O Math 10 2075

Let A–1 = p q , then AA–1 = I
r s

or, a b p q = 1 0
c d r s 0
1
or, ap + br aq + bs = 1 0
cp + dr cq + ds 0
1

i.e. ap + br = 1 ... ... ... (i)
... ... ... (ii)
cp + dr = 0 ... ... ... (iii)
... ... ... (iv)
aq + bs = 0

cq + ds = 1

Solving (i) and (ii) for p and r, Matrices

adp + bdr = d
–cbp +–bdr =–0

(ad – cb)p = d

or, p = ad d bc and r = – ad c bc .
– –

Solving (iii) and (iv) for q and s,

adq + bds = 0
–cbq +–bds =– b

(ad – cb)q = –b

or, q = – ad b bc and s = ad a bc.
– –

d –b
ad – bc
Now, A–1 = p q = ad – bc
r s –c a
ad – bc ad – bc

or, A–1 = ad 1 bc d –b
– – c a

or, A–1 = 1 d –b
|A| – c a

∴ A–1 = 1 d –b
|A| – c a

Example 7: Find the inverse of the matrix B = 1 4 . If B–1 is the inverse of B, prove that BB–1 = I.
7 2
Solution:

Here the given matrix B = 1 4 is
7 2

1 4
| B | = 7 2 = 1 × 2 – 7 × 4 = 2 – 28 = – 26

Determinant / 97

Now, B–1 = 1 d –b = 1 2 – 4 ∴ B–1 = –1 2
|B| – c a – 26 – 7 1 13 13
7
26 –1
26

Now, BB–1 = 1 4 –1 2 –1 + 1143 2 + –132 1 0
7 2 13 13 13 13 0 1
7 = =
26 –1 –7 173 14 –1
26 13 + 13 + 13

\ BB–1 = I

Example 8: If the inverse of the matrix x 2x – 9 is the matrix 3 5 , find the values of x and y.
– y 3 y x
Solution:

Let A = x 2x – 9 and A–1 = 3 5
– y 3 y x

We know that A.A–1 = I

or, x 2x – 9 3 5 = 1 0
– y 3 y x 0 1

or, 3x + (2x – 9)y 5x + x (2x – 9) 1 0
– y × 3 + 3 × y –  y × 5 + 3 × x = 0 1

or, 3x + 2xy – 9y 5x + 2x2 – 9x = 1 0
– 3y + 3y – 5y + 3x 0 1

or, 3x + 2xy – 9y 2x2 – 4x = 1 0
3x – 5y 0 1
0

Equating corresponding elements,

2x2 – 4x = 0

or, 2x(x – 2) = 0

∴ x = 0 or 2

Similarly, 3x – 5y = 1

When x = 0, we have When x = 2, we have
or, 3 × 2 – 5y = 1
or, 3 × 0 – 5y = 1 or, – 5y = 1 – 6
or, – 5y = – 5
or, – 5y = 1 \ y = 1

∴ y = – 1
5

Hence, the required values of x = 0 or 2 and y = – 1 or 1
5

98 / Optional Mathematics Class 10

Some Properties of Inverses

a. The inverse of the product of two non-singular matrices is equal to the product of their inverses taken
in reverse order.

i.e. if A and B are two non-singular matrices of same order, then (AB)–1 = B–1A–1

Let A = 2 5 and B = 3 1 then
1 3 7 2

2 5
| A | = 1 3 = 6 – 5 = 1 and

3 1
| B | = 7 2 = 6 – 7 = – 1

So, A–1 = 1 d – b = 1 3 – 5 = 3 – 5 Matrices
|A| – c a 1 – 1 2 – 1 2

and B–1 = 1 d – b = 1 2 – 1 = – 2 1
|B| – c a – 1 – 7 3 7 – 3

Now, AB = 2 5 3 1 = 41 12
1 3 7 2 24 7

41 12
and | AB | = 24 7 = 287 – 288 = – 1

Hence, (AB)–1 = 1 7 – 12 = 1 7 – 12 = – 7 12
|AB| – 24 41 – 1 – 24 41 24 – 41

Again, B–1A–1 = – 2 1 3 – 5 = – 7 12
7 – 3 – 1 2 24 – 41

Hence, (AB)–1 = B–1A–1.

b. The transpose and inverse of a non-singular matrix is commutative.

i.e. (A–1)T = (AT)–1

Let A = 2 – 3 then AT = 2 7
7 4 – 3 4

2 – 3
Now, | A | = 7 4 = 8 + 21 = 29

2 7 4 3
and | AT| = – 3 4 = 8 + 21 = 29 29 29
– 7
Hence, A–1 = 1 4 3 = 1 4 3 = 29 2
|A| – 7 2 29 – 7 2 29

Hence, (A–1)T = 4 – 7
29 29
3
29 2
29

Determinant / 99

4 – 7
29
and (AT)–1 = 1 4 – 7 = 1 4 – 7 = 29
|AT| 3 2 29 3 2 3 2
29
Hence (A–1)T = (AT)–1. 29

c. If A is a non-singular matrix, then (A–1)–1 = A.

Exercise 3.1A

1. Find the determinants of the following matrices:

a. 4 2 b. 1 –3 3 51 3
1 3 4 3 c. 2 d. 3 8
4
2. Prove that: 4 3 8 9
2

cos A sin A a2 b
a. – sin A cos A = 1 b. = (a – b)(a2 + ab + b2)

a2 + b2 a + b

a – b 2a a + 1 – a
c. – 2b a – b = (a + b)2 d. b 1 – b = 1 + a – b

3. Solve for x:

x 2 2x 8
a. 3 5 = 14 b. x 7 = 12
5
2x 4 3 x =2
c. x 7 = 20 d. 2

4. Find the value of a in the following singular matrices.

a. 4a 3 b. 180 5a
6

a + 3 6 4a – 1 6
c. 6 4 d. 10 a

5. Find the value of k in the following.

3 8 5 2k 3 + k 3 6 k
a. 4 3k = 6 2 b. 7 1 = 2 1

4 1 + k 2 4 3 – k 2 k
c. 6 5 = k 3 d. k – 4 = 3 3

6. a. If P = 2 – 1 and Q = 2 – 6 , find the determinant of PQ.
0 3 4 – 2

b. If P = 1 – 2 and Q = – 3 0 , find the determinant of 5P – 2Q – 3I.
3 4 1 – 2

c. If P = 2 0 and Q = 2 4 , find the determinant of 5P – 1 Q + 2I.
– 1 3 – 6 2 2

100 / Optional Mathematics Class 10

7. Find the inverse of the following matrices:

a. 2 3 b. 7 6 c. 1 0 d. –3 –2
3 5 6 5 0 1 8 5

8. a. If A–1 = 2 –3 , then find the matrix A.
–5 8

b. If B–1 = 5 3 , then find the matrix B.
3 2

Exercise 3.1B

1. a. If A = – 4 5 , B = 4 6 and the determinant of A – B – 5I is 14, calculate the value of x.
7 8 x 3 Matrices

b. If P = 1 2 , Q = x 3 and the determinant of 2P – 3Q is –5, find the value of x.
3 – 4 2 – 1

2. a. If A = 2 3 and B = 0 4 , show that | AB | = | A | | B |.
4 – 1 – 2 1

b. If A = 2 1 and B = 4 3 , show that | AB | = | A | | B |.
4 – 3 7 – 3

3. a. If A = 2 – 4 and B = 3 7 , show that |(A + B)T| = | AT + BT|.
6 5 2 1

b. If M = 4 1 and N = 5 – 6 , show that |(M – N)T| = |MT – NT|.
7 6 7 6

4. a. If the inverse of the matrix x 2x – 9 is the matrix 3 5  . Find the values of of x and y.
y 3 1 y + 3

b. If the inverse of the matrix 2m 7 is the matrix 9 n  , calcu late the value of m and n.
5 9 – 5 4

5. a. If A = 1 2 and B = – 3 2 , then show that (AB)–1 = B–1A–1.
3 4 4 – 1

b. If P = 3 – 1 and Q = – 2 3  , then show that (PQ)–1 = Q–1P–1.
4 2 1 4

6. Find the value of X = x in the following cases:
y

a. 1 4 x = 9 b. 6 – 4 x = 2
2 3 y 8 1 – 1 y – 1

c. 3 4 x = 2 d. csions qq –s cinosqq x = sin q
1 – 1 y 3 y cos q



Determinant / 101

3.2 Solving Linear Equations by Matrix Method

The simultaneous equations in two variables can be solved by different methods, substitution
method, elimination method etc. Here, we deal with the method of solving linear equations of two variables
using matrices.

Let us consider the following linear equations,

a1x + b1y = c1 ... ... (i)
a2x + b2y = c2 ... ... (ii)

Expressing equations (i) and (ii) in the matrix form,

a1 b1 x = c1
a2 b2 y c2

i.e. AX = B

Where A = a1 b1 , X = x and B = c1
a2 b2 y c2

or, A–1(AX) = A–1B [Pre-multiplying both sides by A­–1)

or, (A–1A) X = A–1B

or, IX = A–1B ( A–1A = I)

\ X = A–1B ( IX = X)
Equating the corresponding elements of equal matrices X and A–1B, we get the solution.

Note: If |A| = 0, the given equations represent two parallel lines and the system of equations have no
unique solution.

If |A| ≠ 0, the given equations represent two intersecting lines and the system of equations have
unique solution.

Example 1: If x – 2y = –7 and 3x + 7y = 5 are two simultaneous equations:
i. Express the given equations in the matrix form.
ii. Is there a unique solution of x and y?

iii. Solve the equations.

Solution:

Here, the given equations are x – 2y = – 7 and 3x + 7y = 5.

i. Writing given equations in the matrix form, we get

1 – 2 x = – 7
3 7 y 5

ii. Here, the matrix of the coefficient of variables x and y is 1 – 2 and its determinant is
1 – 2 3 7

| A | = 3 7 = 7 + 6 = 13 ≠ 0

As the determinant A is not equal to 0, the given set of equations have unique solution.

102 / Optional Mathematics Class 10

iii. A–1 = 1 d – b
|A| – c a

= 1 7 2= 7 2
13 – 3 1 13 13
– 3
13 1
13

Now, 1 – 2 x = – 7
3 7 y 5

72

or, x = 13 13 – 7 (³ X = A–1B)
y – 3 1 5

13 13

x – 49 + 10 – 49 + 10 = – 3 Matrices
y 13 13 13 2
or, = =
21 5 21 + 5
13 + 13 13

∴ x = – 3, y = 2

Example 2: Solve by matrix method 4 + 3 = 7 and 3 + 2 = 4.
x y x y

Solution:

Writing the given equations in matrix form,

1

or, 4 3 x =7
3 2 1 4

y

Now, suppose A = 4 3 ,X= 1 ,B= 7
3 2 x 4

1
y

4 3
Here, |A| = 3 2 = 8 – 9 = – 1

A–1 = 1 2 – 3 = 1 2 – 3 = – 2 – 3
|A| – 3 4 – 1 – 3 4 3 4

Using formula,

X = A–1B

1

or, x = – 2 3 7 = – 2 × 7 + 3 × 4 = – 14 + 12 = – 2
1 3 – 4 4 3 × 7 – 4 × 4 21 – 16 5

y

Equating corresponding elements, we get

1x = – 2 i.e. x = – 1 and
2

Solving Linear Equations by Matrix Method / 103

1 = 5 i.e. y = 1
y 5

∴ Values of x and y are – 1 and 15.
2

Example 3: If the cost of 17kg sugar and 4kg tea is Rs.1110 and the cost of 8kg sugar and 2kg tea is

Rs.540, find the cost of sugar per kg and tea per kg by matrix method.

Solution :

Let the cost of sugar per kg = Rs.x

Cost of tea per kg = Rs.y

Then according to question,

1st case, 17x + 4y = 1110

2nd case 8x + 2y = 540

Writing the given equations in the matrix form, we get

187 4 x = 1110
2 y 540

or, AX = B

where A = 17 4 , X = x and B = 1110
8 2 y 540

17 4
Now, |A| = 8 2 = 34 – 32 = 2 ≠ 0

So, the given equations have unique solution.

Now, A–1 = 1 2 – 4 = 1 2 – 4
|A| – 8 17 2 – 8 17

Now, X = A–1B

= 1 2 –4 1110 = 1 2 × 1110 + (– 4) × 540
2 –8 17 540 2 – 8 × 1110 + 17 × 540

= 1 2220 – 2160 = 1 60 = 30
2 – 8880 + 9180 2 300 150

Hence, the cost of sugar per kg is Rs.30 and the cost of tea per kg is Rs.150

Exercise 3.2

1. a. If x + 3y = 5 and 2x – 3y = 1 are two simultaneous equations:
i. Express the given equations in the matrix form.
ii. Is there a unique solution of x and y ?
iii. Solve the equations.
b. If 3x + 5y = 12 and 2x + 3y = 9 are two simultaneous equations:
i. Express the given equations in the matrix form.

104 / Optional Mathematics Class 10

ii. Is there a unique solution of x and y?
iii. Solve the equations.

2. Solve the following pairs of equations by the matrix method and check your answer:

a. 3x + 2y = 1 b. 3x + 5y = 24 c. x + 2y = 8

7x + 5y = 4 5x = 2y + 9 2x + 3y = 11

d. 3x + y = 51 e. 2x + 3y = 5 f. 4x – 3y = 11

4x – 3y = 3 5x – 2y = 3 3x + 7y = – 1

g. 2x + 3y – 18 = 0 h. 3x – 5y = 3 i. 2x – 3y = 1

3x – 2y – 1 = 0 4x + 3y = 4 3x + 4y = 10 Matrices

j. 2x + y = 3 k. x + 3 = 5 l. 2 – 2y = – 1
y x

3x + 2y = 2 4x – 9 = 6 6 + y = 11
y x

m. 2x + 3 = 9 n. y5 = x1 – 1 o. 3x + 2y = 5x + 2y = 4
y 4 5

2x – 3 = 1 5 = 2 – 4 p. x + 1 = y + 3 = x – y
y y x 8 5 4

3. a. The cost of 3 kg of rice and 2 kg of potato is Rs.280. The cost of 4 kg of rice and 3 kg of potato is

Rs.384. Find the cost per kg of each item by using matrix method.

b. The cost of 3 chairs and 2 tables is Rs.27200. The cost of 2 chairs and a table is Rs.14800. Find the cost
of each chair and table using matrix method.



3.3 Cramer’s Rule

Here we introduce a method for solving the system of equations that uses determinants known as
Cramer’s Rule. Cramer’s rule is a viable and efficient method for finding solutions to the system of linear
equations. We will discuss the rule to solve the equations in two variables with the given two equations.
Cramer’s rule will give us the unique solution to a system of equations, if it exists. However, if the system
has no solution or an infinite number of solutions, this will be indicated by a determinant of zero values.

To understand Cramer’s rule, we will connect it with the method of solving system of linear equations
using elimination method.

Let, a1x + b1y = c1 ... ... (1) and
a2x + b2y = c2 ... ... (2) are two equations.

Cramer’s Rule / 105

To eliminate y from these equations, multiplying equation (1) by b2 and equation (2) by b1 and
subtracting equation (2) from equation (1), then

a1b2x + b1b2y = c1b2

–a2b1x +– b2b1y =–c2b1
(a1b2 – a2b1)x = c1b2 – c2b1

or, x = c1b2 – c2b1 and similarly we can get y = a1c2 – a2c1
a1b2 – a2b1 a1b2 – a2b1

These values of x and y can be expressed in the form of determinant as follows:

c1 b1 a1 c1
a2 c2
i.e. x = c1b2 – c2b1 = c2 b2 and y = a1c2 – a2c1 =
a1b2 – a2b1 a1 b1 a1b2 – a2b1 a1 b1
a2 b2
a2 b2

The calculation of the values of variables x and y as indicated above is also known as Cramer's

rule. In the process of solving equations in two variables using Cramer's rule, the different determinants of

co-efficients are represented as D = a1 b1 , Dx = c1 b1 and Dy = a1 c1 . Hence, the solution can
a2 b2 c2 b2 a2 c2

also be represented in the form

c1 b1 a1 c1
a2 c2
x= c2 b2 = DDx and y = = Dy
a1 b1 a1 b1 D
a2 b2
a2 b2

Example 1: Solve using Cramer's rule : 4x + y = 5 and 2x – 3y = 13

Solution:

Given equations are 4x + y = 5 ... ... (1)

and 2x – 3y = 13 ... ... (2)

Comparing these two equations with a1x + b1y = c1 and a2x + b2y = c2 respectively, we have

D= a1 b1 = 4 1 = 4 × (– 3) – 2 × 1 = – 12 – 2 = – 14,
a2 b2 2 – 3

Dx = c1 b1 = 5 1 = 5 × (– 3) – 13 × 1 = – 15 – 13 = – 28 and
c2 b2 13 – 3

Dy = a1 c1 = 4 5 = 4 × 13 – 2 × 5 = 52 – 10 = 42
a2 c2 2 13

Now, x= Dx = – 28 = 2 and y = Dy = 42 = – 3.
D – 14 D – 14

106 / Optional Mathematics Class 10

Example 2: Solve using Cramer's rule : x + 3 = 4 and 2x – 5 = – 9
y 2y

Solution:

Give equations are x + 3 = 4 ... ... (1) and
y
5
2x – 2y = – 9 ... ... (2)

Let z = 1 , then given equations becomes
y

x + 3z = 4 ... ... (3) and

2x – 5 z = – 9 ... ... (4)
2

The determinants of coefficients of equations (3) and (4) are,

D= a1 b1 = 1 3 =1× – 5 – 2 × 3 = – 5 – 6 = – 5 – 12 = – 17 Matrices
a2 b2 2 2 2 2
2 – 5
2

Dx = c1 b1 4 3 =4× – 5 – (– 9) × 3 = – 10 + 27 = 17
c2 b2 = 2
– 5
–  9 2

Dz = a1 c1 = 1 4 = 1 × (– 9) – 2 × 4 = – 9 – 8 = – 17
a2 c2 2 – 9

Now, x = Dx = 17 = 17 × – 2 = – 2 and z = Dz = – 17 = – 17 × – 2 = 2.
D 17 D 17
– 17 – 17
2 2

As we have z = 1 i.e. 2 = 1 and y = 1 .
y y 2

\ x = – 2 and y = 21.

Now consider a pair of equations which represent parallel lines or same line.

2x + 3y = 1 ... ... (1) For 2x + 3y = 1, (-3,5) Y
2x + 3y = 9 ... ... (2) x – 1 2 5

2 3 y 1 – 1 – 3 (0,3)
2 3
Here, D = =2×3–2×3

= 6 – 6 = 0 For 2x + 3y = 9, (-1,1) (3,1)
x – 3 3
Dx = 1 3 =1×3–9×3 y51 0 X' O (2,-1) X
9 3 3
(5,-3)
= 3 – 27 = – 24

Dy = 2 1 =2×9–2×1
2 9
Y'
= 18 – 2 = 16

From this example, we can conclude that if determinant D for given two equations is zero, then the

given equations represent a pair of parallel lines and the equations have no unique solution.

Cramer’s Rule / 107

x + 3y = 2 ... ... (3) For x + 3y = 2, Y
3x + 9y = 6 ... ... (4) x 2 –1 5

Here, D = 1 3 =1×9–3×3 y 0 1 –1 (-4,2)
3 9
(-1,1)
= 9 – 9 = 0
For 2x + 6y = 4, (3,1)
Dx = 2 3 =2×9–6×3 X'
6 9 OX
x –1 5 –4
(5,-1)

= 18 – 18 = 0 y 1 –1 2

Dy = 1 2 =1×6–3×2
3 6

= 6 – 6 = 0 Y'

From this example, we can conclude that if each of determinant D, Dx and Dy for the given two

equations is zero, then the given equations represent a single line and the equations have infinite solutions.

Exercise 3.3

1. If 3x + y = 6 and 4x – y = 1 are two simultaneous equations,

a. Find the determinants D, Dx and Dy of coefficients.
b. Is there unique solution of x and y?

2. If 3x + 5y = 12 and 6x + 10y = 9 are two simultaneous equations,

a. Find the determinants D, Dx and Dy of coefficients.
b. Is there unique solution of x and y?

3. x – 2y = 3 and 3x – 6y = 9 are two simultaneous equations,

a. Find the determinants D, Dx and Dy of coefficients.
b. Is there any solution of x and y?

4. Solve the following pairs of equations by using Cramer's rule:

a. 3x + 4y = 5 b. x + 3y = 1 c. 2x – 5y = 4
4x + y = 30
x + 2y = 3 2x + y = 7 2x + y = 11
y – 3x = 1
d. x – 2y – 4 = 0 e. 2x + 5y – 2 = 0 f.

5y – 3x + 7 = 0 3x – 5y = 3

5. Solve the following pairs of equations by using Cramer's rule:

a. x + 3 = 5 b. 1 + y = 5 c. 1 – 2 = 1
y x x y

3x – 5 = 1 2 – y = 1 2 – 3 = 3
y x x y

d. x + 3 – 7 = 0 e. 2x + 4 = y = 17 – 3x
y 5 4

2xy + 3 = 11y



108 / Optional Mathematics Class 10

4. CO-ORDINAsdaTfsdEfa GEOMETRY

Curriculum Contents Included Periods

a. Angle between two straight lines (Condition of parallel and perpendicular) 20
b. Pair of straight lines
– Lines represented by homogeneous equation of second degree and angle Co-ordinate

between them (Condition of perpendicular and coincident)
c. Conic Section
– Introduction and its types (from the intersection of cone and a plane only)
c. Circle
– Identification of properties of circle and verification with the help of co-

ordinates
– Equations of circle of the form
x2 + y2 = r2
(x – h)2 + (y – k)2 = r2
(x – x1) (x – x2) + (y – y1) (y – y2) = 0 and related problems

Objectives

On the completion of this unit, students will be able to

find the angle between straight lines and state the condition of parallelism and perpendicularity.

find two separate linear equations from the given homogenous equation of second degree.

find the angle between two straight lines given by homogenous equation of second degree.

state the condition of coincident of two straight lines.

find the equation of straight lines passing through the given point and perpendicular to straight
lines represented by given homogeneous equation of second degree.

find the equation of circle in different forms.

find the equation of circle passing through given three points.

find the equation of circle passing through the centre of the given circle and having centre at
given point.

solve the problems related to circles.

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Each of 1 marks Each of 5 marks Quesitons Marks
Co-ordinate Each of 2 marks Each of 4 marks
Goemetry 1 – 2 5
–1

 / 109

4 CO-ORDINATE GEOMETRY

4.1 Angle between Two Straight Lines

Introduction

Two intersecting lines intersect at two angles, which are supplementary to each other. For example, if
the angle between two intersecting lines is 60°, then the other angle will be 180° – 60° = 120°. Generally,
the angle between two straight lines is the acute angle between two straight lines.

We can find the angle between two lines when we know their slopes. Y

Y BD
B P 180° – θ
D
θ
P

θ

θ2 θ1 θ2 θ1

C AX X' C O AX
Y'
X' O

Y'

Let AB and CD be two straight lines having equations y = m1x + c1 and y = m2x + c2 respectively. If
lines AB and CD intersect X-axis at A and C, then BAX = q1 and DCX = q2 where m1 = tan q1 and

m2 = tan q2 are their slopes. Again, if θ be the angle between the lines AB and CD, then q1 = θ + q2.

i.e. θ = q1 – q2

i.e. tan θ = tan (q1 – q2) = tan q1 – tan q2
1 + tan q1 tan q2

i.e. tan θ = m1 m– 1mm2 2 ... ... ... (i)
1+

Again, 180° – θ is the other angle between AB and CD, then

tan (180° – θ) = – tan θ

tan (180° – θ) = – m1 – m2 ... ... ... (ii)
1+ m1 m2

Combining (i) and (ii) we have,

tan q = ± m1 – m2
1+ m1 m2

where θ represent both the angles between AB and CD.

110 / Optional Mathematics class 10

Cor. 1. Condition of Parallelism

Let two lines with slopes m1 and m2 be parallel. i.e. θ = 0°. m1 = slope
m1 – m2 q = 0°
i.e. tan 0° = ± 1+ m1 m2 m2 = slope

i.e. 0 = m1 – m2
1+ m1 m2

i.e. m1 = m2.
Hence, if two lines are parallel then their slopes are equal and vice versa.

Cor. 2. Condition of Perpendicularity

Let two lines with slopes m1 and m2 be perpendicular. i.e. θ = 90°.
m1 – m2
or, tan 90° = ± 1 + m1 m2 m1 = slope
q = 90°
or, csoins 9900°° = ± m1 – m2 m2 = slope
1+ m1 m2

1 m1 – m2 Co-ordinate
or, 0 = ± 1+ m1 m2

or, 1 + m1m2 = 0

∴ m1 . m2 = – 1.

Hence, if two lines are perpendicular then the product of their slopes is – 1 and vice versa.

Cor. 3. Angle between two Straight Lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a1
Let slope of line a1x + b1y + c1 = 0 is m1 = – b1 and that of
a2x + b2y + c2 = 0 is m2 = – ba22,
then the angle between the lines, θ can be calculated by

– a1 – – ba22 –  a1b2 + a2b1 a1x + b1y + c1 = 0
b1 – ba22
tan θ = ± m1 – m2 = ± =± b1b2 = ± a2b1 – a1b2 q = 0°
1+ m1 m2 1 + – ba11 b1b2 + a1a2 a1a2 + b1b2 a2x + b2y + c2 = 0

a1 ba22. b1b2
b1
a. Lines are parallel if =

b. Lines are perpendicular if – a1 × – a2 =–1 i.e. a1 a2 = – 1 a1x + b1y + c1 = 0
b2 b2 b1 b2 q = 90°

i.e. a1a2 + b1b2 = 0. a2x + b2y + c2 = 0

Example 1: Find the acute angle between the lines 2x + 3x = 6 and 4x + y = 7.
Solution:

Here, let 2x + 3y = 6 ... ... (1) and 4x + y = 7 ... ... (2)

From (1), y = – 2x + 6 = – 2 x + 2 and its slope (m1) = – 2
3 3 3

From (2), y = – 4x + 7 and its slope (m2) = – 4

Angle between Two Straight Lines / 111

If q be the angle between the lines, then

m1 – m2 – 2 – (– 4) – 2 + 12
1+ m1 m2 + 3
tan q = ± = ± = ± 3 =± 10 × 3 = ± 10 = ± 0.909
3 11 11
1 – 2 (– 4) 1 + 8
3 3

or, tan q = 0.909 [ Taking positive sign for acute angle]

or, q = tan–1 (0.909) = 42.27°

Hence, the required angle is 42.27°.
Example 2: Find the angle between the lines 3x + y – 7 = 0 and x + 2y + 9 = 0.
Solution:

Let m1 and m2 be the slopes of the given lines 3x + y – 7 = 0 and x + 2y + 9 = 0, then

3x + y – 7 = 0 x + 2y + 9 = 0

or, y = – 3x + 7 19
or, y = – 2 x – 2

∴ m1 = – 3 1
∴ m2 = – 2

If θ be the angle between the given lines.

m1 – m2 –3– – 1 – 3 + 1 –6+1 – 5
1+ m1 m2 2 2 2 2
tan θ = ± = ± = = =
1 3 2+3 5
1 + (–3) – 2 1 + 2 2 2

or, tan θ = ± – 5/2
5/2

or, tan θ = ± 1

or, tan θ = tan 45° or tan 135°

∴ θ = 45° and 135°.

Example 3: Prove that two lines 5x + 3y = 7 and 10x + 6y = 9 are parallel to each other.
Solution:

Here, let 5x + 3y = 7 ... ... (1) and 10x + 6y = 9 ... ... (2)

From (1), y = – 5x + 7 = – 5 x + 7 and its slope (m1) = – 5
3 3 3 3
– 10x + 9 10 9 10 5
From (2), y = 6 = – 6 x + 6 and its slope (m1) = – 6 = – 3

Since, m1 = m2 i.e. the slopes of two lines are equal.

Hence, the given two lines are parallel to each other.

Example 4: If a line 2x + 3y = 6 is perpendicular to the line ax + 4y = 3, find the value of a.
Solution:

The equations of given straight lines are,

2x + 3y = 6 ... ... ... (i) and ax + 4y = 3 ... ... ... (ii)

112 / Optional Mathematics class 10

Slope of the first line (m1) = – 2
3

Slope of the second line (m2) = – a
4

Since, the lines (i) and (ii) are perpendicular to each other,

m1 × m2 = – 1

or, – 2 × – a = – 1
3 4

∴ a = – 6.
Example 5: Show that the points (3, – 1), (0, 2), (1, 3) and (4, 0) are the co-ordinates of the angular

points to a parallelogram and find the angle between diagonals.

Solution: C

Let A(3, – 1), B(0, 2), C(1, 3) and D(4, 0) be the vertices of given parallelogram.

Mid-point of AC is 3 + 1 , –1 + 3 i.e. (2, 1) and B 41° 139°
2 2

Mid-point of BD is 0 + 4 , 2 + 0 i.e. (2, 1) D Co-ordinate
2 2

∴ The diagonal AC and BD of ABCD bisect at (2, 1). Hence, ABCD is a parallelogram.

Slope of diagonal AC is (m1) = 3 + 1 = – 2. A
1 – 3

Slope of diagonal BD is (m2) = 0 – 2 = – 12.
4 – 0

Let θ be the angle between diagonals AC and BD, then

m1 – m2 – 2– – 1 – 3 3 1 3
1+ m1 m2 + (– 2) 2 2 2 2 4
tan θ = ± = ± = =± – × = ±
– 1 2
1 2

∴ tan θ = ± 34.
Taking positive sign, tan θ = 0.75 = tan 41° i.e. θ = 41°

and taking negative sign, tan θ = – 0.75 = tan 139° i.e. θ = 139°

Hence, the angles between the diagonals are 41° and 139°.

Exercise 4.1

1. Find the acute angle between the following pairs of straight lines:

a. 2x = y – 9 and x = y + 21 b. y = 3x + 6 and – 3y = – x + 9

c. 2x – y + 6 = 0 and 3x + y + 4 = 0 d. 6x – 2y + 9 = 0 and 5x – y = 3

2. Find the obtuse angle between the following pairs of lines:

a. 3x – y + 6 = 0 and y + 2x = 5 b. 3x – 4y = 10 and x – 2y + 7 = 0

c. y + 3x = – 7 and y + 5 = 0 d. y – (2 + 3)x = 5 and y – (2 – 3)x = 10

Angle between Two Straight Lines / 113

3. Find the angles between the following pairs of straight lines:

a. x = a + 3y and 3x – y + b = 0 b. 3x + 2y – 7 = 0 and 2x – 3y – 5 = 0

c. 3x – y + 5 = 0 and x – 3y = 5 d. px + qy + r = 0 and px – qy + m = 0

4. Find the angle between the lines given in the figure.

a. C(7,3) b. P(– 2,– 3)

q

q B(6,– 1) R(4,– 4)

A(2,3)

Q(– 1,3)

5. Show that the following pairs of lines are parallel to each other:

a. x + y = – 7 and 2x + 2y = 7 b. 3x + 7 – 4y = 0 and 3x + 9 = 4y

c. 5x – 10y + 3 = 0 and 2x – 4y + 3 = 0 d. ax + by + 4 = 0 and 2ax + 2by + 3 = 0

6. Show that the following pairs of straight lines are perpendicular to each other:

a. x + y = 9 and x – y = 4 b. 4x – 3y = 11 and 3x + 4y = 2

c. 10x + 20y = 13 and 4x – 2y = 15 d. ax + by + c = 0 and bx – ay + d = 0

7. Find the value of k so that the following two straight lines:

a. x + 3y = 7 and 2x + ky + 8 = 0 are parallel.

b. 2x – 3y + 5 = 0 and kx + 3y = 4 are parallel.

c. 4x – ky + 8 = 0 and 6x + 21y – 5 = 0 are perpendicular.

d. 4x + 3y = 7 and kx = 2y + 7 are perpendicular.

e. 2x + 3y + 6 = 0 and ax – 5y + 20 = 0 are perpendicular.

8. a. Show that the line 2x + 4y = 3 is perpendicular to the line passing through the points (7, 3) and
(0, – 11).

b. Show that the line x + 7y = 2 is parallel to the line passing through the points (– 11, 1) and
(3, – 1).

9. a. Show that the line passing through the points (2, 2) and (5, 4) is perpendicular to the line passing
through the points (– 4, 4) and (0, – 2).

b. Show that the line joining points (0, 0) and (2, 3) is parallel to the line joining points (2, – 2) and
(6, 4).

10. a. If the line passing through (2, – 5) and (7, a) is parallel to the line given by the equation
y + 3x + 8 = 0, find the value of a.

b. If the line joining the points (p, 5) and (6, – 7) is perpendicular to the line given by the equation
3x + 4y = 12, find the value of p.

11. a. The angular points of a quadrilateral are (3, 3), (3, 0), (7, – 1) and (8, 5). Find the angle between
the diagonals.

b. Show that the quadrilateral with vertices (4, 3), (2, 1), (6, – 1) and (8, 1) is a parallelogram. Also
find the angle between the diagonals.



114 / Optional Mathematics class 10

4.2 Some Relations between Straight Lines

Introduction

Two straight lines may be intersecting or parallel or perpendicular to each other. We can find the equation
of straight line with its information related to the given straight line.

Equation of a Line Parallel to ax + by + c = 0

The equation of given line is

ax + by + c = 0 ... ... ... (i)

Slope of the line (i) is m1 = – a
b
If the slope of the required straight line be m2 which is parallel to (i), then m1 = m2 = – ba.
The equation of the required straight line is

y = m2x + c1
a
or, y = – b x + c1 ax + by + c = 0 Co-ordinate

or, ax + by – bc1 = 0

Hence, the equation of straight line parallel to ax + by + k = 0

ax + by + c = 0 is of the form ax + by + k = 0 where k = – bc1. only constant term is changed.

Thus, the equation of any line parallel to ax + by + c = 0 is ax + by + k = 0.

Equation of a Line Perpendicular to ax + by + c = 0

The equation of given line is

ax + by + c = 0 ... ... ... (i)

Slope of the line (i) is m1 = – a
b
If the slope of required straight line be m2 which is perpendicular to (i), then m1 × m2 = – 1

or, – a × m2 = – 1
b

b
∴ m2 = a.

The equation of the required straight line is

y = m2x + c1

or, y = b x + c1 ax + by + c = 0
a

or, bx – ay + ac1 = 0 bx – ay + k = 0
Hence, the equation of the straight line perpendicular to
ax + by + c = 0 is of the form bx – ay + k = 0 where k = ac1. co-efficents are interchanged.
sign of one co-efficient is changed.
constant term is changed.

Thus, the equation of any straight line perpendicular to ax + by + c = 0 is bx – ay + k = 0.

Some Relations between Straight Lines / 115

Example 1: Find the equation of a straight line passing through the point (1, – 3) and parallel to the line
2x – 3y = 6.

Solution:

The given equation of a line is 2x – 3y = 6
(1,–3)
2x – 3y = 6 ... ... ... (i)

Equation of straight line parallel to (i) is

2x – 3y + k = 0 ... ... ... (ii)

If the line (ii) passes through the point (1, – 3), then

2 × 1 – 3 × (– 3) + k = 0

i.e. k = – 11

Hence, the equation of required straight line is 2x – 3y – 11 = 0.

Example 2: Find the equation of a straight line perpendicular to 2x + 5y + 3 = 0 and passing through

(– 1, 4).
Solution:

The equation of given line is 2x + 5y + 3 = 0.

i.e. y = – 2 x – 53 ... ... ... (i)
5

The slope of given line (i) is m1 = – 25.
If m2 be the slope of the required straight line, then

m1 × m2 = – 1 Alternately,

i.e. – 2 × m2 = – 1 The given line is 2x + 5y + 3 = 0 ... ... ... (i)
5 Let the line perpendicular to (i) and passing
through (– 1, 4) is
∴ m2 = 5
2

Hence, the equation of required straight line is 5x – 2y + k = 0 ... ... ... (ii)

y – y1 = m2(x – x1) which passes through (– 1A, 4s)line (ii) passes through point (– 1, 4), we get

or, y – 4 = 5 (x + 1) 5 ×(– 1) – 2 × 4 + k = 0
2
or, – 5 – 8 + k = 0
or, 2y – 8 = 5x + 5

\ 5x – 2y + 13 = 0. or, k = 13
Which is the required equation. Hence, the required line is 5x – 2y + 13 = 0.

Example 3: Find the equation of the straight line passing through the point of intersection of the lines
2x – 3y + 5 = 0 and 7x + 4y + 3 = 0 and perpendicular to the line 6x – 7y + 8 = 0.

Solution:

To find the point of intersection of the lines 2x – 3y + 5 = 0 and 7x + 4y + 3 = 0, solve these two
equations.

116 / Optional Mathematics class 10

Multiplying 2x – 3y + 5 = 0 by 4 and the 7x + 4y + 3 = 0 by 3 and adding them, we have

8x – 12y + 20 = 0
21x + 12y + 9 = 0

29x + 29 = 0 Co-ordinate
or, 29x = – 29
∴ x = – 1
Putting the value of x in equation 2x – 3y + 5 = 0, we have
2 × (– 1) – 3y + 5 = 0
or, – 2 – 3y + 5 = 0
or, 3y = 3
or, y = 1
The point of intersection is (– 1, 1).
Now, the equation of straight line perpendicular to 6x – 7y + 8 = 0 is
7x + 6y + k = 0
As it is passing through the point (– 1, 1), we have
or, 7 × (– 1) + 6 × 1 + k = 0
or, – 7 + 6 + k = 0
or, k = 1

Hence, the required equation is 7x + 6y + 1 = 0.

Example 4: Find the equation of the perpendicular bisector of the line segment joining the points
(3, 8) and (– 5, 6).

Solution:

The slope of the line joining given two points A(3, 8) and B(– 5, 6) is

m1 = y2 – y1 = 6–8 = –  2 = 1
x2 – x1 – 5 – 3 –  8 4

If the slope of the perpendicular bisector of AB is m2, then A(3,8) D B(– 5,6)

m1.m2 = – 1

1
or, 4 × m2 = – 1

∴ m2 = – 4 x1 + x2 , y1 + y2
The midpoint D of AB is 2 2

i.e. 3 – 5 , 8 + 6
2 2

i.e. (– 1, 7)
The equation of perpendicular bisector is
y – y1 = m2 (x – x1)
or, y – 7 = – 4(x + 1)
∴ 4x + y – 3 = 0.

Some Relations between Straight Lines / 117

Example 5: Find the equation of the straight line which passes through the point (4, 5) and making an
angle 45° with the straight line 2x + y + 1 = 0.

Solution:

Given line is 2x + y + 1 = 0 = 0
45°
or, y = – 2x – 1 y + 1

2x +

∴ Slope of given line (m1) = – 2. 45° (4,5)

Let slope of the required line be m2 and the angle between the

lines is given to be 45°. i.e. θ = 45°.

or, tan 45° = ± m1 – m2
1 + m1 m2

or, tan 45° = ± – 2 – m2
1 – 2m2

Case I, Taking + ve sign, 1 = – 2 – m2
1 – 2m2

or, – 2 – m2 = 1 – 2m2

∴ m2 = 3.
Hence, the equation of the required straight line is

y – y1 = m2 (x – x1)
i.e. y – 5 = 3 (x – 4)

i.e. 3x – y = 7

Case II, Taking –ve sign, 1 = – –  2 – m2
1 – 2m2

i.e. 2 + m2 = 1 – 2m2

∴ m2 = – 1
3

Hence, the equation of the required straight line is

y – y1 = m2 (x – x1)

i.e. y – 5 = – 1 (x – 4)
3

i.e. 3y – 15 = – x + 4

∴ x + 3y = 19.

Hence, the required straight lines are 3x – y = 7 and x + 3y = 19.

Example 6: Find the equation of the straight line passing through the point that divides the join of
(– 3, – 4) and (7, 1) in the ratio 3:2 and perpendicular to the join.

Solution:

Here given points are A(x1, y1) = A(– 3, – 4) and B(x2, y2) = B(7, 1) and ratio m1:m2 = 3:2

118 / Optional Mathematics class 10

Using formula,

(x, y) = m1x2 + m2x1 , m1y2 + m2y1 The slope of line AB is B(7,1)

or, m1 + m2 m1 + m2 m1 = y2 – y1 (x,y)
or, x2 – x1
or, 3 × 7 + 2 × – 3 , 3 × 1 + 2 × – 4
or, (x, y) = 3+2 3+2 1 – (– 4)
or, m1 = 7 – (– 7)
(x, y) = 21 – 6 , 3 – 8
55 1+4 A(–3,–4)
or, m1 = 7 + 3
15 , – 5
(x, y) = 55 5
or, m1 = 10
(x, y) = (3, – 1)
1
or, m1 = 2

1
If m2 is the slope of the line perpendicular to AB then m1 × m2 = – 1 i.e. 2 × m2 = – 1

Hence, m2 = – 2

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

y – y2 = m2(x – x1) Co-ordinate
or, y – (– 1) = – 2(x – 3)

or, y + 1 = – 2x + 6

or, 2x + y – 5 = 0

Example 7: Find the co-ordinates of the foot of the perpendicualr drawn from (1, 5) to the line
segment joining (1, 1) and (5, 5).

Solution:

Let A(1, 1), B(5, 5) and C(1, 5) are given three points. CD⊥AB and we have to find co-ordinates of point D.

Equation of AB is y – y1 = y2 – y1 (x – x1)
x2 – x1
C(1,5)
5–1
or, y – 1 = 5 – 1 (x – 1)

or, y – 1 = x – 1

or, x – y = 0 … … (i)

Let the equation of CD is … … (ii) A(1,1) D B(5,5)
y + x + k = 0

As CD passes through C(1, 5),

1 + 5 + k = 0

∴ k = – 6

Hence, substituting the value of k in (ii), the equation of CD is

x + y – 6 = 0 … … (iii)

Solving (i) and (iii), we get,

or, x = 3 and y = 3

The required foot of perpendicular is at (3, 3).

Some Relations between Straight Lines / 119

Example 8: If the slope of the base of an equilateral triangle is – 1 and the vertex is (2, 3), find the
equation of remaining sides.

Solution:

Let ABC be equilateral trianlge with vertex A(2, 3) and slope of BC is m1 = – 1. If m2 be the slope of the line

making angle 60° with BC, using formula

or, tan 60° = ± m1 – m2
1 + m1 m2

or, tan 60° = ± – 1 – m2
1 + (– 1)m2

or, 3 = ± – 1 – m2
1 – m2

or, 3(1 – m2) = ± (– 1 – m2)

Taking positive sign, Taking negative sign,

3(1 – m2) = – 1 – m2 3(1 – m2) = 1 + m2

or, 3 – 3m2 = – 1 – m2 or, 3 – 3m2 = 1 + m2

or, m2 – 3m2 = – 1 – 3 or, m2 + 3m2 = 3 – 1

or, – 3m2 + m2 = – 1 – 3 or, m2(1 + 3) = 3 – 1

or, – m2( 3 – 1) = – (1 + 3) or, m2 = 3–1
3+1
or, m2 = 3+1
3–1 3 – 1 3–1
or, m2 = 3 + 1 × 3–1
3 + 1 3+1
or, m2 = 3 – 1 × 3+1 ( 3 – 1)2 3 –2 3+ 1
( 3)2 – 12 3–1
( 3 + 1)2 3 +2 3+ 1 or, m2 = =
( 3)2 – 12 3–1
or, m2 = = 2(2 – 3) = 2 –
2
2(2 + 3) = 2 + \ m2 = 3
2
\ m2 = 3

Let AB has slope m2 = 2 + 3 then its equation is

y – y1 = m2(x – x1)

or, y – 3 = (2 + 3) (x – 2)

or, y – 3 = (2 + 3)x – 2(2 + 3)

or, y = (2 + 3)x – 4 – 2 3 + 3

\ y = (2 + 3)x – 2 3 – 1

If AB has slope m2 = 2 – 3 then its equation is the equation of AC is
y – y1 = m2(x – x1)
or, y – 3 = (2 – 3) (x – 2)

120 / Optional Mathematics class 10

or, y – 3 = (2 – 3) (x – 2)

or, y – 3 = (2 – 3)x – 2(2 – 3)

or, y = (2 – 3)x – 4 + 2 3 + 3

\ y = (2 – 3)x + 2 3 – 1

Example 9: Two opposite vertices of a square are (3, 4) and (1, – 1). Find the co-ordinates of other
vertices.

Solution:

Let the square be with vertices A(3, 4), B(x1, y1), C(1, – 1) and D(x2, y2).

Slope of AC = –1–4 = 52.
1–3

Let m be the slope of a line making angle 45° with AC.

m – 5 D(x2, y2) C(1, – 1)
2
∴ tan 45° = ± 45°
5
1 + m × 2 M Co-ordinate

or, 1 = ± 2m –5 45°
2+ 5m

or, 2 + 5m = ± (2m – 5) A(3, 4) B(x1, y1)
Taking positive sign,
Taking negative sign,

2 + 5m = 2m – 5 and 2 + 5m = – 2m + 5

or, 3m = – 7 or, 7m = 3
∴ m = – 37
∴ m = 3
7

Let the slope of AD (or BC) be – 37, the slope of CD (or AB) will be 37.

Equation of AD is y – 4 = – 7 (x – 3)
3

or, 3y – 12 = – 7x + 21

∴ 7x + 3y – 33 = 0 ... ... ... (i)

Equation of CD is y + 1 = 3 (x – 1)
7
or, 7y + 7 = 3x – 3

∴ 3x – 7y – 10 = 0 ... ... ... (ii)

Solving (i) and (ii), we get

– 30 x 231 = – y 70 = – 1 – 9
– 99 + 49

i.e. – 2x61 = – y = – 1
29 58

i.e. x9 = 1y = 1 i.e. x = 9 , y = 1
2 2 2

Some Relations between Straight Lines / 121

Co-ordinates of D is 9 , 1 .
2 2

The mid-point of median BD is

x1 + 9 , y1 + 1 = 2x1 + 9 , 2y1 + 1
i.e. 2 2 2 2 4 4

The midpoint of median AC is 3 + 1 , 4 + (– 1) = 2, 3
2 2 2

As, M is the mid-point of both medians BD and AC, we have

or, 2x14+ 9 , 2y1 + 1 = 2 , 3
4 2

or, 2x14+ 9 = 2 and 2y1 + 1 = 3
4 2

or, 2x1 + 9 = 8 and 2y1 + 1 = 6

or, 2x1 = – 1 and 2y1 = 5

∴ x1 = – 1 and y1 = 5
2 2

Hence, the co-ordinates of B is – 1 , 5
2 2

Example 10: Find the equation of the straight line which is parallel to 2x + 3y = 11 and such that the

algebraic sum of the intercepts on the axes is 15.

Solution:

Let the equation of line which is the parallel to 2x + 3y – 11 = 0 is

2x + 3y + k = 0

or, 2x + 3y = – k

or, 2x–+k3y = 1
or, – kx/2 y x y –k –k
+ – k/3 = 1 which is of the form a + b = 1 so, 2 and 3 are are the intercepts of

2x + 3y + k = 0.

From question,

–2 k + – k = 15
3

or, – 3k – 2k = 15
6

or, – 5k = 15 × 6

or, k = – 15 × 6
5

or, k = – 18

Hence, the required equation is 2x + 3y – 18 = 0.

122 / Optional Mathematics class 10

Exercise 4.2

1. a. Find the equation of a straight line parallel to 3x – 4y = 12 and passing through the point (2, 2).

b. Find the equation of the line parallel to the line 3x – 2y = 5 and passing through the mid-point of
the line segment joining the points (– 4, 2) and (2, 4).

c. Find the equation of the straight line which is parallel to the line 4x – y + 8 = 0 and passing
through the mid-point of the line joining (1, 5) and (3, 11).

d. Find the equation of straight line parallel to the line joining the points (5, 4) and (2, 2) and
passing through the mid-point of the line segment joining (0, 7) and (8, 7).

2. a. Find the equation of the straight line that passes through point (7, – 3) and perpendicular to the line
5x + 12y + 13 = 0.

b. Find the equation straight line perpendicular to 2x + 3y – 8 = 0 and passing through (9, 2). Co-ordinate

c. Find the equation of the straight line which is perpendicular to the line 2x + y – 8 = 0 and passing
through the mid-point of the line segment joining (0, 7) and (8, 7).

d. Find the euqation of the straight line passing through the point (– 6, 2) and perpendicular to the
line joining the points (– 3, 2) and (5, – 3).

e. M(4, 7) and N(5, – 2) are two points. Find the equation of the straight line perpendicular to MN
and passing through the point (2, 3).

3. a. Find the equation of the straight lines passing through the centroid of the triangle formed by the
vertices (3, – 4), (– 2, 1) and (5, 0) and

(i) parallel to x – 3y = 4 and (ii) perpendicular to 4x – 3y = 5.

b. Find the equation of the straight line which is perpendicular to the line 3x – 4y + 6 = 0 and passes
through the point of intersection of the lines 2x + y = 5 and x – y = 1.

c. Find the equation of the line passing through the point of intersection of the lines x + 2y = 4 and
x = 2y and parallel to the line 6x – 7y + 8 = 0.

4. a. Find the equation of the perpendicular bisector of the line joining the points (– 5, 6) and (0, – 4).

b. The points B(2, 6) and D(0, – 2) are two opposite vertices of a square ABCD, find the equation
of the diagonal AC.

c. A(3, 5) and C(7, 9) are the opposite vertices of a rhombus ABCD. Find the equation of the
diagonal BD.

Some Relations between Straight Lines / 123

d. If the equation of a diagonal AC of given square ABCD is 3x – 4y + 10 = 0 and the co-ordinates
of vertex B is (4, – 5), find the equation of diagonal BD.

5. a. Find the equation of straight lines passing through the point (1, – 4) and making an angle 45°
with the line 2x + 3y + 5 = 0.

b. Find the equation of the straight lines which pass through (2, 3) and making angle 45° with the
line 3x + y – 5 = 0.

6. a. Find the equation of the straight line which passes through the origin and making angle 60° with
the line x + 3y + 3 3 = 0.

b. Find the equation of straight line passing through the point (1, 0) and inclined at an angle 30°
with the line x – 3y = 4.

7. a. The co-ordinates of points P and Q are (2, 5) and (7, – 10) respectively. Find the equation of the
line passing through point M which divides PQ in the ratio 3:2 and perpendicular to PQ.

b. The points A and B have co-ordinates (5, 4) and (– 3, 2) respectively. Find the equation of the line
passing through point P which divides AB in the ratio of 2:1 and perpendicular to AB.

8. a. Find the co-ordinates of the foot of the perpendicular drawn from the point (6, 3) to the line
segment joining (4, 1) and (8, 3).

b. Find the co-ordinates of the foot of perpendicular from (4, 6) to the line through (– 1, 3) and
(7, 1).

9. a. The slope of the base of an equilateral triangle is – 1 and the vertex is the point (2, – 1). Find the
equations of the remaining sides.

b. If the base of an equilateral trianlge is x + y = 3 and the vertex is (2, 3), find the equation of
remaining sides.

10. a. Two opposite vertices of a square are (2, 1) and (– 2, 1). Find the co-ordinates of the other two
vertices.

b. Two opposite vertices of a square are (1, 3) and (7, – 1). Find the co-ordinates of other two
vertices.

11. a. Find the equation of straight line parallel to 2x + 3y = 5 and such that the algebraic sum of the
intercepts on the axes is 10.

b. Find the equation of straight line perpendicular to the line 2x + 3y – 5 = 0 and such that the
algebraic sum of the intercepts on the axes is 5.


124 / Optional Mathematics class 10

4.3 Equation of Pair of Straight Lines

Introduction

Let us consider the equations

2x + y – 1 = 0 ... ... ... (i)

and x + 3y + 2 = 0 ... ... ... (ii)

Let us multiply (i) and (ii),

(2x + y – 1) (x + 3y + 2) = 0 ... ... ... (iii)

The co-ordinates of any point on (i) will satisfy equation (iii). Similarly, every point on (ii) will
satisfy equation (iii). Conversely, any point satisfying (iii), satisfy either (i) or, (ii) or, both. Hence, equation
(iii) is the combined equation of two straight lines (i) and (ii).

Equation (iii) can also be written as

2x2 + 6xy + 4x + xy + 3y2 + 2y – x – 3y – 2 = 0

or, 2x2 + 7xy + 3y2 + 3x – y – 2 = 0 Co-ordinate

Homogeneous Equation of Second Degree in x and y

The general equation of two straight lines through the origin are

a1x + b1y = 0 ... ... ... (i) and a2x + b2y = 0 ... ... ... (ii) where a1, b1, a2 and b2 are constants.
Then their combined equation of (i) and (ii) is

(a1x + b1y) (a2x + b2y) = 0

or, a1a2 x2 + (a1b2 + b1a2) xy + b1b2y2 = 0

Let a1a2 = a, a1b2 + b1a2 = 2h and b1b2 = b then the equation will be

ax2 + 2hxy + by2 = 0 ... ... ... (iii)

Thus, the equation ax2 + 2hxy + by2 = 0 is called homogeneous equation of second degree in x and
y which represents a pair of straight lines passing through origin.

General Equation of Second Degree in x and y

Let us consider two linear equations a1x + b1y + c1 = 0 ... ... ... (i) and
a2x + b2y + c2 = 0 ... ... ... (ii) where a1, b1, c1, a2, b2 and c2 are constants.
Let us multiply (i) and (ii),

(a1x + b1y + c1) (a2x + b2y + c2) = 0 represents two straight lines given by (i) and (ii).
After simplifying above equation, we get

a1a2x2 + (a1b2 + b1a2) xy + b1b2y2 + (a1c2 + c1a2)x + (b1c2 + c1b2) y + c1c2 = 0
Putting, a1a2 = a, b1b2 = b, c1c2 = c, a1b2 + b1a2 = 2h, a1c2 + a2c1 = 2g and b1c2 + b2c1 = 2f
The above equation reduces to ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

Thus, the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is the general equation of second degree in
x and y. It represents a pair of straight lines.

Equation of Pair of Straight Lines / 125

Homogeneous equation of second degree ax2 + 2hxy + by2 = 0

The homogeneous equation of second degree in x and y is ax2 + 2hxy + by2 = 0.

or, by2 + 2hxy + ax2 = 0.

or, by2 + (2hx)y + ax2 = 0.

As, this equation is quadratic in y, it has two roots.

From the formula of solving quadratic equation for y, we have

y = – 2hx ± (2hx)2 – 4.b.ax2
2b

or, y = – 2hx ± 4h2x2 – 4abx2
2b
y = – 2hx ± 2x h2 – ab
or, 2b

or, y = 2(– h ± h2 – ab)x
2b

or, y= –h± h2 – ab x
b

Separating two different equations, we have,

y= –h+ h2 – ab x ... ... (1) and y = – h – h2 – ab x ... ... (2)
b b

Both of these equations are linear in x and y and are satisfied by (0, 0), so they represent two straight

lines passing through the origin.

Homogeneous equation of second degree ax2 + 2hxy + by2 = 0 always represents a pair of straight

lines through the origin.

Condition that the General Equation Represent a Pair of Straight Lines

The general equation of second degree in x and y is

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ... ... (i)

or, ax2 + (2hy + 2g)x + (by2 + 2fy + c) = 0 ... ... (ii)

The equation (ii) is expressed into the form of quadratic in x, so

x = – (2hy + 2g) ± (2hy+ 2g)2 – 4a(by2 + 2fy + c)
2a

= – 2(hy + g) ± 2 (hy+ g)2 – a(by2 + 2fy + c)
2a

= – (hy + g) ± (hy+ g)2 – a(by2 + 2fy + c) ... ... (iii)
a

Let us suppose the equation (iii) gives two linear equations, then

(hy+ g)2 – a(by2 + 2fy + c) is a perfect square.

or, h2y2 + 2hgy + g2 – aby2 – 2afy – ac is a perfect square.

or, (h2 – ab)y2 + (2gh – 2af)y + (g2 – ac) is a perfect square.
or, (2gh – 2af)2 – 4(h2 – ab)(g2 – ac) = 0 [ b2 – 4ac = 0]

or, (gh – af)2 – (h2 – ab)(g2 – ac) = 0

or, g2h2 – 2afgh + a2f 2 – g2h2 + ach2 + abg2 – a2bc = 0

126 / Optional Mathematics class 10

or, – a(abc + 2fgh – af 2 – bg2 – ch2) = 0

or, abc + 2fgh – af 2 – bg2 – ch2 = 0

Thus, the general second degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of
straight lines if abc + 2fgh – af 2 – bg2 – ch2 = 0.
Example 1: Find the combined equation of the lines with equations 2x + y = 0 and x – 4y = 0.

Solution:

The given straight lines are

2x + y = 0 ... ... ... (i)

and x – 4y = 0 ... ... ... (ii)

The single equation of the line (i) and (ii) is

(2x + y) (x – 4y) = 0

or, 2x2 – 8xy + xy – 4y2 = 0

∴ 2x2 – 7xy – 4y2 = 0. This is the required combined equation.
Example 2: Find the equation of the straight lines represented by the equation 3x2 – 8xy + 4y2 = 0.

Solution: Co-ordinate

Given equation is 3x2 – 8xy + 4y2 = 0

or, 3x2 – 6xy – 2xy + 4y2 = 0

or, 3x(x – 2y) – 2y (x – 2y) = 0

or, (x – 2y) (3x – 2y) = 0

Either, x – 2y = 0 ... ... ... (i) or, 3x – 2y = 0 ... ... ... (ii) are required equations.
Example 3: Find the equation of the straight lines represented by the equation x2 + 4y2 + 4xy – 2x – 4y – 15 = 0.
Solution:

Here the given equation is

x2 + 4y2 + 4xy – 2x – 4y – 15 = 0.

or, (x + 2y)2 – 2(x + 2y) – 15 = 0

or, (x + 2y)2 – 5(x + 2y) + 3(x + 2y) – 15 = 0

or, (x + 2y)(x + 2y – 5) + 3(x + 2y – 5) = 0

or, (x + 2y – 5)(x + 2y + 3) = 0

Either x + 2y – 5 = 0 or, x + 2y + 3 = 0

Hence, the required equation of straight lines are

x + 2y – 5 = 0 and x + 2y + 3 = 0.
Example 4: Show that the equation x2 + 3xy + 2y2 + 4x + 5y + 3 = 0 represents a pair of straight lines.
Solution:

Here, the given equation is x2 + 3xy + 2y2 + 4x + 5y + 3 = 0.

Comparing this equation with general second degree equation

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

Equation of Pair of Straight Lines / 127

a = 1, 2h = 3 ⇒ h = 23, b = 2, 2g = 4 ⇒ g = 2, 2f = 5 ⇒ f = 5 and c = 3
2

The given equation represents a pair of straight lines if

abc + 2fgh – af 2 – bg2 – ch2 = 0

5 3 5 2 3 2
2 2 2 2
or, 1 × 2 × 3 + 2 × × 2 × – 1 × – 2 × 22 – 3 × = 0

or, 6 + 15 – 25 – 8 – 27 = 0
4 4

or, 21 – 8 – 25 + 27 =0
4

or, 13 – 52 = 0
4

or, 13 – 13 = 0

or, 0 = 0 which is true.

Hence, the given equation represents a pair of straight lines.

Example 5: Find the equation of the pair of straight lines represented by the equation
x2 + 3xy + 2y2 – 8x – 10y + 12 = 0.
Solution:

Here, the given equation is

x2 + 3xy + 2y2 – 8x – 10y + 12 = 0.

or, x2 + x(3y – 8) + (2y2 – 10y + 12) = 0 which is quadratic in x.

Now, using formula for quadratic equation,

x = – (3y – 8) ± (3y – 8)2 – 4.1(2y2 – 10y + 12)
2.1
= – (3y – 8) ± 9y2 – 48y + 64 – 8y2 + 40y – 48
or, x 2

or, 2x = – (3y – 8) ± y2 – 8y + 16

or, 2x = – (3y – 8) ± (y – 4)2

or, 2x = – (3y – 8) ± (y – 4)

Taking positive sign, Taking negative sign,
2x = – (3y – 8) + (y – 4) 2x = – (3y – 8) – (y – 4)
or, 2x = – 3y + 8 + y – 4 or, 2x = – 3y + 8 – y + 4

or, 2x = – 2y + 4 or, 2x = – 4y + 12

or, 2x + 2y – 4 = 0 or, 2x + 4y – 12 = 0

or, x + y – 2 = 0 or, x + 2y – 6 = 0

Hence, the required equations of straight lines are

x + y – 2 = 0 and x + 2y – 6 = 0.

128 / Optional Mathematics class 10

Example 6: Find the separate equation of the straight lines represented by the equation
x2 + 2xy sec q + y2 = 0.

Solution:

Here, the given equation is

x2 + 2xy sec q + y2 = 0

or, x2 + 2.x.(y sec q) + (y sec q)2 – (y sec q)2 + y2 = 0

or, (x + y sec q)2 – y2sec2q + y2 = 0

or, (x + y sec q)2 – y2(sec2q – 1) = 0

or, (x + y sec q)2 – y2 tan2q = 0

or, (x + y sec q)2 – (y tan q)2 = 0

or, (x + y sec q + y tan q) ( x + y sec q – y tan q) = 0

Either x + y sec q + y tan q = 0 i.e. x + y (sec q + tan q) = 0

or, x + y sec q + y tan q = 0 i.e. x + y (sec q – tan q) = 0

Hence, the required equation are Co-ordinate

x + y (sec q + tan q) = 0 and x + y (sec q – tan q) = 0.
Example 7: Find the area of the triangle formed by the lines 2y2 – 5xy + 2x2 = 0 and y = 2.
Solution:

The given equation is 2y2 – 5xy + 2x2 = 0

or, 2y2 – 4xy – xy + 2x2 = 0

or, 2y(y – 2x) – x(y – 2x) = 0

or, (2y – x)(y – 2x) = 0

Either, 2y – x = 0 ... ... ... (1)

or, y – 2x = 0 ... ... ... (2)

Solving (1) and (2), we get point of intersection is (0, 0).

As y = 2 is one of the side of the triangle, solving (1) with y = 2, we get the point of intersection is (4, 2).

Similarly, solving y = 2 with equation (2), we get the point of intersection (1, 2)

Hence, the vertices of triangle are (0, 0), (4, 2) and (1, 2).

Hence, the area of triangle = 1  |x1y2 – y1x2 + x2y3 – y2x3 + x3y1 – y3x1|
2

= 21 |0 – 0 + 8 – 2 + 0 – 0|

= 12  |6| = 1 × 6
2

= 3 sq. units

Hence, the area of the required triangle is 3 sq. units.

Equation of Pair of Straight Lines / 129

Exercise 4.3A

1. Find the single equation of the following pairs of straight lines.

a. x + y = 0 and x – y = 0 b. x + 2y = 0 and x + y = 0

c 2x + y = 0 and 2x – 3y = 0 d. 11x – 2y = 0 and 3x + 7y = 0

e. ax – by = 0 and bx = –  ay f. (a + b)x + (a – b)y = 0 and (a – b)x + (a + b)y = 0

2. Find the equations of the straight lines represented by the following equations.

a. 3x2 + 7xy + 2y2 = 0 b. x2 – 4xy + 3y2 = 0

c. x2 + xy – 2y2 = 0 d. x2 + 2xy – 3y2 = 0

e. ax2 + (ab + 1)xy + by2 = 0 f. p2x2 – p(q – 1)xy – qy2 = 0

g. ab(x2 – y2) + (a2 – b2)xy = 0 h. ab(x2 + y2) – xy(a2 + b2) = 0

3. Find the equations of the straight lines represented by the following equations.

a. x2 – 4xy + 4y2 – x + 2y = 0 b. x2 + y2 – 2xy + 2x – 2y = 0

c. x2 + 2xy + y2 + 5x + 5y + 6 = 0 d. x2 – 2xy + y2 – 3x + 3y + 2 = 0

Exercise 4.3B

1. Find the single equation of the following pairs of straight lines.

a. x + y + 2 = 0 and 2x + 3y + 4 = 0 b. 2x + y + 2 = 0 and x + 2y + 4 = 0

c. x + y + 3 = 0 and 2x + y – 1 = 0 d. 3x – 2y + 4 = 0 and 2x – y + 8 = 0

e. (sec θ – tan θ) x + y = 0 and (sec θ + tan θ) x + y = 0

2. Show that each of the following equations represents a pair of straight lines.

a. x2 – 2xy + y2 + 5x – 5y + 6 = 0 b. x2 + 2xy + y2 + 6x + 6y + 8 = 0

c. 2x2 – 5xy – 3y2 + 5x + 6y – 3 = 0 d. 2x2 + 7xy + 5y2 – 4x – 2y – 7 = 0

3. Find the equation of straight lines represented by the following equations.

a. x2 + 6xy + 9y2 + 4x + 12y – 5 = 0 b. 2x2 + xy – 3y2 + 10y – 8 = 0

c. 2x2 + 3xy + y2 + 5x + 2y – 3 = 0 d. x2 – 6xy + 8y2­+ 14x – 36y + 40 = 0

4. Find the separate equation of the pair of straight lines represented by the following equations.

a. x2 + 2xy cosec q + y2 = 0 b. x2 – 2xy sec a + y2 = 0

c. y2 + 2xy cot q – x2 = 0 d. (x2 + y2) sin a cos a + xy = 0

5. a. Show that the straight lines x2 + 4xy + y2 = 0 and x – y = 4 form an equilateral triangle.

b. Show that the lines x2 – 4xy + y2 = 0 and x + y = 1 form an equilateral triangle.

c. Find the area of the triangle formed by the lines y2 – 9xy + 18x2 = 0 and y = 9.

d. Show that the equations x2 – 7x + 6 = 0 and y2 – 14y + 40 = 0 form a rectangle. Also find the joint
equation of its diagonals.

e. Show that the 9x2 – 24xy + 16y2 – 12x + 16y – 12 = 0 represents a pair of parallel lines. Find the
distance between the lines.



130 / Optional Mathematics class 10

4.4 Angle between Pair of Straight Lines

Angles between Two Straight Lines Represented by ax2 + 2hxy + by2 = 0

Given equation is ax2 + 2hxy + by2 = 0

or, y2 + 2bh xy + ba x2 = 0 ... ... ... (i)

As, these two lines be y = m1x and y = m2x

i.e. y – m1x = 0 and y – m2x = 0

Product of these two equations

(y – m1x) (y – m2x) = 0
or, y2 – (m1 + m2) xy + m1m2x2 = 0 ... ... ... (ii)

Equations (i) and (ii) are identical, so equating the coefficients of xy and x2, we have

m1 + m2 = – 2h and m1.m2 = a
b b

Let θ be the angle between the lines, we have

tan θ = ± m1 – m2 = ± (m1 – m2)2 =± (m1 + m2)2 – 4m1m2 Co-ordinate
1+ m1m2 1 + m1m2 1 + m1m2

– 2bh 2 × a 4h2 – 4ab 2 h2 – ab
1 b
–4 b2 b
a+b a+b
=± a = =
b
+ bb

2 h2 – ab
=± a+b

∴ q = tan–1 ± 2 h2 – ab
a+b

Cor. 1, If the lines are parallel, or coincident i.e. θ = 0°.

2 h2 – ab
tan 0° = ± a + b

2 h2 – ab
or, 0 = ± a + b

or, 0 = 2 h2 – ab

or, 0 = h2 – ab

or, h2 = ab

Cor. 2, If the lines are perpendicular, i.e. θ = 90°.

2 h2 – ab
tan 90° = ± a + b

or, csoins 9900°° = ± 2 h2 – ab
a+b

or, 10 = ± 2 h2 – ab
a+b

i.e. a + b = 0

Angle between Pair of Straight Lines / 131

Example 1: Find the angle between the straight lines represented by 2x2 – 7xy + 5y2 = 0.
Solution:

Given equation is 2x2 – 7xy + 5y2 = 0

Comparing it with ax2 + 2hxy + by2 = 0

a = 2, 2h = – 7 i.e. h = – 7 and b = 5.
2

If θ be the angle between the straight lines,

7 2 2 49 – 10
2 = tan–1 ± 4
2 h2 – ab 2 – – 2×5
a+b = tan–1 ±
θ = tan–1 ± 2+5 7

2 49 – 40 2 9 2 × 3
= tan–1 ± 4 4 7 2
= tan–1 ± = tan–1 ±
7 7

Hence, θ = tan–1 ± 3 .
7

Example 2: Find the value of c if the line pairs represented by the equation 5x2 – 20xy + cy2 = 0 are

a. perpendicular to each other b. coincident

Solution:

Here given equation is 5x2 – 20xy + cy2 = 0
Comparing it with ax2 + 2hxy + by2 = 0, we get
a = 5, h = – 10, b = c
a. If the given pairs of lines are perpendicular to each other, then
a + b = 0
or, 5 + c = 0
∴ c = – 5
b. If the given pairs of lines are coincident, them
h2 = ab
or, (– 10)2 = 5c
or, 100 = 5c
∴ c = 20
Example 3: Find the equation of pair of lines represented by the equation of x2 – 2xy cosec θ + y2 = 0.

Also find the angle between them.
Solution:

The given equation is x2 – 2xy cosec θ + y2 = 0

or, y2 – 2xy cosec θ + x2 = 0

or, xy 2 – 2 y cosec θ + 1 = 0
x

132 / Optional Mathematics class 10

As this equation is quadratic in yx, it has two solutions, so
a = 1, b = – 2 cosec θ, c = 1, then

yx = –b± b2 – 4ab
2a

or, xy = + 2 cosec q ± (2 cosec q)2 – 4.1.1
2×1

or, xy = 2 cosec q ± 4 cosec2q – 4
2

or, yx = 2cosec q ± 2 cosec2q – 1
2

or, xy = 2 cosec q± 2 cot q
2
or, yx =
2(cosec q ± cot q)
2
or, xy = cosec θ ± cot θ

∴ y = (cosec θ ± cot θ)x Co-ordinate

This is the required equation of the pair of straight lines.

Now, for the angle between the line pair, comparing the given equation with ax2 + 2hxy + by2 = 0, we get

a = 1, h = – cosec θ, b = 1

±2 h2 – ab ± 2 (– cosec q)2 – 1.1
tan φ = a + b =
1+1

= 2 cosec2q – 1 = ± 2 cot q = ± cot θ
2 2

Hence, tan φ = ± tan (90° – θ)

Taking positive sign, we have

tan φ = tan (90° – θ)

∴ φ = 90° – θ

Taking negative sign, we have

tan φ = – tan (90° – θ) = tan {180° – (90° – θ)}

= tan (90° + θ)

∴ φ = 90° + θ

Hence, the required angles are 90° – θ and 90° + θ.

Example 4: Find the separate equation of straight lines represented by the equation
2x2 – 5xy – 3y2 + 5x – y + 2 = 0. Also find the angle between them.
Solution:

Here, the given equation is
2x2 – 5xy – 3y2 + 5x – y + 2 = 0
or, 2x2 + 5x(1 – y) – (3y2 + y – 2) = 0

Angle between Pair of Straight Lines / 133

This is quadratic in x, so {5(1 – y)}2 – 4.2{– (3y2 + y – 2)}
2×2
x = – 5(1 – y) ±

or, 4x = – 5(1 – y) ± 25(1 – 2y + y2) + 24y2 + 8y – 16

or, 4x = – 5(1 – y) ± 25 – 50y + 25y2 + 24y2 + 8y – 16

or, 4x = – 5(1 – y) ± 49y2 – 42y + 9

or, 4x = – 5(1 – y) ± (7y – 3)2 Alternately,

or, 4x = – 5(1 – y) ± (7y – 3) Here, the given equation is
Taking positive sign,
4x = – 5(1 – y) + (7y – 3) 2x2 – 5xy – 3y2 + 5x – y + 2 = 0
or, 4x = – 5 + 5y + 7y – 3
or, 4x = 12y – 8 Taking the equation with homogenous part only,
or, 4x – 12y + 8 = 0
or, x – 3y + 2 = 0 2x2 – 5xy – 3y2 = 0
Taking negative sign,
4x = – 5(1 – y) – (7y – 3) Comparing this equation with ax2 + 2hxy + by2 = 0,
or, 4x = – 5 + 5y – 7y + 3
or, 4x = – 2y – 2 a = 2, h = – 5/2, b = – 3
or, 4x + 2y + 2 = 0
or, 2x + y + 1 = 0 So, tan q = ± 2 h2 – ab .
a+b

=± 2 (–5/2)2 – 2×(– 3)
2–3

= ± – 2 25 + 24
4

= ± 2 × 7 = ± 7
2

\ q = tan–1(± 7)

The required equations are x – 3y + 2 = 0 ... ... (1) and 2x + y + 1 = 0 ... ... (2)

Now slope of line (1), m1 l=in31esa(n1d) slope of line (2), m2= – 2
If q be the angle between and (2), then

m1 – m2 1 – (– 2) 1+6 7 3
1+ m1 m2 3 3 3 1
tan q = ± = ± =± =± ×
1 3–2
1 + 3 × (– 2) 3

or, tan q = ± 7

Hence, the required angle between the lines is

q = tan–1(± 7)

Example 5: Find the equations of the pair of lines passing through the origin and perpendicular to the
lines represented by the equation 3x2 + xy – 10y2 = 0.

Solution:

The given equation is 3x2 + xy – 10y2 = 0.

or, 3x2 + 6xy – 5xy – 10y2 = 0

or, 3x (x + 2y) – 5y (x + 2y) = 0

or, (x + 2y) (3x – 5y) = 0

134 / Optional Mathematics class 10

Either x + 2y = 0 ... ... ... (i)

or, 3x – 5y = 0 ... ... ... (ii)

(i) and (ii) are the equations of two lines represented by given equation. Slope of (i) is m1 = – 12 and slope
of line perpendicular to (i) is m2 = 2.

So equation of line perpendicular to (i) and passes through origin is y = m2x

i.e. y – 2x = 0 ... ... ... (iii)

Slope of (ii) is (m3) = 3 and slope of line perpendicular to (ii) is m4 = – 1 = – 5 .
5 m3 3

So equation of line perpendicular to (ii) passes through origin is y = m4x

or, 3y + 5x = 0 ... ... ... (iv)

Combined equation of (iii) and (iv) is

(y – 2x) (3y + 5x) = 0

or, 3y2 – xy – 10x2 = 0

Which is the required equation. Co-ordinate

Example 6: Find the equations of the pair of lines passing through the (– 2, 3) and parallel to the lines
represented by the equation 2x2 + xy – 6y2 = 0.

Solution:

The given equation is 2x2 + xy – 6y2 = 0
or, 2x2 + 4xy – 3xy – 6y2 = 0

or, 2x(x + 2y) – 3y(x + 2y) = 0

or, (x + 2y)(2x – 3y) = 0

Either x + 2y = 0 ... ... (1) or, 2x – 3y = 0 ... ... (2)

Now, let the equation of any straight line parallel to (1) is x + 2y + k = 0 ... ... (3)
If the line (3) passes through (– 2, 3),
– 2 + 2 × 3 + k = 0

or, – 2 + 6 + k = 0 ⇒ k = – 4
Hence, the equation of straight line through (– 2, 3) and parallel to (1) is
x + 2y – 4 = 0 ... ... (4)

Again let the equation any line parallel to (2) is 2x – 3y + k = 0 ... ... (5)
If the line (5) passes through (– 2, 3),
2×(– 2) – 3 × 3 + k = 0

or, – 4 – 9 + k = 0 ⇒ k = 13

Hence, the equation of straight line through (– 2, 3) and parallel to (2) is 2x – 3y + 13 = 0 ... ... (6)
The combined equation of straight lines (4) and (6) is

(x + 2y – 4) (2x – 3y + 13) = 0

Angle between Pair of Straight Lines / 135

or, 2x2 – 3xy + 13x + 4xy – 6y2 + 26y – 8x + 12y – 52 = 0
or, 2x2 + xy – 6y2 + 5x + 38y – 52 = 0
This is the required equation parallel to 2x2 + xy – 6y2 = 0.
Example 7: Find the straight lines represented by the equation x2 + xy – 6y2 + 7x + 31y – 18 = 0

and find the point of intersection and angle between them.
Solution:
Given equation is x2 + xy – 6y2 + 7x + 31y – 18 = 0
or, x2 + x (y + 7) – (6y2 – 31y + 18) = 0

This is quadratic in x, so in general it has two roots

∴ x= – (y + 7) ± (y + 7)2 + 4(6y2 – 31y +18)
2

or, 2x = – (y + 7) ± y2 + 14y + 49 + 24y2 – 124y + 72

or, 2x = – (y + 7) ± 25y2 –110y + 121

or, 2x = – (y + 7) ± (5y – 11)2

[If expression under square root is not a perfect square then the given equation will not represent two
straight lines.]

2x = – y – 7 ± (5y – 11)

Taking positive sign, Taking negative sign,

Either 2x = – y – 7 + 5y – 11 or, 2x = – y – 7 – 5y + 11

or, 2x – 4y + 18 = 0 or, 2x + 6y – 4 = 0

or, x – 2y + 9 = 0 ... ... ... (i) or, x + 3y – 2 = 0 ... ... ... (ii)

Solving (i) and (ii), the required intersection point is – 523, 11
5
1 1
Slope of (i) m1 = 2 and that of (ii) m2 = – 3

If θ be the angle between (i) and (ii), 3+2
1 1 6
m1 – m2 2 + 3 5 6
1+ m1 m2 6–1 6 5
i.e., tan θ = ± = ± 1 1 =± 6 = ± × = ± 1
2 3
1 + × –

∴ θ = tan–1 (± 1) = 45°, 135°.

Exercise 4.4A

1. Prove that the two lines represented by the following equations are perpendicular to each other.

a. x2 – y2 = 0 b. 11x2 + 2xy – 11y2 = 0

c. x2 – y2 + x + y = 0 d. x2 – y2 + x – y = 0

e. (a – b) x2 + 2abxy + (b – a) y2 = 0

f. (sin2α – cos2β) x2 + 2 cos α sin β xy + (cos2α – sin2β) y2 = 0

136 / Optional Mathematics class 10

2. Prove that the two lines represented by the following equations are coincident.

a. 16x2 – 24xy + 9y2 = 0 b. 4x2 – 20xy + 25y2 = 0
d. 9x2 – 24xy + 16y2 = 0
c. x2 – xy + y2 = 0
4

3. a. Find k so that 5x2 – 8xy + ky2 = 0 may represent two mutually perpendicular lines.

b. If a pair of straight lines represented by an equation mx2 – 5xy – 6y2 = 0 are perpendicular to each
other, find the value of m.

c. If a pair of straight lines represented by 3x2 – 12xy + my2 = 0 are coincident, find the value of m.

d. If a pair of straight lines represented by 2x2 + 6xy + by2 = 0 are coincident, find the value of b.

e. If the equation 2x2 – 2hxy + 3y2 = 0 represents a pair of coincident lines passing through origin,
find the value of h.

Exercise 4.4B

1. Find the angle between the two lines represented by the following equations.

a. 3x2 + 7xy + 2y2 = 0 b. x2 – 4xy + y2 = 0 Co-ordinate

c. x2 – 5xy + 6y2 = 0 d. 2x2 – 3xy – 6y2 = 0

e. x2 + 2xy cot θ + y2 = 0 f. x2 – 2xy sec α + y2 = 0

2. Find the equation of the pair of straight lines represented by the following equations. Also find the
angle between them.

a. x2 + 2xy cosec a + y2 = 0 b. (x2 + y2) sin q cos q – xy = 0

3. a. Find the separate equation of the straight lines represented by the equation 3x2 + 5xy – 2y2­ +
9x + 11y + 6 = 0. Also find the angle between them.

b. Show that the equation x2 – y2 – x + 3y – 2 = 0 represents a pair of straight lines. Find these
lines, and also show that the lines are at right angle.

c. Show that the equation 2x2 – 13xy + 15y2 + 5x – 4y – 3 = 0 represents pairs of straight lines.
Also find the angle between these lines.

4. Find the equations of the pair of straight lines passing through the origin and perpendicular to the
lines represented by the equations.

a. x2 + 2xy – 3y2 = 0 b. 6x2 + 13xy + 6y2 = 0

c. 4x2 + 8xy + 3y2 = 0 d. 6x2 – 7xy – 3y2 = 0

5. a. Show that the two lines x2 – y2 + x + y = 0 are perpendicular to each other.

b. Show that the two straight lines x2 + 6xy + 9y2 + 3x + 9y = 0 are parallel.

6. Find the single equation of the lines through (2, 3) and perpendicular to the lines.

a. 3x2 – 8xy + 5y2 = 0 b. x2 + 3xy – 4y2 = 0

c. 2x2 + 7xy + 6y2 = 0 d. 4x2 – xy – 5y2 = 0

7. Find the equations of the pair of straight lines passing through the point (1, 3) and parallel to the lines
represented by the following equations.

a. x2 – 6xy + 8y2 = 0 b. 4x2 + 8xy – 5y2 = 0

c. 3x2 + 2xy – 8y2 = 0 d. 4x2 + 5xy – 6y2 = 0

Angle between Pair of Straight Lines / 137

8. a. Find the equation of the straight lines passing through the point (2, – 3) and perpendicular to the
lines represented by 3x2 + 5xy – 2y2 + 8x – 5y – 3 = 0

b. Find the equations of the straight lines passing through the point (1, 1) and parallel to the lines
represented by the equation x2 – 5xy + 4y2 + x + 2y – 2 = 0.

9 a. Show that the equation x2 – y2 + 4x – 2y + 3 = 0 represents a pair of straight lines. Find their
point of intersection and the angle between these lines.

b Find the straight lines represented by the equation 2x2 – 7xy – 4y2 + 13x + 38y – 70 = 0. Also
find the point of intersection and angle between them.



4.5 Conic Section (Circle)

Cone Vertex

A cone is a solid generated by rotating a straight line around a fixed Axis Vertex
line making a fixed angle with the fixed line. The revolving line is called the Generator Height angle

generator. The fixed line is called the axis of the cone, the fixed angle is called

the vertical angle and the point of intersection of the generator with the axis

which is fixed is called the vertex of the cone. Here, one end of the generator

is fixed and the other end describes a closed curve in a plane. The closed curve

is called the base of the cone. If the closed curve is a circle, the cone is called

the a right circular cone. The perpendicular distance from the vertex to the Base
base of the right circular cone is called the height of the cone. In a right cone,

the distance from the vertex to any point of the circumference of the base is called the slant height of the cone.

Double Napped Cone

When a straight line intersects a vertical

line at a fixed point, and rotates about that fixed line, Generator Upper
the surface obtained in called a double right circular nappe
cone. A double right circular cone consists of two

cones joined at a fixed point called a Vertex. The line Vertex
that rotates about the line is called the Generator

and the line that remains fixed is called the Axis. A Vertex Lower
right circular cone has a circular base and the axis is angle nappe
perpendicular to the circular base. The perpendicular
Axis

drawn from the vertex to the base of the cone is also

called the height of the cone. The lateral surface of

the right circular cone is called a Nappe. The double Base

right circular cone has two nappes. The nappe above

the vertex is called the Upper nappe and the nappe below the vertex is called the Lower nappe. The angle

between the generator and the axis is called the Vertex angle.

138 / Optional Mathematics class 10

Different Conic Sections

When a plane intersects the double napped cone, we get the intersection portion which is a two
dimentional curves of different types. These curves are called conic sections. In other words. A conic
section (or simple conic) is the intersection of a plane and a double napped cone. We note that the four
different types of conic sections on the basis of the angle made by the plane with the axis. These conic

sections are formed only when the the intersecting plane does not pass throught the vertex of the cone.

Circle Ellipse Parabola Hyperbola

q

q
q

a aa a

q = 90° q>a q=a q<a Co-ordinate

fig. (a) fig. (b) fig. (c) fig. (d)

When the plane is perpendicular, the axis of the cone (q = 90°), the contact of the plane with the cone
is a circle [fig.(a)]. The circle is a closed curve. When the angle between the plane and the axis of the cone
is greater than the vertex angle (q > a), the intersection of the plane and the cone is an ellipse [fig. (b)]. The
ellipse is a closed curve. When the angle between the plane and the axis of cone is equal to the vertex angle
(q = a), the intersection of the plane and the cone is a parabola [fig. (c)]. The parabola is an open curve.
Similarly, when the angle between the plane and the axis of the cone is less than the vertex angle (q < a),
the intersection of the plane and the cone is a hyperbola [fig. (d)]. In this case, the plane intersects both the
nappe as shown, so the hyperbola is two disjoint open curves.

Degenerate Conics

If the plane intersects the double right circular cone at its vertex, the circle and the ellipse become
a point [fig. (e)], the parabola becomes a line [fig. (f)] and the hyperbola becomes two intersecting lines
[fig. (g)]. The figure, so obtained are called degenerate conics.

a a a

fig. (e) fig. (f) fig. (g)

Conic Section (Circle) / 139

Now, we can conclude the following:
⇒ When a plane intersects a double right circular cone, two dimensional curves called conic sections

are formed. There are four types of conic sections : circle, ellipse, parabola and hyperbola.
⇒ When a plane intersects a double right circular cone so that the plane is perpendicular to the axis of

the cone, a circle is obtained.
⇒ When a plane intersects a double right circular cone so that the angle between the palne and the axis

is exactly equal to the vertex angle, a parabola is obtained.
⇒ When a plane intersects a double right circular cone so that the angle between the palne and the axis

is greater than the vertex angle, an ellipse is obtained.
⇒ When a plane intersects a double right circular cone so that the angle between the palne and the axis

is less than the vertex angle, a hyperbola is obtained.
⇒ When a plane intersects a double right circular cone at its vertex the figures formed are called

degenerate conics.

Equation of Circle Circle

When a right circular cone is intersected by a plane par- A O B
allel to the base of the cone, a circle is formed. C



The locus of a moving point which moves so that its distance from a fixed
point is constant is called a circle. The fixed point is known as the centre and the
fixed distance is known as the radius of the circle. Let O be the fixed point and
ABC a curve. Any point on the curve ABC is at equal distance from O i.e. OA =
OB = OC. The curve ABC is a circle. OA is called the radius of the circle and O is
its centre.

Equation of a Circle in Standard Form

The equation of a circle is a locus of a point which moves along the circumference of the circle.
There are the following standard forms of the equation of circle.

Equation of a Circle with Centre at Origin and Radius r

Let O(0, 0) be the radius of a circle. If P(x, y) be any Y
point on the circumference of the circle, then OP = r.
P(x, y)
By using distance formula,
r

OP2 = (x – 0)2 + (y – 0)2 X' O AX

∴ x2 + y2 = r2 ... ... ... (i)

The relation (i) is satisfied by every point on the circle, Y'
hence it represents the equation of circle with centre at origin
and radius r.

140 / Optional Mathematics class 10

Example 1: Find the equation of the circle with centre at origin and radius 5 units.
Solution:

Here, radius (r) = 5 units

The equation of a circle with centre at origin is

x2 + y2 = r2

or, x2 + y2 = 52

∴ x2 + y2 = 25 is the required equation. Y

Equation of a Circle with Centre at (h, k) and Radius r r P(x, y)
C(h, k)
Let C(h, k) be the centre and r be the radius of a circle. If
P(x, y) be any point on the circle, then CP = r.

By using distance formula,
CP2 = (x – h)2 + (y – k)2

or, (x – h)2 + (y – k)2 = r2 ... ... ... (i)

This relation is true for every point on the circle. Hence, it X' O X Co-ordinate
represents a circle with centre at (h, k) and radius r. Y'

Equation (i) can be written as

x2 – 2hx + h2 + y2 – 2ky + k2 – r2 = 0

or, x2 + y2 – 2hx – 2ky + h2 + k2 – r2 = 0

Let – 2h = 2g, – 2k = 2f and h2 + k2 – r2 = c then

or, x2 + y2 + 2gx + 2fy + c = 0 ... ... ... (ii)

The equation (ii) is said to be the General equation of circle. This equation of circle has the following
characteristics.

Note : The equation of the circle is a second degreee equation with equal coefficents of x2 and y2 and no
term containing xy. So, the general equation of the second degree in x and y i.e. ax2 + 2hxy + by2 +
2gx + 2fy + c = 0 represents a circle if a = b an h = 0.

Thus, the general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will represent a circle if
a = b and h = 0.

Example 2: Find the equation of a circle whose centre is at the point (–4, 3) and radius 4 units.
Solution:

Here, centre of circle is (– 4, 3)

i.e. h = – 4, k = 3 and radius r = 4.

The equation of circle with centre at (h, k) and radius r is

(x – h)2 + (y – k)2 = r2

or, (x + 4)2 + (y – 3)2 = 42

or, x2 + 8x + 16 + y2 – 6y + 9 = 16

∴ x2 + y2 + 8x – 6y + 9 = 0 is the required equation.

Equation of Circle / 141

Centre and Radius of a Circle

The general equation of a circle is

x2 + y2 + 2gx + 2fy + c = 0

or, x2 + 2.x.g + g2 + y2 + 2.y.f + f2 – g2 – f2 + c = 0

or, (x + g)2 + (y + f)2­= g2 + f2 – c
or, {x – (–g)}2 + {y – (–f)}2 = g2 + f2 – c 2

Comparing this equation with (x – h)2 + (y – k)2 = r2, we get

h = – g, k = – f and r = g2 + f2 – c

Hence, the centre and the radius of the circle are (– g, – f) and g2 + f2 – c respectively.

Example 3: Find the centre and radius of the centre x2 + y2 – 6x + 4y – 23 = 0.
Solution:

The given equation is
x2 + y2 – 6x + 4y – 23 = 0
or, x2 – 6x + y2 + 4y – 23 = 0
or, x2 – 2.x.3 + 32 + y2 + 2.y.2 + 22 = 23 + 32 + 22
or, (x – 3)2 + (y + 2)2 = 23 + 9 + 4
or, (x – 3)2 + (y + 2)2 = 62.
Comparing it with (x – h)2 + (y – k)2 = r2, we have h = 3, k = – 2 and r = 6
Hence, centre of the circle is (h, k) = (3, – 2) and radius r = 6.

Alternately,

The given equation is
x2 + y2 – 6x + 4y – 23 = 0
Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get
g = – 3, f = 2 and c = – 23
Hence, the centre of the circle (– g, – f) i.e. {– (– 3), – 2} i.e. (3, – 2) and

the radius is r = g2 + f2 – c = g2 + f2 – c

= (–3)2 + 22 – (–23) = 9 + 4 + 23 = 36 = 6

Equation of a Circle in a Diameter Form

Let O be the centre and A(x1, y1) and B(x2, y2) be the ends of diameter AB of the circle. If P(x, y) be
P(x, y)
any point on the circumference of the circle, then

slope of AP (m1) = y – yxyx2121.and B(x2, y2)
yx –– O
slope of BP (m2) = x –
Since, APB is the angle
in the semi-circle APB = 90°. A(x1, y1)

∴ m1 × m2 = – 1

142 / Optional Mathematics class 10

or, xy –– yx11 × y – y2 = – 1
x – x2

or, (y – y1) (y – y2) = – (x – x1) (x – x2)

or, (x – x1) (x – x2) + (y – y1) (y – y2) = 0

Hence, it is the equation of circle in diameter form.

Example 4: Find the equation of a circle whose ends of the diameter are at (– 1, 3) and (5, 7).
Solution:

Here ends of the diameter of the circle are (– 1, 3) and (5, 7).

i.e. x1 = – 1, y1 = 3, x2 = 5 and y2 = 7.
The equation of circle in diameter form is

(x – x1) (x – x2) + (y – y1) (y – y2) = 0
or, (x + 1) (x – 5) + (y – 3) (y – 7) = 0

or, x2 + y2 – 4x – 10y + 16 = 0 is the required equation.

Example 5: Find the equation of the circle passing through point (2, – 1) and which is concentric with Co-ordinate

the circle x2 + y2 + 2x – 6y = 0.

Solution: x2 + y2 + 2x – 6y = 0

Here, the given equation of circle is (2, – 1)

x2 + y2 + 2x – 6y = 0

or, x2 + 2.x.1 + 12 – 12 + y2 – 2.y.3 + 32 – 32 = 0 O
or, (x + 1)2 + (y – 3)2 = 32 + 12
or, (x + 1)2 + (y – 3)2 = 10 Eqn = ?

Comparing this equation with (x – h)2 + (y – k)2 = r2, we get the centre of the circle is (– 1, 3).

Now, we have to find the equation of the circle with centre at (– 1, 3) and passing through point (2, – 1), so
the radius of the required circle is

r = (2 + 1)2 + (–1 – 3)2

or, r = 32 + (–4)2 = 9 + 16

∴ r = 5

The equation of circle is

(x – h)2 + (y – k)2 = r2

or, (x + 1)2 + (y – 3)2 = 52

or, x2 + 2x + 1 + y2 – 6y + 9 = 25

∴ x2 + y2 + 2x – 6y = 15 is the required equation.

Equation of Circle / 143

Example 6: If the co-ordinates of one end of a diameter of the circle having equation x2 + y2 + 4x – 6y
+ 8 = 0 is (0, 2), then find the co-ordinates of the other end of the diameter.

Solution:

Here, the given equation is x2 + y2 + 4x – 6y + 8 = 0. x2 + y2 + 4x – 6y + 8 = 0
Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get (2, – 1)

2g = 4 ⇒ g = 2, 2f = – 6 ⇒ f = – 3, and c = 8 O

Hence, the centre of the circle is at (– g, – f) i.e. (– 2, 3). (?, ?)

As, one end of a diameter is at (0, 2), the centre (– 2, 3) of the circle is the mid-point of the diameter.
If (a, b) is the other end of the diameter using midpoint formula,

(– 2, 3) = 0 + a, 2 + b
2 2

i.e. – 2 = a ⇒ a = – 4, and 3 = 2 + b ⇒ b = 6 – 2 = 4
2 2

Hence, the co-ordinate of the other end of the diameter is (– 4, 4).

Example 7: Find the equation of a circle which passes through the point (1, 4) and the two diameters are
x – y = 1 and 2x + 3y = 7.

Solution:

As the point of intersection of two diameters of a circle is (1, 4) 2x + 3y = 7
the centre of the circle, solving two diameters of the given circle,

x – y = 1 ... ... ... (i) and 2x + 3y = 7 ... ... ... (ii) O
From, (i) and (ii),
2(1 + y) + 3y = 7 x–y=1

or, 5y = 5 ⇒ y = 1

and from (i), x – 1 = 1 i.e. x = 2

∴ Centre of the circle = (2, 1)

Radius of the circle (r) = (1 – 2)2 + (4 – 1)2 = 1 + 9 = 10.

Now, equation of the circle is

(x – h)2 + (y – k)2 = r2

or, (x – 2)2 + (y – 1)2 = ( 10)2

or, x2 – 4x + 4 + y2 – 2y + 1 = 10

∴ x2 + y2 – 4x – 2y – 5 = 0

Which is the required equation.

144 / Optional Mathematics class 10

Exercise 4.5A

Find the equation of the circle in each of the following cases. (Q. No. 1 to 3)

1. a. Centre at origin and radius 6 units

b. Centre at origin and passes through (3, 0)

c. Centre at origin and passes through (4, – 3)

d. Centre at origin and passes through (2, 4)

e. Centre at origin and passes through (a, a)

f. Centre at origin and passes through ( 2a, –a)

2. a. Centre at (– 1, – 3) and radius 3 units

b. Centre at (4, – 2) and diameter 10 units
11
c. Centre at (– 3, 5) and radius 3 units

d. Centre at (3, 4) and radius 5 units

e. Centre at (a, a) and radius 2a units.

f. Centre at (p, q) and radius p2 – q2 units Co-ordinate

3. a. Centre at (4, 0) and passes through (0, 3)

b. Centre at (2, 3) and passes through (5, – 1)

c. Centre at (– 1, 4) and passes through (7, – 2)

d. Centre at (a, b) and passes through (2a, b)

4. Find the equation of the circles when the co-ordinates of the ends of the diameter are

a. (2, – 3) and (– 3, 3) b. (3, 6) and (7, – 2)

c. (3, 2) and (– 1, 6) d. (3, 4) and (2, – 7)

e. (– c, – c) and (c, c) f. (a, 0) and (– a, 0)

5. Find the centre and the radius of the following circles.

a. x2 + y2 = 36 b. (x – 4)2 + y2 = 49

c. (x + 2)2 + (y – 3)2 = 50 d. x2 + y2 – 2mx – 2my = 0

e. x2 + y2 + 6y = 25 f. x2 + y2 – 6x + 2y + 1 = 0

6. Find the equation of the circle shown in the graph given below.

a. Y b. Y c. Y

A(3,3) C(1,3)

X' O X C(0,3) D(-5,1) Centre X
O X'
A(-3,1) X O
X'

Y' Y' Y'

Equation of Circle / 145

Exercise 4.5B

1. Find the centre and the radius of the following circles.

a. x2 + y2 – 3x – y – 13 = 0 b. 2x2 + 2y2 + 4x – 2y + 1 = 0
2 d. 4x2 + 4y2 – 8x + 24y + 32 = 0

c. 5x2 + 5y2 – 50x + 20y + 65 = 0

e. 4(x2 + y2) + 12px – 6py – p2 = 0 f. x2 + y2 – 2ax cos θ – 2ay sin θ = 0

2. a. If the centre of the circle x2 + y2 – ax – by – 12 = 0 is (2, 3), find the value of a and b.

b. If the radius of the circle x2 + y2 + ax + 8y + 9 = 0 is 4 units, find the value of a.

c. If the radius of the circle 4x2 + 4y2 + kx – 24y – 107 = 0 is 6 units, find the value of k.

d. If the circle x2 + y2 – x + 3y + 5 + k = 0 passes through (– 3, – 1), find the centre and radius of the
circle.

e. If the radius of the circle 2x2 + 2y2 – 2x + ky – 3 = 0 is 2 units, find the centre of the circle.

3. a. Find the equation of the circle concentric with the circle x2 + y2 – 6x + 8y – 3 = 0 and having

radius 5 units.

b. Find the equation of the circle concentric with the circle x2 + y2 – 4x – 4y – 8 = 0 and passing

through the point (9, 1). x2 + y2 – 2x – 4y – 44 = 0

c. Find the equation of the circle concentric with the circle x2 + y2 – 2x O
– 6y + 1 = 0 and passing through the point (4, – 1). A(–3,5)

d. In the given figure O is the common centre of both the circle. Find
the equation of the smaller circle.

4. a. Find the equation of the circle having its centre at (4, 6) and passing through the mid-point of the
line segment joining the points (– 1, 3) and (3, 1).

b. Find the equation of the circle having its centre at the mid-point of the line segment joining the
points (3, 7) and (1, – 1) and passing through the point (– 1, 7).

c. Find the equation of the circle with centre at (4, – 2) and passing through the centre of the circle
x2 + y2 – 6x + 10y – 15 = 0.

d. Find the equation of a circle passing through the point (– 3, – 4) and the centre as the point of

division of the line segement joining the points (2, – 3) and A x2 + y2 – 2x + 6y + 1 = 0
(5, 3) in the ratio 1:2.
B

e. In the given figure X and Y are the centres of two equal X(3,5) Y
circles. Find the equation of the circle A.

5. a. If the end of a diameter of the circle x2 + y2 – 4x – 6y – 24 = 0 is (8, 4), show that the co-ordinates
of the other are (– 4, 2).

b. If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), find the co-ordinates of
other end.

146 / Optional Mathematics class 10