Opt. Maths Book for Class-9

UNIT

five Matrix

Estimated Teaching Hours : 20

1. Objectives

SN Level Objectives

To define a matrix.

i Knowledge (K) To define different types of matrices - row, column, zero, unit,

square, diagonal, equal, symmetric, triangular matrices.

To classify given matrices.

ii Understanding (U) To add and subtract matrices of 2×2 order.

To define transpose of a matrix.

iii Application (A) To apply properties of matrix addition.
To apply properties of matrix multiplication.

iv Higher Ability (HA) To solve problems by using properties of addition and
multiplication of matrices.

2. Teaching Materials
Chart paper with different types of matrices.

Chart paper with properties of addition and multiplication of matrices.

3. Teaching Learning Strategies:

Give an practical example of a matrix. (eg. daily commondies and their price) then

define a matrix.

Define different types of matrices with examples. (use chart paper)

Discuss addition and subtraction of matrices.

Discuss properties of matrix addition with illustrated examples.

Define transpose a matrix and its properties with examples.

Give different examples to illustrate the multiplication of two matrices.

eg. a) ab pq
c d×r s

23
b) [1 2] × 4 5

12

245 × 2 3
c) 6 2 3 4 5

Discuss the properties of matrix multiplication with examples.
In each case, questions of corresponding exercises are given to the students of the class

to solve and the teacher gives necessary guidance.

Vedanta Optional Mathematics Teacher's Guide ~ 9 51

Notes:
1. A matrix A of order m×n can be expressed as A = (aij), i = 1, 2, ... ... ..., m and

j = 1, 2, ... ... ..., n.

2. Two matrices of same orders are said to be equal if their corresponding elements are equal.

3. The matrix obtained by interchanging rows and columns of a given matrix A is called

the transpose of a matrix A. It is denoted by A1 ot At ot AT.

4. A matrix which remains same when its rows and columns are interchanged is known

as a symmetric matrix.

5. A square matrix A = (aij) is called skew. Symmetrix matrix if aij = –aji for all i, j and all
diagonal elements are zero.

6. Two matrices A and B are said to be multipliable if the number of columns of A is equal

to the number of rows of matrix B.

7. For three matrices A, B and C of same orders.

i) Commutative property: A + B = B + A

ii) Associative property: (A + B) + C = A + (B + C)

iii) Existence of identity element

A + 0 = 0 + A = A, where 0 is a null matrix of same order of A.

iv) Existence of Additive inverse

A + (–A) = 0

8. Properties of matrix multiplication

Let A, B and C be three square matrices of same order and k be any scalar.

i) Associative property: (AB)C = A(BC)

ii) Left distributive property: A(B + C) = AB + AC

Right distributive property: (B + C)A = BA + CA

iii) Scalar multiplication for a product of matrices

k(A – B) = (kA)B = A(kB)

iv) For identity matrix I,

AI = IA = A

9. Properties of transpose of matrix

i) (A')' = A, where A is any matrix

ii) (A + B)' = A' + B', where A and B are of same orders

iii) (kA)' = kA', where k is a scalar

iv) (AB)' = B'A' Some solved problems

123 b11 b12 b13

1. If 4 6 7 = b21 b22 b23
541 b31 b32 b33
Solution

123 b11 b12 b13

Here, 4 6 7 = b21 b22 b23

541 b31 b32 b33

Equating the corresponding elements of equal matrices, we get

52 Vedanta Optional Mathematics Teacher's Guide ~ 9

b11 = 1, b12 = 2, b13 = 3
b21 = 4, b22 = 6, b23 = 7
b31 = 5, b32 = 4, b33 = 1

2. Construct a 2×2 matrix whose elements aij are given by aij = 4i – 3j.

Solution

Let, A = a11 a12 be the required matrix.
a21 a22

Now a11 = 4×1 – 3×1 = 1, a12 = 4×1 – 3×2 = –2
a21 = 4×2 – 3×1 = 5, a22 = 4×2 – 3×2 = 2

1 –2
 A = 5 2

3. If a matrix has 6 elements in total, what are the possible orders it can have. Give reasons.

Solution
Since we have
Total number of elements in a matrix = no. of rows × no. of columns
Now, total number of elements = no. of rows × no. of columns
i.e. 6 = 6 × 1
= 1 × 6
= 2 × 3
= 3 × 2
Hence the required possible order of the matrix may be 6×1, 1×6, 2×3 or 3×2.

4. If  x–1 p  is an identity matrix, find the values of x, y, p and q.
 q y–1 

Solution x–1 p
q y–1
Here,   is an identity matrix of order 2×2. We can write,
 

 x–1 p  =  1 0 
 q y–1   0 1 

Equating the corresponding elements of equal matrices,

x–1=2 ⇒x=1+2=2
p=0

q = 0, y – 1 = 1 ⇒ y = 2
 p = 0, q = 0, x = 2 and y = 2.

5. For what value of x and y, matrix  2 3x–8  is a scalar.
y–4 2

Solution

Here,  2 3x–8  is a scalar matrix of order 2×2 we can write,
y–4 2

Vedanta Optional Mathematics Teacher's Guide ~ 9 53

 2 3x–8  =  2 0 
y–4 2 0 2

 3x – 8 = 0 ⇒ x = 8
3

and y – 4 = 0 ⇒ y = 4

 x = 8 and y = 4.
3

6. If 6 5 + 4 y = z 6 , find the values of x, y, z and p.
2 x 2 1 p 5

Solution

Here, 65 4y z6
2 x+2 1=p 5

6+4 5+y z 6
or, 2+2 x+1 = p 5

10 5+y z 6
or, 4 x+1 = p 5

Equating the corresponding elements of equal matrices.

z = 10, p = 4, x = + 1 ⇒ x = 4

5 + y = 6 ⇒ y=6–5=1

 p = 4, x = 4, y = 1 and z = 10

236 –4 6 7 –4 –8 –4
7. Let, P = 2 4 1 , Q = 8 2 1 and R = 3 2 1 , verify that:

(P + Q) + R = P + (Q + R)

Solution

Here, P= 2 3 6 ,Q= –4 6 7 and R = –4 –8 –4
2 4 1 8 2 1 3 2 1

P + Q = 2–4 3+6 6+7
2+8 4+2 1+1

= –2 9 13
10 6 2

LHS = (P + Q) + R

= –2 9 13 + –4 –8 –4
10 6 2 3 2 1

= –6 1 9
13 8 3

Also, Q + R

54 Vedanta Optional Mathematics Teacher's Guide ~ 9

= –4 6 7 + –4 –8 –4
8 2 1 3 2 1

= –8 –2 3
11 4 2

RHS = P + (Q + R)

= 2 3 6 + –8 –2 3
2 4 1 11 4 2

= –6 1 9
13 8 3

 LHS = RHS proved.

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

Solution

52
Here, P + Q = 0 9 ... ... ... (i)

35
P – Q = –3 2 ... ... ... (ii)

Adding (i) and (ii), we get

88
2P = –3 11

 P= 4 4
–32
11
2

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

35
Q = P – –3 2

= 4 4 – 3 5
–32 11 –3 2

2

= 4–3 4–6
–23+3 121–2

= 1 –2
3
7
2 2

 P= 4 4 ,Q= 1 –2
–23 11
3 7
2 2 2

Vedanta Optional Mathematics Teacher's Guide ~ 9 55

34
9. If M = 1 5 , show that M + MT is a symmetric matrix.
Solution

34
Here, M = 1 5

31
MT = M = 4 5

34 31
Now, M + MT = M = 1 5 + 4 5

65
= 5 10

Here, a12 = a21= 5
Hence M + MT is a symmetric matrix.

0 2 –45

10. Let A = –2 0 –4
45 4 0

a) Show that A is a skew – symmetric matrix
b) Find the transpose of A.
c) Find A + AT, what type of matrix is formed?

Solution

Here, 0 2 –45

A = –2 0 –4
45 4 0

a) Here, a12 = –a21 = 2
a21 = –a12 = 45
a23 = –a32 = –4
For all i, i, aij = –aji
 A is a skew symmetric matrix.

b) Transpose of A is given by

0 –2 45

AT = 2 0 4
–45 –4 0

c) A + AT –45 0 –2 45

02 –4 + 2 0 4
= –2 0 0 –45 –4 0

45 4

56 Vedanta Optional Mathematics Teacher's Guide ~ 9

000
= 0 0 0

000

 A + AT is a null matrix.

21
11. If P = –2 –1 , then prove that: P2 = P
Solution
Here, P2 = P . P

21 21
= –2 –1 –2 –1

4–2 2–1
= –4+2 –2+1

21
= –2 –1

 P2 = P proved.

1 22 1
12. If = 3 2 and Q = 4 5 , find the value of 3P – 2Q.
Solution
Here, 3P – 2Q

=3 1 2 –2 2 1
3 2 4 5

= 3 6 – 4 2
9 6 8 10

= –1 4
1 –4

11
13. a) If A = 1 1 , show that A2 – 2A = O where O is a null matrix of order 2×2.

Solution
Here, A2 – 2A = A × A – 2A

11 11 11
= 1 1 1 1 – 2 1 1

1+1 1+1 2 2
= 1+1 1+1 – 2 2

22 22 00
= 2 2 – 2 2 = 0 0

= 0
 A2 – 2A = O, where O is a null matrix of order 2×2.

Vedanta Optional Mathematics Teacher's Guide ~ 9 57

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

Solution

3 –5
Here, P = –4 2

3 –5 3 –5
P2 = P . P = –4 2 –4 2

9+20 –15–10
= –12–8 20+4

29 –25
= –20 24

Now, LHS = P2 – 5P 3 –5
29 –25 – 5 –4 2

= –20 24 15 –25
29 –25 – –20 10

= –20 24
14 0

= 0 14
10

= 14 0 1

= 14I
 P2 – 5P = 14I proved.

42
c) If P = –1 1 , then prove that: (P – 2I) (P – 3I) = O. Where O and I are null matrix and

identity matrix of order 2×2.

Solution 2
4 1

Here, P = –1

4 2 10
(P – 3I) = –1 1 –3 0 1
2 30
4 1–0 3
= –1 2
–2
1 2 20
= –1 1–0 2

Also, P – 2I 4
= –1

58 Vedanta Optional Mathematics Teacher's Guide ~ 9

= 2 2
–1 –1

Now, LHS = (P – 2I) (P – 3I) 2 2 1 2
–1 –1 –1 –2

= 2–2 2–2
–2+2 –2+2

= 0 0 =0
0 0

14. If P = 2 4 and Q = –1 5
1 3 2 1 , show that: (PQ)T = QT.PT

Solution

24 –1 5
Here, P = 1 3 and Q = 2 1

21 –1 2
PT = 4 3 , QT = 5 1

2 4 –1 5
PQ = 1 3 2 1

–2+8 10+4
= –1+6 5+3

6 14
= 5 8

65
LHS = (PQ)T = 14 8

–1 2 2 1
Again, RHS QT . PT = 5 1 4 3

–2+8 –1+6
= 10+4 5+3

65
= 14 8

 (PQ)T = QT.PT proved.

1 2 10 y 2
15. a) If x –3 0 1 = 4 –3 , find the values of x and y.

Solution

Here, 12 10 y 2
x –3 0 1 = 4 –3

Vedanta Optional Mathematics Teacher's Guide ~ 9 59

or, 1x 2 = y 2
–3 4 –3

Equating the corresponding elements, we get
 x = 4 and y = 1.

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

Solution

30 a b
Here, A = 0 4 , B = 0 c

Now, AB = A + B

30 ab 30 ab
or, 0 4 0 c = 0 4 + 0 c

3a 3b a+3 b
or, 0 4c = 0 4+c

Equating the corresponding elements, we get,

3a = a + 3 ⇒ a = 3
2

3b = b ⇒ 2b = 0 ⇒ b = 0

4c = 4 + c ⇒ c = 4
3

a = 23, b = 0, c = 34.

c) If  –1 0   x  =  –2 , find the matrix  x 
0 –2 y 4 y

Solution

Here,  –1 0   x  =  –2 
0 –2 y 4

or,  –0x–+2y0  =  –2 
  4 

or,  ––2xy  =  –2 
4

Equating the corresponding elements, we get

x = 2, y = –2

  x  =  2 
y –2

60 Vedanta Optional Mathematics Teacher's Guide ~ 9

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

Solution

ab
Let, R = c d

Then, PR = Q

21 ab 34
i.e. 5 3 c d = 2 4

2a+c 2b+d 34
or, 5a+3c 5b+3d = 2 4

Equating the corresponding elements, we get

2a + c = 3 ... ... ... (i) 2b + d = 4 ... ... ... (ii)

5a + 3c = 2 ... ... ... (iii) 5b + 3d = 4 ... ... ... (iv)

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

a = 7, c = –11

Again, solving equations (ii) and (iv), we get,

b = 8 and d = –12

 a = 7, b = 8, c = –11 and d = –12

78
 R = –11 –12

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

Here, –1 2 –2
2 –2 P = 4

–1 2
As 2 –2 is of order 2×2 and the product matrix is of order 2×1; P must be of order

a
2×1 let P = b

Now, –1 2 a –2
2 –2 b = 4

–a+2b –2
or, 2a–2b = 4

 –a + 2b = –2 ... ... ... (i)

Vedanta Optional Mathematics Teacher's Guide ~ 9 61

2a – 2b = 4 ⇒ a – b = 2 ... ... ... (ii)
Solving equations (i) and (ii), we get,
a = 2, b = 0

 P= 2
0

17. Which matrix pre - multiplies to matrix 1 1 to get 4 5 ?
3 4 6 2

Solution

ab
Let, the required matrix be c d , then by question,

ab 11 45
cd 34=62

a+3b a+4b 45
⇒ a+3d c+4d = 6 2

Equating the corresponding elements, we get

a + 3b = 4 ... ... ... (i)

a + 4b = 5 ... ... ... (ii)

c + 3d = 6 ... ... ... (iii)

c + 4d = 2 ... ... ... (iv)

Solving equation (i) and (ii), we get,

a = 1, b = 1
Solving equation (ii) and (iv), we get,

c = 18, d = –4

ab 11
 c d = 18 –4

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

2a
Here, [a b] 3 = [1 4] b

⇒ [2a + 3b] = [a + 4b]
or, 2a + 3b = = a + 4b
 a = b proved.

62 Vedanta Optional Mathematics Teacher's Guide ~ 9

b) If [m n] m = [13] and m + 1 = n, find the values of m and n.
n

Solution

m
Here, [m n] n = [13]

or, [m2 + n2] = [13]
 m2 + n2= 13 ... ... ... (i)
and m + 1 = n ... ... ... (ii) given
From equation (ii), m = n – 1
put, the value of m in equation (i),
(n – 1)2 + n2 = 13
or, n2 – 2n + 1 + n2 = 13
or, 2n2 – 2n – 12 = 0
or, n2 – n – 6 = 0
or, n2 – 3n + 2n – 6 = 0
or, n(n – 3) + 2(n – 3) = 0
or, (n – 3) (n + 2) = 0
 n = 3, –2
From (ii), when n = 3, m = 2
when n = –2, m = –3
 m = 2, –3 and n = –2, 3
19. 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.

Solution
Here, P(2x + 1) × 2 and Q(3y – 1) × 3
To define the product PQ,
the no. of column of P = the no. of row of Q
i.e. 2 = 3y – 1
 y = 1
Again to define the product QP
the no. of column Q = the no. of row of P
i.e. 3 = 2x + 1
 x = 1
 x = 1 and y = 1.

Vedanta Optional Mathematics Teacher's Guide ~ 9 63

Questions for practice

1. If p q = 2 4 , find the values of p, q, r and s.
r s 6 7

2. Construct a 2×2 matrix whose elements are given by aij = 2i + j.
3. If a matrix has 8 elements in total, what are the possible orders it can have. Give your

reason.

4. If p–4 q is an identity matrix, find the values of p, q, r and s.
r+8 s+3

5. For what value of p and q, the matrix 4 q+4 is a scalar matrix.
p–8 4

6. If 3 3 x +2 1 3 = z –8 , find the values of x, y and z.
0 2 y 2 17 10

7. Find the matrix X and Y, if X + Y = 4 3 and X – Y = 4 5 .
8 6 0 2

8. If A = 1 2 ,B= 2 4 and C = –4 5 , verify that (A + B) + C = A + (B + C).
4 6 3 6 2 7

0 pq

9. If A = –p 0 r , then show that A + A1 is a symmetric matrix.
–q –r 0

12 12
10. If P = 4 6 and Q = 4 6 , verify that: (P . Q)1 = Q1 . P1.

64 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

six Co-ordinate Geometry

Estimated Teaching Hours : 6

1. Objectives

SN Level Objectives

To define locus.

i Knowledge (K) To note steps for finding equation of locus.

To write section formula.
To use distance formula.

ii Understanding (U) To find distance between two points.

To find equation of locus in simple cases.
To find equation of locus under given conditions.
iii Application (A) To find points that divides line joining given two points.

iv Higher Ability (HA) To use and solve very long problems using distance formula
and section formula.

2. Teaching Materials

Graph papers.

Formula list on a chart paper

3. Teaching Learning Strategies:

Review the distance formula and its application.

Define a locus with examples.

Discuss how to write equation of locus.

Discuss problems given in exercise 6.1

Define sections of a line segment.

Derive the section formula and discuss its applications and its special cases (internal,

external mid point etc)

Discuss how to solve problems in exercise 6.2.

Notes:

1) If any point lies on the locus, its coordinates must satisfy the equation of locus.

2) To find the equation of a locus

i) Take any point P(x, y) on the locus

ii) Write the geometric conditions, express it in distance form.

iii) Simplify the algebraic equation.

3. Section formula A(x1,y1) P(x,y) B(x2,y2)
m2
i) Internal division: m1

P(x, y) =  mm1x21 + mm22x1, mm1y12 + mm22y1 
 + + 

ii) External division: m1

P(x, y) =  mm1x21 – mm22x1, mm1y12 – mm22y1  A(x1,y1) B(x2,y2) P(x,y)
 – –  m2

Vedanta Optional Mathematics Teacher's Guide ~ 9 65

4. Mid-point joining two points (x1, y1) and (x2, y2), is P(x, y) =  x1 + x2, y1 + y2
2 2 

5. k - formula k 1
A(x1,y1)
P(x, y) = kxk2++1x1, k2 + y1  P(x,y) B(x2,y2)
k + 1

A(x1,y1)

6. Centroid of a triangle

G(x, y) = x1 + x32 + x3, y1 + y32 + y3 2
G
The centroid of a triangle divides the median in ratio of 2:1. 1

7. Points of trisection: C(x3,y3)
Two points P and Q are called points of trisection if they B(x2,y2)

divide AB is three equal parts.
m1

A(x1,y1) P Q B(x2,y2)

P divides AB in the ratio 1:2 i.e. AP:PB = 1:2

Q divides AB in the ratio of 2:1 i.e. AQMQB = 2:1

P is the mid point of AQ i.e. AP = PQ

Q is the mid point of PB. i.e. PQ = QB.

Some solved problems

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

Solution
Here, x2 + y2 = 25
put the point (4, 3) in x2 + y2 = 25, we get, 16 + 9 = 25 (true)
put the point (3, 2), we get, 9 + 4 = 25 (false)
put the point (5, 0), we get, 25 = 25 (true)
put the point (0, –5), we get 0 + 25 = 25 (true)
put the point (–3, –4), we get, 9 + 16 = 25 (true)
 (4, 3), (5, 0), (0, –5) and (–3, –4) lie on the locus and (3, 2) does not lie on the locus.

2. For what value of λ will the points (2, 3) lie on the locus whose equation is
x2 + y2 + λx + 2y – 30 = 0 ?

Solution
Here, equation of locus is
x2 + y2 + λx + 2y – 30 = 0
put the point (2, 3) in the locus,

66 Vedanta Optional Mathematics Teacher's Guide ~ 9

4 + 9 + 2λ + 6 – 30 = 0
or, 2λ – 11 = 0

 λ = 121.
3. If (2, 3) is a point on locus whose equation is ax + 2y = 16 and also show htat (0, 8) is

another point on the locus.

Solution
Here, ax + 2y = 16 ... ... ... (i)
put (2, 3) lies on the locus (i)
a.2 + 2.3 = 16
or, 2a + 6 = 16
or, 2a = 10
 a = 5
put the value of 'a' in equation (i), we get
5x + 2y = 16
put the point (0, 8) in this equation, we get
5×0 + 2×8 = 16
 16 = 16
 The point (0, 8) also lies on the locus. proved.

4. Find the locus of a point which moves so that it is equidistant from the points (4, 3) and (5, 4).

Solution P(x,y)
Let, P(x, y) be any point in the locus and given points be A(4, 3)
A(4,3) B(5,4)
and B(5, 4).
Then by questions
AP = BP
or, AP2= BP2
or, (x – 4)2 + (y – 3)2 = (x – 5)2 + (y – 4)2
or, x2 – 8x + 16 + y2 – 6y + 9 = x2 – 10x + 25 + y2 – 8y + 16
or, 2x + 2y = 16
 x + y = 8.

5. a) Find the coordinates of the point dividing internally the line joining the points
(4, 5) and (7, 8) in the ratio of 2:3.

Solution
Here, (x1, y1) = (4, 5), (x2, y2) = (7, 8) and m1:m2 = 2:3
Let, (x, y) be the required point that divides the join of given two points in ratio of 2:3
internally then,

(x, y) =  mm1x21 + mm22x1, mm1y21 + mm22y1 
+ +

Vedanta Optional Mathematics Teacher's Guide ~ 9 67

=  2.7 + 33.4, 2.8 + 3.5
2 + 2+3 

= 256, 31 
5

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

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

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

Solution

(a) Let P(x,y) be any on the locus and A(–5,0) and B(5,0) be two points.

Then,
PA 2
PB = 3

or, 9{(x+5)²+(y–0)}= 4{(x–5)²+(y–0)²}
or, 9(x²+10x+25+y²)= 4(x²–10x+25+y²}
or, 9x²+90x+225+9y²= 4x²–40x+100+4y²

or, 5x²+130x+5y²+125=0
or, 5(x²+25x+y²+25)=0

⸫ x²+y²+25x+25=0 which is the required equation.

(b) Let O(x,y) be any point on the locus and A(1,2) and B(5,3) are fixed points.
By questions
PA 4
PB = 5
or, 5PA= PB

or, 5PA²=16PB²
or, 25{(x–1)²+(y–2)²}= 16{(x–5)+(y–3)²}
or, 25(x²–2x+1+y²+4)= 16(x²–10x+25+y²–6y+9)
or, 25x²–50x+25+25y²–100y+100= 16x²–160x+400+16y²–96y+144
⸫ 9x²+9y²+110x–4y–419=0 which is the required equation.

7.(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

Solution
Let p(x,y) be any point on the locus

(i) MP= NP
or, MP²= NP²
or, (x–3)²+(y–2)²= (x–7)²+(y+4)²
or, x²–6x+9+y²–4y+4= x²–14x+49+y²+8y+16
or, 8x–12y–52=0
or, 4(2x–3y–13)=0
⸫ 2x–3y–13=0 which is the required equation.

68 Vedanta Optional Mathematics Teacher's Guide ~ 9

(ii) MP= 2PN
Here, MP²= 4PN²

or, (x–3)²+(y–2)= 4{(x–7)²+(y+4)²]
or, x²–6x+9+y²–4y+4 = 4(x²–14x+49+y²+8y+16)
or, x²–6x+9+y²–4y+4 = 4x²–56x+196+4y²+32y+64
⸫ 3x²+3y²–50x+36y+247=0 is the required equation.

(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

Solution
Let P(x,y) be any points on the locus and A(a.0) and P(–a,0) are two fixed points

Now, PA²+PB²= AB²

(x–a)²+(y–0)²+(x+a)2+(y–0)²= (–a–a)²+(0–0)²
or, x²–2ax+a²+y²+x²+2ax+a²+y²= (2a)²+0
or, 2x²+2y²+2a²= 4a²
or, 2x²+2y²= 2a²
⸫ x²+y²= a² which is the required equation.

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

Solution P (x,y)
Let P(x,y) be an point on the locus.
AB subtend a right angle at P. so we have A(–7,0) (7,0) B
AB²= PA²+PB²
or, (7+7)²+(0–0)²= (x+7)²+(y–0)²+(x–7)²+(y–0)²
or, 196+0= x²+14x+49+y²+x²–14x+49+²

or, 2x²+2y–98=0

or, x²+y²–49=0

⸫ x²+y²=49 which is the required equation.

8. (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

Solution
Let P(x,y) be any point on the locus. A(2,3) and B(6,1) be fixed point.
B given condition,

PA= PB
or, PA²= PB²
or, (x–2)²+(y–3)²= (x–6)²+(y–1)²
or, x²–4x+4+y²–6y+9= x²–12x+36+y²–2y+1
or, 8x–4y+13–37= 0
or, 8x–4y–24= 0
⸫ 2x–y–6= 0 which is the required equation.

Vedanta Optional Mathematics Teacher's Guide ~ 9 69

(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)

Solution
Here,
PA= PB
PA²=PB²
or, (x–a)²+(y–b)²= (x–30)²+(y–3b)²
or, x²–2ax+a²+y²–2by+b²= x²–6ax+9a²+y²–6by+9b²
or, –2ax+6ax–2by+6by= 9a²–a²+9b²–b²
or, 4ax+4by= 8a²+8b²
or, 4(ax+by)= 8(a²+b²)
⸫ ax+by= 2(a²+b²) Proved

9. 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.

Solution
Let P(x,y) be any point on the locus. and A(0,3) and B(0,–2) are fixed point.
By question,
PA²+PB²=6
or, (x–0)²+(y–2)²+(x–0)²+(y+2)²=6
or, x²+y²–4y+4+x²+y²+4y+4=6

or, 2x²+2y²+8= 6
or, 2x²+2y²+2= 0
⸫ x²+y²+1= 0 which is the required equation.

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

Solution
Let P(x,y) be any point on the locus,
By question,
PA²+PB²= 4AB
or, (x–3)²+(y–0)²+(x+3)²+(y–0)²= 4{ (3+3)²+)(0–0)² }

or, x²–6x+9+y²+x²+6x+9+y²= 4×√6²
or, 2x²+2y²+18= 24
or, 2x²+2y²= 24–18
or, 2x²+2y²= 6

⸫ x²+y²= 3 which is the required equation.

(c) The difference of square of distance from the points A(0,1) and B(–4,3) is 16 (PA²–PB²= 16)

Solution
Given, PA²–PB²= 16

70 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, (x–0)²+(y–1)²–{(x+4)+(y–3)²}=16
or, x²+y²–2y+1–x²–8x–16–y²+6y–9=16
or, –8x+4y–24=16
or, 2x–y+10=0

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

Solution XL–ectoPo(r–d1i,n2a54te)=dimvim2dxe11++s mmth12ex2joining A(–3,) and B(2,6) in the ratio of m:n




–1= m2×m(–13+)+mm2 1(2)

or, –m1–m2 = 3m2+3m1

or, 2mmm2 =12 =3m32 1
or,

⸫ m:n= 2:3

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

Solution :
Let the point P(3,–2) divide the join of the points A(1,4) and B(–3,16) in the ratio m:n

Now
X–coordinate= mm2x11++mm21x2

3= m2×m(–13+)+mm2 1×1

oorr,,3mmm121=+3–31m2 = –3m1+m2



⸫ m1:m2 = –1:3

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

Solution
(5,4)= (x,y)

(x1,y1)= (2,1)
(x2,y2)= (7,6)
Ratio(m:n)= ?

Now,
X–coordinate= mm1x12++mm22x1

Vedanta Optional Mathematics Teacher's Guide ~ 9 71

5= m1×m71++mm22×2

or, 5m1+5m2 = 7m1+2m2
or, –2m1 = –3m2
⸫ m1:m2 = 3:2

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

Solution
Let P and Q divide the line joining A(2,–2) and B(–1,4) into three equal parts. Then P divides
the line AB is the ratio 1:2,

The coordinates of P are given by

m1mx12++mm22x1 , mm1y12++mm22y1

1×(–1)+2x2 1×4+2×(–2)

= 1+2 , 1+2

= 30 =(1,0)
3, 3

Here, Q is mid points of PB so coordinates of Q is given by

= x1 + x2 , y1 + y2
2 2
P Q
0+4
= 1+ (–1) 2 A(2,–2) B(–1,4)
2,
=(0,2)

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

Solution
Let A and B divides the line joining P and Q into three equal parts.
Here, A divides PQ in the ratio 1:2.
Coordinate of A,= m1 ×mx1+2+mm2 2x1 , m1×my1+2+mm2 2y1

1×4+2×1 1×2 + 2×2 A B
1+2 , 1+2
P (1,2) Q (4,2)
66
3, 3

=(2,2)

B divides the AB into two equal parts.

Coordinates of B, = x1 + x2 , y1 + y2
2 2

72 Vedanta Optional Mathematics Teacher's Guide ~ 9

= 2+4 2+2
2, 2

=(3,2)

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

Solution A B
The coordinates of A, P (1,–2)

m1mx12++mm22x1 , mm1y21++mm22y1 Q (–3,–4)

=

1(–3)+2×1 1×4 + 2×(–2)
1+2 , 1+2

–1 –0 = –1
3, 3 3 ,0

For the coordinates of B is the mid point of AQ.

= x1 + x2 , y1 + y2
2 2

= –1 +(–3) 0+4
3 2, 2

= –5
3 ,2

(12) (a) If the point (x,14) divide 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.

Solution
Let the point P(x,14) divides the line segment joining the points A(7,11) and B(–18,16) in
the ratio m:n,

The y–coordinates of P= mm1y12++mm22y1
14=m1m×116++mm22×11

or, 14m1+14m2 = 16m1+11m2

⸫or,mm212m=1 = 3m2
3
2 m1mx2+1+mm2x21

The x–coordinate of P=
3×(–18)+2x7
x= 3+2

or, x= 54+14
5

= 54+14
5

Vedanta Optional Mathematics Teacher's Guide ~ 9 73

=–8
⸫ x = –8

(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

Let the ratio be m:n m1xm2+1+mm2x21

Ans The x–coordinate=
or, 5= m1×m81++mm232

or,5m1+5m2 = 7m1+3m2

⸫or,mm3m12 =1 = 2m2
2
3 m1×my2+1+mm2y21

y–coordinate=

y= 2×9+3×7
2+3

y= 18+21 = 39
5 5

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

Solution
Let the points on the x–axis be (a,0) which divides the line segment joining the point the
points A(2,–4) and B(–3,6) in the ratio m:n

y–coordinate= m1×my2+1+mm2y21

or, 0 = m1×6m+m1+2(m–42)

or, 6m1 = 4m2

⸫ mm21 = 2
3

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

Solution
Let the point on axis be (a,0)
y–coordinates= m1my2+1+mm2y21

or 0= m1m6+1m+2m(–23)
or, 6m1 = 3m2

or, mm21 = 1
2

⸫ m1:m2= 1:2

74 Vedanta Optional Mathematics Teacher's Guide ~ 9

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

Solution

Let the point on the Y–axis be (0,a) which divides the line joining A(2,–3) and B(–6,10) in

the ratio m:n
x–coordinates= m1m×x12++mm22x1

or, 0= m1m×(1–+6)m+m2 2(2)
or , 6m1 = 2m2
mm21 1
or, = 3

⸫ m1:m2= 1:3

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

S olution Let the points be (0,a) and the ratio be m:n
m1mx21++mm22x1
x–coordinates= x–coordinates=

or, 0= m1m(–1+2)m+2m2(5)
or , mm212m=1 = 5m2
or, 5
2

⸫ m1:m2= 2:5
(e) Find the ratio in which the line joining (3,–2) and (–3,6) is cut by the axes of the coordinates.

Solution
Let the point on the x–axis be (a,0) which cut the line segment in the ratio m:n

y–coordinate= y–coordinate= m1y2m+1m+2my12

or, 0 = m1×m5+1+mm2(2–2)

or, 6m1 = 2m2
⸫ m1:m2 = 1:3

Again, Let the point on the y–axis be (0,b) which cuts the line segment in the ratio m:n

ratio m:n m1x2m+1m+2mx12
x–coordinate=

or, 0= m1×(m–31)++mm22×3

or, 3m1 = 3m2
mm12
or, = 1
1

⸫ m1: m2 = 1:1

14. (a) Prove that the following points represents the vertices of a parallelogram.

Vedanta Optional Mathematics Teacher's Guide ~ 9 75

(i) (–2,–1), (1,0), (4,3) and (1,2)

Solution
Let A(–2,–1), B(1,0), C(4,3) and D(1,2) are the vertices of a quadrilateral.

Here, the mid point of AC

= x1 + x2 , y1 + y2
2 2
D (1,2) C (4,3)
–2+4 –1+3
= 2,2

= (1,1)

The mid point of BD

= x1 + x2 , y1 + y2 A (–2,–1) B (1,0)
2 2

= 1+1 0+2
2,2

= (1,1)

Therefore, the diagonals of quadrilateral have same mid points, they bisects each other.

Hence, given points are vertices of the parallelogram.

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

Solution
Let the fourth vertix be S(x,y). Diagonals of parallelogram biset each other. They have
common mid point.
Mid point of PR = x1 + x2 y1 + y2 S (x,y) R (4,0)
2 2
,

= 1+4 2+0 O
2,2

= 5 P (1,2) Q (1,0)
2 ,1

The coordinate of mid point of QS= 5
2 ,1

= x1 + x2 , y1 + y2 = 5
2 2 2 ,1

= x+1 y+0 = 5
2,2 2 ,1

= x+1 y+0 = 5
2,2 2 ,1

x+1 5 y+0
∴ 2 = 2 2 =1

or, x=4 or, y=2

The coordinates of S are (4,2)

76 Vedanta Optional Mathematics Teacher's Guide ~ 9

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

Solution
Let P(3,7), Q(5,–7), R(–2,5) and S(x,y) are the four vertices of a parm.
Mid point of PR= Mid point of QS

or, 3+(–2) 7+5 = 5+x –7+y
2, 2 2, 2

⇒ 3–2 5+x 7+5 –7+y
2= 2 2= 2
or, x=–4
or, y=19

Therefore, the fourth vertex is (–4,19)

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

Solution
Let A(–1,3), B(1,–1) and C(5,1) are the verties of a triangle CD be the median. D is mid point
of AB.
The coordinates of D= x1 + x2 , y1 + y2 C (5,1)
2 2

= –1+1 3–1
2, 2

= (0,1)

The length of CD= (x2–x1)²+(y2–y1)² A (–1,3) D(0,1) B (1,–1)

= (x2–x1)²+(y2–y1)²
= 5²+0²

=5 units

(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.

Solution
Let A(x1,y1), B(x2,y2), and C(x3,y3) are the vertices of triangle ABC
Here, P(2,–1), Q(–2,2) and R(–1,4) are the mid points of AB, BC and CA respectively.

Mid point of AB= x1 + x2 , y1 + y2 C
2 2 Q (–2,2)

(2,–1)= x1 + x2 , y1 + y2
2 2

x1+2 x2 = 2 y1+2 y2 = –1 R (–1,4)

x1+x2 =4 ........(i) or, y1+y2 = –2 ........(ii) B

Mid point of BC= x2 + x3 , y2 + y3 P (2,–1)
2 2

(–2,2)= x2 + x3 , y2 + y3 A
2 2

Vedanta Optional Mathematics Teacher's Guide ~ 9 77

or, x2+x3 =–4 ........(iii) or, y2+y3 = 4 ........(iv)

Mid point of CA= x1+ x3 , y1+2 y3
2

(–1,4)= x1+2 x3 , y1+2 y3

or, x1+x3 =–2 ........(v) or, y2+y3 = 8 ........(vi)

Adding equation (i), (iii) and (v) Adding (ii), (iv) and (vi)

2(x1+x 2+x3) = –2 2(y1+y2+y3) = 10

x1+x 2+x3 = –1 (y1+y2+y3) = 5

From equation (i), substituting x1+x2 = 4

4+ x3 = –1 –2+y3 = 5

x3 =–5 y3 = 7

Similarly,

x1= 3 y1 = 1
x2= 1 y2 = –3

The vertices are, (3,1), (1,–3) and (–5,7)

(16) (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 P.

Solution:
PQRS is a parallelogram

Mid points of PR= Mid Points of QS

5+2 y+3 = 0+3 –7+7
2, 2 2, 2

x+5 3 y+3 5
∴ 2 =2 and 2 = 2
or, x=–2
or, y=2

∴ P(x,y)= P(–2,2)

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

Solution are S(x2,y2) are vertices of parallelogram ) is the mid point
Let P(1,1), Q(3,6), R(x1,y1) the O(7,
of PR and QS.

Mid point of PR, S (x2,y2) R (x1,y1)

7, 7 = 1+2x1 , 1+2y1
2

∴ x1=13, y1=6 O 7, 7
Mid point of QS, 2

P (1,1) Q (3,6)

7, 7 = 3+x 6+2y2
2 2,

78 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, x2 =11, y2= 1
The remaining vertices are (13,6) and (11,1)

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

Solution :

Let A(1,1), B(1,7) and C(7,1) are the vertices of triangle.

AB= (1–1)²+(7–1)² =6 B(1,7)

BC= (7–1)²+(1–7)² = 6²+6² =6 2

CA= (7–1)²+(1–1)² = 6² =6 M
Here longest side is BC, Hence BC is hypotenuse.

Mid point of BC= M 1+7 7+1 =M(4,4) A(1,1) C(7,1)
2, 2
Distance of mid points and vertex A= (4–1)²+(4–1)²

i.e A.M.= 3²+3²

=3 2 units

distance of mid points and vertex C= (4–7)²+(4–1)²

i.e C.M. =3 2 units

Distance of the vertex B from M = (4–1)²+(4–7)²

⸫ AM = BM = CM i.e BM = 9 + 9 = 3 2
Hence mid point of hypotenuse is equal distance from the vertices. Proved

Questions for practice

1. For what value of k will the point (1, –2) lie on the curve x2 + y2 + kx – 4y – 15 = 0.
2. Show that the points (0, 0), (0, 4), (4, 0) are the vertices of an isoceles right angled triangle.
3. Find the coordinates of a pooint dividing the line joining the points (–5, 4) and (11, –4)

internally in the ratio of 5:3.
4. A(20, 19) and B(16, 23) are the positions of two sources of drinking water. A water supplier

wants to supply water in a city P(2, 0). From which source should the supplier supply the
water so that the corporation may choose only a minimum distance (take 1 unit = 0.5km)
5. Find the equation of locus of a point which moves so that its distance from the point (2, 1) is 5.
6. Show that A(3, 5), B(1, 1) C(5, 3) and D(7, 7) are the vertices of a rhombus.
7. Determine the ratio in which 2y – x + 2 = 0 divides the line joining (3, –1) and (8, 9).
8. The point (2, 8) divides the line segment joining the points (4, –8) and (m, n) in the ratio
of 1:2 internally. Find the values of m and n.
9. Find the centroid of triangle PQR wiht vertices P(4, 5), Q(0, –2) and R(5, 9).
10. Find the equation of locus of the points such that the sum of its distances from two
fixed points (0, 3) and (0, –3) is 8.

Vedanta Optional Mathematics Teacher's Guide ~ 9 79

UNIT

seven Equations of Straight Lines

Estimated Teaching Hours : 8

1. Objectives

SN Level Objectives

To tell equation of straight line in the form of

parallel to X-axis.

parallel to Y-axis.

i Knowledge (K) slope intercept form
double intercept form.

perpendicular form

slope point form.

two points form.

To write formula to find area of triangle

To find equation of straight line in above forms.

ii Understanding (U) To find distance between two parallel lines.
To find distance of a point from a line.

To calculate area of a triangle and a quadrilateral when

coordinates of vertices are given.

iii Application (A) To find the equation of straight lines of verbal
questions stated as in (1).

iv Higher Ability (HA) To derive equation of straight lines in standard forms.
To derive formulae of area of triangle.

2. Teaching Materials

Chart papers with diagrams of equations of straight lines.

3. Teaching Learning Strategies:

Derive the equations of straight lines in the following form :-

– parallel to X-axis.

– parallel to Y-axis.

– slope intercept form

– double intercept form

– perpendicular form

– slope point form

– two points form

In every formula derivation, special cases (if exists) are explained with at least one example of each.

– some questions are given to students to solve from exercise, necessary guidances are given.

– derive formula to find distance from a point to a line.

– derive formula to calculate are of triangles.

– review the concept of collinearity and explain it in case of area of triangle.

– solve some problems of given exercise as examples to illustrate the formulas stated in above.

80 Vedanta Optional Mathematics Teacher's Guide ~ 9

Notes:
1) Slope of a line joining two points.

m = xy22 – yx11
–

2) Three points A, B and C are collinear. Then
slope of AB = slope of BC.
3) Equation of straight line in slope intercept form
y = mx + c
If it passes through the origin, equation of the line is y = mx.
4) Equation of a straight line
i) parallel to y-axis x = a
ii) parallel to x-axis y = b
5) Three standard equations of straight lines are
i) slope - intercept form, y = mx + c

ii) double - intercept form , x + y =1
a b

iii) perpendicular form, xcosα + ysinα = p

6) Equations of straight lines in special cases are

i) slope - point form, y – y1= m(x – x1)

ii) two points form, y– y1 = xy22 – yx11 (x – x1)
–

7) Slope of the line Ax + By + C = 0 is (m) = – Coefficient of x
Coefficient of y

8) The perpendicualr distance of a line Ax + By + C = 0 from point P(x1, y1) is given by
d = Ax1 + By1 + C

A2 + B2

9) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC, then area of triangle is given by

∆ = 1 x1 x2 x3 x1
2 y1 y2 y3 y1

= 1 [(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y1 – x1y3)] .
2

10) The vertices of triangle/quadrilateral are always taken as anticlockwise direction to
make the area of triangle/quadrilateral positive.

11) Area of quadrilateral with vertices A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) is given by

∆ = 1 x1 x2 x3 x4 x1
2 y1 y2 y3 y4 y1

= 1 [(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y4 – x4y3) + (x4y1 – x1y4)]
2

Vedanta Optional Mathematics Teacher's Guide ~ 9 81

Some solved problems

1. Find the equation of a straight line parallel to X-axis and passing through the point (–4, 5).

Solution
Here, We know that equation of a straight line parallel to X-axis is y = b ... ... ... (i)
It passes through the point (–4, 5)
 5 = b
put the value of b in equation (i)
y = 5 whcih is the required equation.

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

Solution 3
Here, y-intercept (c) = 4
inclination (θ) = 60°
slope (m) = tan60° =

By using formula,
y = mx + c

or, y = 3x + 4

 y = 3x + 4 is the required equation.

3. Find the equation given straight line PQ.

Solution y
Here, from the given figure, we have
X-intercept (a) = 4 Q
Y-intercept (b) = –5
Equation of given straight line is (4,0)

x + y = 1 x' O x
a b

or, x + y = 1
4 –5

 5x – 4y = 20 which is the required equation. (0,–5)

4. Find the intercepts of the given equation. y'
4x – 3y = 1. P

Solution
Here, 4x – 3y = 1

xy
1 + = 1
4 –13

82 Vedanta Optional Mathematics Teacher's Guide ~ 9

comparing it with x + y = 1
a b

x-intercept (a) = 1
4

y-intercept (b) = –31
Alternative Method

Here, 4x – 3y = 1

put y = 0, then x = 1
4

put x = 0, then y = –31

The given line cuts X-axis and Y-axis at 14, 0 and 0, –31

 x-intercept (a) = 1
4

y-intercept (b) = –13

5. Find the equation of straight line given in the figure.

Solution y
In the given figure perpendicular distance V

of the line from origin 45˚
p = 3 p=3
inclination (α) = 90° + 45° = 135°
By using formula, O
x cosα + y sinα = p
or, x cos 135° + y sin 135° = 3 y'

or, x – 1  + y . 1 = 3
2 2
x' x

or, –x + y = 3 2 U

 x – y + 3 2 = 0 is the required equation.

6. Write the equation 4x – 3y + 12 = 0 into slope
intercept form.

Solution

Here, 4x – 3y + 12 = 0

or, 3y = 4x + 12

 y = 34x + 4

which is in the form of y = mx + c

where slope (m) = 4
3

y - intercept (c) = 4 units.

Vedanta Optional Mathematics Teacher's Guide ~ 9 83

7. Write the following equations in double intercept form:
4x – 3y – 36 = 0

Solution

Here, 4x – 3y – 36 = 0

or, 4x – 3y = 36

dividing both sides by 36, we get

x – y = 1
9 12

Which is in the form of x + y = 1
a b

Where, x - intercept (a) = 9

y - intercept (b) = –12

8. Reduce the given equation in normal form

x – y – 2 2 = 0
Solution
Here, x – y – 2 2 = 0

or, 12 + (–1)2 = 1 + 1 = 2

Dividing the given equation by 2 on both sides,

1 x + – 1  y = 2 2
2 2 2

or, 1 x – 1 y = 2
2 2

 x cos315° + y sin315° = 2
Which is in the form of x cosα + y sinα = p
Where, α = 315°, p = 2 units.

9. Find the equation of the straight line making an angle of 225° with X-axis and passing
through (–2, –4).

Solution
Inclination (θ) = 225°, slope (m) = tan225° = 1
(x1, y1) = (–2, –4)
By using formula,
y – y1 = m(x – x1)
or, y + 4 = 1(x + 2)
or, x – y = 2
 x – y = 2 is the required equation.

10. Find the perpendicular distance from (4, 5) to the line 2x + 3y = 25.

Solution
Here, 2x + 3y = 25

84 Vedanta Optional Mathematics Teacher's Guide ~ 9

i.e. 2x + 3y – 25 = 0
comparing it with ax + by + c = 0, we get
a = 2, b = 3 and c = –25
(x1, y1) = (4, 5)
The perpendicular distance of above line is given by

d = ax1 + by1 + c
a2 + b2

= 2.4 + 3.5 – 25 = –2 = 2 units
22 + 32 13 13
11. Find the distance between the parallel lines

4x + 5y = 20 and 4x + 5y = 40

Solution
Here, 4x + 5y = 20 ... ... ... (i)
4x + 5y = 40 ... ... ... (ii)
From equation (i), when y = 0, then x = 5
(5, 0) is a point on the line (i)
The perpendicular distance of the line (ii) from a point (5, 0) on line (i) is given by,

d = ax1 + by1 + c
a2 + b2

= 4×5 + 5×0 – 40 = 20 units
42 + 52 41

Alternative Method

Given equations of lines are

4x + 5y = 20 ... ... ... (i)

4x + 5y = 40 ... ... ... (ii)

Writing each of equations in the form of

y = mx + c

From (i) y = –54 x + 4

From (ii) y = –45 x + 8
Slope (m) = –54, c1 = 5, c2 = 8

Perpendicular distance between two parallel lines are given by

d = c1 – c2
1 + m2

= 4–8 = (–4) × 5 = 20 units
41 41
1 +  –45 2


Vedanta Optional Mathematics Teacher's Guide ~ 9 85

12. Find the area of ∆ABC of A(2, 2), B(4, 5) and C(4, 8)

Solution
Here, To find area of ∆ABC

∆ = 1 2 4 4 2
2 2 5 8 2

= 1 [(10 – 8) + (32 – 20) + (8 – 16)]
2

= 1 [2 + 12 – 8]
2

= 1 × 6 = 3 sq. unit.
2

13. Using area of triangle show that the following points are collinear

(a, b + c), (b, c + a), (c, c + b)

Solution
Let us find the area of triangle formed from given vertices.

∆ = 1 a b c a
2 b+c c+a a+b b+c

= 1 [ca + a2 – b2 – bc + ab + b2 – c2 – ca + bc + c2 – a2 – ab]
2

= 1 × 0 = =0
2

Since the area is zero, the given points are collinear.

14. Find the area of the quadrilateral whose vertices are (4, 6), (8, 4), (8, 8) and (–4, 2).

Solution
Here, The area of given quadrilateral is given by

∆ = 1 4 8 8 –4 4
2 6 4 8 2 6

= 1 [(16 – 48) + (64 – 32) + (16 + 32) + (–24 – 8)]
2

= 1 × 8 = 4 sq. units.
2

15. Find the value of k, if the points (k, 2 – 2k), (1 – k, 2k) and (–4, 6 – 2k) are collinear.

Solution
Since the given three points (k, 2 – 2k), (1 – k, 2k) and (–4, 6 – 2k) are collinear area of
triangle formed is zero.

∆ = 1 k 1–k –4 k
2 2–2k 2k 6–2k 2–2k

or, 0 = (2k2 – 2 + 2k + 2k – 2k2) + (6 – 2k – 6k + 2k2 + 8k) + (–8 + 8k – 6k + 2k2)

86 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, 0 = 4k2 + 6k – 4
or, 2k2 + 4k – k – 2 = 0
or, 2k(k + 2) – 1(k + 2) = 0
or, (k + 2) (2k – 1) = 0
Either k + 2 = 0 ⇒ k = – 2

or, 2k – 1 = 0 ⇒ k = 1
2
 k = –2, 1
2

Alternative Method

Let the given collinear points be A(k, 2 – 2k), B(1 – k, 2k) and C(–4, 6 – 2k)

then slope of AB = slope of BC

2k – 2 + 2k = 6 – 2k – 2k
1–k–k –4 – 1 +k

or, 4k – 2 = 6 – 4k
1 – 2k k–5

or, 4k2 – 2k – 20k + 10 = 6 – 4k – 12k + 8k2
or, 4k2 + 6k – 4 = 0
or, 2k2 + 3k – 2 =
or, 2k2 + 4k – k – 2 = 0
or, 2k(k + 2) – 1(k + 2) = 0
or, (k + 2) (2k – 1) = 0
Either k + 2 = 0 ⇒ k = – 2

or, 2k – 1 = 0 ⇒ k = 1
2

 k = –2, 1
2

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

Solution
Given points P(–4,3), Q(1,2) and R(9,10)

Slope of PQ= xy22––yy11 = 2+3 = 5 =1
1+4 5

slopeSolofpPeQoafnQdRs=lopxy22e––oyyf11 10–2 8 points are equal. Hence, they are collinear.
Here = 9–2 = 8 =1
QR with a common

(b) If (a,0), (0,b) and (1,1) are collinear then prove that: 1 1 =1
a+ b

Solution
Let the given points are A(a,0), B(0,b) and C(1,1)

Slope of AB= b–0 = –b
0–a a

Slope of BC= 1–b = 1–b
1–a

Vedanta Optional Mathematics Teacher's Guide ~ 9 87

Slope of AB= Slope of BC

or, –b
a = 1–b

or, –b=a–ab

or, ab= a+b

or, 1= 1 + 1
ab ab

or, 1 1 = 1 proved.
a+ b

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

Solution
The given points A(a,0), B(0,b) and C(x,y)

Slope of AB= Slope of BC

b–0 = y–b
0–a x–0
or, bx= –ay+ab

or, bx+ay= ab

or, 1= bx ay
ab + ab
x y
or, a+ b =1

(d) If (x,y), (0,c) and ( ,0) are collinear prove that y= mx+c

Solution :

Given points A(x,y), B(0,c), C(– ,0)

Slope of AB= Slope of BC

c–y = 0–c
0–x –c–0

or, c–y = ––cc×m
–x
or, c–y= –mx

∴ y= mx+c Proved

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

Solution
Let x–intercept(a)= k

y–intercept(b) = k [∴ intercept are equal]

The equation of the line

xy
a+ b =1

or, xy
k+ k =1
or, x+y= k ............(i)

88 Vedanta Optional Mathematics Teacher's Guide ~ 9

But the line passes through (2,4)
∴ 2+4= k
i.e. k=6
Required equation of line, x+y=6

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

Solution
Let the x–intercept(a)= k

y–intercept(b) =–k

Equation of line is xy
a+ b =1

or, xy
k + –k = 1

or, x–y= k .............(i)
But it passes through (–4,–3), so we have

(–4)–(–3)= k
or, –4+3= k
⸫ k= –1
The required equation is x–y= k

or, x–y= –1
⸫ x–y+1 = 0

(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

Solution
Let x–intercept (a)= k

y–intercept (b) = 2k

Equation of line is

xy
a+ b =1

or, xy
k + 2k = 1

or, 2x+y= 2k

If passes through (4,5), so we have

2×4+5 = 2k

or, 8+5= 2k

or, k= 13
2
13
Required equation, 2x+y= 2k = 2× 2

∴ 2x+y= 13

Vedanta Optional Mathematics Teacher's Guide ~ 9 89

(d) Find the equation of a straight line which passes through the points (–1,3) and makes
intercepts on x–axis thrice as long as that on y–axis.

Solution
Let x–intercept(a)= 3k

y–intercept(b) = k

Equation of line is

xy
a+ b =1

or, xy
3k + k = 1

or, x+3y= 3k ..........(i)

It passes through (–1,3), so we have

(–1)+3(3)= 3k

or, k= 8
3
8
Required equation of line is x+3y= 3× 3

⸫ x+3y= 8

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

Solution
Let the line cuts of x–axis at (a,0) and Y–axis at (0,b)

Equation of straight line is x y =1 (o,b)
a+ b
Here (2.3) is the mid point of line joining (a,0) and (0,b)

∴ (2,3)= a+0 0+b (2,3)
2, 2

or, a=4, b=6

Required equation of line

x y = 1 (a,o)
4+ 6

⸫ 3x+2y= 12

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

Solution
Let the line cuts the x–axis at (a,0) and y–axis at (0,b)
(4,4) is the mid points of line joining (a,0) and (0,b)

∴ (x,y)= x1 + x2 , y1 + y2
2 2

(4,4)= a+0 0+b
∴a=8, b=8 2, 2

Equation of line is

x y = 1
a+ b

90 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, x y = 1
8+ 8
⸫ x+y=8

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

Solution
Let the line cut the x–axis at A(a,0) and y–axis at B(0,b). The points C(1,–5) divides the line
joining AB in the ratio 1:3.

Here,

(x,y)= m1mx12++mm22x1 , mm1y12++mm22y1

or, (1,–5) = 1×0+3×1 1×b + 3×0 y

1+3 , 1+3

or, (1,–5) = 3a b
4, 4
Now we have, A (a,o)
b Ox
1= 3a –5 = 4 x′
4 C (1,5) 
B
or,= 4 or, b = –20
3 y′ (o,b) 
The equation of line is

xy
a+ b =1

or, xy
4 + –20 = 1

or, 3x y
4 – 20 = 1

⸫ 15x–y=20

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

Solution (o,b) y
Let the line cuts x–axis at A(a,0) and y–axis at B(0,b) B
Here, C(–2,3) divides the line in the ratio 3:4.

Now, (x,y)= m1mx12++mm22x1 , mm1y21++mm22y1 4

or, (–2,3) = 3×0+4×a 3×b + 4×0
3+4 , 3+4

or, (–2,3) = 4a 3b C (–2,3) 
7, 7 3

∴ –2= 4a 3b x′ O x
7 3= 7 A (a,o) y′ 91
–7
or, a= –14 = 2 or b=7
4

Vedanta Optional Mathematics Teacher's Guide ~ 9

The equation of the line is

x y = 1
a+ b
x y
or, –7 + 7 = 1

2

or, –2x y = 1
7+ 7

⸫ 2x–y+7=0

19. Find the area of triangles formed by the line with the coordinates axis.
(a) 4x+5y–12=0

(b) 8x+5+40=0

Solution
(a) 4x+5y–12=0

or, 4x+5y= 12

or, 4x 5y
12 + 12 = 1

or, x + y =1
3 12

15 xy
a+b
Comparing with =1, we get

x–intercept(a)= 3

y–intercept(b)= 12
Area of triangle 5
= 1 ×a×b
2

= 1 ×3× 12
2 5

= 18
5

= 3.6 sq. unit

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

or, 8x+5y= –40

or, 8x + , 5y =1
–40 –40
–x y
or, 5+ –8 =1

Comparing with xy
a + b =1

x–intercept(a)= –5, y–intercept(b)= –8

Area of triangle = 1 ×a×b
2

= 1 ×(–5)×(–8)
2

= 20 sq. units

92 Vedanta Optional Mathematics Teacher's Guide ~ 9

(20) Find the lengths of line intercepted between the coordinates axis of the following line
x y
(a) 3x–4y= 24 (b) 4+ 5 =1

Solution

(a) 3x–4y=24

or, 3x – 4y =1 a²+b²
24 24 a
b
⸫ x – y =1 x y a²+b²
8 6 a+ b

Comparing with = 1 , we get

x–intercept(a)= 8, y–intercept(b)= –6

The length of the intercepted segment between the axis =

= 8²+6² = 10 units

(b) xy
4 + 5 =1

Solution
x–intercept= 4, –intercept= 5

The length of the line segment = a²+b² = 4²+5² = 41 unit.

(21) Find the equation of a straight line

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

Solution
Let x–intercept(a)= k

∴ y–intercept(b) = 9–k [∴ a+b=9]

Equation of line,

xy =1
a+b

or, x y =1 = 1 ..............(i)
k + 9–k

But the line passes through (4,1), So,

41 =1
k + 9–k

or, 36–4k+k =1
k(9–k)

or, 36–3k= 9k–k²

or, k²–12k+36= 0

or, (x–6)2

∴ x= 6

The required equation of the line is

x y =1
6 + 9–6
⸫ x+2y= 6

Vedanta Optional Mathematics Teacher's Guide ~ 9 93

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

Solution
Let x–intercept(a)= k

y–intercept(b) = k–4 [⸪ a–b=4]

Equation of the line,

xy =1
a+b
xy
or k + k–4 =1

The line passes through (3,1), we get

xy =1
k + k–4
or, 3k–12+k= k(k–4)

or, 4k–12= k²–4k
or, k²–8k+12= 0

or, k²–6k–2k+12= 0

or, k(k–6)–2(k–6)= 0
or, (k–6) (k–6)= 0

∴ k= 2,6

When k= 2

The equation of line,

xy =1
2 + 2–4
or, x–y= 2

When k=6,

The equation of the line is

xy =1
6 + 6–4
⸫ x+3y= 6

22. (a) If the perpendicular distance of a straight line from the origin is 8 units and the
inclination of the perpendicular with x–axis is 210°. Find the equation of the line.

Solution
Here p=8

α=210°
Equation of the line is

xcosα +ysinα =p
or, xcos210°+ysin210°= 8

or, x × –√3 +y –1 =8
2 2

or, –√3x – y = 8
2 2

⸫ √3x+y+16= 0

94 Vedanta Optional Mathematics Teacher's Guide ~ 9

(b) The length of perpendicular distance of a straight line from the origin is 2√3 units and
the slope of inclination of perpendicular is –√3. Find the equation of the line.

Solution
Here, p=2√3

tanα = –√3
tanα = tan120° or tan300°

α =120° or 300°
when α =120°

The equation of line,

xcosα +ysinα = p
or, xcos120°+ysin120°= 2√3

or,x –1 +y √3 = 2√3
2 2

or, –x+√3y = 2√3
2
or, –x+√3y= 4√3

or, x–√3y+4√3= 0

when α =300°
The equation of line,

xcosα +ysinα = p

xcos300°+ysin300°= 2√3

or, x × 1 + y –√3 = 2√3
2 2

⸫ x–√3y= 4√3

(c) A line whose perpendicular distance from the origin is 4 units and the slope of

perpendicular is 2 . Find equation of the line.
3

Solution
Here, p=4
slope= tanα = p 2 3 3
b= 3 2²+3² = 13
b
cosα = h= 3 =
p²+b²
22
or, sinα = p 2²+3² = 13
h=
The equation of the line,

xcosα +ysinα = p

or, x. 3 2 =4
√13 + y. √13

⸫ 3x+2y= 4√13

23. (a) The length of perpendicular drawn from the origin on a straight line is 3units and the
perpendicular is inclined at an angle of 120° to the x–axis. Find the equation of straight

Vedanta Optional Mathematics Teacher's Guide ~ 9 95

line. Also prove the line passes through the point (–3, √3).

Solution
Here, p= 3units

α = 120°

Equation of the line is

xcosα +ysinα = p

or, x.cos120°+ysin120°= 3

or, x –1 +y. √3 = 3
2 2

or, –x+√3y= 6

or, x–√3y+6=0

putting the point (–3,√3) on the equation.

–3–√3.√3+6= 0

or, –3–3+6= 0

⸫ 0=0 (true)

Hence the line passes through the point(–3,√3).

(b) Find the equitation of a straight line whose length of perpendicular drawn from the
origin on the straight line is 4units and the perpendicular is inclined at an angle of 60°
with x–axis. Also prove that it passes through the point(5,√3).

Solution
Here, p= 4units

α = 60°

Equation of the line

xcosα +ysinα = p

or, x.cos60°+ysin60°= p

or, x× 1 + y. √3
2 2
or, x+√3y= 8

Putting the point (5,√3) on the equation,

x+√3= 8

or, 5+√3.√8= 8

or, 5+3=8

or, 8=8 true)

Hence the line passes through the point (5,√3).

24. Reduce the following equation in slope intercept from and hence find the slope and
y–intercept.

(a) 13x+65y= 130

Solution
Here, 65y= –13x+130

96 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, y= –13x 130
or, y = – 5165x+2+ 65

which is in the form of y= mx+c [slope intercept form]

∴ slope(m)= –1
5
y–intercept(c)= 2

(b) 3x–2y–8= 0

Solution
Here, –2y= –3x+8

or, 2y= 3x–8

or, y= 3x 8
or, y= 2 –2
3x –4
2
which is in the form of y=mn+c

⸫ m= 3 , c= –4
2

25. Reduce the following equations in double intercept form and hence find the x–intercept
and y–intercept.

(a) 3x+9y= 45

Solution xy
2 + 2–4
Here, =1

or, xy =1
15 + 5
xy
which is in the form of a+b =1 [Double intercept form]

⸫ x–intercept(a)= 15

y–intercept(b)= 5

(b) √3x+y–12= 0

Solution
Here, √3x+y= 12

or, √3x y =1
12 + 12
√x y
or, 12 + 12 =1 which is in the form of
12 × √√33 = 4√3
⸫ x–intercept(a)= 12 = √3
√3
y–intercept(b)= 12

26. Reduce the following equations into perpendicular form and hence find p and α.

(a) x+y+1= 0 (b) x–y–1= 0 (c) √3x+y= 8

(d) x+√3y= 4 (e) x–y= 3√2 (f) x–y= 5

Vedanta Optional Mathematics Teacher's Guide ~ 9 97

(a) x+y+1= 0

Solution
Here, comparing it with Ax+By+c= 0

A=1, B=1, c=1

A²+B² = 1²+1² = 2

Dividing both sides of equation by √2, we get

x + y + 1 = 0
√2 √2 √2
1 –1
or, x. 1 + y. √2 = √2
√2

or, x –1 + y. –1 = 1 [ p must be positive]
√2 √2 √2

or, x.cos225°+y.sin225°= 1
√2
which is in the form of xcosα +ysinα = p

∴ α = 225°

p= 1
√2

(b) x–y–1= 0

Solution
Here, x–y= 1

(coeff of x)2+(coeff of y)² = 1²+(–1)² = 2

Dividing both sides by √2,

x – y = 1
√2 √2 √2

or, x . 1 + y. –1 = 1
√2 √2 √2
√12, Which is in the form of xcosα + ysinα = p
or, x.cos315°+y.sin315°=

∴ α = 315°, p= 1
√2

(c) √3x+y= 8

Solution

Here, (√3)2+1² = 3+1 = 2

Dividing both sides by 2

√3x + 2y= 8
2 2

or, √23. x + y. 1 =4
2

or, x.cos30°+y.sin30°= 4, which is in the form of xcosα + ysinα = p

∴ α = 30, p= 4

(d) x+√3y= 4

98 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution 12+(√3)² = 1+3 = 2

Here, Dividing both sides by

x + √3 = 4
2 2 2
1 √3
or, x. 2 +y. 2 =2

⸫ x.cos60°+ysin60²= 2, which is in the form of xcosα + ysinα = p
α =60², p=2

(e) x–y= 3√2

Solution Dividing both sides by 1²+(–1)² = √2
x y 3√2
√2 – √2 = √2

or, x. 1 +y. –1 =3
√2 √2

or, x.cos315°+y.sin315°= 3, which is in the form of xcosα + ysinα = p
∴α =315°, p=3

(f) x–y= 5

Solution
Dividing both sides by 1²+1² =√2
x y 5
√2 – √2 = √2

or, x. 1 +y. –1 = 5
√2 √2 √2
5
or, x.cos315°+y.sin315°= √2 , which is in the form of xcosα + ysinα = p

∴ α = 315°, p= 5
√2
xy x y
27. Transform the equation a+ b = 1 into normal form and hence show that a² + b² =
1 distance of the line from origin)
p² (where p is the perpendicular

Solution Given, equation xy
a + b =1
comparing with Ax+By+C=0, we get

A= 1 , B= 1 , C=–1
a b

A²+B² = 1 2 1 2 1 1
a + b a2 b2
= +

Dividing both sides by 1 + 1
a2 b2

x y –1

a 1 + 1 + 1 + 1 = 1 + 1
a2 b2 b a2 b2 a2 b2

Vedanta Optional Mathematics Teacher's Guide ~ 9 99

or, – x – y = 1
1 1 1 1 1 1
a a2 + b2 b a2 + b2 a2 + b2

which is in the form of x.cosα +y.sinα = p

∴ p= 1

1 + 1
a2 b2

or, 11 1 proved.
p² = a² + b²

28. The straight line 4x+5y–20= 0 cuts x–axis and A and y–axis at B. Find the x–intercept
and y–intercept. Also find the area of ∆AOB.
Y

Solution B(0,4)
Given, equation of the line O A(5,0)

4x+5y–20= 0 Y'

4x+5y= 20

or, 4x 5y =1
+204y+=210
or, x X' X
5
∴ x–intercept(a)= 5

y–intercept(b)= 4

∴ Area of triangle= 1 a×b = 5×4
2
=10sq. units

29. Find the value of k such that the line 2x+3y–k= 0 forms a triangle with coordinate axis
whose area is 12sq. units

Solution
Given equation,

2x+3y–k= 0

or, 2x+3y= k

or, 2x + 3 = 1
k k
xoAr–r,ienatkeoxr/f2cterp+iatkn(a3/g3)l=e==2k112 ab
y–intercept(b)= k
3



12= 12× 2k× k
3
or, 144 = k²

⸫ k= ±12

30) If P and Q are two points on the line x–y+1= 0, and are at a distance of 5units from the
origin. Find the area of ∆POQ.

100 Vedanta Optional Mathematics Teacher's Guide ~ 9