Vedanta Opt. Math 9 Final (2078)

vedanta Excel In Opt. Mathematics - Book 9 Matrix

5.4 Transpose of Matrix

a bc
Let A = d e f be a square matrix. Interchange the rows and columns of the matrix and

ghi

adg
denote it by A' = b e h . The matrix A' is called the transpose of matrix A.

cf i

Definition : Let A be a given matrix. Then the matrix obtained by interchanging the rows

and columns of the given matrix A is called the transpose of matrix A. It is denoted by A' or
AT or At.

For example : Let A = 146 then its transpose is given by A' = 1 2
231 4 3 . Here, the order of
matrix A is 2 × 3 and order of A' is 3 × 2. 6
1

Properties of Transpose of Matrix

(a) The transpose of a matrix is the matrix itself. i.e. (AT)T = A.

Let A = a b , then AT = a c and (AT)T = a b . ? (AT)T =A
c d b d c d

(b) The transpose of the sum of two matrices is the sum of their transposes.

i.e. (A + B)T = AT + BT

Let A = 1 2 and B = 6 4
3 4 5 2

Then, A + B = 1 2 + 6 4
3 4 5 2

= 1+6 2+4 = 7 6
3+5 4+2 8 6

(A + B)T = 7 8
6 6

Again, AT = 1 3 and BT = 6 5 .
2 4 4 2

Then AT +BT = 1 3 + 6 5 = 7 8
2 4 4 2 6 6

? (A + B)T = AT + BT

(c) If A is any matrix and k is any scalar (real number), then (kA)T = kAT

Let A= 123
456

Take k = 3, kA = 3 123 = 3 6 9
456 12 15 18

3 12
(kA)T = 6 15

9 18

1 4 3 12
Again kAT = 3 2 5 = 6 15

3 6 9 18

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Hence (kA)T = kAT
i.e. (3A)T = 3AT

Worked out Examples

Example 1. If P = 1 2 3 , find PT and verify that (PT)T = P
Solution : 7 8 10

Example 2. Here, P = 1 2 3
Solution : 7 8 10
Then interchanging the rows and columns of the matrix P, we get,

17
PT = 2 8

3 10

Also, (PT)T = 1 2 3 ? (AT)T = A Proved.
7 8 10

If P = 1 2 and Q = 2 4 , verify that (P + Q)T = PT + QT
4 6 6 9

Here, P = 1 2 and Q = 2 4
4 6 6 9

Then PT = 1 4 and QT = 2 6
2 6 4 9

P+Q= 1 2 + 2 4 = 3 6
4 6 6 9 10 15

(P + Q)T = 3 10
6 15

and PT + QT = 1 4 + 2 6 = 3 10
2 6 4 9 6 15

? (P + Q)T = PT + QT Proved.

Exercise 5.4

Very Short Questions

1. (a) Define transpose of a matrix with an example.

(b) State the properties of transpose of a matrix.

2. Find the transpose of the following matrices:

(a) P = [1 2 3], 7
(b) Q = 8

9

(c) R= 2 4 6 123
8 2 3 (d) M = 4 5 6

789

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Short Questions :

3. (a) If A = 2 6 8 , then final AT and verify that (AT)T = A.
3 2 5

(b) If Q = 1 7 8 , verify that (QT)T
2 3 7

4. If P = 6 7 8 and Q = 3 2 4 , then verity that
2 3 4 6 2 1

(i) (P + Q)T = PT + QT

(ii) (PT)T + (QT)T = P + Q

Long Questions

5. (a) If M = 3 4 , show that M + MT is a symmetric matrix.
1 5

(b) If P = 1 3 , show that P + PT is a symmetric matrix.
2 4

0 2 –45
6. Let A = –2 0 –4 , then

45 4 0

(a) Show that A is a skew - symmetric matrix.

(b) Find transpose of A.

(c) Find A + AT. What type of matrix is formed?

7. Let P = 2 3 for k = 4, verify that (kP)T = kPT.
4 5

0 1 –4
8. If R = –1 0 –5 , then show that RT = –R

450

1 (b) [7 8 9] 28 147
2. (a) 2 (c) 4 2 (d) 2 5 8

3 63 369

0 –2 45 103
6. (b) 2 0 4

45 –4 0

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5.5 Matrix Multiplication

(I) Multiplication of a Matrix by a scalar

Let A = 2 3 , then we can write,
5 4

A+A = 2 3 + 2 3
5 4 5 4

= 2+2 3+3 = 4 6
3+5 4+2 10 8

Also, we write 2A = 2 2 3 = 2×2 2×3 = 4 6
5 4 2–5 2×4 10 8

Similarly, we can find the values of 3A, 4A and 5A.

Definition : Let A be any matrix and k be a scalar, then the matrix obtained by multiplying
each element of A by k is denoted by kA and it is called scalar multiplication of matrix A by
k. The order of matrix kA is same order as that of A.

Example 1. Let A = 2 4 6 , then find the value of 4A.
Solution : 3 2 1

Example 2. Here, A = 2 4 6 , 4A =4 2 4 6 = 4×2 4×4 4×6
Solution : 3 2 1 3 2 1 4×3 4×2 4×1

Example 3. = 8 16 24
Solution : 12 8 4

If P = 3 2 ,Q= 4 5 , then find the value of 3P + 2Q.
4 5 2 1

Here, P = 32 and Q = 4 5
45 2 1

Now, 3P + 2Q = 3 3 2 +2 4 5
4 5 2 1

= 9 6 + 8 10 = 17 16
12 15 4 2 16 17

Find the matrix X if A = 2 3 , B= 2 3 , and X + 5A = 4B
4 6 3 –1

Here, A = 2 3 ,B= 2 3
4 6 3 –1

Now, X + 5A = 4B

or, X = 4B – 5A = 4 2 3 –5 2 3
3 –1 4 6

= 8 12 – 10 15 = –2 –3
12 –4 20 30 –8 –34

(II) Multiplication of Matrices :

Mr Gyan bought 20 packets of tooth paste, 30 bottles of mineral water, and 25 packets of
tasty biscuits at the rate of Rs. 75, Rs. 20 and Rs. 45 respectively. How much money did he
spend? Can we use matrix method to solve those problem ?

Yes, we can solve it by use of matrix. Let, matrix of toothpaste Mineral water, Biscuit be M.

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i.e., M = [20 30 25] be matrix of goods.

75 Tooth paste
N = 20 water

45 Biscuits

be matrix of rates.

75
Now, MN = [20 30 25] 20

45

= [20 × 75 + 30 × 20 + 25 × 45]

= [1500 + 600 + 1125] = [3225]

Hence, Mr. Gyan spent total amount of Rs. 3225 on purchasing the goods.

Definition : Two matrices are said to be multiplicable if the number of columns of the first
matrix is equal to the number of rows of the second matrix. An element of the product
matrix is obtained by adding the products of the corresponding elements of the first matrix
matrix and column of the second matrix.

Let there be Am ×n and Bn×P . Then, the product AB is defined as the number of the column
of the first matrix A is equal to the number of the rows of the second matrix. The product
matrix AB is of order m × p.

Examples : A = 1 2 and B = 4 2 , then AB is defined.
2 4 3 2

Solution : Now, AB = 1 2 42
2 4 32

= 1×4+2×3 1×2+2×2 = 10 6
2×4+4×3 2×2+4×2 20 12

Process of Matrix Multiplication :

Multiplying a Row by a Column

Let us find the product

x
[a b c] y

z
we have to multiply a 1 × 3 matrix by a 3 × 1 matrix.

The number of the column in the first matrix is the same as the number of rows in the
second matrix; so, they are compatible.

The product is given by,

x
[a b c] y = [ax + by + cz]

z
Multiplying Larger Matrices

We know that if two matrices are multiplicable, we multiply a row by a column, multiplying
each element in the ith row of the first matrix by the corresponding element in the jth column
of the second matrix and adding the result.

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Let us take

A= 1 2 3 3 2
2 3 4 and B = 3 1
5
4
Here A and B matrices are of orders 2 × 3 and 3 × 2 respectively.

The product matrix is of order 2 × 2.

Let AB = 1 2 3 3 2 = e11 e12
2 3 4 3 1 e21 e22
4 5
To get e11, multiply Row 1 of the first matrix by Column 1 of the second matrix.

3
e11 = [1 2 3] 3 = 1 × 3 + 2 × 3 + 3 × 4 = 21

4
To get e12, multi Row 1 of the first matrix by column 2 of the second matrix.

2
e12 = [1 2 3] 1 = 1 × 2 + 2 + 1 + 3 × 5 = 19

5
To get e21, multiply Row 2 of the first matrix by Column 1 of the second matrix.

3
e21 = [2 3 4] 3 = 2 × 3 + 3 × 3 + 4 × 4 = 31

4
To get e22, multiply Row 2 of the first matrix by column 2 of the second matrix.

2
e22 = [2 3 4] 1 = 2 × 2 + 3 × 1 + 4 × 5 = 27

5
writing the product matrix, we get

e11 e12 = 21 19
e21 e22 31 27
Therefore, we have,

At a gl21an23ce,34we334wri215te,= 21 19
31 27

123 32 = 1×3+2×3+3×4 1×2+2×1+3×5
234 31 2×3+3×3+4×4 2×2+3×1+4×4×5
45
21 19
= 31 27

Properties of Matrix Multiplication :

(a) In general matrix multiplication is not commuatative

Let A and B be two matrices such that AB and BA are defined, then AB z BA.

Let A = 2 3 and B = 4 5
3 2 2 3

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Now, AB = 2 3 45
3 2 23

= 2×4+3×2 2×5+3×3 = 14 19
3×4+2×2 3×5+2×3 16 21

Again, BA = 4 5 23
2 3 32

= 4×2+5×3 4×3+5×2 = 23 22
2×2+3×3 2×3+3×2 13 12

? AB z BA

Hence, the commutative property does not hold in matrix multiplication

(b) Associative property of Matrix Multiplication
Let A, B and C be three matrices such that AB and BC are defined Then (AB)C= A(BC)

Let A = 1 0 ,B= 2 1 ,C= –1 2
1 2 1 1 3 2

Now, AB = 1 0 2 1 = 2+0 1+0 = 2 1
1 2 1 1 2+2 1+2 4 3

(AB)C = 2 1 –1 2
4 3 32

= –2 + 3 4+2 = 1 6
–4 + 9 8+6 5 14

Again, BC = 2 1 –1 2
1 1 32

= –2 + 3 4+2 = 1 6
–1 + 3 2+2 2 4

A(BC) = 1 0 16
1 2 24

= 1+0 6+0 = 16
1+4 6+8 5 14

? (AB)C = A(BC)

Hence, associative property holds in matrix multiplication.

(c) Distributive Property :
Let A,B, and C be three matrices such that AB and AC are defined. Then,

A(B + C) = AB + AC.

Let A = 1 2 ,B= 2 1 and C = 2 1
0 3 3 2 4 3

Now, B+C= 2 1 + 2 1 = 4 2
3 2 4 3 7 5

A(B + C) = 1 2 42 = 4 + 14 2 + 10 = 18 12
0 3 75 0 + 21 0 + 15 21 15

Again AB + AC = 1 2 21 + 1 2 21
0 3 32 0 3 43

= 2+6 1+4 + 2+8 1+6
0+9 0+6 0 + 12 0+9

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= 8 5 + 10 7 = 18 12
9 6 12 9 21 15

? A(B + C) = AB + BC

Hence, distributive property holds for matrix multiplication.

(d) Identity Property (or Identity Element for Multiplication)
Let A be a square matrix and I be an identity matrix of the same order as that of A, then

A×I=A=I×A

Let A = 5 6 ,I= 1 0
7 8 0 1

Now, AI = 5 6 1 0 = 5+0 0+6 = 5 6
7 8 0 1 7+0 0+8 7 8

Again, IA = 1 0 56 = 5+0 6+0 = 5 6
0 1 78 0+7 0+8 7 8

? A.I = A = IA

(e) Transpose of Product Property :

Let A = 2 1 and B = 1 0 , then
3 4 2 3

(AB)T = BTAT

Now, AB = 2 1 1 0 = 2+2 0+3 = 4 3
3 4 2 3 3+8 0 + 12 11 12

(AB)T = 4 11
3 12

AT = 2 3 and BT = 1 2
1 4 0 3

Again, BTAT = 1 2 2 3 = 2+2 3+8 = 4 11
0 3 1 4 0+3 0 + 12 3 12

? (AB)T = BTAT

Hence, the transpose of product of two matrices is equal to the product of transpose of
the matrices in reverse order.

12

Example 4. Let A = –3 4 and B = 1 2 , then find AB and BA if possible.
Solution : 21
–2 4
1 2
Here, A = –3 4 and B = 1 2 . As A3×2 and B2×2, AB is defined
1 –2 4
2
and B2×2 and A3×2 and A3×2, BA is not defined.

Now, AB 1 2 12
= –3 4 –2 4
1
2

1–4 2+8 –3 10
= –3 – 8
–6 + 16 = –11 10
2–2
4+4 08

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

Example 5. If 5 X= 20 15 , find the matrix X.
Solution : 2 8 6

Since 5 is of order 2 × 1 and 20 15 is of order 2 × 2, hence X must be of
2 8 6

order 1 × 2. Let X = [x y]

Now, 5 [x y] = 20 15
or, 2 8 6

5x 5y = 20 15
2x 2y 8 6

Equating the corresponding elements, we get,

5x = 20 or, x = 4

and 5y = 15 or, y = 3

Example 6. If P = 1 2 then, prove that P2 – 2P – 5I = O, where I and 0 are unit matrix
Solution: 3 1

Example 7. and null matrix of order 2 × 2 respectively.
Solution:
Here, P = 1 2
3 1

P2 = P.P = 1 2 12
3 1 31

= 1+6 2+2 = 7 4
3+3 6+1 6 7

2P = 2 1 2 = 2 4
3 1 6 2

5I = 5 1 0 = 5 0
0 1 0 5

Now, P2 – 2P – 5I = 7 4 – 2 4 – 5 0
6 7 6 2 0 5

= 7 4 – 2+5 4+9
6 7 6+0 2+5

= 7 4 – 7 4
6 7 6 7

= 7–7 4–4 = 0 0 =O
6–6 7–7 0 0

? P2 – 2P – 5I = O Proved.

If P + Q = 5 2 and P – Q = 3 6 , find the matrices P and Q.
0 9 0 –1

Here, P + Q = 5 2 ................ (i)
0 9

and P–Q= 3 6 ............ (ii)
0 –1

adding equations (i) and (ii), we get,

2P = 8 8
0 8

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

or, P = 1 8 8 (multiplying both sides by 1 )
2 0 8 2

? P= 4 4
0 4

put the value of P in equation (i), we get,

Q = 5 2 –P= 5 2 – 4 4
0 9 0 9 0 4

= 1 –2
0 5

Example 8. If P = 1 2 and Q = 2 1 , then show that (PQ)T = QTPT.
Solution: 3 4 4 2

Example 9. Here, P = 1 2 and Q = 2 1 , then
Solution: 3 4 4 1

PT = 1 3 and QT = 2 4
2 4 1 2

Now, PQ = 1 2 . 2 1 = 2+8 1+4 = 10 5
3 4 4 1 6 + 16 3+8 22 11

LHS = (PQ)T = 10 22
5 11

RHS = QTPT = 2 4 . 1 3 = 2+8 6 + 16 = 10 22
1 2 2 4 1+4 3+8 5 11

? LHS = RHS Proved.

What matrix pre–multiplies –1 3 to get (–9 5)?
7 1

Here, –1 3 is a square matrix of order 2 × 2 and the product matrix is
71

[–9 5], the required matrix must be of order 1 × 2.

Let [x y] be the required matrix.

Now, [x y] –1 3 = [–9 5]
7 1

or, [–x + 7y 3x + y] = [–9 5]

Equating the corresponding elements, we get,

– x + 7y = –9 .............. (i)

3x + y = 5 ................. (ii)

From equation (i) x = 7y + 9

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

3(7y + 9) + y = 5

or, 21y + 27 + y = 5

or, 22y = – 22

? y = –1

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

Put the value of y in equation (i), we get
x = 7 × (–1) + 9 = 2

? x = 2, y = –1

Exercise 5.5

Very Short Questions

1. (a) Under what condition two matrices are multiplicable ?

(b) If Am×p and Bp×n and AB = C, what is the order of product matrix C ?

(c) Are A2×3 and B3×2, multiplicable ?

(d) If A2×3 and B3×3 are AB and BA defined ? Give reasons.

(e) For what types of matrices, A2, A3, A4, are defined ?

(f) If A = 2 4 , find (i) 5A, 1 A (ii) 3 A.
7 5 2 2

2. Find the product of the matrices given below.
2 1
(a) [1 2] 5 (b) 2 [4 5]

2 (d) 23 24
(c) [4 3 2] 1 45 62

3

Short Questions :

3. Find the product of given matrices if possible:

(a) A= 1 2 ,B= 3 2 find AB.
2 4 2 5

(b) P= 1 2 3 ,Q= 1 2
4 5 6 4 3 , find PQ.
2
1

(c) M= 2 1 , N = [2 3], find MN.
3 0

(d) P= 1 2 3 ,Q= 3
4 5 6 1 , find PQ.

2

4. (a) If A = 1 –2 1 and B = 2 1 find AB and BA.
2 1 3 3 2
1 1

3

(b) If A = [1 2 3] and B = 2 , find AB and BA.

1

5. (a) If P = 2 1 , then prove that : P2 = P.
–2 –1

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(b) If P = 1 0 , prove that : P2 = P
0 1

6. If P = 1 2 and Q = 2 1 , find the values of :
3 2 4 5

(i) 2P + 3Q (ii) 3P – 2Q.

Long Questions

7. (a) If A = 1 1 show that A2 – 2A = O, where O is a null matrix of order of 2 × 2.
1 1

(b) If P = 3 –5 , proved that P2 – 5P = 14I where I is a unit matrix of order 2 × 2.
–4 2

(c) If A = 3 1 , prove that A2 – 5A + 7I = O, where I and O are unit and null
–1 2

matrices of order 2 × 2.

(d) If P = 4 2 , then prove that (P – 2I) (P – 3I) = O, where O is a null matrix of
–1 1

order 2 × 2 and I unit matrix.

8. (a) If A + 2B = 3 20 and 2A + B = 3 1 3 , find the matrices A and B.
3 35 0 3 7

(b) If 2P + 3Q = 2 7 12 and P + 2Q = 1 0 –1 . Find the matrices P and Q.
13 12 23 –4 –1 –6

9. (a) If A = 2 4 and B = 2 3 , show that : (AB)T = BTAT
3 1 0 4

(b) If P = 2 4 and Q = –1 5 , show that : (PQ)T = QTPT
1 3 2 1

10. (a) If A = 10 and B = 10 show that : AB = BA
02 04

(b) If A = 53 and B = –1 3 show that AB = BA = I where I is a unit matrix of
21 2 –5
order 2 × 2.

11. (a) If 1 2 1 0 = y 2 , find the values of x and y.
x –3 0 1 4 –3

(b) If 4 1 1 –1 = x 1 , then find the values of x and y.
7 –3 13 4 y

(c) If A = 3 0 ,B= a b and AB = A + B, find the values of a, b and c.
0 4 0 c

(d) If –1 0 x = –2 , find the matrix x .
0 –2 y 4 y

12. (a) If P = 2 1 and Q = 3 4 and PR = Q, find the matrix R.
5 3 2 4

(b) If –1 2 ×P= –2 , find the matrix P.
2 –2 4

13. Which matrix pre–multiplies to matrix 1 1 to get 4 5 ?
3 4 6 2

14. (a) If [a b] 2 = [1 4] a , then prove that : a = b.
3 b

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

(b) If [m n] m = [1 3] and m + 1 = n, find the values of m and n.
n
15. (a) If P is a matrix of order (2x + 1) × 2 and Q is another matrix of order (3y – 1) × 3.

If PQ and QP both are defined, find the values of x and y.

(b) Matrix M has x-rows and (11 – x) columns. Matrix N has y-rows and (y + 5)
columns. If MN and NM both are defined, find the values of x and y.

16. If A = 1 2 ,B= 2 4 ,C= 5 2 ,I= 1 0 then verity the following :
3 0 3 5 3 4 0 1

(i) AB z BA (ii) (AB)C = A(BC) (iii) A(B + C) = AB + AC

(iv) IA = AI = A (v) (BC)T = CTAT.

Project work
17. Prepare a report on "operations on matrices" and present in your classroom.

1. (a) No. of column of 1st matrix = No. of rows of 2nd matrix. (b) m × n

(c) Yes (d) AB defined but BA not defined (e) square matrix

10 20 12 36
25 35 (iii) 21 15
(f) (i) (ii) 7 5
22
22
(c) [17]
2. (a) [12] (b) 45 (d) 22 14
8 10 38 26

3. (a) 7 12 (b) 15 11 (c) MN is not defined (d) 11
14 24 36 29 29

4. (a) –3 –2 , 4 –3 53 6 9 6. (i) 8 7 (ii) –1 4
10 7 7 –4 9 (b) [10], 2 4 6 18 19 1 –4
3 –1 2 3
41

8. (a) A = 1 0 2 ,B= 1 1 –1 (b) P= 1 14 27 ,Q= 0 –7 –14
–1 1 3 2 1 1 38 27 64 –21 –14 –35

11. (a) x = 4, y = 1 (b) x = 5, y = – 16 (c) a = 3 , b = 0, c = 4
2 3

(d) x = 2, y = – 2 12. (a) 7 8 (b) 2 13. 11
–11 –12 0 18 –4

14. (b) m = 2, –3, n = 3, –2 15. (a) x = 1, y = 1 (b) x = 8, y = 3

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vedanta Excel In Opt. Mathematics - Book 9 Coordinate Geometry

Coordinate Geometry 6

6.0 Review

Group discuss the following questions :
(a) Name the four quardrants in the given graph.

(b) Write down the signs of coordinates in each quadrant.

(c) What are the coordinates of any points in the X-axis and Y-axis?

(d) Write the coordinates of the four points P, Q, R and S in the given graph.

(e) Write the distance formula to find the distance between points P(x1, y1) and Q(x2, y2).
(f) From the given graph, find the distance between the points P and R.

Y

7
6
5
Q 4 P

3
2
1
X' -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 X
-2
-3
-4
R -5
S

Y'

6.1 Locus Bus stop Road

Let us study the following activities : House
(a) Sima draws a map to show the path from her house to AP B

nearest bus stop. d
M Fixed Line N
Sima has to cover fixed distance from her house to the
bus stop to go her school.

(b) Let a point P move such that its distance from a fixed
line is always equal to d. The point P traces out a
straight line AB parallel to the fixed line MN. The
point P moves in a straight line.

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vedanta Excel In Opt. Mathematics - Book 9 Coordinate Geometry

(c) Let a point P move in a plane such that its distance P
from a fixed point O is equal to r. The point P traces r
out a circle with centre at O and radius r.
O
(d) Let us consider a set of points in the X-axis. In the
X-axis, all the y-coordinates one zero. Hence y = 0 Y
represents X-axis.

All the above activities are related to path traced by O y=0 X
a point. The path traced by a point under certain X'

condition is a locus. Y'

Definition : The locus of a point is defined as the path traced out by the moving point under
given geometrical conditions. It may also be defined as a locus is the set of points which
satisfy the given geometrical conditions.

The following points should be taken in mind while dealing with a locus:
(i) Every point in the locus must satisfies the given geometrical conditions.

(ii) A point which does not satisfy the given geometrical conditions cannot be on the locus.

(iii) Every point which lies in the locus satisfies the equation of the locus.

(iv) The points which do not lie in the locus do not satisfy the equation of the locus.

(v) To find the locus of a moving point, plot some points satisfying the given geometrical
conditions, and then join them.

Procedure of finding the equation of a locus

To find the equation of a locus with given geometrical conditions, the following are some
general rules:

(i) Draw a figure according to given geometrical condition and take a point P(x, y) on the locus.

(ii) Write down the geometrical conditions according to the points move.

(iii) Express the geometrical conditions in terms of the coordinates (x, y) by using distance
formula and simplify it. The simplest form of the given condition is the required
equation of the locus.

Worked out Examples

Example 1. Does the point (2, 3) lie in the locus x2 + y2 = 13 ?
Solution : Here, x2 + y2 = 13
Put the point (2, 3) i.e. x = 2, y = 3 in the given equation, we get,

22 + 32 = 13
or, 4 + 9 = 13
? 13 = 13 (True).
Hence, the point (2, 3) lies in equation of locus x2 + y2 = 13.

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vedanta Excel In Opt. Mathematics - Book 9 Coordinate Geometry

Example 2. Find the value of k if the point (4, 3) lies in the locus x2 + y2 + kx + 3y – 16 = 0.
Solution :
The given equation of locus is x2 + y2 + kx + 3y – 16 = 0
Example 3.
Solution : The point (4, 3) lies in the locus.

Example 4. So, put x = 4 and y = 3 in the above equation.
Solution :
42 + 32 + k.4 + 3.3 – 16 = 0

or, 16 + 9 + 4k + 9 – 16 = 0

or, 4k = – 18

? k = – 9
2

If the point (2, 3) lies in the locus whose equation is 5x + by = 16. Show that
(0, 8) also lies in the locus.

Here, 5x + by = 16 .... (i)

The point (2, 3) lies in the locus

5.2 + b.3 = 16

or, 3b = 16 – 10

or, 3b = 6

? b=2

Put the value of b in (i), it becomes,

5x + 2y = 16

Put the point (0, 8) in the locus.

5.0 + 2.8 = 16

? 16 = 16 (True)

Hence the point (0, 8) also lies in the locus.

Find the equation of locus of points which move at equidistant from the

points A(2, 3) and B(4, 5). Y

Let P(x, y) be any point in the locus which is at 6 B(4, 5)
equidistant from the points 5
4
A(3, 2) and B(4, 5) 3 P(x, y)

2 A(3, 2)
1
Then, AP = BP
X' -4 -3 -2 -1-O1 1 2 3 4 5 X

or, AP2 = B2 (Squaring on both sides.) -2
-3

or, (x – 3)2 + (y – 2)2 = (x – 4)2 + (y – 5)2 Y'

or, x2 – 6x + 9 + y2 – 4y + 4 = x2 – 8x + 16 + y2 – 10y + 25

or, 2x + 6y = 28

or, x + 3y = 14

? x + 3y = 14 is the required equation of locus.

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Example 5. Find the equation of locus of a point which moves so that its distance from
Solution:
X–axis is always 6 units. Y
Example 6.
Solution: Let P(x, y) be any point in the locus P(x, y)
and A(x, 0) be any point in the X–axis.
Example 7.
Solution: Then, PA = 6 6

Example 8. or, PA2 = 36 (Squaring on both sides) X' A (x, 0) X
Solution: or, (x – x)2 + (y – 0)2 = 36

or, y2 = 36
? y = ± 6 is the required equation of the locus. Y'

Find the equation of locus of point which moves such that its distance from
X–axis is half of its distance from the origin.

Let P(x, y) be any point in the locus and A(x, 0) be a point in X–axis.

Then by question, Y P(x, y)

PA = 1 OP
2

or, 4PA2 = OP2 (Squaring on both sides.)

or, 4PA2 = OP2 X' O A(x, 0) X
Y'
or, 4{(x – x)2 + (y – 0)2} = (x – 0)2 + (y – 0)2

or, 4y2 = x2 + y2

or, 3y2 = x2

or, x2 = 3y2

? x2 = 3y2 is the required equation of the locus.

Let A(1, 2) and B(5, 3) are two fixed points. Find the equation of locus of

point P so that PA : PB = 3:2. Y

Let P(x, y) be any point on the locus and A(1, 2) P(x, y)

and B(5, 3) are two fixed points. 3 A(5, 3)
2 B(1, 2)
Then, PA = 3 1
PB 2 X' O 1 2 3 4 5 6 X

2PA = 3PB

Squaring on both sides, Y'
4{(x – 1)2 + (y – 2)2} = 9{(x – 5)2 + (y – 3)2}

or, 4{(x2 – 2x + 1) + (y2 – 4y + 4)} = 9{(x2 – 10x + 25) + (y2 –6y + 9)}

or, 4x2 + 4y2 – 8x – 16y + 20 = 9x2 + 9y2 – 90x – 54y + 306

? 5x2 + 5y2 – 82x – 38y + 286 = 0 is the required equation.

Let A(3, 0) and B(–3, 0) be two fixed points. Find the locus of point P which
moves so that the sum of square of distance is 25.

Let P(x, y) be any point in the locus

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and A(3, 0) and B(–3, 0) are two fixed points. Y

Then, by questions, AP2 + BP2 = 25 P(x, y)

or, (x – 3)2 + (y – 0)2 + (x + 3)2 + (y – 0)2 = 25

or, x2 – 6x + 9 + y2 + x2 + 6x + 9 + y2 = 25

? 2x2 + 2y2 = 7 is the required equation. X' B 0 A X
(–3, 0) (3, 0)

Y'

Exercise 6.1

Very Short Questions

1. (a) Write down the formula to find the distance between two points P(x1, y1) and Q(x2, y2).
(b) Define locus of a point.

2. Find the equation of locus of a point which moves so that

(a) Its distance from X-axis is always 4 units.

(b) Its distance from X-axis is always 5 units

(c) Its ordinate is always – 6

(d) Its abscissa is always 8.

Short Questions

3. Find the distance between the following pair of points:

(a) (4, 5) and (6, 8) (b) (–2, –2) and (–4, –4)

(c) (a + b, a – b) and (a – b, a + b) (d) (p – q, p + q) and (p + q, p – q)

4. (a) Show that the points (4, 5), (5, –2) and (1, 1) are the vertices an isosceles triangle.

(b) Show that the points (1, 1), (–1, –1) and (– 3, 3 ) are the vertices of an equilateral
triangle.

5. (a) Does the point (2, 4) lies in the locus whose equation is

(i) x + 2y = 10 (ii) x2 + y2 = 20

(b) Do the points (3, 2), (4, 3), (5, 0), (0, –5), and (–3, –4) lie on the locus whose
equation is x2 + y2 = 25?

6. (a) For what value of k will the point (2, 1) lie in the locus whose equation is

x2 + y2 – 4x + 3y + k = 13 ?

(b) For what value of O will the point (2, 3) lie on the locus whose equation is

x2 + y2+ Ox + 2y – 30 = 0 ?

7. (a) If (4, 4) is a point on the locus whose equation is y2 = ax, prove that (16, 8) is
another point on the locus.

(b) If (2, 3) is a point on locus whose equation is ax + 2y = 16 and also show that
(0, 8) is another point on the locus.

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8. Find the locus of a point which moves so that

(a) Its distance is 4 units from the point (4, –2)

(b) Its distance from (1, 2) and (2, –3) are equal.

(c) It is at equidistant from the points (4, 3) and (5, 4)

9. Find the locus of a point which moves so that

(a) Its distance from X-axis is twice its distance from Y-axis.

(b) Its distance from the point (2, 1) is double its distance from (1, 2)

(c) Its distance from Y-axis is half of its distance from the origin.

(d) Its distance from the point (0, 2) is one third of its distance from (3, 0).

Long Questions

10. Find the locus of a point which moves so that

(a) The ratio of its distance from the point (–5, 0) and (5, 0) is 2:3.

(b) The ratio of its distance from the points (1, 2) and (5, 3) is 4:5.

11. (a) Let M(3, 2) and N(7, –4) are two fixed points and P(x, y) be a variable point in the
locus, then find the equation of locus under the following conditions

(i) MP = NP (ii) MP = 2PN

(b) Let A(a, 0) and B(–a, 0) be two fixed points. Find the locus of a point which moves
such that PA2 + PB2 = AB2.

(c) Let A(– 7, 0) and B(7, 0) be two fixed points on a circle. Find the locus of a moving
point P at which AB subtend a right angle.

12. (a) If a point (x, y) is equidistant from the point (2, 3) and (6, 1). Show that the
equation of locus is given by 2x – y = 6.

(b) Let A(a, b) and B(3a, 3b) are two fixed points and P(x, y) is a point such that
PA = PB, then prove that the equation of locus is given by ax + by = 2(a2 + b2).

13. Find the equation of the locus of a point such that

(a) The sum of square of its distance from (0, 2) and (0, –2) is 6.

(b) The sum of the squares of its distances from A(3, 0) and B(–3, 0) is four times the
distance from A and B.

(c) The difference of square of distance from the points A(0, 1) and B(–4, 3) is 16.

(i.e. PA2 – PB2 = 16)

Project Work

14. In which field of our daily life do we use locus? Group discuss in the class and prepare
a report on it.

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1. (a) PQ = (x2 – x1)2 + (y2 – y1)2 2. (a) y = ±4 (b) y = ±5
3. (a) 13 (b) 2 2
(c) y = –6 (d) x = 8

(c) 2 2 b (d) 2 2 q 5.(a) (i) Yes (ii) Yes
(b) x – 5y = 4 (c) x + y = 8
6. (a) k = 13 (b) O = 11
2

8. (a) x2 + y2 – 8x + 4y + 4 = 0

9. (a) y = 2x (b) 3x2 + 3y2 – 4x – 14y + 15 = 0

(c) 3x2 – y2 = 0 (d) 8x2 + 8y2 + 6x – 36y + 27 = 0

10. (a) x2 + y2 + 26x + 23 = 0 (b) 9x2 + 9y2 + 110x – 4y = 419

11. (a) (i) 2x – 3y – 13 = 0 (ii) 3x2 + 3y2 – 50x + 36y + 247 = 0 (b) x2 + y2 = a2

(c) x2 + y2 = 49 13.(a) x2 + y2 + 1 = 0 (b) x2 + y2 – 3 = 0 (c) 2x – y + 10 = 0

6.2 Section Formula

Review

(a) Discuss the meaning of ratio with examples. B
(b) Find the ratio of AP : PB in the given figure. 3cm

4cm P 3cm B
Fig. (i)
Internal Division : Let AB be a given line segment and
P be a point on it. If P lies within AB, P is said to divide A
AB internally in the ratio of AP : PB. If means that P is an
P
internal point of AB.
6cm
To find the coordinates of a point dividing the line
segment joining two given points internally in the given A Fig. (ii)

ratio.

Let A(x1, y1) and B(x2, y2) be two given points. Let P(x,y) be any point on the line joining A
and B. Let P divide AB in the ratio of m1 : m2 internally. The we write AP : PB = m1 : m2.

Perpendiculars AL, PN, and BM are drawn from the point A,P, and B respectively on the
X-axis. Y

B(x2, y2)

Again perpendiculars AQ, PR are drawn from the

points A and B to PN and BM respectively. P(x, y) R

Then, AQ = LN = ON – OL = x – x1 A(x1, y1) Q
PR = NM = OM – ON = x2 – x1

QP = NP – NQ = NP – LA = y – y1 X' O LN MX

BR = MB – MR = MB – NP =y2 – y Y'

Since QP and BR are parallel, triangles PAQ and BPR are similar.

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The corresponding sides of similar triangles are proportional.

? AP = APRQ = QRBP
BP
or, mm12 xx2––xx1 yy2––yy1

Taking the first two ratio, we get,

mm12 xx2––xx1
or, m1x2 – m1x1 = m2x – m2x1

or, m1x2 + m2x1 = m2x – m2x1

or, m1x2 + m2x1 = x(m1 + m2)

? x = m1x2 + m2x1
m1 + m2

Again, taking the first and third ratios, we get,

mm12 yy2––yy1
or, m1y2 – m1y = m2y – m2y1

or, m1y2 + m2y1 = (m1 + m2) y

? y = m1y2 + m2y1
m1 + m2
m1x2 + m2x1 m1y2 + m2y1
Hence, the coordinates of P are m1 + m2 , m1 + m2

Special Cases

(i) If P(x, y) is the mid point of line segment joining A(x1, y1) and B(x2, y2) then AP = BP
and AP : BP = 1:1.

Then, (x, y) = m1x2 + m2x1 , m1y2 + m2y1 = x1 + x2 , y1 + y2
m1 + m2 m1 + m2 2 2

(ii) If P(x, y) is an external point of the line segment joining P(x, y)
B(x2, y2)
the point A(x1, y1) and B(x2, y2), then the coordinates of
P are given by

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

A(x1, y1)

(iii) If the point P(x1, y1) divides the line segment joining the points A(x1, y1) and B(x2, y2) in
the ratio of k : 1 internally, then the coordinates of P are given by,

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

For the problems in which the ratio is to be calculated by which a point divides the line
segment, the formula is more convenient.

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Centroid Formula

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of triangle A(x1,y1)
∆ABC. Let D,E, and F be the mid points of the sides BC, CA and

AB respectively. Then the medians AD, BE and CF are drawn, The 2

medians intersect at G which is called the centroid of the triangle F 1G E
B(x2,y2) D (x3,y3)
ABC. Since D is the mid-point of BC, the coordinates of D are
x2 + x3 y2 + y3
2 , 2

From plane geometry, we know that the centroid of a triangle

divides the median in the ratio 2:1. i.e. G divides AD in 2:1 ratio.

2 × x2 + x3 + 1 × x1 2 × y2 + y3 + 1 × y1
2 + 1 2 + 1
G (x, y) =G 2 , 2

=G x1 + x2 + x3 , y1 + y2 + y3
3 3

Definition : The point of intersection of the medians of a triangle is called centroid of the

triangle.

Worked out Examples

Example 1. Find the coordinates of the points which divide the line segment joining
Solution:
(3, 1) and (6, 7) in the ratio of 2:3 B(6, 7)
Example 2.
Solution : Here, (x1, y1) = (3, 1), (x2, y2) = (6. 7) 3
m1 : m2 = 2 : 3 P(x, y)
2
Let the required point be P(x, y) A(3, 1)

Then, by using formula,

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

= 2 × 6+3 × 3, 2 × 7+3 × 1
2+3 2+3

= 12 + 9, 14 + 3
5 5

? (x, y) = 21 , 17
5 5

Find the coordinates of the point which divides the line segment joining

(2,1) and (8, 2) in ratio of 3:2 externally. P(x, y)

Let given two points be A(2, 1) and B(8, 2) 2
Then, A(x1, y1) = A(2, 1), B(x2 y2) = B(8, 2)
B(8, 2)
AP : BP = m1 : m2 = 3:2 3
Let the required external point be P(x, y)

Then, by using formula, A(2, 1)

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P(x, y) = m1x2 – m2x1 , m1y2 – m2y1 = 3 × 8 – 2 × 2 , 3×2–2×1
m1 – m2 m1 – m2 3 – 2 3–2

= (24 – 4, 6 – 2) = (20, 4)

Example 3. Find the coordinates of the mid-point of the line segment joining the point
Solution: (4, 6) and (–2, –4).

Example 4. Here, (x1, y1) = (4, 6) and (x2, y2) = (–2, –4).
Solution: Let the required mid-point be (x, y).

Then by using formula,

(x, y) = x1 + x2 , y1 + y2 = 4 – 2 , 6–4
2 2 2 2

= 2 , 2 = (1, 1)
2 2

In what ratio does the point (3, 2) divide the line segment joining the points
(1, 4) and (–2, 8).

Here, (x1, y1) = (1, 4)

(x2, y2) = (–2, 8)

Let (3, 2) divides the line segment joining the points (1, 4) and (–2, 8) in

ratio of m1:m2.

Now, x = m1x2 + m2x1
m1 + m2

or, 3 = m1(–2) + m2 . 1 or, 3m1 + 3m2 = –2m1 + m2
m1 + m2
m1 2
or, 5m1 = – 2m1 or, m2 = – 5

? m1 : m2 = – 2:5

Hence, the point (1, 4) divides the join of the given points in 2:5 ratio
externally.

Example 5. The point (2, 4) divides the line segment joining the points (4, –6) and (p, q)
Solution: in the ratio of 1:2 internally. Find the values of p and q.

Here, (x1, y1) = (4, –6) and (x2, y2) = (p, q) m1:m2 = 1:2

(x, y) = (2, 4)

Then, x = m1x2 + m2x1
m1 + m2

or, 2 = 1 . p + 2 . 4 or, 2 = p + 8
1 + 2 3

or, 6 = p + 8 ? p = –2

and y = m1y2 + m2y1
m1 + m2

or, 4 = 1 . q+2. (–6)
1+2

or, 12 = q – 12 ? q = 24

Hence, (p, q) = (–2, 24)

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Example 6. In what ratio is the line joining the points (2, –3) and (5, 8) divided by X-axis
Solution : and Y-axis?

Let the X-axis divide the line segment joining the points (2, –3) and (5, 8) in

ratio of m1:m2 Y (5, 8)
Here, (x1, y1) = (2, –3) 8
7
(x2, y1) = (5, 8) 6
At X–axis, y-coordinate = 0 5
4
3
2
y = m1y2 + m2y1 1 X
m1 + m2 X' -6 -5 -4 -3 -2 -1-1O 1 2 3 4 5 6
-2
or, 0 = m1 . 8 + m2 . (–3) -3 (2, -3)
m1 + m2
or, 8m1 – 3m2 = 0 Y'

or, 8m1 = 3m2

? m1 = 3
m2 8

Hence, X-axis divides the joining of given points in 3:8 internally.

Again, let Y-axis divide the join of given points in m:n ratio, At Y-axis
X-coordinate = 0

x = mx2 + nx1
m + n

or, 0 = m .5 + n . 2 or, 0 = m × 5 + n × 2
m + n

or, m : n = – 2 : 5

Hence, Y–axis divides the joining of given points in 2:5 externally in ratio of 2:5.

Example 7. Find the coordinates of the points which trisect the join of (2, 4) and (8, 9)
Solution: equal parts.

Let given points be A(2, 4) and B(8, 9), P(x', y') and Q(x", y") be the required

points which divide AB in three equal parts. P divides AB is 1:2 ratio,

m1:m2 = 1:2 (x1, y1) = (2, 4), (x2, y2) = (8, 9)

Now, P(x', y') = m1x2 + m2x1 , m1y2 + m2y1 B(8, 9)
m1 + m2 m1 + m2 Q(x", y")

= 1 × 8 + 2 × 2 , 1×9+2×4
1 + 2 1+2

= 12 , 17 = 4, 17 P(x', y')
3 3 3

Q(x", y") is he mid-point of PB. (x1, y1) = 4, 17 ,
3
A(2, 4)
x1 + x2 y1 + y2
(x", y") = 2 , 2 (x2, y2) = (8, 9)

= 4 + 8 , 17 + 9 = 6, 22
2 3 3

2

? P 4, 17 and Q 6, 22 are the required points.
3 3

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Example 8. Show that the points A(1, 2), B(3, 0), C(7, 4), and D(5, 6) are the vertices of a
Solution :
parallelogram. A(1, 2) D(5, 6)
Example 9.
Solution: Here, A(1, 2), B(3, 0), C(7, 4), and D(5, 6) are the
vertices of quadrilateral ABCD.

A quadrilateral is a parallelogram if the diagonals

bisect each other at a point. B(3, 0) C(7, 4)

Mid-point of diagonal BD = 3 + 5 , 0+6 = (4, 3)
2 2

Mid-point of diagonal AC = 1 + 7 , 2+4 = (4, 3)
2 2

Hence, the mid-point of diagonal BD = the mid-point of diagonal AC

? ABCD is a parallelogram.

If P(8, 5), Q(–7, –5), and R(–5, 5) are the three vertices of a parallelogram
PQRS in order. Find the fourth vertex.

Here, P(8,5), Q(–7, –5) R(–5, 5), and S(x', y') are the vertices of a parallelogram.

Now, mid-point of diagonal PR. P(8, 5) S(x', y')

= 8 – 5 , 5+5 = 3 , 5
2 2 2

Q(–7,–5) R(–5, 5)

Again, mid-point of QS = – 7 + x' , – 5+ y'
2 2

since PQRS is a parallelogram,

mid-point of diagonal PR = mid-point of QS

or, 3 , 5 = – 7+ x' , – 5 + y'
2 2 2

? 3 = –7 + x' and 5 = – 5+ y'
2 2 2

or, 3 = – 7 + x' or, 10 = – 5 + y'

? x' = 10 ? y' = 15

Hence, the fourth vertex of the parallelogram is (10, 15).

Example 10. The vertices of a triangle are (3, –5), (–7, 4), and (10, –2). Find the centroid of

the triangle. A(3, –5)

Solution: Let the vertices of triangle be A(3, –5), B(–7, 4) and C(10, –2)

Now, A(x1, y1) = (3, –5) E
C(10, –2)
B(x2, y2) = (–7, 4) G

C(x3, y3) = (10, –2) B(–7, 4)

Let G(x, y) be the centroid of the triangle then by using formula,

G(x, y) = x1 + x2 + x3 , y1 + y2 + y2
3 3

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= 3 + (–7) + 10 , – 5+4+ (–2)
3 3

? G(x, y) = (2, – 1)

Hence, the centroid of triangle is (2, –1).

Exercise 6.2

Very Short Questions

1. Let A(x1, y1) and B(x2, y2) be two given points and P divides AB in the ratio of m1 : m2.
Then, write the following formula.

(a) Section formula (i) Internal division (ii) External division
(b) Mid-point formula
A(x1,y1)

2. In the adjoining figure of triangle F E
(a) What are AD, BE and CF called ? B(x2,y2) G
(b) What is the point G called ? D (x3,y3)
(c) Write down the coordinates of G.

3. (a) Write down the coordinates of the mid-point of the line segment joining A(4, 5) and
B(8, 10).

(b) Find the coordinates of the point which divides the join of (5, 4) and (8, 9) in the
ratio of 1:2 internally.

(c) Find the coordinates of the point that divide the join of (1, 3) and (2, 7) externally
in the ratio of 3:4.

(d) Find the coordinates of centroid of the triangle PQR whose vertices are P(1, 2),
Q(4, 5) and C(4, 2).

Short Questions

4. (a) Find the coordinates of a point dividing internally the line joining the points.

(i) (4, 5) and (7, 8) in the ratio of 2:3

(ii) (–3, 6) and (1, –2) in the ratio of 4:5.

(b) Find the coordinates of a point dividing externally the line joining the points.

(i) (1, 4) and (–2, 16) in the ratio of 1:3.

(ii) (5, –2) and (9, 6) in the ratio of 3:1.

5. (a) A line segment having an end (5, –2) is divided by the point (8, 4) in the ratio of
3:1, find the another end.

(b) If A(2, k) is the mid-point of the line segment PQ joining the points P(4, 3) and
Q(0, 2), find the value of k.

(c) If the coordinates of the mid-point of the line joining the point is (3, 5) and (x, y) is
(1, –1). Find the values of x and y.

(d) The point (–2, –9) divides the line segment joining (a. b) and (2, 7) in the ratio of
3:4, find the values of a and b.

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6. (a) Find the coordinates of the centre of circle from the given circles.

(i) (ii)

A(6, –7) O B(6, 7) P(2, 4) O Q(4, 8)

(b) Find the coordinates of unknown vertices of the given parallelograms.

(i) A(8,5) D(p,q) (ii) P(–2, –1) S(1, 2)

(1,3) (1, 1)

B(9,2) C(x,y) Q(a, b) R(x, y)

7. (a) If the centroid of the triangle is (1, 3), its two vertices are (2, 3) and (–3, 4), find the
third vertex.

(b) The medians of a triangle meet at(1, 1) whose two vertices are (5, 9) and (–4, 1).
Find the third vertex.

Long Question

8. (a) In what ratio does the point –1, 24 divide the line joining the points (–3, 4) and
(2, 6)? 5

(b) In what ratio does the point (3, –2) divide the join of the points (1, 4) and (–3, 16) ?

(c) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1)
and (7, 6) ?

9. (a) Find the coordinates of the points of trisection of the line segment joining the
points (2, –2) and (–1, 4).

(b) Find the coordinates of the points of trisection of the line segment joining the
points P(1, 2) and Q(4, 2).

(c) Find the coordinates of the points of trisection of the line segment joining the
points (1, –2) and (–3, 4).

10. (a) If the point (x, 14) divides the line segment joining the points (7, 11) and (–18, 16).
In which ratio the point divides the line and hence find the value of x.

(b) The point (5, y) divides the line segment joining the points (3, 7) and (8, 9) in a
ratio. Find the ratio and value of y.

11. (a) In what ratio does X-axis divide the line, segment joining the points (2, –4) and
(–3, 6)?

(b) In what ratio does X-axis divide the line segment joining the points (2, –3) and
(5, 6)?

(c) In what ratio does Y-axis divide the line segment joining the points (2, –3) and
(–6, 10) ?

(d) In what ratio does Y-axis divide the line segment joining the point (5, 4) and (–2, –2)?

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(e) Find the ratios in which the line joining (3, –2) and (–3, 6) is cut by the axes of
coordinates .

12. (a) Prove that the following points represent the vertices of a parallelogram.

(i) (–2, –1), (1, 0), (4, 3) and (1, 2)

(ii) (2, –2), (8, 4), (5, 7) and (–1, 1)

(iii) (10, –6), (2, –6), (–4, –2) and (4, –2)

(b) Three consecutive vertices of a parallelogram are respectively P(1, 2), Q(1,0) and
R(4, 0). Find the fourth vertex S.

(c) If (3, 7), (5, –7) and (–2, 5) are the vertices of parallelogram, find the coordinates of
the fourth vertex opposite to (5, –7).

13. (a) If the points (–1, 3), (1, –1) and (5, 1) are the vertices of a triangle, then find the
length of the median drawn through (5, 1).

(b) If P(2, –1), Q(–2, 2) and R(–1, 4) are the mid points of the sides of ∆ABC, find the
vertices of the triangle.

14. (a) P(x, y), Q(0, –2), R(5, 3) and S(3, 7) are the vertices of a parallelogram PQRS. Find
the coordinates of point P.

(b) The two consecutive vertices of a parallelogram are (1, 1) and (3, 6). The diagonals
of the parallelogram cut each other at (7, 7/2). Find the remaining vertices of the
parallelogram.

(c) A right angled triangle has the vertices (1, 1), (1, 7) and (7, 1), prove that the mid
point of the hypotenuse lies at equal distance from each of its vertices.

Project Work

15. List the application of section formula and mid-point formula in our daily life and
discuss it on your class room.

3. (a) 6, 15 (b) 6, 17 (c) (–2, –9) (d) (3, 3)
2 3
(ii) (11, 10)
4. (a) (i) 26 , 31 (ii) – 11 , 22 (b) (i) 5 , –2 (d) a = –5, b = –21
5 5 9 9 2 (ii) D(p, q) = (–7, 4)
5
5. (a) (9, 6) (b) k = 2 (c) x = –1, y = –7

6. (a) (i) (6, 0) (ii) (3, 6) (b) (i) C(x, y) = (–6, 1)

7. (a) (4, 2) (b) (2, –7) 8.(a) 2 : 3 (b) 1 : 3 externally (c) 3 : 2

9. (a) (1, 0), (0, 2) (b) (2, 2), (3, 2) (c) – 1 , 0 and –5 , 2
3 3
39
10. (a) 3 : 2, x = –8 (b) 2 : 3, y = 5 11.(a) 2 : 3 (b) 1 : 2
(c) 1 : 3
(d) 5 : 2 (e) 1 : 3, 1 : 1
12. (b) (4, 2)
14. (a) (–2, 2) (c) (–4, 19) 13.(a) 5 (b) (3, 1), (1, –3), (–5, 7)

(b) (13, 6), (11, 1)

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Equations of Straight 7
Lines

7.0 Review

Group discuss the following questions.
(i) What is difference between a straight line and a curve ?

(ii) Draw a diagram joining the following points on a graph paper, is it a curve or straight line?

x 2 34 5

y 4 6 8 10

(iii) Draw a graph of x – y = 1. Does it represent a straight line.

7.1.1 Equations of Straight Lines Parallel to Coordinate Axes

(i) Equations of straight Lines Parallel to Y-axis. Y B
a x=a
In the given figure, AB is a line parallel to Y-axis. The X'
distance between the Y-axis and the line AB is 'a' units. Y' X
Hence every point on line has x-coordinate a. Hence, the A
equation of the line AB is x = a. If the line AB is right of
Y-axis, then value of a is positive and if it is in the left of
Y-axis, the value of a is negative.

Y

P 5 N
(-7, 3) 4 (7, 4)
(-7, 2) 3 (7, 3)
(-7, 1) 2 (7, 2)
(7, 1)
(-7, 0) 1 (7, 0)
1 2 3 4(75, -61)7 8
X' -8 -7 -6 -5 -4 -3 -2 -1 O X

-1
-2
-3 (7, -3)

-4
-5
-6
Q -7 (7, -6) M

Y'

For example, in the above graph line MN is parallel Y-axis and the points on the line MN are
(7, 0), (7, 1), (7, 2), (7, 4), (7, 1), (7, –3), (7, –6), and so on. We note that x-coordinate is 7 at
every point of the line; hence, the equation of MN is x = 7.

Again, the line PQ is parallel to Y-axis and every point on the line, x-coordinate is –7. Hence,
the equation of the line PQ is x = –7 or x + 7 = 0

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Example 1. Find the equation of a straight line parallel to Y-axis and at 4 units right
Solution: Y
from the Y-axis. Show the line in the figure. 5 N
Here, equation of any line parallel to Y-axis is x = a. 4
3
The distance of the line from Y-axis = a = 4. 2 x=4

Hence, the required equation of straight line is 1 X
X' -2 -1-1O 1 2 3 4 5
-2 N
x=a i.e. x = 4.
In the figure, MN is a straight line parallel to Y-axis, whose equaYti'on is x = 4.

Example 2. Find the equation of the straight line parallel to Y-axis and passing through
Solution : the point (–4, 6).

Here, the equation of a line parallel to Y-axis is x= a

This line passes through the point (–4, 6)

Hence, a = –4 i.e. x = –4 or x + 4 = 0 is the required equation of a straight line.

(ii) Equation of Straight Line Parallel to X-axis A Y B
X' y=b X
In the figure AB is parallel to X-axis. The distance between
AB and X-axis is b units. Hence, every point on the line has b
y-coordinate b. Hence, the equation of the line AB is y = b. O

Y

A y=4 B

X' O X Y'

C y = -3 D

Y'

In the above graph, AB and CD are parallel to X-axis. Every point on line AB has y-coordinate
4. It means that AB is at 4 units above X-axis. Hence, its equation is y = 4. Again, every
point on line CD has y-coordinate –3. It means that CD is at 3 units below X-axis. Hence, its
equation is y = –3 or y + 3 = 0.

Example 3. Find the equation of a straight line parallel to X-axis at 5 units below X-axis.
Solution :
Show it in figure. Y
Example 4.
Solution : Here, equation of a line parallel to X-axis is y = b. X' O X
The line is 5 units below X-axis, hence b = –5.

Hence, the required equation of straight line is y = –5

y = –5 or y + 5 = 0. Y'

Find the equation of straight line parallel to X-axis and passing through the

following points: (i) (7, 5) and (ii) (–5, –6)

(i) Let the equation of a line parallel to X-axis be y = b, as the line passes
through the point (7, 5), b= 5. Hence, the required equation of line is y = 5.

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(ii) Let the equation of straight line parallel to X– axis be y = b, as the line
passes through the point (–5, –6), b= –6. Hence, the required equation
of straight line is y = –6 or y + 6 = 0.

7.1.2 Slope or Gradient of a straight Line.

Let the angle made by a line AB with X-axis in anti-clockwise Y B
be T. It is called inclination of the line AB. The tangent of the X
inclination of a line is called slope of the line. It is denoted by T
m and given by m = tan T. O
Y'
For example : If inclination of a line is (T) = 45°, X'
then, slope (m) = tan 45° = 1
A
Definition : Let T be the inclination of a straight line then the
tangent of the inclination is called its slope. it is also known
as gradient and denoted by m.

? slope (m) = tan T.

Note :
(i) Slope of a line is positive if the inclination T is acute. (0°<T<90°)

(ii) Slope of a line is negative if the inclination T is obtuse, (90° < T < 180°).

(iii) If line is parallel to X–axis, then slope is zero. i.e. m= tan T° = 0

(iv) If line is parallel to Y–axis, then slope is i.e. m = tan 90° = ∞.

Example 5. Find the slope of the given lines in the figures.

(a) Y (b) Y X
B N
135°
135°
X' P O

X' CO X
A Y'
Y' M

Solution : (a) Here, ‘XCB + ‘BCX' = 180° (Linear pair angles)
‘XCB = 180° – 135° adjacent angles.

Inclination = ‘ XCB = 45°

Hence, the slope of AB = tan 45° = 1.
(b) Here slope of MN (T) = ‘XON = 135°

Hence, the slope of line MN = tan 135 = –1

Example 6. In the given figure of right angled isoceles triangle, the hypotenuse BC is
Solution : parallel to X-axis. Find the slopes of the sides of the triangle.

Here, ABC is an isoceles right angled triangle.

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So, ‘ABC = ‘ACB = 45° Y
Now, BC is parallel to X-axis. A
Inclination of BC = 0c.

Hence slope of BC = tan 0c = 0. B 45° 45° C 135°
Inclination of BA = 45° 45°
X' X
O

Hence slope of BA = tan 45° = 1. Y'

Inclination of CA = 180° – 45° = 135°

Hence the slope of CA = tan 135° = –1

? Slope of BA = 1, slope of CA = –1, slope of BC = 0.

(b) Slope a line passing through two points :

Let AB be a given line passing through points A(x1, y1) and B(x2, y2). AL, BM
perpendiculars are drawn through A and B on
X-axis. Again, AN is drawn perpendicular to BM. Y B(x2, y2)

Now, OL = x1, OM = x2, A(x 1, y 1)
LM = OM – OL = x2 – x1 T
LA = y1 = MN, MB = y2, N
NB = MB – MN = y2 – y1 X' T L MX
QO
Let the inclination of the line AB be T .
Y'

‘BAN = ‘AQM = T (? corresponding angles an AN//XOX')

Then, slope of AB is given by,

m = tan T = rise = NB = y2 – y1
run ON x2 – x1
y2 – y1
Hence, the slope of a line passing through two points (x1, y1) and (x2, y2) is m = x2 – x1

Example 6. Find the slope of a line passing through the points A(4, 2) and B(6, –7).

Solution : Here, (x1, y1) = (4, 2)
Example 7.
(x2, y2) = (6, –7)

? slope of the line (m) = y2 – y1 = –7 – 2 =– 9
x2 – x1 6–4 2

If the slope of a line passing through the points (6, 7) and (9, k) is 3 , find
the value of k. 2

Solution: Here, (x1, y1) = (6, 7)

(x2, y2) = (9, k)

slope (m) = 3
2
y2 – y1 3 k –7
slope (m) = x–27– x1 or, 2 = 9 –6
or,
3 = k 3 or, 9k = 2k – 14
2

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or, 7k = –14 or, k = – 2
? k = –2

Collinear Points

A set of points lying on the same straight line are called collinear points. C
In the adjoint figure, A, B, and C points are on the same straight line.
Hence, they are collinear points. If A, B, and C are collinear points, then B
slope of AB = slope of BC = slope of AC. A

Hence, the slopes of the line segments of the same line are Y C(8, 9)
equal to each other. 9
8
7 B(6, 7)
For examples : 6
5
4
In the adjoining graph A(1, 2), B(6, 7), and C(8, 9) are 3
collinear points. 2 A(1, 2)
1
Slope of AB = 7–2 = 5 =1 X' -3 -2 -1-1O 1 2 3 4 5 6 7 8 9 X
6–1 5
9 – 7 2 -2
Slope of BC = 8 – 6 = 2 =1 Y'

Slope of AC = 9 – 2 = 7 =1
8 – 1 7
Hence, slope of AB = slope of BC = slope of AC.

Example 8. Show that the points A(–3, –2), B(1, 2), and C(9, 10) are collinear.
Solution :
Given points are A(–3, –2), B(1, 2) and C(9, 10),

we have, slope (m) = y2 – y1
x2 – x1

slope of AB (m1) = 2+2 = 4 = 1.
1+3 4
8
Slope of BC (m2) = 10 – 2 = 8 = 1.
9–1
10+2 12
Slope of AC (m3) = 9+3 = 12 = 1.

? Slope of AB = slope of BC = slope of AC.

Hence, the points A, B and C are collinear.

Intercepts Y
B(0, b)
Let MN be a straight line. It cuts X-axis at A and Y-axis at B. N
Then, O
Y'
(i) OA is called x-intercept of line MN and its is denoted by a.

? x-intercept of MN = OA = a X' A(a, 0)
X
(ii) OB is called y-intercept of MN and it is denoted by b.
M

? y-intercept of MN = OB = b

(iii) OA and OB are called intercepts of the line MN.

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Note :
(i) x-intercept is considered positive if it is measured to the right of origin and negative

if it is measured to the left of origin.

(ii) y-intercept is considered positive if it is measured above the origin and negative if it
is measured below the origin.

(iii) A horizontal line has no y-intercept and a vertical line has no y-intercept. only the
oblique line or inclined line has both x-intercept and y-intercept.

Example 9. Find the intercepts of the following straight lines as shown in the figures.

(a) Y (b) Y
X' B (0, 6) X'
O X
(4, 0) A(3, 0)
OA X
B(0, -5)

Y' Y'

(c) Y (d) Y

B(0, 3) X' O X
A(-4, 0)

X' A(-3, 0) O X B(0, -5)
Y'
Y' y-intercept (OB) = b = 6
y-intercept (OB) = b = –5
Solution : (a) x-intercept (OA) = a = 4 y-intercept (OB) = b = 3
y-intercept (OB) = b = –5
(b) x-intercept (OA) = a = OA = 3

(c) x-intercept (OA) = a = –2

(d) x-intercept (OA) = a = –4

7.1.3 Equation of Straight Line Y P(x, y) B
In Slope Intercept Form (y=mx + c)

To find the equation of straight line in form of y = mx + c. CT N
Let AB be a straight line which makes an intercept OC on Y-axis. MX
O
Then, let OC = c, and let T be the inclination of AB. Y'

Then, slope of the line AB(m) = tan T. X' Q T
Let P(x, y) be any point on the line AB. A
PM is drawn perpendicular to X-axis from P.

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Then, OC = MN = c, MP = y

OM = CN = x,

NP = MP – MN = y – c.

Since CN is parallel to X-axis,

‘PCN = ‘CQM = T.

From right angled triangle PCN,

tan T = PN
cCN
or, m = y –
x
or, mx = y – c

? y = mx + c

? y = mx + c is the required equation of straight line in slope intercept form.

Note :

If the line AB passes through the origin and y-intercept c = 0, the equation of the straight
line passing the origin is y = mx.

Example 10. Find the equation of straight line whose y-intercept is 4 and slope is 3 .
4
Solution: Here, y-intercept (c) = 4

slope (m) = 3 .
4
Now, equation of the straight line is

y = mx + c

or, y= 3 x+4
4
or, 4y = 3x + 16

? 3x – 4y + 16 = 0 is the required equation of the straight line.

Example 11. Find the slope of y-intercept of line y = 3x + 8.
Solution: Here, comparing equation y = 3x + 8 to y = mx + c,

y-intercept (c) = 8
slope (m) = 3

Example 12. Find the equation of straight line passing through the origin which inclination
is 60°.

Solution: Here, angle of inclination (T) = 60°
slope (m) = tan 60° = 3

Equation of the straight line passing through the origin is,
y = mx

or, y = 3x
? y = 3x is the required equation of the straight line.

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Example 13. Find the equation of straight lines equally inclined to the axes and passing
through the point (0, –8).

Solution: There may be two lines l1 and l2 equally inclined to the axes as shown in the

figure. Y

For line l1, angle of inclination (T1) = 45°

Slope (m) = tan 45° = 1 X' 135° 45° 45° X
y-intercept (c) = –8 45° O

Equation of the line l1 is 45° 45°

y = mx + c (0, –8)

or, y = 1 . x – 8 l1 l2
or, y = x – 8 Y'

? x–y–8=0

For the line l2, the inclination (T2) = 135°, slope (m) = tan 135° = –1
Equation of straight line is y-intercept (c) = –8

y = mx + c

or, y = – 1 . x – 8

? x+y+8=0

Hence, the required equations of the lines are x – y – 8 = 0 and x + y + 8 = 0

Exercise 7.1

Very Short Questions
1. (a) Define slope of a line.

(b) Define x-intercept and y-intercept of a line.
(c) Write down the equation of straight line in slope -intercept form.
(d) What is the equation of straight line passing through the origin and having

slope m?
2. Find the equation of straight line from the following figures:

(a) Y (b) Y

M 4 N X' O X
X' O X P –5 Q
Y'
Y'

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(c) Y A (d) YE
5 X'

X' O X O (4, 0) X

Y' B Y' F

(e) Q Y (f) Y
(-3, 4) B

X' O X X' O 45° X

P Y' A
Y'

(g) Y (h) Y

X' A 135° X X' O AX

O

B (0, –3) 45°
Y' B (0, –6)

Y'

3. Find the equation of straight lines parallel to Y-axis passing through the following point:

(a) (2, 4) (b) (–6, 3) (c) (–4, 5) (d) (–2, –4)

4. Find the equation of straight lines parallel to X-axis and passing through the following
points.

(a) (–4, 5) (b) (–6, –6) (c) (2, 7) (d) (8, 3)

5. (a) Find the equation of straight lines parallel to X-axis and at a distance of-

(i) 4 units above X-axis (ii) 7 units below X-axis

(iii) 7 units above X-axis
3

(b) Find the equation of straight lines parallel to Y-axis and at a distance of-

(i) 6 units right of Y-axis (ii) 8 units left of Y-axis

(iii) 4 units left of Y-axis
3

6. Find the slope of lines passing through the points:

(a) (4, 5) and (6, 7) (b) (2, 4) and (–2, –4)

(c) (–6, –4) and (4, 2)

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Short Questions
7. Fill slope and y-intercept of given lines in the table:

Equations Slope (m) y-intercept (c)
(a) y = 2x + 1
(b) y = – 4x + 5
(c) x + y = 2
(d) x – y = 4
(e) 2x + 3y = 7
(f) 9x + 3y = 12
(g) y = 3x + 4
(h) y = – 3x + 2

8. Find the equation of straight line under the given conditions:

Slope (m) y-intercept (c)

(a) 1 4

(b) – 1 2

(c) 4 –6
3

(d) 3 0 (passing through origin)

(e) – 3 4
4

(f) – 2 –4

(g) 1 –1
6 –5

(h) tan 60°

(i) tan 120° –6

9. (a) Find the equation of a straight line cutting off the y-intercept 4 from the axis of y
and inclined to 60° with the positive direction of X-axis.

(b) Find the equation of a straight line making an angle of 120° with the positive
direction of X-axis and cutting of an intercept 8 from the negative Y-axis.

(c) Find the equation of a straight line passing through the point (0, –5) and inclined
at an angle of 150° with positive direction of X-axis.

Y A

10. (a) The side BC of an equilateral triangle ABC is parallel

to X-axis find the slopes of sides AB, AC, and BC.

B C
X' O X

Y'

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Y

(b) The sides PQ and RS of a rectangle PQRS is parallel P Q
R
to X-axis, calculate the slopes of PQ and PS. S
X
X' O
P
Y'
S
(c) In the figure alongside, PQRS is a square in which Y

the diagonal QS is parallel to X-axis. Then, find the

slopes of PQ, PS and RP. Q

Long Questions: X' O RX
Y'

11. (a) If three points are P(–4, –3), Q(1, 2), and R(9, 10), then prove that they are collinear.

(b) If (a, 0), (0, b) and (1, 1) are collinear, prove that 1 + 1 = 1
a b
x y
(c) If (a, 0), (0, b) and (x, y) are collinear, prove that a + b = 1

(d) If (x, y), (0, c) and – c , 0 are collinear, prove that y = mx + c.
m
12. Derive the equation of straight lines in the slope intercept form.

Project Work

13. Prepare a report on "uses of slopes in our daily life" and present in your class.

2. (a) y = 4 (b) y + 5 = 0 (c) x = 5 (d) x = 4

(e) x + 3 = 0 (f) y = x (g) x + y + 3 = 0 (h) x – y = 6

3. (a) x = 2 (b) x + 6 = 0 (c) x + 4 = 0 (d) x + 2 = 0

4. (a) y = 5 (b) y + 6 = 0 (c) y = 7 (d) y = 3

5. (a) (i) y = 4 (ii) y + 7 = 0 (iii) 3y – 7 = 0

(b) (i) x = 6 (ii) x + 8 = 0 (iii) 3x + 4 = 0

6. (a) 1 (b) 2 (c) 3
5

7. (a) m = 2, c = 1 (b) m = –4, c = 5 (c) m = –1, c = 2 (d) m = 1, c = –4

(e) m = – 2 , c = 7 (f) m = – 3, c = 4 (g) m = 3, c = 4 (h) m = – 3, c = 2
3 3 (c) 4x – 3y – 18 = 0 (d) y = 3x

8. (a) y = x + 4 (b) x + y = 2

(e) 3x + 4y = 16 (f) 2x + y + 4 = 0 (g) x – 6y = 6 (h) y = 3x – 5

(i) 3x + y + 6 = 0 9.(a) y = 3x + 4 (b) 3x + y + 8 = 0

(c) x + 3y + 5 3 = 0 10.(a) 3, – 3, 0 (b) 0, f (c) 1, – 1, f

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

7.2 Equation of a Straight Line in Intercepts form : x + y = 1
a b

To find the equation of a straight line in the form of x + y = 1. Y
a b
N
Let MN be a straight line which cuts X-axis at A
B(0, b)
and Y-axis at B such that,

x-intercept = OA = a P(x, y)

y-intercept = OB = b

Then, the coordinates of A and B are respectively X' O A(a, 0) X
Y' M
(a, 0) and (0, b).

Let P(x, y) be any point on the line AB. Then

slope of AP = y – 0 = x y a
x – a –
b – y b–y
slope of PB = 0 – x = –x

Since, A, P and B are collinear, we have

slope of AP = slope of PB

or, x y a = b– y
– –x

or, – xy = bx – xy – ab + ay

or, ay + bx = ab

Dividing both sides by ab, we get

ay + bx = ab
ab ab
ay bx
or, ab + ab = 1

? x + y = 1 is the required equation of the straight in double intercepts form.
a b

Alternative Method

Let MN be a straight line, which cuts X-axis at A and Y-axis at B such that its

x-intercept = OA = a

y-intercept = OB = b Y
N
Let P(x, y) be any point on AB. PQ and PR are drawn
perpendiculars from P to OX and OY. B

Then PQ = y

PR = OQ = x bR P(x, y)

Now, area of 'OBA = Area of 'BPO + Area of 'POA A
X
or, 1 OA × OB = 1 PR × OB + 1 PQ × OA X' O Q
2 2 2 a M

or, 1 a × b = 1 x × b + 1 y × a Y'
2 2 2

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Dividing by 1 ab on both sides, we get,
2
x y
1 = a + b

? x + y = 1, which is the required equation of the line in intercepts form.
a b

Worked out Examples

Example 1. Find the equation of a straight line whose x-intercept and y-intercepts are
Solution: respectively 4 and 5.

Example 2. Here, x-intercept (a) = 4
Solution:
y-intercept (b) = 5
Example 3.
Solution: Equation of required straight line is

x + y = 1
a b
x y
or, 4 + 5 = 1

or, 5x + 4y = 20

? 5x + 4y = 20 is the required equation.

Find the x-intercept and y-intercept of a straight line 4x + 3y = 12.

Here, 4x + 3y = 12

or, 4x + 3y = 12
12 12 12
x y
or, 3 + 4 = 1

Comparing it to x + y = 1, we get
a b

x-intercept (a) = 3

y-intercept (b) = 4

Find the equation of a straight line which passes through the point (3, –5)
and cuts off intercepts on the axes equal but opposite in sign.

Let x-intercept = a and y-intercept = b

But a = – b

The equation of the straight line is

x + y = 1
a b
x y
or, –b + b = 1

or, – x + y = b

or, x – y – b = 0 .......................... (i)

It passes through the point (3, –5)

So, 3 – (–5) – b = 0

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

or, 3 + 5 = b
? b=8
Substituting the value of b in equation (i), we get,

x–y–8=0
? x – y = 8 is the required equation.

Example 4. Find the equation of straight line which passes through the point (–2, 3) such
Solution: that its portion between the axis is divided by the point in the ratio 3 : 4.

Example 5. Let the required line be MN whose x-intercept is a and y-intercept is b, as
Solution:
shown in the figure. The point P(–2, 3) divides the intercepted portion

AB is ratio 3 : 4. Then, Y

x-coordinate of P = m1x2 + m2x1 = 3 × 0 + 4 × a N
m1 + m2 3 + 4 B(0, b)
4a
or, – 2 = 7 3
4
? a = – 7 P(–2, 3)
2
m1y2 + m2y1
and y-coordinate of P = m1 + m2
or, 0
3 = 3 × b + 4× X' A(a, 0) O X
3 + 4 M Y'
3b
or, 3 = 7

? b=7

The required equation is,

x + y = 1
a b
or, –x27+ 7y= 1
or, – 2x + y = 7

? 2x – y + 7 = 0 is the required equation.

Find the area of triangle formed by the line 3x – 4y – 12 = 0 with the

coordinate axis. Y

We have to find intercepts of the given line

3x – 4y – 12 = 0 X' O X
A(4, 0)
Here, 3x – 4y = 12

or, x – y = 1 B(0, -3)
4 3 Y'
x y
or, 4 + –3 = 1

Comparing it with x + y = 1, we get,
a b

x-intercept (a) = 4

y-intercept (b) = –3

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Area of triangle formed by the line with coordinate axis is given by,

Area of triangle = 1 |OA| × |OB|
2
1 1
= 2 × |4| × |–3| = 2 × 4 × 3 = 6 sq. units.

Note :

Since area cannot be in negative, we take positive value.

Example 6. Find the length of a line intercepted between the coordinate axes if the
Solution:
equation of the line is 4x + 3y = 24. Y
Example 7.
Solution: Here, equation of the given line is 4x + 3y = 24 B(0, 8)

Now, 4x + 3y = 24
24 24
x y
or, 6 + 8 = 1

Comparing it with x + y = 1
a b

x-intercept (a) = OA = 6 X' X
A(6, 0)
y-intercept (b) = OB = 8 O

The coordinates of A and B are (6, 0) and (0, 8). Y'

Now, AB = (x2 – x1)2 + (y2 – y1)2 = (0 – 6)2 + (8 – 0)2
= 36 + 64 = 100 = 10 units.

Find the equation of straight lines which pass through the point (3, 4) and
have intercepts on the axes such that their sum is 14.

Let x-intercept = a

and y-intercept = b

Then, a + b = 14

or, a = 14 – b ........................... (i)

Equation in intercepts form is

x + y = 1
a b
x y
or, 14 – b + b = 1

It passes through the point (3, 4).

3 b + 4 = 1
14 – b

or, 3b + 56 – 4b = b(14 – b)

or, 3b + 56 – 4b – 14b + b2 = 0

or, b2 – 15b + 56 = 0

or, b2 – 8b – 7b + 56 = 0

or, b(b – 8) – 7(b – 8) = 0

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

or, (b – 8) (b – 7) = 0

Either, b – 8 = 0 or b – 7 = 0

? b=8 ? b=7

when b = 7, a = 14 – 7 = 7 [From (i)]

when b = 8, a = 14 – 8 = 6

Required equation of straight line, when a = 7, b = 7 is

x + y = 1
a b
x y
or, 7 + 7 = 1

? x+y=7

Again, required equation of the line when a = 6, b = 8

x + y = 1
6 8
4x + 3y
or, 24 = 1

or, 4x + 3y = 24

? The required equations are x + y = 7 and 4x + 3y = 24.

Exercise 7.2

Very Short Questions

1. (a) Write down the equation of a straight line in double intercepts form.

(b) In equation x + y = 1 , what do a and b represent?
a b

2. (a) In an equation 4x + 3y = 12, put x = 0 and y = 0 successively, what do the values

obtained of x and y represent?

(b) Find the intercepts of given line 6x + 8y = 48.

(c) Write down the equation 6x – 5y – 30 = 0 in the form of x + y = 1.
a b

Short Questions

3. Find the equation of straight lines from the following figures:

(a) Y (b) Y

B(0, 6) X' O A(4, 0) X

X' A(4, 0) X B(0, –5)
Y'
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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

(c) Y (d) Y
X' B(0, 7)
A(–4, 0)
O X

B(0, –5) X' X

A(–5, 0) O

Y' Y'
4. Find the equation of straight lines having intercepts as given below in table:

x-intercept (a) y-intercept (b)

(a) 4 5

(b) – 4 6

(c) 7 8

(d) – 4 –5

(e) – 6 8
5 5

(f) 4 3
2
3

5. Find the intercepts of the following equations:

(a) 4x – 3y =1 (b) 8x + 5y – 40 = 0

(c) 9x + 2y – 27 = 0 (d) 2x + 3y – 18 = 0

(e) 4x + 5y = 20

Long Questions

6. (a) Find the equation of a straight line which cuts equal intercepts on the axes and
passes through the point (2, 4).

(b) Find the equation of straight line which cuts equal intercepts on the axes equal in
magnitude but opposite in sign and passes through the point (–4, –3).

(c) Find the equation of a straight line which passes through the points (4, 5) and
makes y-intercept twice as long as that on x-intercept.

(d) Find the equation of a straight line which passes through the point (–1, 3) and
makes intercepts on the X-axis thrice as long that on Y-axis.

7. (a) Find the equation of the straight line whose portion intercepted is bisected at (2, 3).

(b) Find the equation of a straight line which passes through the point (4, 4) and the
portion of the line intercepted between the axis is bisected at the point.

(c) Find the equation of a straight line passing through the point (1, –5) and the
portion between the axis is divided by this point in the ratio of 1 : 3.

(d) Find the equation of a straight line which passes through the point (–2, 3) and the
intercepted portion of the line is divided by this point in the ratio of 3 : 4.

8. Find the area of triangle formed by the line with the coordinate axis.

(a) 4x + 5y – 12 = 0

(b) 8x + 5y + 40 = 0

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

9. Find the lengths of line intercepted between the coordinates axis of the following lines.

(a) 3x – 4y = 24

(b) x + y = 1
4 5

10. Find the equation of a straight line

(a) passing through the point (4, 1) and makes intercepts on the axis, the sum of
whose lengths is 9.

(b) passing through the point (3, 1) and makes intercepts on the axis, the difference of
whose lengths of intercepts is 4.

11. Derive the equation of a straight line in double intercepts form.

Project Work

12. Draw the graphs of the following straight lines and find their slopes, intercepts. Interpret
them geometrically.

(a) 4x + 5y = 20 (b) 5x + 6y = 30 (c) y = x

1. (a) x + y = 1 (b) x-intercept = a, y-intercept = b
a b (b) x-intercept = 8, y-intercept = 6
x y
2. (a) y = 4, x = 3 (c) 5 – 6 = 1

3. (a) 3x + 2y = 12 (b) 5x – 4y = 20 (c) 5x + 4y + 20 = 0

(d) 7x – 5y + 35 = 0 4.(a) 5x + 4y = 20 (b) 3x – 2y + 12 = 0

(c) 8x + 7y = 56 (d) 5x + 4y + 20 = 0 (e) 20x – 15y + 24 = 0
1 1
(f) 9x + 8y = 12 5.(a) x-intercept = 4 , y-intercept = – 3

(b) x-intercept = 5, y-intercept = 8 (c) x-intercept = 3, y-intercept = 27
2

(d) x-intercept = 9, y-intercept = 6 (e) x-intercept = 5, y-intercept = 4

6. (a) x + y = 6 (b) x – y + 1 = 0 (c) 2x + y = 13 (d) x + 3y = 8

7. (a) 3x + 2y = 12 (b) x + y = 8 (c) 15x – y = 20 (d) 2x – y + 7 = 0

8. (a) 3.6 sq. units (b) 20 sq. units 9. (a) 10 units (b) 41 units

10. (a) x + 2y = 6 (b) x + 3y = 6, x – y = 2

Standard Forms of Equations of Straight Line

The following three forms of equations of a straight line are called standard forms.

(a) Slope intercept form : y = mx + c

(b) Intercepts form : x + y = 1
a b

(c) Normal form : x cos D + y sin D = p

we have already derived equations of a straight line in slope -intercept form and double
intercepts form. Here, we derive equation of a straight line in normal or perpendicular form.

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

7.3 Equation of a straight line in Normal or Perpendicular form
(x cos D + y sin D = p)

To find the equation of a straight line in the form x cos D + y sin D = p

Let a straight line MN cut the X-axis at A and Y-axis at B such that

x-intercept = OA

y-intercept = OB

Let OQ = p be perpendicular distance the line Y
MN from the origin and ‘AOQ = D. NB

Then, ‘OBQ = 90° – ‘BOQ = 90° – (90° – D) = D

From right angled 'OAQ, Q
p
sec D = OA = OA
OQ p
X' D AX
? OA = OQ sec D = p sec D O M

Again, from right angled triangle OBQ, Y'

cosec D = OB OpB
OQ

? OB = OQ cosec D = p cosec D

Now, equation of the straight line MN is given by,

x + y = 1
OA OB
x y
or, p sec D + p cosec D = 1

or, x cos D + y sin D = 1
p p

? x cosD + y sinD = p is the required equation of a straight line in normal / perpendicular

form.

Alternative Method
Let MN be a straight line which cuts X-axis at A and Y-axis at B such that

x-intercept = OA Y
y-intercept = OB. NB

Draw OQ perpendicular to MN from the origin.

Let D be the angle made by OQ with OX in positive direction. Q90°-D

i.e. ‘AOQ = D and let OQ = p X' p AX
Then, ‘QOB = 90° – D D M
O
147
From right angled 'OQA, Y'

cos D = OQ = p
OA OA
p
or, OA = cos D

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Again from right angled 'OBQ,

cos (90° – D) = OQ
OB
p
or, sin D = OB

or, OB = p D
sin

Now,the equation of MN in double intercepts form is

x + y = 1
OA OB
x y
or, p + p =1

cos D sin D

? x cos D + y sin D = p is the required equation of the straight line MN in

perpendicular form / normal form.

Note :

(i) If the length of the perpendicular p from the origin to the line is zero, equation of the
line will be x cos D + y sin D = 0

(ii) When D = 0°, then x = p which is equation of a line parallel to Y-axis.
(iii) When D = 90°, then y = p which is equation of a line parallel to X-axis.
(iv) The value of p is always positive.
(v) Angle D is always taken with positive X-axis.

Worked out Examples

Example 1. The perpendicular distance of a straight line from the origin is 4 units and
Solution:
the angle made by the perpendicular with X-axis is 60°. Find the equation of
Example 2.
Solution: the straight line. Y

Here, length of the perpendicular (p) = 4 B
The angle made by perpendicular with
Q
X-axis (D) = 60° 4
Now, the equation of straight line is
X' D AX
x cos D + y sin D = p O
or, x cos 60° + y sin 60° = 4

or, x × 1 + y × 3 = 4 Y'
2 2

or, x + 3y = 8

? x + 3y = 8 is the required equation of straight line.

A straight line passes through the point (4, 4) and the inclination of
perpendicular is 60°. Find the equation of the straight line.

Here, length of perpendicular (p) is required.

The inclination of the perpendicular (D) = 60°

Now, the equation of required straight line is

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

x cos D + y sin D = p Y

or, x cos 60° + y sin 60° = p B
Q
or, x × 1 + y × 3 = p .............. (i)
2 2 p

Since it passes through the point (4, 4), 60° AX
X' O
4 × 1 + 4 × 3 = p
2 2 Y'

or, 2 + 2 3 = p

? p = 2 + 2 3 = 2(1 + 3)

Substituting the value of p in equation (i), we get

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

? x + y 3 = 4(1 + 3) is the required equation.

Example 3. The length of perpendicular from the origin to the line is 2 3 and the
Solution:
inclination of the perpendicular is 150°. Find the equation of the straight

line. Prove that the line passes through the point (2, 6). Y

Here, length of perpendicular (p) = 2 3

Inclination of the perpendicular (D) = 150° B

Now, the equation of straight line is,

x cos D + y sin D = p Q
2 3 150°
or, x cos 150° + y sin 150° = 2 3 X' X
AO
or, x× – 3 + y × 1 = 2 3
2 2
Y'
or, – 3x + y = 4 3

? 3x – y + 4 3 = 0 ..................... (i)
which is the required equation.
Put the point (2, 6 3) i.e. x = 2, y = 6 3 in equation (i), we get

3×2–6 3+4 3=0

or, 2 3 – 6 3 + 4 3 = 0
or, 6 3 – 6 3 = 0, (which is true.)
Hence, the line (i) passes through the point (2, 6 3).

Exercise 7.3

Very Short Questions

1. (a) Write down the equation of a straight line in normal form or perpendicular form.

(b) In normal form of equation of a straight line x cos D + y sin D = p, what are the
meanings of D and p?

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

(c) What are the equations of a straight line in three standard forms? List them.

(d) If a straight line passes through the origin and makes an angle of 120° with X-axis,
what will be its equation?

2. Write down the equations of the straight lines from the following figures.

(a) Y (b) Y
N Q

P 120°

p=5 X' O X
p=4
X' 45° X
O M

Y' P
Y'

(c) S Y (d) Y

X' 45° OX V

P P

T p=3 45°
Y'
X' U O X

Y'

Short Questions

3. If p is the perpendicular distance of a line from the origin and D the angle of inclination
with X-axis. Find the equation of straight under the following conditions.

p D
45°
(a) 1 120°
150°
(b) 2 210°
135°
(c) 2
120°
(d) 5

(e) 7
5

(f) 3
2

(g) 3 1 tan–1 1
2 2

(h) 6 tan–1 2
7 3

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