Vedanta Opt. Maths Book 10 Final (2078)

Short Questions

1. Find the adjoint matrices of the following matrices:

(a) A= 2 3 (b) B= 6 7 (c) C= 10 5
4 5 3 2 23

2. Which of the following matrices have their inverse?

(a) A= 4 6 (b) B= 4 10
2 3 2 5

(c) C= 2 4 2 24
6 7 (d) D = 2 8

3

3. Find the inverse matrices of the following matrices if possible:

(a) A= 2 1 (b B= 6 2 (c) C= 1 2
7 4 9 3 4 7
cosecT cotT
(d) D= 2 7 (e) E= cosT - sinT (f) F = cotT
4 14 sinT cosT cosecT

4. Show that inverse of each of the following matrix is the matrix itself:

(a) 1 0 (b) 01 (c) 0 -1 (d) -1 0
0 1 10 -1 0 0 -1

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 139

vedanta Excel In Opt. Mathematics - Book 10

5. In each of the following, prove that each matrix is an inverse of other:

(a) 3 1 and 2 -1 (b) 2 3 and 8 -3
5 2 -5 3 5 8 -5 2

(c) 5 3 and -4 -3
-7 -4 7 5

Long Questions

6. (a) If P = x 1 and Q = 2 -1 are inverse to each other find the value of x.
5 2 -5 x

(b) If M = x -1 and N = 2 -1 are inverse to each other, find the values of x
and y. 5 -2 y -3

(c) If -1 2 and q 2 are inverse to each other, find the value of o and q.
2p -7 4 1

7. (a) If P = 2 0 and Q = 4 1 ,
5 3 7 2

(i) Find P–1 and Q–1 (ii) Find (PQ)-1 (iii) Show that (PQ)-1 = Q-1P-1

(b) If A 5 3 and B = 3 2 , show that (AB)-1 = B-1A-1.
3 2 4 3

(c) If M = 2 3 and N = 4 5 , show that (MN)-1 = N-1M-1
4 5 6 7

53

8. (a) If A = 3 2 , then, show following:

(i) (A-1)-1 = A (ii) (A-1)T = (AT)-1 (iii) A-1 A = AA-1 I

(b) If B = 4 1 , then, show the following:
7 2

(i) (B-1)-1 = B (ii) (B-1)T = (BT)-1 (iii) BB-1 = B-1B = I

9. For what value of x the product matrix 3 2 2 -1 does not have its inverse
matrix. x 4 3 2

1. (a) 5 -3 (b) 2 -7 (c) 3 -5
-4 2 -3 6 -2 10

2. C 3.(a) 4 -1 (b) B-1 does not exist.
-7 2

(c) -7 2 (d) D-1 does not exist
4 -1

(e) cosT sinT (f) cosecT - cotT
-sinT cosT -cotT cosecT

6. (a) 3 (b) x = 3, y = 5 (c) p = 2, q = 7
9. x = 6
7. (a) (i) P-1 = 1/2 0 , Q-1 = 2 -1
-5/6 1/3 -7 4

140 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

7.3 Solution of a system of linear Equation of in two variables by

Inverse Matrix Method

Let us consider two linear equations:

a1x + b1y = c1 ................. (i)

a2x + b2y = c2 ................. (ii)

Writing the above equations in matrix form,

a1x + b1y = c1
c2
( ( ( (a2x + b2y
( ( ( ( ( (or,a1 b1 x
a2 b2 y = c1
c2
or, AX = C ........... (iii)

a1( (where, A =b1 c1( ( ( (C = x
a2 b2 , c2 ,X= y

If A is a non - singular matrix, pre-multiplying (iii) by A-1, we get

A-1(AX) = A-1C

or, (A-1A) X = A-1C

or, I X = A-1C
or, X = A-1C,
? X = A-1 C
which gives the solution of above system of linear equations.
Note :
(a) A unique solution of given system of linear equations is possible if |A| ≠ 0.

If |A| = 0, A-1 does not exist and the solution of above system of equations is not
possible.
(b) Write the equation in the form of ax + by = c.

Steps in solving simultaneous equations of two variables by matrix method:
(a) Write the given equations in the form of ax + by = c , where a, b, and c are constants.

If coefficient of any variable is absent write it zero.
(b) Denote the coefficients of x and y by matrix A and constants in RHS by C and write in

the form AX = C.
(c) If |A| ≠ 0, find the inverse matrix of A., i.e. A-1
(d) Multiply by A-1 on both sides, we get, X = A–1C
Equate the corresponding elements in LHS and RHS which give the values of x and y.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 141

vedanta Excel In Opt. Mathematics - Book 10

Example : Solve by matrix method.
Solution:
4x - 3y = 11

3x + 7y = - 1

Writing the given equations in the matrix form.

4 -3 x 11
37 y = -1

or, AX = C

or, X = A-1C ..........(i)

Where, A = 4 -3 ,X= x 11
3 7 y = -1

|A| = 4 -3 = 28 + 9 = 37 ≠ 0.
3 7
Since |A| ≠ 0, A-1 exists and given system of linear equations have unique

solution.

Adj.A = 7 3 A-1 = 1 Adj.A = 1 7 3
-3 4 |A| 37 -3 4

From (i), we get,

X = 1 7 3 11
37 -3 4 -1

= 1 77 – 3 = 1 74 = 2
37 –33 – 4 37 -37 -1

? x = 2 ? x = 2, y = –1
y -1

Worked out Examples

Example 1. Write the given system of linear equation in matrix form:
Solution:
(a) 4x + 3y = 4 and 2x + 7y = 5

(b) 3y - x + 7 = 0 and 2x - 4 = y

(a) Here, given equation are

4x + 3y = 4

and 2x + 7y = 5

writing the above equations in the matrix form,

43 x = 4
27 y 5

(b) Here, given equation are

3y - x + 7 = 0

2x - 4 = y

142 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

These equation must be written in the form of ax + by = c

-x + 3y = -7

or, x - 3y = 7 .......... (i)

and 2x - y = 4 ........... (ii)

Writing the equations (i) and (ii) in the matrix form

1 -3 x = 7
2 -1 y 4

Example 2. Factorise 2x +3y
Solution: 4x - 7y
Example 3.
Here, 2x +3y = 23 x
Solution: 4x - 7y 4 -7 y

Solve by matrix method :

4x + 8y = 2

2x + 4y = 2

Given equations can be written as

4x + 8y= 2 or 2x + 4y = 1 ......(i)

and 2x + 4y = 2 or, x + 2y = 1 ..... (ii)

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

24 x = 1
12 y 1

or, AX = C ........... (i)

Where A = 2 4 ,C= 1 ,X= x
1 2 1 y

|A| = 2 4 =4-4=0
1 2

Since |A| = 0, A-1 does not exist and given system of equations do not have

unique solution.

Example 4. Solve by matrix method :
Solution:
3 + 4y = 9 2 +y= 8
x x 3

Writing the equations in the matrix form.

34 19
21 x
y = 8
3

or, AX = C ..............(i)

3 4 1 9
2 1 x 8
where, A = , X= y , C= 3

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 143

vedanta Excel In Opt. Mathematics - Book 10

|A| = 3 4 = 3 - 8 = -5 ≠ 0
2 1

Since |A| ≠ 0, A-1 exists and the given system of linear equations have unique

solution.

Adj. A = 1 -4
-2 3

A-1 = 1 Adj.A = 1 1 -4
|A| -5 -2 3

Now from (i), X = A-1 C = 1 1 -4 9
-5 -2 3 8
3

1 9 - 32 1 27-32
-5 3 -5 3
= =
-18 +234 -18 + 8

= 1 -5 1
-5 3 =3

-10 2

11
? x =3

y2

i.e. 1 = 1 and y = 2 ? x = 3 and y = 2
x3
Example 5.
Solution Solve by matrix method.

2 + 5 =1 and 3 + 2 = 19
x y x y 20

Writing the given equations in matrix form

25 1 1
32 x = 19
1
y 20

or, AX = C .............. (i)

where, A = 2 5 , X= 1 , C= 1
|A| = 3 2 x 19
1 20
y

2 5 = 4- 15 = - 11 ≠ 0
3 2

Since |A| ≠ 0, A-1 exists and the given system of linear equations have unique
solution.

Adj. A = 2 -5 , A-1 = 1 Adj. A. = 1 2 -5
-3 2 |A| -11 -3 2

144 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

Now, from (i)

X = A-1C = 1 2 -5 1
-11 -3 2 19
20

1 2 - 19 1 -141 1
11 4 11 -1110 4
=- =- = 1
-3 + 19 10
10
11
x 4
? 1 = 1

1 y 10
x 1
i.e. = 4 or, x=4
or, y = 10
and 1 = 1
y 10

? x = 4 and y = 10

Exercise 7.3 (b) B = 2x-4y

Short Questions
1. Factorize the given matrices:

(a) A = 4x+3y

x-y 5x + y

2. If A = a b , then answer the following questions:
c d

(a) Find determinant of matrix A..

(b) Under what condition A does not have its inverse.

3. If A, C, and X are three square matrices such that AX = C, write X in terms A-1 and C.

4. (a) If 1 0 X= 4 , find the value of X.
0 1 5

(b) If 32 X= 13 , find the value of X.
23 12

5. Check whether the system of linear equations have unique solution or not.

(a) x + y = 4 (b) 4x + 2y = 8 (c) 3x + 11y = 7

x-y=3 x-y=1 6x + 22y = 5

(d) x - 2y = 4 (e) 3x + 5y = 2 (f) 3 + 2 =1
x y
5 3
3x - 5y - 7 = 0 5y - 3x = 3 x - y =9

6. Solve for x and y in each of the following.

(a) 1 -8 x = -7 (b) 1 -2 x = -7
23 y 5 37 y 5

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 145

vedanta Excel In Opt. Mathematics - Book 10

(c) 57 x = 1 (d) 4 -3 x = 11
14 y -5 37 y -1

Long Questions

7. Solve the following system of linear equations by matrix method:

(a) x+ y = 5 (b) 9x - 8y = 12 (c) 3x + 5y = 21

x-y=3 2x + 3y =17 2x + 3y = 13

(d) x - 2y = -7 (e) 3x - 2y = 5 (f) 3x + y = 3

3x + 7y = 5 x+y=5 3x + 2y = 2

(g) 2x - 3y = 1 (h) 4x - 3y = 11 (i) x = 2y - 1

4y + 3x = 10 3x = 5 - y y = 2x

(j) 2x = 1 - y (k) 3x + 2y = 9 (l) 3x + 2y - 3 = 0

3x + 2y = 1 2x + 3y = 11 6x + 5y - 9 = 0

8. Solve the following equation by matrix method:

(a) x + y =4 (b) x + y = 7 (c) 5x + y = 5
3 3 2 2 2
x y 4
5x - 4y = - 3 2 + 3 = 3 3 x - y = -2

(d) x= 2 y (e) x - 2y = -7 (f) 3x + 2y = 1
3 2
3x 7y xy

4x - 3y = 1 5 + 5 =1 3 - 3 =1

9. Solve the following equations by matrix method:

(a) 4 + 3 =1 (b) 3 + 2 = 13 (c) 4 + 2y = 8
x y x y x
3 2 1 5 3 2
x - y = 24 x - y = 9 x + 3y = 10

(d) 3x + 4 = 10 (e) 3 + 1 = 5 (f) x + y =1
y x y 6 4
y
-2x + 3 = - 1 2 - 1 = 5 x - 4 =1
y x y 2
2 1 1
(g) x + y =2 (h) x + 2y =8 (i) 2x + 3y = 10xy
3 4 =5
y 1 1 5 4
x+ 2x - y = -1 x + y = 13

(j) 3y + 4x = 2xy (k) 3x + 5y = 7x + 3y = 4
4 5
7y + 5x = 29xy

(l) 2x + 4 =y= 40 - 3x (m) 2x + 5y = 1 = 3x - 2y
5 4 5
10. Equations of pair of lines are 2x - y = 5 and x - 2y = 1. Do the following:

(a) Write the equations in matrix form

(b) Is there unique solution of above given equations ?

(c) Solve the equations.

(d) check the solutions to show that the values x and y so obtained are true.

146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

1. (a) 43 x (b) 2 -4 x
1 -1 y 51 y

2. (a) ad - bc (b) |A| = 0 3. X = A-1C

4. (a) 4 (b) 3
5 2

5. (c) It does not have unique solution, others have solutions.

6. (a) (1, 1) (b) (-3, 2) (c) (3, -2) (d) (2, -1)
(d) (-3, 2)
7. (a) (4, 1) (b) (4, 3) (c) (2, 3) (h) (2, -1)
(l) (-1, 3)
(e) (3, 2) (f) 4 , -1 (g) (2, 1) (d) (-2, -3)
3
1 2
(i) 3 , 3 (j) (1, -1) (k) (1, 3)

8. (a) (5, 7) (b) (-10, 24) (c) 18 , 70
23 23

(e) (-3, 2) (f) (2, -1)

9. (a) (8, 6) (b) 1 , 1 (c) (2, 3) (d) (2, 1)
3 2 (g) (3, 2)
1 (k) (2, 2) 1 1
(e) 2 , -1 (f) (3, 2) (h) 6 , 4

(i) 1 , - 2 (j) 13 , - 13 (l) (8, 4)
7 11 106 73

(m) 15 , 13 10.(b) (3, 1)
19 19

7.4 Cramer’s Rule

We discuss the method of solving simultaneous equations consisting of two variables with
the help of determinants. It is also used to solve simultaneous equations with three variables.
But we limit it in solving simultaneous equations of two variables only.

Cramer's rule is the process of finding solution of simultaneous equations by determinant
method. It is also called determinant method of solving simultaneous equations of two or
three variables.

Solution of simultaneous equations with two variables by Cramer's Rule:
Let, a1x + b1y = d1 ........ (i)
and a2x + b2y = d2 ........ (ii)
be two linear equations, where x and y are variables and the remainings are constants.

Now, multiplying equation (i) by b2 and equation (ii) by –b1 and adding them, we get,
a1b2x + b1b2y = b2d1
–a2b1x – b1b2y = - b1d2

Adding, (a1b2 – a2b1)x = b2d1 – b1d2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147

vedanta Excel In Opt. Mathematics - Book 10

or, x = d1b2 – d2b1
a1b2 – a2b1
This equation can be put in the form of determinant

x= dd12 bb21 = D1
a1 b1 D

a2 b2

Where, D1 = add1 21 bbb121 , than D z 0, where D2 = aa21 dd21
Similarly, D= DDa22 , b2
y= provided

Note :

(i) D1 is also written as '1 or Dx

(ii) D2 is also written as '2 or Dy

(iii) If D = 0, there is no solution of given system of equations.

Steps to be used in Cramer's rule

The following steps are to be used in finding the solution of simultaneous linear equations

of two (or three) variables by using Cramer's rule:

(a) Arrange the given system of equations such that the terms containing the variables x
and y on left hand side and constant on the right hand side.

(b) Arrange the system of equations such that the term containing the variable x in the
first, then y.

(c) If any variable is missing in the equation, write the variable with coefficient zero.

(d) Collect the coefficients of the variables x and y, and the constant term and put them in
order.

(e) Write the determinant D with the elements of the coefficients of x and y.

(f) Write tfhirestdvetaerrimabilnea)nbtyDth1 ferocmonDstabnyt replacing the first column of D (i.e., the coefficient
of the column.

Similarly write determinant D2 replacing the second column of D by the constant column.

(g) The, use the formula:

x= D1 , y = D2
D D

Provided that D z 0.

Note :
If D = 0, there is no solution of given system of equations.

148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

Worked out Examples

Example 1: Solve the following simultaneous equations by using Cramer's rule:
Solution: x+y=5
x-y=3
Here, x + y = 5

x-y=3
Writing the coefficients of x, y and constants.

coefficient of x coefficient of y constant
5
11 3

1 -1

Now, D = 1 1 = -1 – 1 = -2
1 -1
5 1
D1 = 3 -1 = -5 – 3 = -8

D2 = 1 5 = 3 – 5 = -2
1 3

Using Cramer's rule,

x = D1 = -8 = 4

D -2
DD2 -2
y= = -2 =1

? x = 4 and y = 1

Example 2: Solve the given equations by using Cramer's rule:
Solution:
2x – 3y = 3 4x – y = 11

Here,

coefficient of x coefficient of y constant
3
2 -3 11

4 -1

Now, D = 2 –3 = –2 + 12 = 10
4 –1

D1 = 3 –3 = – 3 + 33 = 30
11 –1

D2 = 2 3 = 22 – 12 = 10
4 11

By using Cramer's rule,

x= DD1 = 30 =3
10

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149

vedanta Excel In Opt. Mathematics - Book 10

y= D2 = 10 =1
D 10

? x = 3 and y = 1.

Example 3: Solve the given simultaneous equations by using determinant method.
Solution: 2(x + y) = 3x
7x = 2(3y + 7)
From the first equation,

2x + 2y = 3x
or, 2x – 3x + 2y = 0
? –x + 2y = 0 ........... (i)
From the second equation,

7x = 6y + 14
or, 7x – 6y = 14 ............ (ii)
Now,

coefficient of x coefficient of y constant
-1 2 0
7 -6 14

–1 2

Now, D = 7 –6 = 6 – 14 = –8
0 2
D1 = 14 –6 = 0 – 28 = –28

D2 = –1 0 = –14 – 0 = –14
7 14

By using Cramer's rule, we get

x= DD1 = –28 = 7
–8 2
D2 –14 7
y= D = –8 = 4

? x = 7 and y = 74.
2

Example 4: Solve the given simultaneous equations by using cramer's rule.
Solution: 3x + 4y – 4 = 0
2x + 5y + 7 = 0
From the first equation,

3x + 4y = 4
and from the second equation

2x + 5y = –7

150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

Now,

coefficient of x coefficient of y constant

344

2 5 -7

Now, D = 3 4 = 15 – 8 = 7
2 5
4 4
D1 = –7 5 = 20 + 28 = 48

D2 = 3 4 = –21 – 8 = –29
2 –7

By using Cramer's rule,

x= D1 = 48
D 7
D2
y= D = –29
7
48 -29
? x= 7 and y = 7

Example 5: A marketing boy sales 10 kg of rice, 12 kg of tea at cost Rs 1460 on Sunday.
Solution: On Monday he sales 5 kg of rice and 7 kg of sugar at cost Rs 810. Find the
costs of each kg of rice and sugar. (Use Cramer's rule)

Let Rs x and Rs y be the prices of per kg of rice and sugar respectively.

Then, we have,

10x + 12y = 1460
or, 5x + 6y = 730 ........ (i)
and 5x + 7y = 810 ........ (ii)
Now, to solve above equations using Cramer's rule,

coefficient of x coefficient of y constant

5 6 730

5 7 810

Now, D = 5 6 = 35 – 30 = 5
5 7
730 6
D1 = 810 7 = 5110 – 4860 = 250

D2 = 5 730 = 4050 – 3650 = 400
5 810

By using Cramer's rule,

x= D1 = 250 = 50
D 5
D2 400
y= D = 5 = 80

? The costs of per kg of rice and sugar are Rs 50 and Rs 80 respectively.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 151

vedanta Excel In Opt. Mathematics - Book 10

Exercise 7.4

Very Short Questions :

1. (a) In equations a1x + b1y = c1 and a2x + b2y = C2, what are the determinants
representing D, Dx and Dy?

(b) If D = 20, Dx = 20, Dy = 40, write the values of x and y.

(c) If D = 4, Dx = 1 , Dy = 1 , find the values of x and y.
2 4

Long Questions :

2. Solve the following system of equations by using Cramer's rule:

(a) 3x – 2y = 1 and –x + 4y = 3

(b) 3x – 5y = 2 and 2x + y = 4

(c) 2x – 3y = 2 and 4x – y = 1

(d) 5x – 4y = –3 and 7x + 2y = 6

(e) 2x + 3y = 17 and 9x – 8y = 12

(f) 2x – y = 1 and 3x + y = 9

(g) 2x – 3y = 3 and 4x – y = 11

(h) 3x + 5y = 21 and 2x + 3y = 13

(i) 5x + 8y = 340 and 7x + 6y = 320

(j) 5x – 3y = 8 and 9x + 7y = 126

3. Solve the following system of equations by using Cramer's rule:

(a) 3 + 2 = 1 and 4 + 3 = 17
x y x y 6
3 5 4 3 29
(b) x + y = 1 and x + y = 30

(c) x – y = 2 and x + y = 59
3 4 8 3 24
2 5 3 6
(d) x + y = 30 and x + y = 27

(e) x – y = –6 and 3x – 1 = y
6 4

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

(a) 2x – 5 = 28 and 4x + 3 = –9
y y
10 4
(b) x – 2y = –1 and x + 3y = 11

(c) 8 – 9 = 1 and 10 + 6 = 7
x y x y

(d) 2(3x – y) = 5(x – 2) and 3(x + 4y) = 2(y – 3)

(e) 7(x – y) = x + y and 5(x + y) = 35 (x – y)

152 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

5. A helicopter has 4 seats for passengers. Those willing to pay first class fares can take 60
kg of baggage each, but tourist class passengers are restricted to 20 kg each. It can carry
only 120 kg of baggage altogether. To find the number of passengers of each kind, use
Cramer's rule.

6. If the total cost of 6 kg potatoes and 4 kg of tomatoes is Rs. 200 and the total cost of 1 kg
of potatoes and 1 kg tomatoes is 42. Find the price of per kg of each vegetable by using
determinant.

Project Work

7. Write two simultaneous equations related to cost of two daily used commodities and
solve them by using Cramer's rule.

1. (a) D = aa12 bb12 , Dx = cc12 bb21 , Dy = aa12 cc12 (b) (1, 2) (c) 1 , 1
8 16

2. (a) (1, 1) (b) 22 , 8 (c) 1 , - 3 (d) 9 , 51
13 13 10 5 19 38

(e) (4, 3) (f) (2, 3) (g) (3, 1) (h) (2, 3)

(i) (20, 30) (j) (7, 9) 3.(a) - 3 , 2 (b) (6, 10)
(c) (9, 4) (d) - 1 , 1 8 9

(e) (12, 8)

4. (a) 3 , - 1 15 12 (c) (2, 3) (d) -7, 3
2 5 (b) (2, 3) 6. (16, 26) 2
5. (1, 3)
(e) No solution

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 153

vedanta Excel In Opt. Mathematics - Book 10 8

Co-ordinate Geometry

8.0 Review B(x2,y2)

Let us review some formulae of coordinate geometry which we studied
in previous class.

(i) Distance Formula :
The distance between two points A (x1, y1) and B(x2, y2) is A(x1,y1)
given by

d = AB = (x2 - x1)2 + (y2 - y1)2 B(x2, y2)
(ii) Section Formula:
m2

(a) Internal Division : If P divides the join of A(x1, y1) and m1 P(x, y)
B(x2, y2), in ratio of m1 : m2 internally, then, coordinates
of P are given by

( (P(x,y) = P A(x1, y1)
m1x2 +m2x1 , m1y2 +m2y1
m1 + m2 m1 + m2

( (If the ratio is k:1, then, P(x,y) = P
kx2 +x1 , ky2 +y1
k+ 1 k+ 1

(b) External Division : If P divides the joining of A(x1, y1) and B(x2, y2) in m1 : m2 ratio,
externally, then, the coordinates of P are given by,

( (P (x, y) = P P(x , y)
m1x2 - m2x1 , m1y2 - m2y1 m2
m1 - m2 m1 - m2

If the ratio is k:1, them m1 B(x1, y1)

( (P (x, y) = P
kx2 - x1 , ky2 - y1
k-1 k-1

(c) Mid point formula : If P divides AB in two equal A(x1, y1)
parts, then, P is the midpoint of AB, the coordinates of P are given by,

( (P(x, y) = P
x1 + x2 , y1 + y2 A(x1, y1) P(x, y) B(x2, y2)
2 2

154 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

(iii) Slope formula : The slope of line segment joining two B(x2,y2)

points A(x1, y1) and B(x2,y2) is given by m = y2 - y1
If AB makes an angle T with positive X-axis, x2 - x1

A(x1,y1) A(x1, y1)

then (m) = tanT F 2E

G

(iv) Centroid of triangle : The point of intersection of B(x2, y2) 1
medians of triangle is called centroid. In the given D C(x3, y3)
figure,G is the centroid of triangle
Y P(x, y)
x1 + x2 + x3 y1 + y2 + y3
G(x,y) = G 3 , 3

(vi) Equation of straight line in slope- intercept. BT

c

In the figure, straight line AB has, y-intercept = OB = c X T X’
AO

Y’

Equation of the line is y = mx + c. Q Y
B(0, b)

(vii) Equation of straight line in double intercept form.

b

x-intercept of PQ =OA = a A(a, 0)
X
y - intercept of PQ = OB = b X' O a P

Equation of PQ is x + y = 1. Y' Y
a b M B

(viii) Equation of straight line in normal form P
p
perpendicular distance of MN from origin = OP = p
X' O D A X
Angle made by OP with OX is ‘AOP = D Y' N
Equation of straight line MN is x cosD + y sinD = p
Y N

(ix) Equations of straight line in point slope form P (x1, y1)
slope of MN = (m) = tanT
MN passes through the point P(x1, y1) X' T X
Equation of MN is y - y1 = m (x - x1) N
O

M

Y'

Y

(x) Equation of straight line in two points form Q(x2, y2)
P (x1, y1)
Equation of straight line joining two points P(x, y) and X'
Q(x2, y2) is M OX
Y'
y - y1 = y2 - y1 (x - x1)
x2 - x1 155

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

(xi) Perpendicular distance of a point from a line. P(x1, y1)

The perpendicular distance of the P(x1, y1) from the line A Q B

AB with equation ax + by+ c = 0 is d = ax1 + by1 + c straight line of the form
a2 + b2

(xii) Discuss how to change the linear equation of
ax + by + c = 0 into following three standard forms:

(a) y = mx + c (b) x + y = 1 (c) xcosD + ysinD = p
a b

(xiii) Area of triangle : Let A(x1, y1), B(x2, y2) and C(x3, y3), be three vertices of ∆ABC. Then area
of ∆ABC is given by,

∆= 1 |x1y2 - x2y1+ (x2y3 - x3y2) + (x3y1- x1y3)|
2

8.1 Angle between two straight lines

Let us consider the following three case of two straight line l1 and l2.
(i) l1

O
T

l2
Here, two straight lines l1 and l2 intersect at O and four angles are formed.
(ii) l1

O l2

Here, two straight lines l1 and l2 are perpendicular to each other.
(iii) l1

l2

Two lines l1 and l2 are parallel to each other. Y

The angle between them is zero. QN

To find the angle between two lines y = m1x + c1 y = m2x + c2 S S–T
T y = m1x + c1
and y = m2x + c2
Let MN and PQ be two straight lines with equations X' T T X
y = m1x + c1 and y = m2x + c2. OA B

Let the lines intersect at S. The lines MN and PQ cut MP
OX at A and B respectively making angles T2, and T
with the positive direction of X-axis. Y'

156 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

Then, slope of line MN (m1) = tanT, slope of line PQ (m2) = tanT2
Let the one angle between MN and PQ be T, i.e. and ‘MSP = T

Then, by plane geometry,

T1 = T + T2 [ ? Exterior angle of a triangle is sum of two opposite interior angles]
or, T = T1 - T2
? tanT = tan(T1 - T2)

tanT1 - tanT2
? tanT = 1 + tanT1.tanT2m1 - m2

( (‘MSP = T = tan-1 1 + m1. m2 ...... (i)

Also, ‘PSN is angle between the lines

‘PSN = 180° - T

? tan (‘PSN) = tan (180° - T) = - tanT = - m1 - m2
m1 - m2
( ( ‘PSN = tan-1 - 1 + m1.m2 ....... (ii) 1 + m1.m2

combining above (i) and (ii), the angle between

( (m1 - m2

MN and PQ = tan-1 ± 1 + m1.m2
This relation gives an angle between the two lines. Another angle is its supplement.

m1 - m2
If 1 + m1.m2 is positive, the acute angle between the two lines is obtained. Otherwise

it gives the obtuse angle.

?

( (Note :
T = tan-1 m1 - m2
± 1 + m1.m2

(a) If the value of tanT is positive, T is acute angle.

(b) If the value of tanT is negative, T is obtuse angle.

(c) If the value of tanT is zero, the lines is parallel or coincident.

(d) If the value of tanT is undefined (f) when the value of T is 90°.

To find the condition that two lines y = m1x + c1 and y = m2x + c2 are parallel.
If two lines y = m1x + c1 and y = m2x + c2 are parallel, to each other then angle between
them is °.

i.e. T = 0°
or, 0ta=n01° m=+11m-m+m11m1m-2m1Ÿm1 2m1 - m2 = 0
? m1 = m2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 157

vedanta Excel In Opt. Mathematics - Book 10

Therefore, two lines are parallel if their slopes are equal or if the lines are parallel their
slopes are equal.

To find the condition that two lines y = m1x + c1 and y = m2x + c2 are perpendicular
If two lines are perpendicular, the angle between them is 90°

i.e, T = 90°

or, tcsaoinns999000°°°==1m1+m1+m-1 mm-1mm11m21 2 or, 1 = m1 - m1
or, 1 + m1 . m2 = 0 0 1+ m1m2

or, m1 . m2 = –1

or, m1m2 = –1,

Therefore, two lines are at right angles if the product of their slopes is equal to –1.

To find the angle the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Equations of two lines are:

a1x + b1y + c1 = 0......... (i)

a2x + b2y + c2 = 0 ........ (ii)

slope of line (i), m1 = – a1
slope of line (ii), m2 = - ba12
b2
Let T be the angle between the two lines,

( ( ( ((( ((tanTm1 - m2 - a1 – - a2
+ m1.m2 b1 b2

then = ± 1 =± a1 a2
? b1 b2
1+ - -

( ( ( (= ±
a1b2 – a2b1 =± a1b2 – a2b1
b1b2 + a1a2 a1a2 + b1b2
( (T tan-1
± a1b2 - a2b1
a1a2 + b1b2
( (Therefore, the angle between two lines is tan–1
± a1b2 - a2b1 .
a1a2 + b1b2

To find the condition for parallelism of two lines a1x + b1y + c1 = 0

and a2x + b2y +c2 = 0
a1b2 - a2b1
We have, tanT = a1a2 + b1b2

For parallelism, T = 0°, tan0° = a1b2 - a2b1
a1a2 + b1b2
a1 b1
or, a1b2 - a2b1 = 0 ? a2 = b2

158 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

or, a1b2 = a2b1
Therefore, the two lines are parallel if the coefficients of x and y are proportional.

To find the condition for perpendicularity or orthogonality of two lines

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
a1b2 - a2b1
We have, tanT = a1a2 + b1b2

cotT = a1a2 + b1b2
a2b1 - a1b2
If T = 90° then cot90° = 0

or, cot90° = a1a2 + b1b2 or, 0= a1a2 + b1b2
a2b1 - a1b2 a2b1 - a1b2
? a1a2 + b1b2 = 0

Therefore, two lines are at right angles if the sum of the product of coefficients of x and
product of the coefficient of y is 0.

To find the equation of a line parallel to ax + by + c = 0

The given equation of line is ax + by + c = 0 ......... (i)

Slope of line (i), m1 = - coefficient of x = - a
coefficient of y b

Let the equation of a line parallel to (i) be y = mx + c ........ (ii)

Slope of line (ii), m2 = m

Since two lines (i) and (ii) are parallel m1 = m2

or, a = m
b
Putting the value of m in equation (ii),

y = - a x + C
b
or, by + ax + (-bc) = 0

or, ax + by + k = 0, where k = -bc.

Hence, the required equation of straight line parallel to (i) is ax + by + k = 0

Therefore, we change only the constant term in the given equation to get the equation of any
line parallel to given line.

Equation of a straight line perpendicular or orthogonal to ax + by + c = 0.

The equation of given line is ax + by + c = 0......... (i)

Its slope (m1) = - coefficient of x =- a
coefficient of y b

Let the equation of a straight line perpendicular to (i) be y = mx + c........ (ii)

Its slope (m2) = m
Since two lines (i) and (ii) are perpendicular m1.m2 = -1

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 159

vedanta Excel In Opt. Mathematics - Book 10

( (or, - a m=- 1 ? m = b
b of m in equation y= a
Putting the b
value (ii), we get a x+c

or, ay = bx + ac

or, bx - ay + ac = 0

or, bx - ay + k = 0 Where k = ac

Therefore, the required equation of straight line perpendicular to (i) is bx - ay + k = 0.

Therefore, we interchange the coefficients of x and y in given equation and sign of any one
of them is also changed. The constant term of given equation is also changed.

Worked out Examples

Example 1. Find the acute angle between the lines:
Solution:
y - (2 + 3 ) x - 5 = 0 ................(i)
Example 2.
Solution: y - (2 - 3 ) x - 2 = 0 ................. (ii)

160 Slope of line (i) m1 = - coefficient of x
coefficient of y

= (-) (2 + 3 ) =2+ 3
1 3
coefficient of x
Slope of line (ii), m2 = - coefficient of y = 2 -

Let T be the angle between given lines. Then

tanT = ± (m1 - m2)
1+ m1m2

2+ 3 -2+ 3
= 1+ (2 + 3 ) (2 - 3 )

= ± 2 3 =± 2 3 =± 3
1+ 4 -3 2

For acute angle we take positive sign only, we get

tanT = 3

or, tanT = tan60°

? T = 60°

Therefore, the required angle is 60°.

Find the obtuse angle between the given lines. ( i.e., y = mx form)
y = 3 x and 3 x + y + 3 = 0
Given equations of the lines are,

y = 3 x ........(i)
3 x + y + 3 = 0.......... (ii)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

slope of line (i), m1 = 3

slope of line (ii), m2 = - 3 =- 3
1
Let T be the angle between the given lines. Then

tanT = ± (m1 - m2) =± 3 +3 3) =± 2 3
1+ m1m2 1+( 3 ) (- 1-3

=±2 3 =±( 3)
-2
For obtuse angle, we take negative sign,

tanT = - 3

or, tanT = tan 120° ? T =120°

Therefore, the required angle is 120°

Example 3. Show that the lines 4x - 8y + 7 = 0 and x - 2y + 8 = 0 are parallel to each
Solution: other.

Example 4. The given equations of lines are 4x - 8y + 7 = 0 ......... (i)
Solution:
x - 2y + 8 = 0 ......... (ii)
Example 5.
Solution: Slope of line (i), m1 = - coefficient of x = - 4 = 1
coefficient of y -8 2

Slope of line (ii), m2 = - coefficient of x = - 1 = 1
coefficient of y -2 2

? m1 = m2

Therefore, the given lines are parallel to each other.

Show that the lines 2x + y + 2 = 0 and x - 2y + 3 = 0 are perpendicular to
each other.

The equations of given lines are

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

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

Slope of line (i), m1 = - coefficient of x = – 2 = –2
coefficient of y 1

Slope of line (ii), m2 = - coefficient of x = – 1 = 1
coefficient of y -2 2
1
Here, m1.m2 = - 2 × 2 =-1

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

If the lines 2x + 3y = 5 and kx + y = 2 are parallel to each other. Find the
value of k.

The equation of given lines are :

2x + 3y = 5 ............ (i)

kx + y = 2 ............ (ii)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 161

vedanta Excel In Opt. Mathematics - Book 10

slope of line (i), m1 = - coefficient of x = 2
coefficient of y 3
coefficient of x k
slope of line (ii), m2 = - coefficient of y = – 1 =-k

Since, the lines (i) and (ii) are parallel, we have

m1 = m2

or, - 2 =-k
3
2
or, k = 3

Therefore, the value of k is 2 .
3

Example 6. If the lines 3x + 4y - 7 = 0 and kx - 3y + 5 = 0 are perpendicular to each
Solution: other, find the value of k.

Example 7. The equations of given lines are,
Solution:
3x + 4y + 7 = 0 .......(i)
Example 8.
Solution: kx - 3y + 5 = 0 ......... (ii)

slope of line (i), m1 = - coefficient of x =- 3
coefficient of y 4
coefficient of x k k
slope of line (ii), m2 = - coefficient of y = - -3 = 3

since the lines (i) and (ii) are perpendicular to each other, m1.m2 = -1

or, - 3 × k =-1 or, k = 4
4 3
Therefore, the value of k is 4.

If the line joining the points (-1, k), and (5,6) is perpendicular to
2x + y + 7 = 0, find the value of k.

The equation of given line is 2x + y + 7 = 0

Its slope (m1) = - coefficient of x = - 2 = -2
coefficient of y 1

slope of line joining points (-1, k) and (5,6)

(m2) = y2 - y1 = 6-k = 6-k
x2 - x1 5+1 6
Since the lines are perpendicular, m1.m2 = -1

or, -2 × 6-k =-1 or, 6 - k = 0
2
? k=6

Find the equation of straight line which is parallel to the line
4x -3y + 12 = 0 and passing through the point (1, 2).

Equation of given straight line is 4x - 3y + 12 = 0......... (i)

Its slope is (m1) = - coefficient of x = - 4 = 4
coefficient of y -3 3

Equation of the line passing through point (1, 2) is y - y1 = m2(x - x1), where

162 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

m2 is slope.

or, y - 2 = m (x - 1)........... (ii)

Since line (ii) is parallel to (i), their slopes are equal m1 = m2 = 4
3
Putting the value of m2 is equation (ii)

y - 2 = 4 (x - 1)
3
or, 3y - 6 = 4x - 4

? 4x - 3y + 2 = 0

Therefore the required equation is 4x - 3y + 2 = 0.

Alternative Method

The equation of given straight line is 4x - 3y + 12 = 0 .............(i)

The equation of any straight line parallel to (i) is 4x - 3y + k = 0....... (ii)

The line (ii) passes through the point (1, 2)

4.1 - 3.2 + k = 0

or, 4 - 6 + k = 0

? k=2

Putting the value of k in equation (ii) we get, 4x - 3y + 2 = 0

Therefore, the required equation is 4x - 3y + 2 = 0.

Example 9. Find the equation of straight line perpendicular to the line
Solution: 2x - 3y + 4 = 0 passing through the point (2, 1).

The equation of given straight line is 2x - 3y + 4 = 0 .......... (i)

Its slope (m1) = - coefficient of x
coefficient of y
2
= - 2 = 3
(-3)
The equation a line passing through (2, 1) is y - y1 = m2(x - x1),

where m2 is slope

or, y - 1 = m2 (x - 2) ............... (ii)

Since the line (i) is perpendicular to (ii), we get

m1m2 = -1

or, 2 m2 = - 1
3
3
? m2 = - 2

Putting the value of m1 in equation (ii), we get

y-1=- 3 (x - 2)
2
or, 2y - 2 = - 3x + 6

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 163

vedanta Excel In Opt. Mathematics - Book 10

or, 3x + 2y = 8

Therefore, the required equation is 3x + 2y = 8.

Alternative Method:

The equation of given line is 2x - 3y + 4 = 0 ........ (i)

Equation of any line perpendicular to (i) is 3x + 2y + k = 0 .......... (ii)

where k is a constant to be determined.

The line (ii) passes through the point (2, 1)

3.2 + 2.1 + k = 0

or, 6 + 2 = - k ? k=-8

Putting the value of k in equation (ii), we get,

3x + 2y - 8 = 0

or, 3x + 2y = 8

Therefore, the required equation is 3x + 2y = 8.

Example 10. Find the equation of a straight line passing through the point (-2, 5) and
perpendicular to the line joining (3, 4) and (2, 8).

Solution: The equation of a line passing through the point (-2, 5) is

y - 5 = m1 (x + 2)......... (i)

where m1 is the slope of line (i)

slope of line joining the points (3, 4) and (2, 8) is m2 = 8-4 = 4 =-4
2-3 -1
Since the two lines are perpendicular to each other,

m1m2 = - 1

or, m1 (-4) = -1

or, m1 = 1
4
Putting the value of m1 in equation (i) , we get

y-5= 1 (x + 2)
4
or, 4y - 20 = x + 2

or, x - 4y + 22 = 0

? x - 4y + 22 = 0

Therefore, the required equation is x - 4y + 22 = 0.

Example 11. Find the equation of the straight line parallel to the line 2x - 3y + 4 = 0 and
passing through the mid-point of the line joining two points (3, 5) and (7, 3)

Solution: Let given line be PQ whose equation is 2x - 3y + 4 = 0 .......... (i)

Let the given points be A(3,5) and B(7, 3).

164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

( (Mid - point of AB = M 2x – 3y + 4 = 0
x1 + x2 , y1 + y2 P A(3, 5) Q
2 2 S
( (= M
3+7 , 5+ 3 R
2 2 = M (5, 4)
B (7, 3)
Equation of any line parallel to (i) is 2x - 3y + k = 0 ........ (ii)

The line (ii) passes through the point M(5, 4).

So. putting the point M(5, 4) in equation (ii),

we get, 2.5 - 3.4 + k = 0

or, 10 - 12 + k = 0 or, -2 + k = 0

or, k = 2

Put the value of k in equation (ii), we get,

? 2x - 3y + 2 = 0 which is the required equation.

Example 12. Find the equation of the perpendicular bisector of the line joining two points
(5, 7) and (4, 5).

Solution: Let A and B be given two points with coordinates (5, 7) and (4,5) respectively.

Let D be the mid points of AB.

( ( ( (Then, the coordinates of D are 5+4 , 7+ 5 i.e. 9 ,6
2 2 2

Let CD be the perpendicular bisector of AB. C

Slope of AB (m1) = y2 - y1 = 5-7 = -2 = 2
x2 - x1 4-5 -1

Let slope of CD be m2

Then m1.m2 = -1 A D (4, 5) B
or, 2.m2 = –1 (5, 7)

or, m2 = – 1
2
Now, equation of CD is y – y1 = m (x – x1)

or, y – 6 = – 1 x – 9
2 2
2x – 9
or, 2y – 12 = – 2 or, 4y – 24 = –2x + 9

? 2x + 4y = 33 which is the required equation.

Example 13. Find the equation of the straight lines passing through the point (3, -2) and
making an angle of 45° with 6x - 7y - 5 =0

Solution: The equation of straight line passing through point (3, -2) is given by

y - y1 = m(x - x1)
or, y + 2 = m1 (x - 3) ....... (i)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 165

vedanta Excel In Opt. Mathematics - Book 10

Its slope is (m) = m1

Again, the equation of given line is 6x - 7y - 5 = 0 ....... (ii) P(3, -2)

Its slope is (m2) = - coefficient of x = - 6 = 6 45° 45°
coefficient of y -7 7 A 6x-7y-5=0B

Angle between the lines (i) and (ii) is 45°

tanT = ± (m1 - m2)
1 + m1m2

m1 - 6
7
or, tan45° = ±
1 + m1 6
or, 7
or, (7m1 - 6) 7
1= ± 7 × 7 + 6m1

7 + 6m1 = ± (7m1 - 6)

Taking positive sign, we get

7 + 6m1 = 7m1, -6 ? m1 = 13
or, -m1 = -13
Taking negative sign, we get

7 + 6m1, = - 7m1 + 6

or, 13m1 = -1 ? m1 = -1
13
Putting the rules of m1 in equation (i),

Putting the values of m1 in equation (i),

Now, for m1 = -1
13
y – y1 = m1(x - x1)

or, y + 2 = -1 (x – 3)
13
or, 13y + 26 = –x + 3

or, x + 13y + 23 = 0

For m1 = 13, equation of the straight line is y + 2 = 13 (x – 3)
or, 13x – y = 41.

Hence, the required equation of lines are x + 13y + 23 = 0 and

13x – y = 41.

Exercise 8.1

Very Short Questions
1. (a) Write the formula to find the angle between two lines y = m1x + c1 and

y = m2x + c2

166 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

(b) Write the conditions of parallelism and perpendicularity of the two lines

y = m1x + c1 and y = m2x + c2
(c) If two lines with equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel,

show that a1b2 = a2b1
(d) If two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are perpendicular to each

other, show that a1a2 + b1b2 = 0
Short Questions

2. Find the acute angle between two lines in each of the following pair of straight lines:

(a) y - (2 - 3 ) x = 5 and y = (2 + 3 ) x + 8.

(b) 2x + 4y = 7 and x + 2y = 5

(c) x = 3 y + 2 and 3 x - y = 5

(d) xcosD + ysinD = p and x sinD - y cosD = q

3. Find the obtuse angle between the lines in each of the following pair of straight lines:

(a) y + 3 x + 8 = 0 and y + 20 = 0

(b) 2x + y = 3 and 3x + 2y = -1

(c) y = 3 x + 8 and 3 x + y + 5 = 0

(d) 3x + 2y = 1 and 2x + 3y + 7 = 0.

4. (a) Show that the lines 3x+4y=10 and 6x + 8y + 10 = 0 are parallel to each other.

(b) Show the lines x + y = 2 and 2x + 2y = 3 are parallel to each other.

(c) Show that the line x-y + 2 = 0 and the line jointing the points (4,6) and (10,12)
are parallel to each other.

(d) Show that the line 3x+y+5 = 0 and the line joining the points (7,0) and (6, 3) are
parallel to each other.

5. (a) Show that the lines 3 x + y = 9 and -x + 3 y + 3 = 0 are perpendicular to
each other.

(b) Show that lines 3x-y- 2 = 0 and x + 3y + 5 = 0 are perpendicular to each other.

(c) Show that the lines joining the points (7, -5) and (3, 4) is perpendicular to the line
4x - 9y + 7 = 0

6. Find the value of k when-

(a) The pair of straight lines kx + 3y - 8 = 0 and 3x - 2y + 11 = 0 are parallel.

(b) The pair of straight lines 2x + 5y + 2 = 0 and kx + y + 6 = 0 are parallel.

(c) The pair of lines 3 x + 2y = 9 and kx + 3 y + 3 = 0 are perpendicular to each
other.

(d) The lines represented by equations kx + 3y + 5 = 0 and 4x - 3y + 10 = 0 are
perpendicular to each other.

(e) The line passing through the points (-5, 7) and (1, -2) is perpendicular to the lines
joining the points (1, -3) and (4, k)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 167

vedanta Excel In Opt. Mathematics - Book 10

(f) The line joining the points (8,2) and (4,1) is parallel to the line joining the points
(6,4) and (2, k)

7. Find the equations of straight lines which-

(a) ppppaaaasssssssseieensssgttthhhthrrrooorouuuugggghhhhttthhhtheeeeppppooooiiinnninttt t(((-42(24,,,,535-))3)aa)annanddnddpppeaparrerpaarleplllneeeldnltidtocoiuctthluhaelrealrtliointnoetehthehhaaelvivnliiinnenggehsahsllvaooivppnieegng-s43lso23.lpoepe6545..
(b)
(c)
(d)

8. Find the equation the straight line which-

(a) passes through the point (2,3) and parallel to the line x - 2y - 2 = 0

(b) passes through the point (4, 3) and parallel to the line 2x + 3y + 12 = 0

(c) passes through the point (-1, -2) and perpendicular to the line 6x + 7y = 42.

(d) passes through the point (-1, -2) and perpendicular to the line 5x+4y+ 20 = 0

(e) passes through the point of intersection of the straight line 3x + 4y = 7 and

5x - 2y = 3 and perpendicular to the line 2x + 3y = 5

(f) passes through the point of intersection of 2x - 3y + 1 = 0 and x + 2y = 3 and
parallel to the line 4x + 3y = 12

(g) divides the line joining the points P(-1, -4) and Q (7, 1) in the ratio of 3:2 and
perpendicular to it.

(h) passes through the centroid of ∆ABC with vertices A(4,5), B(-4, -5) and C(1,2) and
parallel to 7x + 5y = 35.

9. Find the equation of the altitude AD of 'ABC with vertices A(2, 3), B(-4, 1) and C(2, 0)
drawn from the vertex A.

10. If the angle between the lines (a2 - b2) x - (p + q) y = 0 and (p2 - q2) x + (a + b) y = 0 is
90°, prove that (a - b) (p - q) = 1.

Long Questions

11. (a) From the point P(-2, 4), perpendicular PQ is drawn to the line AB: 7x - 4y + 15 = 0
Find the equation of PQ.

(b) Find the equation of the altitude PQ drawn from the vertex P to QR in ∆PQR with
P(2,3), Q(-4, 1) and R(2,0).

12. Find the equation of the perpendicular bisector to the join of the following two points.

(a) P(-3, 5) and Q(-6, 7) (b) M(5,6) and N(7, 10) (c) R(2, 4) and S(-2, -4)

13. (a) In rhombus PQRS, P(2, 4) and R(8, 10) are the opposite vertices. Find the equation
of diagonal QS.

(b) M(5,1) and P(-3, 3) are two opposite vertices of square MNPQ. Find the equation
of diagonal QN.

(c) In a square PQRS, P(2,3) and R(-6, 5) are the end points of diagonal PR. Find the

168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

equation of the other diagonal.

14. (a) Determine the equation of the straight lines through (1, -4) that make an angle of
45° with the straight line 2x + 3y + 7 = 0

(b) Find the equations of the straight lines passing through the point (3, 2) and making
angle of 45° with the line x - 2y - 3 = 0.

(c) Find the equation of two lines passing through the point (1, - 4) and making an
angle of 45° with the lines 2x - 7y +5 = 0

(d) Find the equation of the straight line passing through the point (3 - 2) and making
an angle of 60° with the line 3x + y - 1 = 0.

(e) Determine the value of m so that 3x - my - 8 = 0 makes an angle of 45° with the
line 3x + 5y - 17 = 0

15. (a) Find the equation of the sides of an equilateral triangle whose vertex is (1, 2) and
base is y = 0.

(b) Find the equation of the sides of right angled isosceles triangle whose vertex is at
(-2, -3) and whose base is x = 0

(c) Find the equation of the sides of right angled isosceles triangle whose vertex is
(2, 4) and the equation of base is x = 0.

1. (a) tanT = ± m1 - m2 (b) m1 = m2, m1 . m2 = -1
1 + m1 . m2

2. (a) 60° (b) 0° (c) 30° (d) 90°

3. (a) 120° (b) 172.87° (c) 120° (d) 172.87°
(c) -2
6. (a) - 9 (b) 2 (d) 9
2 5 4

(e) -1 (f) 3

7. (a) 2x + 3y = 13 (b) 3x - 4y + 8 = 0 (c) 5x + 4y + 22 = 0

(d) 5x + 6y = 50 8.(a) x - 2y + 4 = 0 (b) 2x + 3y = 17 (c) 7x - 6y = 5

(d) 4x - 5y = 6 (e) 3x - 2y - 1 = 0 (f) 4x + 3y = 7

(g) 40x + 25y = 127 (h) 21x + 15y = 17 9. 6x - y = 9

11. (a) 4x + 7y = 20 (b) 6x - y = 9 12.(a) 6x - 4y + 51 = 0

(b) x + 2y = 22 (c) x + 2y = 0 13.(a) x + y = 12 (b) 4x - y = 2

(c) 4x - y + 12 = 0 14.(a) 5x + y = 1, x - 5y = 21

(b) 3x - y = 7, x + 3y = 9 (c) 9x - 5y = 29, 5x + 9y + 31 = 0

(d) y + 2 = 0, 3 x - y = 2 + 3 3 (e) 12, - 3
4

15. (a) 3 x - y + 2 = 3, 3 x + y = 2 + 3 (b) x - y = 1, x + y + 5 = 0

(c) x + y = 6, x - y + 2 = 0

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 169

vedanta Excel In Opt. Mathematics - Book 10

8.2 Pair of straight Lines

Introduction :

Let us consider equations of two straight lines:
2x + 3y + 5 = 0 .............. (i)
x + y + 6 = 0 .............. (ii)

Above equations can be multiplied to find a single equation representing them.
(2x + 3y + 5) (x + y + 6) = 0

or, 2x2 + xy + 12x + 3xy + 3y2 + 18y + 5x + 5y + 30 = 0
or, 2x2 + 3y2 + 5xy + 17x + 23y + 30 = 0
or, 2x2 + 5xy + 3y2 + 17x + 23y + 30 = 0 ..... (iii)
Equation (iii) represents a pair of lines representing lines having equation (i) and (ii)

General Equation of second degree.

An equation of the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is called the general equation
of the second degree in x and y.

Example : 2x2 + 3xy + 5y2 + 3x + 4y + 15 = 0

This equation is in the form of ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,

where a = 2, h = 3 , b = 5, g = 3 f = 2, c = 15.
2 2

Let us consider two general equations of straight lines

a1x +b1y + c1 = 0 ........... (i)

and a2x + b2y + c2 = 0 .......... (ii)

Combining above two equations, we get (a1x + b1y + c1) (a2x + b2y + c2) = 0 on simplification,

We write, a1a2 x2 + (a1b2 + a2b1) xy + b1b2y2 + (b1c2 + b2c1) y + (c1a2 + c2a1) x + c1c2 = 0

or, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ........... (iii)

Where, a = a1a2, a1b2 + a2b1 =2h, b1b2 = b, c1a2 + c2a1 = 2g, b1c2 + b2c1 = 2f, c1c2 = c

The above equation (iii) is called general equation of second degree in x and y. It represents
a pair of lines if it can be factorized into two linear factors otherwise it represents a curve.

Homogeneous Equations :

If the sum of indices of variables x and y in each term of an equation is same, such type of
an equation is called homogeneous equation in x and y.

Examples :
(a) x2 + 2xy + y2 = 0 is a homogeneous equation of second degree in x and y.

(b) x3 + 3x2y + 4xy2 + y3 = 0 is a homogeneous equation of the third degree in x and y.

170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

General equation of the second degree that represents a pair of lines through the origin.
The general equation of the lines passing through the origin are

a1x + b1y = 0 ..... (i)
a2x + b2y = 0 ...... (ii)
Both of these equations are linear in x and y pass through the origin are in the form of y = mx.

Combining these equations, we get,

(a1x + b1y) (a2x + b2y) = 0
or, a1a2x2 + (a1b2 + a2b1) xy + b1b2y2 = 0
or, ax2 + 2hxy + by2 = 0 ............. (iii)

where a = a1a2, 2h = a1b2 + a2b1 , b = b1b2 .

The above equation (iii) is called homogenous equation of the second degree in x and y
which always represents a pair of lines through the origin.

Example : x2 + 3xy + 2y2 = 0

This equation is in the from of ax2 + 2hxy + by2 = 0

we can write,

x2 + 2xy + xy + 2y2 = 0

or, x(x + 2y) + (y(x + 2y) = 0

or, (x + 2y) (x + 6) = 0

Here, x + 2y = 0 and x + y = 0 are equation of two line passing through the origin.

To show that the homogeneous equation of degree two ax2 + 2hxy + by2 = 0 always
represents a pair of lines through the origin.

Proof : The homogeneous equation of second degree in x and y is

ax2 + 2hxy + by2 = 0

or, y2 + 2h . y + a = 0
x2 b x b
y 2+ 2h y a
or, x b x + b = 0

This equation is a quadratic in y and hence it has two roots.
x

Let the roots of the equations be m1 and m2.

Then, we write,

m1 = y and m2 = y
x x

or, y = m1x and y = m2x

Both of these equations are linear in x and y pass through the origin and are in the form of
y = mx.

Hence, ax2 + 2hxy + by2 = 0 always represents a pair of lines through the origin.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 171

vedanta Excel In Opt. Mathematics - Book 10

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

It can be written as ax2 + (2hy)x + by2 = 0

This equation is quadratic in x in the form of ax2 + bx + c = 0

x = -2hy ± (2hy)2 - 4.a.by2 = 2y(-h± h2 - ab )
2a 2a
( (?
= - h ± h2 - ab y
a

Taking positive sign and negative sign, we get

ax + (h - h2 - ab ) y = 0 ........... (i)

ax + (h + h2 - ab ) y = 0 .......... (ii)

Each of these equations is linear in x and y. Hence they represents straight lines. Moreover,
they are in the form of Ax + By = 0. Hence the lines (i) and (ii) are equations of the lines
passing through the origin.

Note :
If equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of lines,

then ax2 + 2hxy + by2 = 0 represents a pair of lines through the origin parallel to above
pair of lines.

If ax2 + 2hxy + by2 = 0 represents a pair of lines y = m1x and y = m2x

Then, (y - m2x) (y - m2x) = 0

y2 - (m1 + m2) xy + m1 m2x2 = 0 ........ (i)

Also, ax2 + 2hxy + by2 = 0

or, y2 + 2h xy + a y2 = 0 ........... (ii)
b b
2h a
The above equations (i) and (ii) are identical if (m1 + m2) =- b , m1m2 = b

To find angle between the two straight lines represented by ax2 + 2hxy + by2 = 0.

Given equation is

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

We know that equation (i) represents a pair of lines passing through the origin.

Let the lines represented by equation (i) be Y 0

y = m1x ............. (ii) =

x

y = m2x .............. (iii) m
1

– y – m1x = 0

y

Then, m1m2 = a –2h X0 X’
b , m1 + m2 = b

slope of line (ii) = m1 Y’

slope of line (iii) = m2

172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

Let T be the angle between the two lines (ii) and (iii), then,

tanT = ± m1 - m2 =± (m1 + m2)2 - 4m1m2
1 + m1m2 1 + m1m2

( (= ± -2h 2 - 4 a 4h2 a
( (= ± 2 b b b2 b
- 4
=±
a a+b
1+ b
b

h2 - ab ± 2 h2 - ab
a + b , T = tan-1 a+b

Condition for perpendicularity Y

The two lines represented by ax2 + 2hxy + by2 = 0

are perpendicular if T = 90°, y = m 2x

tan90° = ± 2 h2 - ab X y = m 1x X'
a+b
a+b O
or, cot90° = ±
2 h2 - ab Y'

or, 0 = a + b

? a+b=0

Therefore, when a + b = 0, the lines are perpendicular each other.

Condition for Coincident

Two lines represented by ax2 + 2hxy + by2 = 0 are coincident if T = 0°,

tan0° = ± 2 h2 - ab
a+b
or, h2 - ab = 0

? h2 = ab

Therefore, when a + b = 0 then the lines are coincident.

Worked Out Examples

Example 1. Find the single equation of the lines with equations:
Solution:
(a) x - 3y = 0 and x + 3y = 0 (b) x + 2y + 3 = 0 and x + y = 0

(a) The equations of the lines are

x - 3y = 0 ..........(i)

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

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

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 173

vedanta Excel In Opt. Mathematics - Book 10

(x - 3y) (x + 3y) = 0
? x2 - 9y2 = 0 is the required equation.
(b) The equations of the lines is,

x + 2y + 3 = 0 ....... (i)
x + y = 0 ........... (ii)
The single equation of above the lines is
(x + 2y + 3) (x + y) = 0
or, x (x + 2y + 3) + y(x + 2y + 3) = 0
or, x2 + 2xy + 3x + xy + 2y2 + 3y = 0
? x2 + 3xy + 2y2 + 3x + 3y = 0 is the required equation.

Example 2. Find the separate equations of two lines represented by the equation.
Solution:
(a) 6x2 + 5xy - 6y2 = 0 (b) 2x2 - 5xy - 3y2 + 3x + 19y - 20 = 0

(a) Here, 6x2 + 5xy - 6y2 - 0

or, 6x2 + 9xy - 4xy - 6y2 = 0

or, 3x (2x + 3y) - 2y (2x + 3y) = 0

or, (2x + 3y) (3x - 2y) = 0

Either 2x + 3y = 0 ........... (i)

or, 3x - 2y = 0 ............ (ii)

3x - 2y = 0 and 2x + 3y = 0 are the required equations

(b) Here, 2x2 - 5xy - 3y2 + 3x + 19y - 20 = 0

It can be written 2x2 - x (5y - 3) + 19y - 3y2 - 20 = 0

Which is in the form of ax2 + bx + c = 0, where,

a = 2, b = 3 - 5y, c = 19y - 3y2 - 20

Now, x = - b± b2 -4ac
2a
(5y - 3) ± (3 - 5y)2 - 4.2.(19y - 3y2 - 20)
= 2.2

= (5y - 3) ± 9-30y+25y2 - 152y + 24y2 + 160
4

= (5y - 3) ± 49y2 - 182y + 169
4

= (5y - 3) ± (7y - 13)2
4
or, 4x = (5y - 3) ± (7y - 13)

Taking positive sign, we get,

4x = 5y - 3 + 7y - 13

174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur