origin to a line meets it at the point (–2, 9), nd the equation of the
line.

Answer. The slope of the line joining the origin (0, 0) and point (–2, 9) is m
9−0 9
1 = = −
−2−0 2

Accordingly, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is
1 1 2
m2 = − = − =
m1 9 9
(− )
2

Now, the equation of the line passing through point (–2, 9) and having a slope m2 is

2
(y − 9) = (x + 2)
9

9y − 81 = 2x + 4

i.e., 2x − 9y + 85 = 0

Page : 220 , Block Name : Exercise 10.2

Q16 The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C.
In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Answer. It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the value of L is
125.134.
Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and (110,
125.134) in the XY plane.
Therefore, the linear relation between L and C is the equation of the line passing through points (20,
124.942) and (110, 125.134).
125.134−124.942
(L − 124.942) = (C − 20)
110−20

0.192
L − 124.942 = (C − 20)
90

0.192
i.e... L = (C − 20) + 124.942, which is the required linear relation
90

Page : 220 , Block Name : Exercise 10.2

Q17 The owner of a milk store nds that,
he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre.
Assuming a linear relationship between selling price and demand,
how many litres could he sell weekly at Rs 17/litre?

Answer. The relationship between selling price and demand is linear.
Assuming selling price per litre along the x-axis and demand along the y-axis,
we have two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship
between selling price and demand.
Therefore, the linear relationship between selling price per litre and demand is the equation of the line
passing through points (14, 980) and (16, 1220).
1220−980
y − 980 = (x − 14)
16−14

240
y − 980 = (x − 14)
2

y − 980 = 120(x − 14)

i.e., y = 120(x − 14) + 980

When x = Rs 17 litre,

y = 120(17 − 14) + 980

⇒ y = 120 × 3 + 980 = 360 + 980 = 1340

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

Page : 220 , Block Name : Exercise 10.2

Q18 P (a, b) is the mid-point of a line segment between axes. Show that equation

of the line is .
x y
+ = 2
a b

Answer. Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.
Since P (a, b) is the mid-point of AB,
0+x y+0
( , ) = (a, b)
2 2

x y
⇒ ( , ) = (a, b)
2 2

x y
⇒ = a and = b
2 2

∴ x = 2a and y = 2b

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).
The equation of the line passing through points (0, 2b) and (2a, 0) is
(0−2b)
(y − 2b) = (x − 0)
(2a−0)

−2b
y − 2b = (x)
2a

a(y − 2b) = −bx

ay − 2ab = −bx

i.e., bx + ay = 2ab

On dividing both sides by ab, we obtain
bx ay 2ab
+ =
ab ab ab

x y
⇒ + = 2
a b

x y
⇒ + = 2
a b

x y
Thus, the equation of the line is + = 2
a b

Page : 220 , Block Name : Exercise 10.2

Q19 Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find bequation of the line.

Answer. Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio 1: 2.

Let the respective coordinates of A and B be (x, 0) and (0, y).
Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,
1×0+2×x 1×y+2×0
(h, k) = ( , )
1+2 1+2

2x y
⇒ (h, k) = ( , )
3 3

2x y
⇒ h = and k =
3 3

3h
⇒ x = and y = 3k
2

Therefore, the respective coordinates of A and B are

3h
( , 0)
2
and (0,3k)

Now, the equation of line AB passing through points

3h
( , 0)
2

(0, 3k) is

3k−0 3h
(y − 0) = (x − )
3h 2
0−
2

2k 3h
y = − (x − )
h 2

hy = −2kx + 3hk

i.e., 2kx + hy = 3hk

Thus,the required equation of the line is 2kx + hy = 3hk.

Page : 220 , Block Name : Exercise 10.2

Q20 By using the concept of equation of a line, prove that the three points (3, 0),
(– 2, – 2) and (8, 2) are collinear.

Answer. In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear,
it suf ces to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8,
2).
The equation of the line passing through points (3, 0) and (–2, –2) is
(−2−0)
(y − 0) = (x − 3)
(−2−3)

−2
y = (x − 3)
−5

5y = 2x − 6

i.e., 2x − 5y = 6

It is observed that at x = 8 and y = 2,
L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.

Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). Hence,
points (3, 0), (–2, –2), and (8, 2) are collinear.

Page : 220 , Block Name : Exercise 10.2

Exercise 10.3

Q1 Reduce the following equations into slope - intercept form and nd their slopes
and the y - intercepts.
(i) x + 7y = 0, (ii) 6x + 3y – 5 = 0, (iii) y = 0.

Answer. (i) The given equation is x + 7y = 0. It can be written as
1
y = − x + 0 … (1)
7

This equation is of the form y = mx + c, where.
1
m = − and c = 0
7

Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are − and 0
1

7

respectively.
(ii) The given equation is 6x + 3y – 5 = 0.
It can be written as
1
y = (−6x + 5)
3

5
y = −2x + . . . (2)
3

5
This equation is of the form y = mx + c, where m = −2 and c =
3

Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are–2 and 5

3

respectively.
(iii) The given equation is y = 0.
It can be written as
y = 0.x + 0 … (3)
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0
respectively.

Page : 227 , Block Name : Exercise 10.3

Q2 Reduce the following equations into intercept form and nd their intercepts on
the axes.
(i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6, (iii) 3y + 2 = 0.

Answer. (i) The given equation is 3x + 2y – 12 = 0.
It can be written as
3x + 2y = 12

3x 2y
+ = 1
12 12

x y
i.e., + = 1 . . . (1)
4 6

This equation is of the form , where a = 4 and b = 6.
x y
+ = 1
a b

Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6
respectively.

(ii) The given equation is 4x – 3y = 6.
It can be written as
4x 3y
− = 1
6 6

2x y
− = 1
3 2

x y
i.e., + = 1 … (2)
3 (−2)
( )
2

y
This equation is of the form , where a = and b = -2.
x 3
+ = 1
a b 2

Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are and –2
3

2

respectively.
(iii) The given equation is 3y + 2 = 0.
It can be written as
3y = −2
y
i.e., = 1 . . . (3)
2
(− )
3

x

¯
¯¯a
This equation is of the form a = 1

2
= −
3

Therefore, equation (3) is in the intercept form, where
2
−
the intercept on the y-axis is 3
and it has no intercept

on the x -axis.

Page : 227 , Block Name : Exercise 10.3

Q3 Reduce the following equations into normal form. Find their perpendicular distances
from the origin and angle between perpendicular and the positive x-axis.
(i) x − √3y + 8 = 0, (ii) y – 2 = 0, (iii) x – y = 4.

Answer.
(i) The given equation is x − √3y + 8 = 0

It can be reduced as:

x − √3y = −8

⇒ −x + √3y = 8

2 2
On dividing both sides by √(−1) + (√3) = √4 = 2

obtain

x √3 8
− + y =
2 2 2

1 √3
⇒ (− )x + ( )y = 4
2 2

∘ ∘
⇒ × cos 120 + y sin 120 = 4 . . . (1)

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line
x cos ω + y sin ω = p, we obtain ω = 120° and p = 4.
Thus, the perpendicular distance of the line from the origin is 4, while the angle between the
perpendicular and the positive x-axis is 120°.
(ii) The given equation is y – 2 = 0.
It can be reduced as 0.x + 1.y = 2
On dividing both sides by √0 + 1 = 1 , we obtain 0.x + 1.y = 2
2 2

⇒ x cos 90° + y sin 90° = 2 … (1)
Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line
x cos ω + y sin ω = p, we obtain ω = 90° and p = 2.
Thus, the perpendicular distance of the line from the origin is 2, while the angle between the
perpendicular and the positive x-axis is 90°.
(iii) The given equation is x – y = 4.
It can be reduced as 1.x + (–1) y = 4

Page : 227 , Block Name : Exercise 10.3

Q4 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Answer. The given equation of the line is 12(x + 6) = 5(y – 2).
⇒ 12x + 72 = 5y – 10
⇒12x – 5y + 82 = 0 … (1)
On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 12, B = –5, and C
= 82.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
|Ax1 +By1 +C|
d = .
√A2 +B2

The given point is (x1, y1) = (–1, 1).
Therefore, the distance of point (–1, 1) from the given line
|12(−1)+(−5)(1)+82| |−12−5+82| 65|
= units = units = units = 5
√(12)2 +(−5)2 √169 13

Page : 227 , Block Name : Exercise 10.3

y
Q5 Find the points on the x-axis, whose distances from the line are 4 units.
x
+ = 1
3 4

Answer.
The given equation of line is

x y
+ = 1
3 4

or, 4x + 3y − 12 = 0 . . . (1)

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C =
–12.
Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x , y ) is given by 1 1

|Ax1 +By1 +C|
d = .
√A2 +B2

Therefore,

|4a+3×0−12|
4 =
√ 2 2
4 +3

|4a−12|
⇒ 4 =
5

⇒ |4a − 12| = 20

⇒ ±(4a − 12) = 20

⇒ (4a − 12) = 20 or − (4a − 12) = 20

⇒ 4a = 20 + 12 or 4a = −20 + 12

⇒ a = 8 or − 2

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

Page : 227 , Block Name : Exercise 10.3

Q6 Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0.

Answer.
It is known that the distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 =

|C1 −C2 |
d =
√A2 +B2

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Here, A = 15, B = 8, C1 = –34, and C2 = 31.
Therefore, the distance between the parallel lines is
|C1 −C2 | |−34−31| |−65| 65
d = = units = units = unit
2
√A +B 2 2
√(15) +(8) 2 17 17

(ii) The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.
lx + ly + p = 0 and lx + ly – r = 0
Here, A = l, B = l, C1 = p, and C2 = –r.
Therefore, the distance between the parallel lines is
|C1 −C2 | |p+r| |p+r| |p+r| 1 p+r
d = = units = units = units = ∣ ∣ unit
2 2 2 2 2
∣ l
∣
√A +B √l +l √2l l√2 √2

Page : 227 , Block Name : Exercise 10.3

Q7 Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through
the point (–2, 3).

Answer. The equation of the given line is
3x − 4y + 2 = 0

3x 2
or y =
4
+
4
, which is of the form y = mx + c
3 1
or y = x +
4 2
3
∴ Slope of the given line =
4

It is known that parallel lines have the same slope.
3
m=
∴ Slope of the other line = 4

Now, the equation of the line that has a slope of and passes through the point (–2, 3) is

3
(y − 3) = {x − (−2)}
4

4y − 12 = 3x + 6

i.e., 3x − 4y + 18 = 0

Page : 228 , Block Name : Exercise 10.3

Q8 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Answer. The given equation of line is x-7 y+5=0.
or ,y = x + , which is of the form y = mx + c.
1 5

7 7

∴Slope of the given line = 1

7
1
m=− =−7

The slope of the line perpendicular to the line having a slope of
1
1 ( )
is 7

7

The equation of the line with slope –7 and x-intercept 3 is given by
y = m (x – d)
⇒ y = –7 (x – 3)
⇒ y = –7x + 21
⇒ 7x + y = 21

Page : 228 , Block Name : Exercise 10.3

Q9 Find angles between the lines √3x + y = 1 and x + √3y = 1.

Answer.
The given lines are √3x + y = 1 and x + √3y = 1

1 1
y = −√3x + 1 … (1) and y = − x + … (2)
√3 √3
1
m2 =−
The slope of line (1) is m1 = −√3 , while the slope of line (2) is √3

The acute angle i.e., θ between the two lines is given by

∣ m1 − m2 ∣
tan θ =∣ ∣
∣ 1 + m1 m2 ∣

1
∣ − √3 + ∣
√3
∣ ∣
tan θ = ∣ ∣
1
∣ 1 + (−√3) (− )∣
∣ √3 ∣

Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

Page : 228 , Block Name : Exercise 10.3

Q10 The line through the points (h, 3) and (4, 1) intersects the line 7 x-9 y-19=0
at right angle. Find the value of h.

Answer. The slope of the line passing through points (h, 3) and (4, 1) is
1−3 −2
m1 = =
4−h 4−h

The slope of line 7x – 9y – 19 = 0 or y =
7 19 7
x − is m2 =
9 9 9

It is given that the two lines are perpendicular.

∴ m1 × m2 = −1

−2 7
⇒ ( ) × ( ) = −1
4−h 9

−14
⇒ = −1
36−9h

⇒ 14 = 36 − 9h

⇒ hh = 36 − 14

22
⇒ h =
9

Thus, the value of h is
22

9

Page : 228 , Block Name : Exercise 10.3

Q11 Prove that the line through the point (x , y ) and parallel to the line Ax + By + C = 0 is
1 1

A (x − x ) + B (y − y ) = 0.
1 1

Answer. The slope of line Ax + By + C = 0 or y = (
−A −C A
)x + ( ) m = −
B B B
is

It is known that parallel lines have the same slope.
∴ Slope of the other line = m = − A

B

The equation of the line passing through point (x1, y1) and having a slope m = − is
A

B

y − y1 = m (x − x1 )

A
y − y1 = − (x − x1 )
B

B (y − y1 ) = −A (x − x1 )

A (x − x1 ) + B (y − y1 ) = 0

Hence, the line through point (x1, y1) and parallel to line Ax + By + C = 0 is
A (x –x1) + B (y – y1) = 0

Page : 228 , Block Name : Exercise 10.3

Q12 Two lines passing through the point (2, 3) intersects each other at an angle of 60 ∘

If slope of one line is 2, nd equation of the other line.

Answer. It is given that the slope of the rst line, m1 = 2.
Let the slope of the other line be m2.
The angle between the two lines is 60°.
∘ m −m
= ∣ ∣
1 2
∴ tan 60
∣ 1+m m ∣
1 2

2−m2
∣ ∣
⇒ √3 = ∣ ∣
1+2m2

2−m2
⇒ √3 = ± ( )
1+2m2

2−m2 2−m2
⇒ √3 = or √3 = − ( )
1+2m2 1+2m2

⇒ √3 (1 + 2m2 ) = 2 − m2 or √3 (1 + 2m2 ) = − (2 − m2 )

⇒ √3 + 2√3m2 + m2 = 2 or √3 + 2√3m2 − m2 = −2

⇒ √3 + (2√3 + 1)m2 = 2 or √3 + (2√3 − 1)m2 = −2

2−√3 −(2+√3)
⇒ m2 = or m2 =
(2√3+1) (2√3−1)

2−√3
Case I : m 2 = ( )
2√3+1

(2−√3)
The equation of the line passing through point (2, 3) and having a slope of
(2√3+1)
is

2−√3
(y − 3) = (x − 2)
2√3+1

(2√3 + 1)y − 3(2√3 + 1) = (2 − √3)x − 2(2 − √3)

(√3 − 2)x + (2√3 + 1)y = −4 + 2√3 + 6√3 + 3

(√3 − 2)x + (2√3 + 1)y = −1 + 8√3

In this case, the equation of the other line is (√3 − 2)x + (2√3 + 1)y = −1 + 8√3
−(2+√3)
Case II : m 2 =
(2√3−1)

−(2+√3)
The equation of the line passing through point (2, 3) and having a slope of is
(2√3−1)

−(2+√3)
(y − 3) = (x − 2)
(2√3−1)

(2√3 − 1)y − 3(2√3 − 1) = −(2 + √3)x + 2(2 + √3)

(2√3 − 1)y + (2 + √3)x = 4 + 2√3 + 6√3 − 3

(2 + √3)x + (2√3 − 1)y = 1 + 8√3

In this case, the equation of the other line is. (2 + √3)x + (2√3 − 1)y = 1 + 8√3
Thus, the required equation of the other line is (2 + √3)x + (2√3 − 1)y = −1 + 8√3 or
(2 + √3)x + (2√3 − 1)y = 1 + 8√3

Page : 228 , Block Name : Exercise 10.3

Q13 Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Answer. The right bisector of a line segment bisects the line segment at 90°.
The end-points of the line segment are given as A (3, 4) and B (–1, 2).
Accordingly, mid-point of AB = (
3−1 4+2
, ) = (1, 3)
2 2

Slope of AB =
2−4 −2 1
= =
−1−3 −4 2

∴Slope of the line perpendicular to AB = −
1
= −2
1
( )
2

The equation of the line passing through (1, 3) and having a slope of –2 is
(y – 3) = –2 (x – 1)
y – 3 = –2x + 2
2x + y = 5
Thus, the required equation of the line is 2x + y = 5.

Page : 228 , Block Name : Exercise 10.3

Q14 Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Answer. Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x
– 4y – 16 = 0.

Slope of the line joining (–1, 3) and (a, b), m1 =
b−3

a+1

Slope of the line 3x – 4y – 16 = 0 or y = 3

4
x − 4, m2 =
3

4

Since these two lines are perpendicular, m1m2 = –1
b−3 3
∴ ( ) × ( ) = −1
a+1 4

3b−9
⇒ = −1
4a+4

⇒ 3b − 9 = −4a − 4

⇒ 4a + 3b = 5 . . . (1)

Point (a, b) lies on line 3x – 4y = 16.
∴3a – 4b = 16 … (2)
On solving equations (1) and (2), we obtain
68 49
a = and b = −
25 25

Thus, the required coordinates of the foot of the perpendicular are ( 68

25
,−
49

25
) .

Page : 228 , Block Name : Exercise 10.3

Q15 The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2).
Find the values of m and c.

Answer. The given equation of line is y = mx + c.
It is given that the perpendicular from the origin meets the given line at (–1, 2).
Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.
∴Slope of the line joining (0, 0) and (–1, 2) =
2
= −2
−1

The slope of the given line is m.
∴ m × −2 = −1 [The two lines are perpendicular]

1
⇒ m =
2

Since point (–1, 2) lies on the given line, it satis es the equation y = mx + c.
∴ 2 = m(−1) + c

1
⇒ 2 = (−1) + c
2

1 5
⇒ c = 2 + =
2 2

Thus, the respective values of m and c are 1 5
and
2 2

Page : 228 , Block Name : Exercise 10.3

16. If p and q are the lengths of perpendiculars from the origin to the lines cosθ − sin θ = kyx 2cos θ and x
sec θ + y cosec θ = k,respectively,
prove that p + 4q = k . 2 2 2

Answer. The equations of given lines are
x cos θ – y sinθ = k cos 2θ … (1)
x secθ + y cosec θ= k … (2)
|Ax1 +By1 +C|
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d =
√A2 +B2

.
On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain A = cosθ, B = –
sinθ, and C = –k cos 2θ.
It is given that p is the length of the perpendicular from (0, 0) to line (1).
|A(0)+B(0)+C| |C| |−k cos 2θ|
∴ p = = = = | − k cos 2θ| . . . (3)
√A2 +B2 √A2 +B2 √cos2 θ+sin2 θ

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain A = secθ, B =
cosecθ, and C = –k.
It is given that q is the length of the perpendicular from (0, 0) to line (2).
|A(0)+B(0)+C| |C| |−k|
∴ q = = = . . . (4)
√A2 +B2 √A2 +B2 √sec2 θ+csc2 θ

From (3) and (4), we have
2
|−k|
2 2 2
p + 4q = (| − k cos 2θ|) + 4( )
√sec2 θ+csc2 θ

2
2 2 4k
= k cos 2θ + 2 2
(sec θ+csc θ)

2
2 2 4k 1
= k cos 2θ + 2
+ 2
)
(cos θ sin θ
2
2 2 4k
= k cos 2θ + 2 2
sin θ+cos θ
( )
sin2 θ cos2 θ

2
2 2 4k
= k cos 2θ +
1
( )
2 2
sin θ cos θ

2 2 2 2 2
= k cos 2θ + 4k sin θ cos θ
2 2 2 2
= k cos 2θ + k (2 sin θ cos θ)

2 2 2 2
= k cos 2θ + k (2 sin θ cos θ)

2 2 2
= k (cos 2θ + sin 2θ)

2
= k

Hence, we proved that p 2
+ 4q
2
= k
2
.

Page : 228 , Block Name : Exercise 10.3

Q17 In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), nd the equation and length of
altitude from the vertex A.

Answer. Let AD be the altitude of triangle ABC from vertex A.
Accordingly, AD⊥BC

The equation of the line passing through point (2, 3) and having a slope of 1 is
(y – 3) = 1(x – 2)
⇒x–y+1=0
⇒y–x=1
Therefore, equation of the altitude from vertex A = y – x = 1.
Length of AD = Length of the perpendicular from A (2, 3) to BC
The equation of BC is
2+1
(y + 1) = (x − 4)
1−4

⇒ (y + 1) = −1(x − 4)

⇒ y + 1 = −x + 4

⇒ x + y − 3 = 0 . . . (1)

The perpendicular distance (d) of a line Ax + By + C = 0
|Ax1 +By1 +C|
from a point (x1, y1) is given by d = .
√A2 +B2

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 1, B = 1, and C =
–3.
∴Length of AD
|1×2+1×3−3| |2|
2
= units = units = units = √2
2 2
√1 +1 √2 √2

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and √2 units respectively.

Page : 228 , Block Name : Exercise 10.3

Q18 If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and
b,
then show that = +
1
2
. 1
2
1
2
p a b

Answer. It is known that the equation of a line whose intercepts on the axes are a and b is
x y
+ = 1
a b

or bx + ay = ab

or bx + ay − ab = 0 . . . (1)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
|Ax1 +By1 +C|
d =
√A2 +B2

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C =
–ab.
Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line (1), we obtain

|A(0)+B(0)−ab|
p =
√b2 +a2

|−ab|
⇒ p =
√a2 +b2

On squaring both sides, we obtain
2
(−ab)
2
p = 2 2
a +b

2 2 2 2 2
⇒ p (a + b ) = a b

2 2
a +b 1
⇒ 2 2
=
a b p2

1 1 1
⇒ = 2
+ 2
p2 a b

Hence, we showed that p2
1
=
a
1
2
+
1

b
2
.

Page : 228 , Block Name : Exercise 10.3

Miscellaneous Exercise

Q1 Find the values of k for which the line (k − 3)x − (4 − k ) y + k 2 2
− 7k + 6 = 0

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

Answer. The given equation of line is
2 2
(k − 3)x − (4 − k ) y + k − 7k + 6 = 0 … (1)

(a) If the given line is parallel to the x-axis, then Slope of the given line = Slope of the x-axis, The given
line can be written as
2 2
(4 − k ) y = (k − 3)x + k − 7k + 6 = 0

(k−3) 2
k −7k+6
y = 2
x + 2
, which is of the form y = mx + c.
(4−k ) (4−k )

2
∴ Slope of the given line = (4 − k )

Slope of the x-axis = 0
(k−3)
∴ 2
= 0
(4−k )

⇒ k − 3 = 0

⇒ k = 3

Thus, if the given line is parallel to the x-axis, then the value of k is 3.
(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be unde ned.
(k−3)
The slope of the given line is (4−k )
2

(k−3)
Now, is unde ned at k = 4
2

2
(4−k )

=4
2
k

⇒ k = ±2
Thus, if the given line is parallel to the y-axis, then the value of k is ±2.
(c) If the given line is passing through the origin, then point (0, 0) satis es the
given equation of line.

2 2
(k − 3)(0) − (4 − k ) (0) + k − 7k + 6 = 0

2
k − 7k + 6 = 0

2
k − 6k − k + 6 = 0

(k − 6)(k − 1) = 0

k = 1 or 6

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Page : 233 , Block Name : Miscellaneous Exercise

Q2 Find the values of θ and p, if the equation x cos θ + y sinθ = p is the normal form of the line
√3x + y + 2 = 0.

Answer. The equation of the given line is √3x + y + 2 = 0.
This equation can be reduced as
√3x + y + 2 = 0

−√3x + y = 2

On dividing both sides by √(−√3) 2
+ (−1)
2
= 2 , we obtain
√3 1 2
− x − y =
2 2 2

√3 1
⇒ (− ) x + (− )y = 1 . . . (1)
2 2

On comparing equation (1) to x cos θ + y sin θ = p , we obtain
√3 1
cos θ = − , sin θ = − , and p = 1
2 2

Since the values of sin θ and cos θ are negative θ = π + π

6
=
7π

6
,
Thus, the respective values of θand p are and 1
7π

6

Page : 233 , Block Name : Miscellaneous Exercise

Q3 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1
and – 6, respectively.

Answer. Let the intercepts cut by the given lines on the axes be a and b.
It is given that
a + b = 1 … (1)
ab = –6 … (2)
On solving equations (1) and (2), we obtain
a = 3 and b = –2 or a = –2 and b = 3
It is known that the equation of the line whose intercepts on the axes are a and b is
x y
+ = 1 or bx + ay − ab = 0
a b

Case I: a = 3 and b = –2
In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6.
Case II: a = –2 and b = 3
In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6.
Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.

Page : 233 , Block Name : Miscellaneous Exercise

Q4 What are the points on the y-axis whose distance from the line is 4 units.
x y
+ = 1
3 4

y
Answer. Let (0, b) be the point on the y-axis whose distance from line x

3
+
4
= 1 is 4 units.
The given line can be written as 4x + 3y – 12 = 0 … (1)
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C =
–12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
|Ax1 +By1 +C|
d = .
√A2 +B2
y
Therefore, if (0, b) is the point on the y-axis whose distance from line x

3
+
4
= 1 is 4 units, then :
|4(0)+3(b)−12|
4 =
√42 +32

|3b−12|
⇒ 4 =
5

⇒ 20 = |3b − 12|

⇒ 20 = ±(3b − 12)

⇒ 30 = (3b − 12) or 20 = −(3b − 12)

⇒ bb = 20 + 12 or 30 = −20 + 12

32 8
⇒ b = or b = −
3 3

Thus, the required points are (0, .
32 8
) (0, − )
3 3
and

Page : 233 , Block Name : Miscellaneous Exercise

Q5 Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) And
(cos ϕ, sin ϕ).

Answer. The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) are given by
sin ϕ−sin θ
y − sin θ = (x − cos θ)
cos ϕ−cos θ

y(cos ϕ − cos θ) − sin θ(cos ϕ − cos θ) = x(sin ϕ − sin θ) − cos θ(sin ϕ − sin θ)

x(sin θ − sin ϕ) − y(cos ϕ − cos θ) + cos θ sin ϕ − cos θ sin θ − sin θ cos ϕ + sin θ cos θ = 0

x(sin θ − sin ϕ) + y(cos ϕ − cos θ) + sin(ϕ − θ) = 0

Ax + By + C = 0, where A = sin θ − sin ϕ, B = cos ϕ − cos θ, and C = sin(ϕ − θ)

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
|Ax1 +By1 +C|
d = .
√A2 +B2

Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is
|(sin θ−sin ϕ)(0)+(cos ϕ−cos θ)(0)+sin(ϕ−θ)|
d =
√(sin θ−sin ϕ)2 +(cos ϕ−cos θ)2

| sin(ϕ−θ)|
=
2 2 2 2
√sin θ+sin ϕ−2 sin θ sin ϕ+cos ϕ+cos θ−2 cos ϕ cos θ

| sin(ϕ−θ)|
=
2 2 2 2
√(sin θ+cos θ)+(sin ϕ+cos ϕ)−2(sin θ sin ϕ+cos θ cos ϕ)

| sin(ϕ − θ)|
=
√1 + 1 − 2(cos(ϕ − θ))

| sin(ϕ − θ)
=
√2(1 − cos(ϕ − θ))

| sin(ϕ − θ)|
=

2 ϕ−θ
√2 (2 sin ( )
2

| sin(ϕ − θ)|
=
ϕ−θ
|2 sin( )
2

Page : 233 , Block Name : Miscellaneous Exercise

Q6 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the
lines x – 7y + 5 = 0 and 3x + y = 0.

Answer. The equation of any line parallel to the y-axis is of the form
x = a … (1)
The two given lines are
x – 7y + 5 = 0 … (2)
3x + y = 0 … (3)
On solving equations (2) and (3), we obtain x = − and y = . 5

22
15

22

Therefore, (− 5

22
,
15

22
) is the point of intersection of lines (2) and (3).

Since line x = a passes through point (− 5

22
,
15

22
,
) a = −
5

22
.

Thus, the required equation of the line is a = − .
5

22

Page : 233 , Block Name : Miscellaneous Exercise

y
Q7 Find the equation of a line drawn perpendicular to the line through the point, where it
x
+ = 1
4 6

meets the y-axis.

Answer. The equation of the given line is .
x y
+ = 1
4 6

This equation can also be written as 3x + 2y – 12 = 0
x + 6, which is of the form y = mx + c
−3
y =
2

∴Slope of the given line = − 3

2

∴Slope of line perpendicular to the given line = − 1
=
2

3
3
(− )
2

Let the given line intersect the y-axis at (0, y).
y
On substituting x with 0 in the equation of the given line, we obtain 6
= 1 ⇒ y = 6

∴The given line intersects the y-axis at (0, 6).
The equation of the line that has a slope of and passes through point (0, 6) is
2

3

2
(y − 6) = (x − 0)
3

3y − 18 = 2x

2x − 3y + 18 = 0

Thus, the required equation of the line is 2x − 3y + 18 = 0.

Page : 233 , Block Name : Miscellaneous Exercise

Q8 Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Answer. The equations of the given lines are
y – x = 0 … (1)
x + y = 0 … (2)
x – k = 0 … (3)
The point of intersection of lines (1) and (2) is given by x = 0 and y = 0
The point of intersection of lines (2) and (3) is given by x = k and y = –k
The point of intersection of lines (3) and (1) is given by x = k and y = k
Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (k, k).
We know that the area of a triangle whose vertices are (x , y ) , (x , y ) , and (x , y ) is.
1 1 2 2 3 3

1
|x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 )|
2

Therefore, area of the triangle formed by the three given lines
1
= |0(−k − k) + k(k − 0) + k(0 + k)| square units
2
1
2 2
= ∣k + k ∣
∣ ∣ square units
2

1
2
= ∣2k ∣ square units
∣ ∣
2
2
= k square units

Page : 233 , Block Name : Miscellaneous Exercise

Q9 Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and 2x – y – 3 = 0 may
intersect at one point.

Answer. The equations of the given lines are
3x + y – 2 = 0 … (1)
px + 2y – 3 = 0 … (2)
2x – y – 3 = 0 … (3)
On solving equations (1) and (3), we obtain
x = 1 and y = –1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also
satisfy line (2).
p (1) + 2 (–1) – 3 = 0
p–2–3=0
p=5
Thus, the required value of p is 5.

Page : 233 , Block Name : Miscellaneous Exercise

Q10 If three lines whose equations are y = m1 x + c1 , y = mx x + c2 and y = m3 x + c3 are

concurrent, then show that m1 (c2 − c3 ) + m2 (c3 − c1 ) + m3 (c1 − c2 ) = 0

Answer. The equations of the given lines are
y = m1x + c1 … (1)
y = m2x + c2 … (2)
y = m3x + c3 … (3)
On subtracting equation (1) from (2), we obtain
0 = (m2 − m1 ) x + (c2 − c1 )

⇒ (m1 − m2 ) x = c2 − c1

c2 −c1
⇒ x =
m1 −m2

On substituting this value of x in (1), we obtain
c2 − c1
y = m1 ( ) + c1
m1 − m2

m1 c2 − m1 c1
y = + c1
m1 − m2
m1 c2 − m1 c1 + m1 c1 − m2 c1
y =
m1 − m2
m1 c2 − m2 c1
y =
m1 − m2

is the point of intersection of lines (1) and (2).?
c2 −c1 m1 c2 −m2 c1
∴ ( , )
m1 −m2 m1 −m2

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2)
will also satisfy equation (3).
m1 c2 −m2 c1 c2 −c1
= m3 ( ) + c3
m1 −m2 m1 −m2

m1 c2 −m2 c1 m3 c2 −m3 c1 +c3 m1 −c3 m2
=
m1 −m2 m1 −m2

m1 c2 − m2 c1 − m3 c2 + m3 c1 − c3 m1 + c3 m2 = 0

m1 (c2 − c3 ) + m2 (c3 − c1 ) + m3 (c1 − c2 ) = 0

Hence, m1 (c2 − c3 ) + m2 (c3 − c1 ) + m3 (c1 − c2 ) = 0

Page : 233 , Block Name : Miscellaneous Exercise

Q11 Find the equation of the lines through the point (3, 2) which make an angle of 45 with the line x –
∘

2y = 3.

Answer. Let the slope of the required line be m1.
The given line can be written as y = x − , which is of the form y = mx + c
1 3

2 2

∴Slope of the given line = m 2 =
1

2

It is given that the angle between the required line and line x – 2y = 3 is 45°.
We know that if θisthe acute angle between lines I and I with slopes m and m respectively, then
1 2 1 2

.
m2 −m1
∣ ∣
tan θ = ∣ ∣
1+m1 m2

∘ |m1 −m2 |
∴ tan 45 =
1+m1 m2

1
∣ −m1 ∣
2
⇒ 1 = ∣ m ∣
1
∣ 1+ ∣
2

1−2m
1
( )
2

⇒ 1 = | |
2+m
1

2

1−2m1
∣ ∣
⇒ 1 = ∣ ∣
2+m1

1−2m1
⇒ 1 = ±( )
2+m1

1−2m1 1−2m1
⇒ 1 = or 1 = − ( )
2+m1 2+m1

⇒ 2 + m1 = 1 − 2m1 or 2 + m1 = −1 + 2m1

1
⇒ m1 = − or m1 = 3
3

Case I: m1 = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = (x – 3)
y – 2 = 3x – 9
3x – y = 7
Case II: m1 = − 1

3

The equation of the line passing through (3, 2) and having a slope of − is: 1

3
1
y − 2 = − (x − 3)
3

3y − 6 = −x + 3

x + 3y = 9

Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

Page : 233 , Block Name : Miscellaneous Exercise

Q12 Find the equation of the line passing through the point of intersection of the lines
4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Answer. Let the equation of the line having equal intercepts on the axes be
x y
+ = 1
a a

Or x + y = a . . . (1)

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain x = .
1 5
and y =
13 13

is the point of intersection of the two given lines.
1 5
∴ ( , )
13 13

Since equation (1) passes through point ( ,
1 5
, )
13 13

1 5
+ = a
13 13

6
⇒ a =
13

6
∴ Equation (1) becomes x + y = , i.e., 13x + 13y = 6
13

Thus, the required equation of the line is 3x + 13y = 6

Page : 233 , Block Name : Miscellaneous Exercise

Q13 Show that the equation of the line passing through the origin and making an angle
y
θ with the line y mx c is y = mx + c is = x
m±tan

1∓m tan

Answer. Let the equation of the line passing through the origin be y = m x. 1

If this line makes an angle of θ with line y = mx + c, then angle θ is given by
m1 −m
∣ ∣
∴ tan θ = ∣
1+m1 m ∣

y
∣ − m ∣
x
⇒ tan θ = ∣ ∣
∣ y ∣
∣ 1 + m ∣
x
y
−m
x
⇒ tan θ = ± ( y
)
1+ m
x

y y
−m −m
x x
⇒ tan θ = y
or tan θ = − ( y
)
1+ m 1+ m
x x

y
−m
x
tan θ = y
1+ m
x

Case I :
y
−m
x
tan θ = y
1+
x

y y
⇒ tan θ + m tan θ = − m
x x

y
⇒ m + tan θ = (1 − m tan θ)
x

y m+tan θ
⇒ =
x 1−m tan θ
y
−m
x
tan θ = − ( y
)
1+ m
x

Case II :
y
−m
x
tan θ = − ( y
)
1+ m
x

y y
⇒ tan θ + m tan θ = − + m
x x

y
⇒ (1 + m tan θ) = m − tan θ
x

y m−tan θ
⇒ =
x 1+m tan θ

Therefore, the required line is given by
y m±tan θ
=
x 1∓m tan θ

Page : 234 , Block Name : Miscellaneous Exercise

Q14 In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Answer. The equation of the line joining the points (–1, 1) and (5, 7) is given by
7−1
y − 1 = (x + 1)
5+1

6
y − 1 = (x + 1)
6

x − y + 2 = 0 . . . (1)

The equation of the given line is
x + y – 4 = 0 … (2)
The point of intersection of lines (1) and (2) is given by
x = 1 and y = 3

Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k. Accordingly, by section
formula,
k(−1)+1(5) k(1)+1(7)
(1, 3) = ( , )
1+k 1+k

−k+5 k+7
⇒ (1, 3) = ( , )
1+k 1+k

−k+5 k+7
⇒ = 1, = 3
1+k 1+k

−k+5
∴ = 1
1+k

⇒ −k + 5 = 1 + k

⇒ k = 4

Thus, the line joining the points (–1, 1) and (5, 7) is divided by line
x + y = 4 in the ratio 1:2.

Page : 234 , Block Name : Miscellaneous Exercise

Q15 Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line
2x – y = 0.

Answer. The given lines are
2x – y = 0 … (1)
4x + 7y + 5 = 0 … (2)
A (1, 2) is a point on line (1).
Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain x = .
−5 −5
and y =
18 9

∴Coordinates of point B are ( .
−5 −5
, )
18 9

By using distance formula, the distance between points A and B can be obtained as

2 2
5 5
AB = √(1 + ) + (2 + ) units
18 9

2 2
23 23
= √( ) + ( ) units
18 9

2 2
23 23
= √( ) + ( ) units
2×9 9

2 2 2
23 1 23
= √( ) ( ) + ( ) units
9 2 9

2
23 1
= √( ) ( + 1) units
9 4

23 5
= √ units
9 4

23 √5
= × units
9 2

23√5
= units
18
23√5
Thus, the required distance is 18
units.

Page : 234 , Block Name : Miscellaneous Exercise

Q16 Find the direction in which a straight line must be drawn through the point (–1, 2)
so that its point of intersection with the line x + y= 4 may be at a distance of 3 units from this point.

Answer. Let y = mx + c be the line through point (–1, 2).
Accordingly, 2 = m (–1) + c.
⇒ 2 = –m + c
⇒c=m+2
∴ y = mx + m + 2 … (1)
The given line is
x + y = 4 … (2)
On solving equations (1) and (2), we obtain
2−m 5m+2
x = and y =
m+1 m+1

2−m 5m+2
∴ ( , )
m+1 m+1
is the point of intersection of lines (1) and (2)

Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,
2 2
2−m 5m+2
√( + 1) + ( − 2) = 3
m+1 m+1

2 2
2−m+m+1 5m+2−2m−2 2
⇒ ( ) + ( ) = 3
m+1 m+1

2
9 9m
⇒ 2
+ 2
= 9
(m+1) (m+1)

2
⇒ 1 + m = 1

2 2
⇒ 1 + m = m + 1 + 2m

⇒ m = 0

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

Page : 234 , Block Name : Miscellaneous Exercise

Q17 The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1).
Find an equation of the legs (perpendicular sides) of the triangle.

Answer. Let A(1,3) and B(−4,1) be the coordinates of the end points of the hypotenuse.
Now, plotting the line segment joining the points A(1,3) and B(−4,1) on the coordinate plane,
we will get two right triangles with AB as the hypotenuse.

Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle
having AB as the hypotenuse can be either P or Q.

CASE 1: When ∠ APB is taken. The perpendicular sides in ∠ APB are AP and PB. Now, side PB is parallel
to x-axis and at a distance of 1 units above x-axis. So, equation of PB is, y=1 or y−1=0. The side AP is
parallel to y-axis and at a distance of 1 units on the right of y-axis. So, equation of AP is x=1 or x−1=0.
CASE 2: When ∠AQB is taken. The perpendicular sides in ∠ AQB are AQ and QB. Now, side AQ is parallel
to x-axis and at a distance of 3 units above x-axis. So, equation of AQ is, y=3 or y−3=0. The side QB is
parallel to y-axis and at a distance of 4 units on the left of y-axis. So, equation of QB is x=−4 or x+4=0.
Hence, the equation of the legs are : x=1, y=1 or x=−4, y=3

Page : 234 , Block Name : Miscellaneous Exercise

Q18 Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane
mirror.

Answer. The equation of the given line is
x + 3y = 7 … (1)
Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.

b−8 1
AB = , while the slope of line (1) = −
a−3 3

Since line (1) is perpendicular to AB,
b−8 1
( ) × (− ) = −1
a−3 3

b−8
⇒ = 1
3a−9

⇒ b − 8 = 3a − 9

⇒ 3b − b = 1 . . . (2)

a+3 b+8
Mid-point of AB = ( , )
2 2

The mid-point of line segment AB will also satisfy line (1).
Hence, from equation (1), we have
a+3 b+8
( ) + 3( ) = 7
2 2

⇒ a + 3 + 3b + 24 = 14

⇒ a + 3b = −13 . . . (3)

On solving equations (2) and (3), we obtain a = –1 and b = –4.
Thus, the image of the given point with respect to the given line is (–1, –4).

Page : 234 , Block Name : Miscellaneous Exercise

Q19 If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, nd the value of m.

Answer. The equations of the given lines are
y = 3x + 1 … (1)
2y = x + 3 … (2)
y = mx + 4 … (3)
Slope of line (1),m = 3 1

Slope of line (2), m = 2
1

2

Slope of line (3),m = m 3

It is given that lines (1) and (2) are equally inclined to line (3). This means that
the angle between lines (1) and (3) equals the angle between lines (2) and (3).
m −m m −m
∴ ∣ ∣ = ∣ ∣
1 3 2 3

∣ 1+m m ∣ ∣ 1+m m ∣
1 3 2 3

1

3−m ∣ 2 −m ∣
⇒ ∣
∣
∣ =
∣ ∣ ∣
1+3m 1
∣ 1+ m ∣
2

3−m 1−2m
⇒ ∣
∣
∣ = ∣ ∣
1+3m ∣ ∣ m+2 ∣

3−m 1−2m
⇒ = ±( )
1+3m m+2

3−m 1−2m 3−m 1−2m
⇒ = or = −( )
1+3m m+2 1+3m m+2

3−m 1−2m
If = , then
1+3m m+2

(3 − m)(m + 2) = (1 − 2m)(1 + 3m)

2 2
⇒ −m + m + 6 = 1 + m − 6m

2
⇒ 5m + 5 = 0

2
⇒ (m + 1) = 0

⇒ m = √−1, which is not real

Hence, this case is not posible.

3−m 1−2m
If = −( ) , then
1+3m m+2

⇒ (3 − m)(m + 2) = −(1 − 2m)(1 + 3m)

2 2
⇒ −m + m + 6 = − (1 + m − 6m )

2
⇒ 7m − 2m − 7 = 0

2±√4−4(7)(−7)
⇒ m =
2(7)

2±2√1+49
⇒ m =
14

1±5√2
⇒ m =
7

1±5√2
Thus, the required value of m is
7

Page : 234 , Block Name : Miscellaneous Exercise

Q20 If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x –
2y +7 = 0 is always 10.
Show that P must move on a line.

Answer. The equations of the given lines are
x + y – 5 = 0 … (1)
3x – 2y + 7 = 0 … (2)
The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by
|x+y−5| |3x−2y+7|
d1 = and d2 =
√(1)2 +(1)2 √(3)2 +(−2)2

|x+y−5| |3x−2y+7|
i.e., d1 = and d2 =
√2 √13

It is given that d1 + d2 = 10

|x+y−5| |3x−2y+7|
∴ + = 10
√2 √13

⇒ √13|x + y − 5| + √2|3x − 2y + 7| − 10√26 = 0

⇒ √13(x + y − 5) + √2(3x − 2y + 7) − 10√26 = 0

Assuming (x + y − 5) and (3x − 2y + 7) are positive ]

⇒ √13x + √13y − 5√13 + 3√2x − 2√2y + 7√2 − 10√26 = 0
, which is the equation of a line.
⇒ x(√13 + 3√2) + y(√13 − 2√2) + (7√2 − 5√13 − 10√26) = 0

Similarly, we can obtain the equation of line for any signs of (x + y − 5) and (3x − 2y + 7).
Thus, point P must move on a line.

Page : 234 , Block Name : Miscellaneous Exercise

Q21 Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Answer. The equations of the given lines are
9x + 6y – 7 = 0 … (1)
3x + 2y + 6 = 0 … (2)
Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of
P (h, k) from line (1) is given by
|9h+6k−7| |9h+6k−7| |9h+6k−7|
d1 = = =
2 2
(9) +(6) √117 3√13

The perpendicular distance of P (h, k) from line (2) is given by
|3h+2k+6| |3h+2k+6|
d2 = =
2 2
√(3) +(2) √13

Since P (h, k) is equidistant from lines (1) and (2),d 1 = d2

|9h+6k−7| |3h+2k+6|
∴ =
3√13 √13

⇒ |9h + 6k − 7| = 3|3h + 2k + 6|

⇒ |9h + 6k − 7| = ±3(3h + 2k + 6)

⇒ 9h + 6k − 7 = 3(3h + 2k + 6) or 9h + 6k − 7 = −3(3h + 2k + 6)

The case 9h + 6k − 7 = 3(3h + 2k + 6) is not possible as

9h + 6k − 7 = 3(3h + 2k + 6) ⇒ −7 = 18 (which is absurd)

∴ 9h + 6k − 7 = −3(3h + 2k + 6)

9h + 6k – 7 = – 9h – 6k – 18
⇒ 18h + 12k + 11 = 0
Thus, the required equation of the line is 18x + 12y + 11 = 0.

Page : 234 , Block Name : Miscellaneous Exercise

Q22 A ray of light passing through the point (1, 2) re ects on the x-axis at point A and the re ected ray
passes through the point (5, 3).
Find the coordinates of A.

Answer.

Let the coordinates of point A be (a, 0).
Draw a line (AL) perpendicular to the x-axis.
We know that angle of incidence is equal to angle of re ection. Hence, let
∠BAL = ∠CAL = Φ
Let ∠CAX = θ
∴∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]
= 180° – θ – 180° + 2θ
=θ
∴∠BAX = 180° – θ

3−0
Now, slope of line AC =
5−a

3
⇒ tan θ = . . . (1)
5−a

2−0
Slope of line AB =
1−a

∘ 2
⇒ tan(180 − θ) =
1−a

2
⇒ − tan θ =
1−a

2
⇒ tan θ = . . . (2)
a−1

From equations (1) and (2), we obtain
3 2
=
5−a a−1

⇒ 3a − 3 = 10 − 2a

13
⇒ a =
5

Thus, the coordinates of point A are ( 13

5
, 0)

Page : 234 , Block Name : Miscellaneous Exercise

Q23 Prove that the product of the lengths of the perpendiculars drawn from the
Points (√a x y
2 2 2 2 2
− b , 0) and (−√a − b , 0) to the line cos θ + sin θ = 1 is b
a b

Answer.
The equation of the given line is

x y
cos θ + sin θ = 1
a b

Or, bx cos θ + ay sin θ − ab = 0 … (1)

2 2
Length of the perpendicular from point (√a − b , 0)
to line (1) is

2 2
b cos θ(√a −b )+a sin θ(0)−ab| ∣b cos θ√a2 −b2 −ab∣
∣ ∣
p1 = =
√ 2 2 2 2 √ 2 2 2 2
b cos θ+a sin θ b cos θ+a sin θ

2 2
Length of the perpendicular from point (−√a − b , 0)
to line (2) is

2 2
b cos θ(−√a −b )+a sin θ(0)−ab ∣b cos θ√a2 −b2 +ab∣
∣ ∣
p2 = = … (3)
√ 2 2 2 2 √ 2 2 2 2
b cos θ+a sin θ b cos θ+a sin θ

On multiplying equations (2) and (3), we obtain

|b cos θ√a2 −b2 −ab∥(b cos θ√a2 −b2 +ab)

p1 p2 =
2
2
(√b2 cos2 θ+a2 sin θ)

∣(b cos θ√a2 −b2 −ab)(b cos θ√a2 −b2 +ab)|
∣

=
2 2 2 2
(b cos θ+a sin θ)

2
2 2 2
(b cos θ√a − b ) − (ab)

=
2 2 2
(b cos θ + a sin θ)

2 2 2 2 2 2
b cos θ (a − b ) − a b |
=
2 2 2 2
(b cos θ + a sin θ)

2 2 2 4 2 2 2
∣
∣a b cos θ − b cos θ − a b ∣
∣
=
2 2 2 2
b cos θ + a sin θ
2 2 2 2 2 2
b ∣
∣a cos θ−b cos θ−a ∣
∣
= 2
2 2 2
b cos θ+a sin θ

2 2 2 2 2 2 2 2 2
b ∣
∣a cos θ−b cos θ−a sin θ−a cos θ∣
∣ 2 2
= [sin θ + cos θ = 1]
2
b2 cos2 θ+a2 sin θ

2 2 2 2 2
b | − (b cos θ + a sin θ)
=
2 2 2 2
b cos θ + a sin θ
2 2 2 2 2
b (b cos θ + a sin θ)
=
2 2 2 2
(b cos θ + a sin θ)

2
= b

Hence, proved

Page : 234 , Block Name : Miscellaneous Exercise

Q24 A person standing at the junction (crossing) of two straight paths represented bythe equations 2x –
3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach
the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should
follow.

Answer. The equations of the given lines are
2x – 3y + 4 = 0 … (1)
3x + 4y – 5 = 0 … (2)
6x – 7y + 8 = 0 … (3)
The person is standing at the junction of the paths represented by lines (1) and (2).
On solving equations (1) and (2), we obtain x = − .
1 22
and y =
17 17

Thus, the person is standing at point (− .
1 22
, )
17 17

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point
(−
1

17
,
22

17
) .

Slope of the line(3) =
6

7

∴Slope of the line perpendicular to line (3)
The equation of the line passing through (− 1

17
,
22

17
) and having a slope of − 7

6
is given by
22 7 1
(y − ) = − (x + )
17 6 17

6(17y − 22) = −7(17x + 1)

102y − 132 = −119x − 7

119x + 102y = 125

Hence, the path that the person should follow is 119 x+102 y=125.

Page : 234 , Block Name : Miscellaneous Exercise