Optional Math 9

iv) IImff aAPt=r=ixSSSSRSTSSS<.–30122 12–F343an––d263BWWXWWWVWWW = <1 1F , prove that A + B is a symmetrical matrix.
v) 2 3 prove that 3P – Q is lower
2 9 –9
and Q = 2 8 –18 triangular
–7 –10 1

6. If A = <3 2F , B = <–1 3F and C = 2 4 prove that the followings.
1 –2 4 1 –1 3

i) A + B = B + A

ii) A + (B + C) = (A + B) + C

iii) A + 0 = 0 + A = A

iii) A + (–A) = 0

v) 2(A + B) = 2A + 2B

7. i) If AAAT===<<1313SSSSSSSTRS334 –2––121222FF, B–,12f1i=XWVWWWWWWWn,d<f13iAnTd–.43AF. find AT + BT.
ii) If
iii) If

iv) If A = <3 2 F and B = <1 –3F , find (A + B)T.
1 –1 34

v) If A = <2 1F , B = <–1 2F , find (2A – B)T.
3 2 1 2

8. PRIME more creative questions.

a. i) If A = <3 2F and B = <1 5F , prove that (A + B)T = AT + BT.
1 –2 –4 1

ii) If A = < 3 2F and (A + B)T = <6 –1F , find the matrix B.
–1 4 21

iii) If A = <4 3F and A + B = <6 5F , find B and prove that it is symmetrical.
2 1 4 3

iv) If A + B = <6 5F , A – B = <2 1F , find the matrices A & B.
4 3 0 –1

v) If P + Q = <4 7F & P–Q= <2 –3F , find the matrices P and Q.
–3 –1 5 –3

b. i) If = x +1 2yq+–24G is an identity matrix, find the value of x, y, p and q.
3p – 6

PRIME Opt. Maths Book - IX 97

ii) If <x –1 –4F is the additive inverse of <–1 4F , find the value of x and y.
y +3 5 –4 –5

iii) If <5 6F + <1 yF = <6 8F , find ‘x’ and ‘y’.
7 x 0 1 7 4

iv) If A = <x 3F , B = =4 + x y 2 G , C = <8 5F and A + B = C, find the value of ‘x’ and ‘y’ .
2 1 –1 – 2 1 1

v) If A = <2x 15F , B = < 6 5 F and AT = B, find the value of ‘x’ and ‘y’.
5 3y 15 12

Answer

1. Show to your teacher.
ii) SSTSRSSSSS204 064 320WWWXWWVWWW iii) <56 –25F
2. i) <3 1 0F
–1 5 2

iv) <–22 1 1F v) <53 44F
1 3

3. i) <4 1F ii) <55 55F iii) <–42 –31F
3 4

iv) <54 9F v) <163 18 1–15F
15

4. i) <10 3F ii) <15 –77F iii) <175 –11F
4 –3

iv) <–11 8F v) <–02 –1111F
–8

5. Proved 6. Proved
ii) SSSSSSRTSS124 –321WXWWWWVWWW iii) <–41 34F
7. i) <1 3F
2 –2

iv) <–41 4F v) <50 52F
3

8.a. i) Proved ii) <3 0F iii) <22 22F
0 –3

iv) A = <4 3F , B = <22 22F v) P= <3 2F , Q = <1 5F
2 1 1 –2 –4 1

b. i) x = 0, y = – 1, p = 2, q = 2 ii) x = 2, y = 1
v) x = 3, y = 4
iii) x = 3, y = 2 iv) x = 2, y = 2

98 PRIME Opt. Maths Book - IX

3.3 Multiplication of matrices:

Let us take an example,

Pranisha bought 2 kg of potatoes at the rate of Rs. 20, 3kg of tomatoes at the rate of Rs. 10
and 4kg of soyabean at the rate of Rs. 15. How much she had paid for each items.
Taking the informations in matrix form as,

A = 62 3 4@1×3 [Items in kg]

B = TRSSSSSSSS112500WWWVWWWWXW3×1 [Rate of cost of items respectively]

Then,
Total cost paid by Pranisha can be taken as,
= 2 × 20 + 3 × 10 + 4 × 15

= 40 + 30 + 60
= Rs. 130

This operation can be taken in matrix form as,
Total cost = Item (kg) × Rate
T = A[622××3B204@+SSSSTSRSSS3112050×WWWWWVXWWW10
= + 4 × 15]
=

= [40 + 30 + 60]

= [130]

Let us consider another example,
A company wants to establish three types of factories A, B, C in a place and needs male
and female workers in different months as below.

Factories A B C
Workers

Male 8 12 6

Female 324

Required workers in the factories. Jestha
Months Baishakh
3
Factories 4
A2 2
B3
C4

PRIME Opt. Maths Book - IX 99

How many workers are needed in each months in total?
HereP, T=he<38in1f2o2rm64aFt,i onsQin=mSSSRSSSTSS324atr324ixWWVWWWWWWX form:

Baishakh (Male) = 8 × 2 + 12 × 3 + 6 × 4

= 16 + 36 + 24

= 76

Baishakh (Female) =3×2+2×3+4×4

= 6 + 6 + 16

= 28

Jestha (Male) = 8 × 3 + 12 × 4 + 6 × 2

= 24 + 48 + 12

= 84

Jestha (Female) =3×3+2×4+4×2

=9+8+8

= 25

Number of workers in respective months can be written in bracket as,

<76 84F
28 25

Here, Row 1 shows the male and
Row 2 shows the female workers,

Wheres,

Column 1 shows the Baishakh and
Column 2 shows the Jestha

It can b=e<e38xp12r2es64seF2d×3inTSSSSSRSSS324ma324tVWWXWWWWWWr3i×x2 as,
P.Q

= =83××22++122××33++46××44 83××33++122××44++46××22G 2×2

= <76 84F
28 25 2×2

100 PRIME Opt. Maths Book - IX

3.3.1 Conclusion from the above examples for multiplication of the
matrices (Ways of matrix multiplication)

When number of columns of first matrix is equal to the number
of rows of second matrix, they can be multiplied where all the
elements of the rows of first matrix should be multiplied with
the corresponding element of columns of second matrix and
added to each other for the multiplication of the matrices.

i.e. For the matrices A and B having order m × p and p × n respectively.

Am×p × Bp×n = (AB)m×n

equal

(AB exists)

For the multiplication taking an example of any two matrices A and B having order 2 × 2.

A= <a bF , B = <p qF
c d r s

Then,

AB = <a bF <p qF 1st step
c d 2×2 r s 2×2

= =a×p + b×r ........ G 2nd step
....... .........

= =a×p + b×r a×q + b×sG 3rd step
...... ......

= =a×p + b×r a×q + b×sG 4th step
c×p + d×r ......

= =a×p + b×r a×q + b×sG 5th step
c×p + d×r c×q + d×s

= =ap + br aq + bsG Result of AB
cp + dr cq + ds

PRIME Opt. Maths Book - IX 101

L et uBT IA sfhABcAeo n=n,sSSTSSSSRSS====i134dRSSSSSSSSRTSSSSSSSSTSSSSSSSTRS<eD1132133444––re32×57×4×2am333––WVWXWWWWWWn132o22––+++5oa12WWWWXWWWWVFtn32(23WWWWWVWWXWh××–d322××e×2244Br)<SSSSSSSRSTe×13344=x4a<––34m13213224×p×FVXWWWWWWWW×–3211l11××2e22F+:++ (23–××2(()––×22())–2)WWXVWWWWWW

Here, the no. of column of first matrix are not equal to the no. of rows of second
matrix.

i.e. 2 ≠ 3
\ BA does not exists.

3.3.2 Conditions of matrix multiplications:

i) Number of columns of first matrix should equal to the number of rows of second

matrix.
ii) Elements of rows of first matrix should multiply with corresponding elements of

second matrix successively.
iii) Order of new matrix will be the number of rows of first and columns of second.
.
i.e. A3×2 B2×4 = (AB)3×4

3.3.3 Properties of matrix multiplication :
1. Commutative property does not hold.

AB ≠ BA

Taking A = <2 3F and B = <–1 2F
Then, 1 4 3 2

AB = <2 3F <–1 2F
1 4 2×2 3 2 2×2

= =2× (–1) + 3×3 2×2 + 3×2G
1× (–1) + 4×3 1×2 + 4×2

102 PRIME Opt. Maths Book - IX

= <7 10F
11 10

BA = <–1 2F <2 3F
3 2 2×2 1 4 2×2

= =(–1) ×2 + 2×1 (3–×13)+×32×+42×4G
3×2 + 2×1

= <0 5F
8 17

\ AB ≠ BA

2. Associative property:

(AB)C = A(BC)

Taking, A = <2 1F , B = <3 1F and C = <2 4F
–1 3 2 –1 1 3

L.H.S. = (AB)C

= (< 2 1F<3 1 F2 <2 4F
–1 32 –1 1 3

= <2 4F =2×3 + 1×2 2×1 + 1 (–1) G
1 3 (–1) ×3 + 3×2 (–1) ×1 + 3 (–1)

= <8 1F <2 4F
3 –4 1 3

= =8×2 + 1×1 ×1 8×4 + 1×3 G
3×2 + (–4) 3×4 + (–4) 3

= <17 35F
2 0

R.H.S. = A(BC)

= <2 1F =3×2 + 1×1 ×1 3×4 + 1×3 G
–1 3 2×2 + (–1) 2×4 + (–1) ×3

= <2 1F <7 15F
–1 3 3 5

= =2×7 + 1×3 22××415++(–11×)5×3G
(–1) ×7 + 3×3

= <17 35F
2 0

\ L.H.S. = R.H.S. proved

PRIME Opt. Maths Book - IX 103

3. Distributive property over addition.

A(B + C) = AB + AC

Taking A = <4 2F , B = <–2 3F , C = <3 2F
3 1 –1 1 1 4

L.H.S. = A(B + C)

= <4 2F (<–2 3F + <3 2F2
3 1 –1 11 4

= <4 2F =–2 + 3 3 + 2G
3 1 –1 + 1 1 + 1

= =4×1 + 2×0 4×5 + 2×5G
3×1 + 1×0 3×5 + 1×5

= <4 30F
3 20

R.H.S. = AB + AC

= <4 2F <–2 3F + <4 2F <3 2F
3 1 –1 1 3 11 4

= =4× (–2) 2 (–1) 4×3 + 2×1G + =4×3 + 2×1 4×2 + 2×4G
3× (–2) + 1 (–1) 3×3 + 1×1 3×3 + 1×1 3×2 + 1×4

= <–10 14F + <14 16F
–7 10 10 10

= =–10 + 14 17 + 16G
–7 + 10 10 + 10

= <4 30F
3 20

\ L.H.S. = R.H.S. proved

4. Multiplicative identity
For any matrix A, ∃ a matrix I such that AI = IA = A. I is called the multiplicative identity.

A×I=I×A=A

Taking A = <5 4F , I = <1 0F
Then, 2 1 0 1

AI = <5 4F <1 0F
2 1 0 1

= =5×1 + 4×0 5×0 + 4×1G
2×1 + 1×0 2×0 + 1×1

= <5 4F
2 1

= A

104 PRIME Opt. Maths Book - IX

IA = <1 0F <5 4F
0 1 2 1

= =1×5 + 0×2 1×4 + 0×1G
0×2 + 1×2 0×4 + 1×1

= <5 4F
2 1

= A

5. Multiplication over transpose:

(AB)T = BTAT

Taking, A = <3 2F and B = <–2 –3F
4 1 1 4

L.H.S. = (AB)T
3 2 –2 –3 T

= (<4 1F< 1 4 F2

= =3× (–2) + 2×1 3 (–3) + 2×4GT
4× (–2) + 1×1 4 (–3) + 1×4

= <–4 –1FT
–7 –8

= <–4 –7F
–1 –8

R.H.S. = BTAT

= <–2 –3FT <3 2FT
1 4 4 1

= <–2 1F <3 4F
–3 4 2 1

= =(–2) ×3 + 1×2 (–2) ×4 + 1×1G
(–3) ×3 + 4×3 (–3) ×4 + 4×1

= <–4 –7F
–1 –8

\ L.H.S. = R.H.S. proved

PRIME Opt. Maths Book - IX 105

Worked out Examples

1. If A = <13 –22F and B = <–42 1 43F find AB.
2

Solution :

AB = <3 2F<4 1 3F
1 –2 –2 2 4

= =13××4(–+42) +(–(2–)2) (–2) 3×1 + 2×2 3×3 + 2×4 G
1×1 + (–2) ×2 1×3 + (–2) ×4

= <8 7 17F
8 –3 –5

2. Which matrix pre- multiplies to <01 32F gives [2 13]?
Solution :

Let,

Am × z <1 2F = [2 13]1 × 2
0 3 2×2

\ m = 1

Here,

m×2 2×2=1×2
Let, The required matrix be [ab]
Then,

[a b] <1 2F = [2 13]
0 3

or, [a × 1 + b × 0 a × 2 + b × 3] = [2 13]

or, [a 2a + 3b] = [2 13]

by equating the corresponding elements,

a = 2

and 2a + 3b = 13

or, 2 × 2 + 3b = 13

or, 3b = 9

\ b = 3
\ The required matrix is [2 3]

3. If A = <12 –32F , find the value of A2 – 3A + 2I.

Solution :

A2 – 3A + 2I = <2 3 F <2 3F – 3<12 3F + 2<10 0F
1 –2 1 –2 –2 1

= =2×2 + 3×1 ×1 2×3 + 3× (–2) G
1×2 + (–2) 1×3 + (–2) (–2)

= <7 0F – <6 9F + <2 0F
0 73 –6 0 2

106 PRIME Opt. Maths Book - IX

= =70 – 6 + 2 0 – 9 + 02G
– 3 + 0 7 + 6 +

= <–23 1–95F

4. Which matrix post multiplies to <43 12F gives <––74 ––18F ?
Solution :

Here, AB = C

Taking order : 2 × 2 2 × m = 2 × 2

∴ m = 2 and order of required matrix is also 2 × 2.

Let, the matrix be B = <ac dbF
Then,

<34 12F <ac dbF = <––74 ––18F

or, =34××aa + 2×c 3×b + 2×dG = <––47 ––18F
+ 1×c 4×b + 1×d

or, =34aa + 2c 3b + 2ddG = <––47 ––18F
+ c 4b +

By equating the corresponding elements,

4a + c = – 7 ⇒ c = – 4a – 7 .............................. (i)

3a + 2c = – 4 ⇒ c= –3a – 4 .............................. (ii)
2

4b = d = – 8 ⇒ d = –4b – 8.............................. (iii)

3b + 2d = – 1 ⇒ d= –3b – 1 .............................. (iv)
2

Solving equation (i) and (ii)

–4a – 7 = –3a – 4
2

or, –8a – 14 = –3a – 4

or, –5a = 10

∴ a = –2

Solving equation (iii) and (iv)

–4b – 8 = –3b – 1
2

or, –8b – 16 = –3b – 1

or, –5b = 15

∴ b = –3

PRIME Opt. Maths Book - IX 107

Putting the value of ‘a’ and ‘b’ in equation (i) and (iii) respectively.

c = –4(–2) – 7 = 1

d = –4(–3) – 8 = 4

\ The required matrix is <–2 –3F
1 4

5. If P = <04 05F and Q = <0x zyF and PQ = P + Q, find x, y and z.

Solution :
PQ = P + Q

or, <04 0F<x yF = <4 0F + <x yF
50 z 0 50 z

or, =4 × x + 0×0 4 × y + 0× zG = =4 + x 0 + yG
0 × x + 5×0 0 × y + 5× z 0 + 0 5 + z

or, <40x 4yF = =4 +x 0 + yG
5z 0 +0 5 + z

By equating the corresponding elements,

4x = 4 + x ⇒ x= 4
3

4y = y ⇒ y= 1
4

5z = 5 + z ⇒ z= 5
4

108 PRIME Opt. Maths Book - IX

Exercise 3.3

1. i) Write down the conditions of matrix multiplication.
ii) Write down distributive property of multiplication over addition.
iii) Am×p is a matrix and Bp×n is another matrix can A and B be multiplication?
iv) In what condition AB = BA = A?
v) Write down the associative property of multiplication.

2. vWiii))i ) h icPCAh===ofRSSSSSSTSSSSSSSSTSSR<531232t34hVXWWWWWWWWe,––122Q1f1oWWWWWVXWWW=l,l12oD6Fw1,=Bin3SSSRTSSSSS=g1322mSSSTSSSSSR@–13212a 1XVWWWWWWWWt r324icWWWWWXWWWVes can multiply? Write down with reason.

ii) A = <1 3 2F , B = <2 4 1F
4 1 –2 –1 3 2

iv) M = SSSTSRSSSS134 2 –121XVWWWWWWWW , N = RSSSSSSSTS324 –122WVXWWWWWWW
2
1

3. Write down the order of the following matrix multiplication by multiplying them.

i) MN where, M = <2 –1F , N = <4 1F
3 2 2 –1

ii) AB where, A = <3 2F , B = <1 2 1F
1 4 3 1 4

iii) QP where, P = SRTSSSSSSS<<32–1311212–41131–12–22241FFXWWWWWWWVW,, QB, Q===SSSSSSSTRSSSSSSTRSSS13––4[2221–1211311WWWWWXWWWVXVWWWWWWWW 3]
iv) PQ where, P =
v) BA where, A =

4. If A = <3 2F , B = <2 4F and C = <1 3F , prove that the followings:
1 –2 1 3 2 4

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

iii) (AB)T = BTAT iv) BC ≠ CB

v) AI = IA = A where I is identity matrix of order 2 × 2

PRIME Opt. Maths Book - IX 109

5. i) Which matrix pre-multiplies to <2 0F gives 65 9@ ?
1 3

ii) Which matrix post-mulitplies to <1 2F gives <8F ?
0 3 9

iii) Which matrix pre-multiplies to <3 –1F gives <7 –1F ?
2 0 7 –3

iv) Which matrix post multiplies to <–1 2F gives <–1 6F ?
–2 1 –5 0

v) If <–4 0F A = <4F , find the matrix A.
5 7 9

6. i) If A = <–4 4F and B = <2 –5F , prove that : AB is a null matrix.
–7 7 2 –5

ii) If M = <4 7F and N = <–5 7F , prove that : MN is an identity matrix.
3 5 3 –4

iii) If P = <7 –5F and Q = <–2 5F , prove that : PQ = QP = I
3 –2 –3 7

iv) If A = <4 1F , I and O are the identity and zero matrix of order 2 × 2.
–1 2

Prove that : A2 – 6A + 9I = 0.

v) If I is an identity matrix of order 2 × 2 and M = <4 2F .
Prove that : (M – 2I)(M – 3I) = 0 –1 1

7. PRIME more creative questions:

a. i) If A = <3 –5F and I is the unit matrix of order 2 × 2, prove that A2 = 5A + 14I.
–4 2

ii) If P = <2 0F , Q = <p qF and PQ = P + Q, find p, q and r.
0 3 0 r

iii) If AT = <2 –3F and BT = <1 –2F , prove that (AB)T = BTAT.
1 2 3 4

iv) If ASSSSSSSRST23y=xz–++<ac283WVXWWWWWWWdb=F , B= <1 6 7F and C = <2 1 4F and AB = C, find the matrix ‘A’.
v) If SSSTSSRSSS–523 ––2423WWWWVWXWWW1 4 find x, y 167
<3F , and z.
2

b. i) If A = <a 0F , B = <4 0F and A2 = B, find the value of ‘a’ and ‘b’.
–3 b –9 1

ii) M = < 1 xF , N = <–3 8F and M2 = N, find the value of ‘x’ and ‘y’.
–2 y –8 5

iii) If <2 1F<3 1F = <8 cF find the value of ‘a’, ‘b’ and ‘c’.
–1 ab –1 3 –4

110 PRIME Opt. Maths Book - IX

iv) If A = <3 4F and AB is an identity matrix of order 2 × 2, find the matrix B.
2 3

v) Collects the marks obtained by 9 students from Roll No. 1 to 9 of your class in
first terminal examination and present such marks in matrix form. Discuss the

possible matrices of different order. Is square matrix formed?

8. Project work
Collects the types of matrices and operations on matrices in a chart paper and

present it in your classroom.

Answer

1. Show to your teacher.

2. Show to your teacher.

3. i) <6 3F ii) < 9 8 11F iii) 63 10 7@1×3
16 1 2×2 v) RSSSSSSTSS11583––7796 1–6427WWWWWVXWWW32××33
1F
iv) <137 5 2×2
4. Proved

5. i) 61 3@ ii) <32F iii) <13 –21F

iv) <13 2F v) <–21F
4

6. Proved

7.a. ii) p = 2, q = 0, r = 3 iv) <10 1F v) x = 3, y = 4, z = –3
2 0 iii) a = 3, b = 2, c = 1

b. i) a = ±2, b = ±1 ii) x = 2, y = 3

iv) B = <3 –4F v) Show to your subject teacher.
–2 3

PRIME Opt. Maths Book - IX 111

Matrices

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. What is scalar matrix? Write down with example.

2. a. If A = <1 2F and B = <3 –1F , find 3A – 2B.
–2 3 2 1

b. Which matrix pre-multiplies to <2 0F results [7 9]?
1 3

c. If A = <2 1F , find A2 – 3A + 5I.
3 4

3. a. If A + B = <4 1F , and A – B = <–2 3F find A and B.
0 4 –4 2

b. If <1 xF + <a –1F = <2a–2 1F , find the value of a, b, x and y.
–2 y b 1 3b–6 4

4. If A = <–2 4F , B = <3 1F and C = <4 1F Prove that A(B + C) = AB + AC.
3 1 –2 –1 3 2

112 PRIME Opt. Maths Book - IX

Unit 4 Co-ordinate Geometry

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 2 2 1 1 6 15 30
Weight 2 4 4 5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to find the distance formula and its application.
• Students are able to find the locus point and its equation in different

conditions.
• Students are able to find the Section formula (internal, external, mid-point

and centroid of a triangle).
• Students are able to find the slope of a straight line in different condition
• Student are able to find the equation of straight line in standard conditions.
• Students are able to reduce the linear equation into standard form of

equation.
• Students are able to find the perpendicular distance of straight line from a

point and distance between two parallel lines.
• Students are able to find the area of triangle and quadrilateral using co-

ordinates.

Materials Required:
• Chart paper
• Chart of formulae used in co-ordinate
• Chart of standard formula with derivation.
• Graph paper.

PRIME Opt. Maths Book - IX 113

4.1 Distance between any two points (Review only)

DLeist;ta‘dn’cbeebtehtewdeiesntatnhceepboeitnwtseAen(xt1h, ey1p)oainndt AB((xx12,, y2). B(x2, y2)
Draw, AM^OX, BN^OX and AC^BN. y1) and

Then, = Mxx12N,, AB=MNO==Nyy–21 Y (x2, y2)
OM = (x1, y1) B
ON = OM = x2 – x1 d
AC C
X
BC = BN – CN = BN – AM = y2 – y1
AB = d N

In right angled DABC, A

h2 = p2 + b2

or, AB2 = AC2 + BC2

or, d2 = (x2 – x1)2 + (y2 – y1)2 OM
\ d = (x2 – x1)2 + (y2 – y1)2

It is the distance formula.

Some important informations: (Things to remember)
• Origin point is O(0, 0)
• A point on x – axis is (x, 0)
• A point on y – axis is (0, y)
• To prove right angled triangle : h2 = p2 + b2
• To prove isosceles triangle : Any two sides should be equal.
• To prove equilateral triangle : All side should be equal.
• To prove rhombus : All 4 sides should be equal.
• To prove square : All 4 sides should be equal and h2 = p2 + b2 for a diagonal
or diagonals are equal.
• To prove parallelogram.
Opposite sides should be equal. (mid points of diagonals should be equal)
• To prove rectangle : Opposite sides should be equal and h2 = p2 + b2 for a
diagonal. (i.e. diagonals are equal.)
• To prove the points A, B, C in circumference of a circle,
OA = OB = OC where O is centre of circle.

Worked out Examples

1. Plot the points A(1, 3) and B(4, 7) in graph paper and find the distance between the
points A and B by drawing perpendicular lines to the axes using run & rise.

Solution:
The given points A(1, 3) and B(4, 7) in graph are;
Here, AC = MN = ON – OM = 4 – 1 = 3 = run
BC = BN – CN = BN – AM = 7 – 3 = 4 = rise

114 PRIME Opt. Maths Book - IX

\ AB = (run)² + (rise)² Y
B
= 32 + 42
AC
= 25

= 5 units

2. Find the distance between the points (2, –3) and (5, 1). X’ O M X
Solution: N
The given points are:
Y’
A(2, –3) = (x1, y1)
B(5, 1) = (x2, y2)
Using distance formula,

d = (x2 – x1)2 + (y2 – y1)2
d(AB) = (5 – 2)2 + (1 + 3)2

= 32 + 42
= 5 units.

3. Prove that the line joining the points A(3, 4), B(7, 7) and C(11, 10) are collinear
points.

Solution :
The given points are A(3, 4), B(7, 7) & C(11, 10)
Using distance formula,

d = (x2 – x1)2 + (y2 – y1)2

d(AB) = (7 – 3)2 + (7 – 4)2 = 5 units

d(BC) = (11 – 7)2 + (10 – 7)2 = 5 units

d(AC) = (11 – 3)2 + (10 – 4)2 = 10 units
Here, AC = AB + BC
or, 10 = 5 + 5
or, 10 = 10
Hence, They are collinear points.

4. Prove that the points A(–2, 3), B(–2, –4), C(5, –4) and D(5, 3) are the vertices of a

square. A(–2, 3) D(5, 3)
Solution :
The given points are
A(–2, 3), B(–2, –4), C(5, –4), D(5, 3)

Using distance formula

d = (x2 – x1)2 + (y2 – y1)2

d(AB) = (–2 + 2)2 + (–4 – 3)2= 7units

d(BC) = (52 + 2)2 + (–4 + 4)2 = 7units

d(CD) = (5 + 5)2 + (3 + 4)2 = 7units B(–2, –4) C(5, –4)

PRIME Opt. Maths Book - IX 115

d(AD) = (5 + 2)2 + (3 – 3)2 = 7units

d(AC) = (5 + 2)2 + (–4 – 3)2= 7 2 units
Here, AB = BC = CD = DA
Also, AB2 + BC2 = AC2

or, 72 + 72 = ( 7 2 )2

or, 98 = 98
Hence, They are the vertices of a square.

5. Find the co-ordiante of a point on x-axis which is 5 units distance from a point (5, 4).

Solution : Let, the point on x-axis be A(x, 0)
The given point is B(5, 4)

Using distance formula, (5, 4)
d = 5 units
d2 = (x2 – x1)2 + (y2 – y1)2

or (5)2 = (x – 5)2 + (0 – 4)2

or, 25 = x2 – 10x + 25 + 16

or, x2 – 10x + 16 = 0

or, x2 – (8 + 2)x + 16 = 0 (x, 0)
or, x2 – 8x – 2x + 16 = 0

or, x(x – 8) – 2(x – 8) = 0

or, (x – 8) (x – 2) = 0

Either, Or

x – 8 = 0 x–2=0

\ x = 8 \ x=2

\ The required point is (8, 0) or (2, 0)

Exercise 4.1

1. i) Write down the co-ordinate of a point lies in x-axis & y-axis.
ii) If run = 3 units, rise = 4units of a line segment AB, find the length of AB.
iii) Write down the distance formula of a point P(x, y) from the origin.
iv) In what condition the DABC will be right angled at B?
v) Write down the condition of the quadrilateral being a square.

2. Find the distance between the points given below.

i) (3, –7) and (9, 1) ii) (–5, 6) and (10, –2)

iii) (8, –3) and (–4, 2) iv) (a + b, a – b) and (b – a, b + a)

v) (aSina, aCosa) and (bCosa, – bSina)

3. i) Prove that the point (1, 2) is equidistance from the points (4, – 2), (5, –1) and (–2, 6).
ii) Prove that the point on x-axis (–3, 0) is equidistance from the points (3, 8) and (5, –6).
iii) Prove that the points (3, 4), (–4, 3) and (0, –5) are the points of circumference of a

circle of centre origin.

116 PRIME Opt. Maths Book - IX

iv) Prove that the point on y-axis (0, 4) is equidistance from the points (15, –4) and (–8, –11).
v) Prove that the point (–4, 4) is the mid point of line joining the points (–3, 7) and (–5, 1)

4. i) Prove that the points (0, 4), (3, –2) and (5, –6) are collinear.
ii) Prove that the points (3, 4), (7, 7) and (11, 10) are collinear.
iii) Prove that the points (2, 3), (5, 8), (0, 5) and (–3, 0) are the vertices of a rhombus.
iv) Prove that the points (7, 10), (–2, 5) and (3, –4) are the vertices of isosceles right

angled triangle.
v) Prove that the vertices (4, 8), (0, 2), (3, 0) and (7, 6) are of a rectangle.

5. i) Find the co-ordinate of a point on x-axis which is 5 units distant from the point
(–2, 3)

ii) Find the co-ordinate of a point on y-axis which is 13 units distance from the
point (2, 5)

iii) Find the value of ‘m’ where (m, –2) is 10 units distance from the point (–2, 6).
iv) Find the co-ordinate of a point on x-axis which is equidistance from the points

(1, 3) and (–6, 4)
v) Find the co-ordinate of a point on y-axis which is equidistance from the points

(6, 10) and (–8,8).

6. PRIME more creative questions
i) Find the co-ordinate of a point which is equidistance from the points (2, –1), (1,

0) and (2, 7)
ii) Find the cirum-centre of a triangle having vertices (–5, –2), (–3, –4) and (9, 12).
iii) The two vertices of an equilateral triangle are (–2, 0) and (–8, 0), find the co-

ordinate of 3rd vertix lies in second quadrent.
iv) If a point p(x, y) is equidistance from the points (3, –2) and (2, 1) prove that

x – 3y – 4 = 0.
v) Prove that the points (3, –3), (3 3, 3 3) and (–3 3, –3 3) are the vertices of

an isosceles triangle.

1. Show to your teacher. Answer iii) 13 units
2. i) 10 units iii) – 8 or 4
iv) 2 a2 + b2 units ii) 17 units iii) (–5, 3 3)
5. i) (2, 0) or (–6, 0) v) a2 + b2 units
iv) (–3, 0) ii) (0, 2) or (0, 8)
v) (0, 2)
6. i) (5, 3) ii) (3, 4)

PRIME Opt. Maths Book - IX 117

4.2 Equation of Locus of a point

Let us consider a moving point P moves in different conditions w.r.t. a fixed point as given

in diagrams. P A P1
P1 P4 P2 B
P3
P2
P3 P4

P1

A P2 P4

P3 B P1

P1 P6 P3 P2

P2 P5

P4 P4

P3 P5 P6

In the above diagrams, the moving point P moves in different form of geometrical shapes

like circular, straight, curve, parabola, hyperhola etc according to the given geometrical

conditions. Such forms are the path traced out by the moving point while moving from

one place to another place called locus of the point.

The path traced out by a moving point while moving from one
place to another place under any geometrical conditions is called

locus of the point.

The paths traced out by a moving point can be expressed in the form of equation according
to the given conditions which is called equation of locus of a point.

Example : If a moving point P(x, y) makes a distance of 3 units always from the origin O(0, 0)
Then, OP = 3
OP2 = 9

using distance formula

d2 = (x2 – x1)2 + (y2 – y1)2
or, OP2 =9

or, (x – 0)2 + (y – 0)2 = 9

or, x2 + y2 = 9
It is the equation of locus of a point P.

118 PRIME Opt. Maths Book - IX

Worked out Examples

1. Find the locus of a point which moves so that its abscissa is always 4.
Solution :
Let, P(x, y) be a moving point
By the equation,
abscissa is always 4.
or, x - component = 4
or, x = 4
\ The required equation of locus is x = 4.

2. Prove that the points (1, 2) and (–3, 10) always lie in the locus of equation 2x + y – 4 = 0
Solution :
The given equation of locus is,
2x + y – 4 = 0
At point (1, 2)
2 × 1 + 2 – 4 = 0
or, 4 – 4 = 0
or, 0 = 0 (true)

Again,
At point (–3, 10)
2(–3) + 10 – 4 = 0
or, 0 = 0 (true)
Hence, (1, 2) and (–3, 10) always lie in the given locus proved.

3. Find the equation of locus of a point which is equi distance from the points (2, –1)

and (1, 2)

Solution : P(x, y)
Let, P(x, y) be a moving point.
The given fixed points are A(2, –1) and B(1, 2).

By the question. A(2, –1) B(1, 2)
AP = BP
or, AP2 = BP2 [\ squaring on both sides?]

using distance formula,

d2 = (x2 – x1)2 + (y2 – y1)2

Then taking
AP2 = BP2

or, (x – 2)2 + (y + 1)2 = (x – 1)2 + (y – 2)2

or, x2 – 4x + 4 + y2 + 2y + 1 = x2 – 2x + 1 + y2 – 4y + 4

or, –2x + 6y = 0

or, x – 3y = 0

\ x – 3y = 0 is the required equation.

PRIME Opt. Maths Book - IX 119

4. Find the equation of locus of a point which moves so that its distance from the point

(–2, 1) is double its distance from the point (2, 3).

Solution : P(x, y)
Let, P(x, y) be a moving point.
The fixed points are A(–2, 1) and B(2, 3)

by the question, B(2, 3)
AP = 2BP
or, AP2 = 4BP2

using distance formula A(–2, 1)

d2 = =(x42B–Px21)2 + (y2 – y1)2
or, AP2

or, (x + 2)2 + (y – 1)2 = 4{(x – 2)2 + (y – 3)2}

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

or, 3x2 + 3y2 – 20x – 22y + 47 = 0

It is the required equation of locus.

5. Find the equation of locus of a point ‘P’ which moves so that from the points A(–a, 0)
and B(a, 0) under the condition AP2 + BP2 = AB2.

Solution:
Let, P(x, y) be a moving point.
The fixed points are A(–a, 0) and B(a, 0)
By the question,
PA2 + PB2 = AB2
using distance formula

d2 = (x2 – x1)2 + (y2 – y1)2

Taking,
PA2 + PB2 = AB2

or, (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 = 4a2

or, 2x2 + 2y2 = 2a2

or, x2 + y2 = a2

It is the required equation of locus.

120 PRIME Opt. Maths Book - IX

Exercise 4.2

1. i) What is locus?
ii) Which type of locus represented by the equation x + y = 3?
iii) Which type of locus represented by the equation x2 + y2 = a2?
iv) Which type of locus represented by the equation x = 4?
v) Which type of locus represented by the equation y = –2?

2. i) Which of the points (2, 2), (1, 5), (–2, 10) and (3, 0) lie in the locus having

equation 2x + y – 6 = 0
ii) Prove that the points (3, 4), (–4, –3) and (–5, 0) lie in the locus of equation x2 +

y2 = 25
iii) If (3, 2) lies in the locus of equation 3x + my – 17 = 0, find the value of ‘m’.
iv) Prove that the point (2, –3) lies in the locus having equations 4x – y – 11 = 0 and

x2 + y2 = 13.
v) If a point (1, –2) lies in the locus having equation x2 + y2 + 2x + py – 3 = 0, find

the value of ‘p’.

3. Find the equation of locus of a point which moves so that under the following
conditions.

i) Making abscissa always 3
ii) Making ordinate always – 4
iii) Making distance from the origin always 5 units.
iv) Making sum of abscissa and ordinate always 7.
v) Making distance from a point (1, 2) is always 4 units.

4. Find the equation of locus of a point under the conditions given below.
i) It moves such that equidistance from the points (3, –1) and (1, 3).
ii) It moves making equal distance from the points (2, 1) and (–3, 2)
iii) It moves making distance from x-axis is double the distance from y-axis.
iv) It moves such that its distance from (1, 2) is double the distance from (1, –1).
v) It moves such that its distance from the origin is thrice that distance from (–1, 2)

5. Find the equation of locus of a point ‘P’ under the following conditions.
i) It moves such that distance from y-axis is double the distance from x-axis.
ii) It moves such that from the points A(–2, 0) and B(2, 0) under the condition AP2

+ BP2 = AB2
iii) It moves such that from the points A(1, –2) and B(2, 1) under the condition of

2AP – BP = 0.
iv) It moves such that sum of the square of distance from A(2, –3) and B(3, 4) is

always 40 units.

v) It moves such that the difference of distance from the point (3, – 2) and (2, 1) is

always zero.

PRIME Opt. Maths Book - IX 121

6. PRIME more creative questions.
i) Find the equation of locus of a point so that ratio of its distance from (2, 1) and

from the point (1, –2) is always 2:3.

ii) Find the equation of locus of a point so that it moves from the circumference of
a circle having equation x2 + y2 = k where one of the point in the circumference

is (–3, 0)
iii) If a point (4, 4) lies in the equation of locus y2 = ax. Find the value of ‘a’. Also

prove that another point (16, 8) lies in that locus.
iv) Find the equation of locus of a point P which moves so that it makes right angle

at point ‘P’ from the fixed points A(–m, 0) and B(m, 0).
v) Find the equation of locus of a point which moves so, that its distance fro (1, –2)

is half of the distance from (2, 1).

Answer

1. Show to your teacher.

2. i) (2, 2), (–2, 10), (3, 0) iii) 4 v) 2

3. i) x – 3 = 0 ii) y + 4 = 0 iii) x2 + y2 = 25
iv) x + y = 7
v) x2 + y2 – 2x – 4y – 11 = 0

4. i) x – 2y = 0 ii) 5x – y + 4 = 0 iii) 2x – y = 0

iv) x2 + y2 – 2x + 4y + 1 =0

v) 8x2 + 8y2 + 18x – 36y + 45 = 0

5. i) x – 2y = 0 ii) x2 + y2 = 4

iii) 3x2 + 3y2 – 4x + 18y + 15 = 0

iv) x2 + y2 – 5x – y – 1 = 0

v) x – 3y – 4 = 0

6. i) 5x2 + 5y2 – 28x – 34y + 25 = 0
ii) x2 + y2 = 9
iii) a = 4
iv) x2 + y2 = m2
v) 3x2 + 3y2 – 4x + 18y + 15 = 0

122 PRIME Opt. Maths Book - IX

4.3 Section Point

Let us consider a point P cuts a line segment AB at a point where the point divides the
line segment in two parts AP and BP. Hence the point P is called the section point of a line
segment AB as shown in diagram. Such two sections AP and PB may be equal or may be

in a certain ratio. A m1 P m2 B

Internal Section

The point which may cut in equal A m Pm B
section is called mid-point. B
The point which may cut in unequal Mid Section
sections externally or internally is
P m1 A m
called section point.

4.3.1 Centroid of a triangle m2
External Section

The intersecting point of medians of a triangle is called the
centroid of the triangle. It cuts the medians in the ratio 2:1.

Here, In DABC, P, Q and R are the mid-point of sides of DABC where A
AP, BQ and CR are called medians.
The intersecting point of the medians AP, BQ and CR is ‘G’ which is R Q
G
called the centroid of DABC. B C
P

4.3.2 Co-ordinate of section point of a line segment.

i. When a point cuts internally to a line Y
segment joining the points (x1, y1) & (x2, y2).
Let, a point C(x, y) cuts the line joining the B(x2, y2)

points A(x1, y1) and B(x2, y2) internally in the C(x, y)
Drartaiowntmhe1:pme2r.pendiculars, m2
S

AM^OX, BN^OX, m1 Q
CP^OX, AQ^CP,
A
CS^BN, (x1, y1)
Then, OM
CABCSQQS ====PBCMNNPP=––=QOSONPNP==––CBOOPNPM–=–=ACxMx2P––==xxyy.12.––yy1.. P NX




AC : CB = m1:m2

PRIME Opt. Maths Book - IX 123

In DACQ and DCBS
i) \ AQC = \ BSC → Both being 90°
ii) \ QAC = \ SCB → Being corresponding angles.
iii) \ ACQ = \ CBS → Being remaining angles of Ds.
\ DACQ ∼ DCBS by AAA axiom.

\ AQ = CQ = AC
CS BS CB

or, xx2––xx1 = y – y1 = m1
y2 – y m2

Taking,

or, xx2––xx1 = m1
m2

or, m2x – m2x1 = m1x2 – m1x

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

\ x= m1 x2 + m2 x1
m1 + m2

Again,

y – y1 = m1
y2 – y m2

or, m2y – m2y1 = m1y2 – m1y
or, y(m1 + m2) = m2y1 + m1y2

\ y= m1 y2 + m2 y1
m1 + m2

\ Co-ordiante of section point is
m1 y2 + m2 y1
C(x, y) = ( m1 x2 + m2 x1 , m1 + m2 )
m1 + m2

ii) Mid-point of a line segment
If point ‘C’ be the mid-point of the line segment AB in the above dliinagerasemg,mme1n=t
Then the section point so considered is called mid-point of the AmB2.,

where, m1 y2 + m2 y1
m1 + m2
(x, y) = ( m1 x2 + m2 x1 , )
m1 + m2

If m1 = m2 for mid-point , m2 (y1 + y2)
or, (x, y) = m1 (x1 + x2) 2m2

2m1

\ (x, y) = x1 + x2 , y1 + y2
2 2

It is the co-ordinate of mid-point.

124 PRIME Opt. Maths Book - IX

iii. If the section point cuts a line segment externally.
Let, A point P(x, y) cuts the line joining the
points A(x1, y1) and B(x2, y2) externally in the Y

rDartaiowmth1:emp2e. rpendiculars, P(x, y)
AM^OX, BN^OX and PQ^OX
AR^PQ and BS^PQ , y )

2 2
B(x m2
m1 S

Then,
AR ===:=BPNPMPQQQQ=––==mSROOQ1QQ:Qm==–=2PPOQOQNM––=B=AxNMx–=–=xyx2y1––yy21
BS A R
PR (x1, y1) N QX
PS OM
AP

Now, In DARP and DBSP,
\ R = \ S
→ Both being 90°

\ A = \ B → Corresponding angles
\ P = \ P → Common angle
\ DARP ∼ DBSP → By AAA axiom.

\ AR = PR = AP
BS PS BP

or, xx –x1 = y –y1 = m1
– x2 y – y2 m2

Taking,

x –x1 = m1
x – x2 m2

or, m2x – m2x1 = m1x – m1x2
or, m1x2 – m2x1 = x(m1 – m2)

\ x= m1 x2 – m2 x1
m1 – m2

Again,

y –y1 = m1
y – y2 m2

or, m2y – m2y2 = m1y – m1y2
or, m1y2 – m2y1 = y(m1 – m2)

\ y= m1 y2 – m2 y1
m1 – m2

\ co-ordinate of section point externally is,
m1 x2 – m2 x1 m1 y2 – m2 y1
\ (x, y) = ( m1 – m2 , m1 – m2 )

It is the co-ordinate of section point externally.

PRIME Opt. Maths Book - IX 125

iv. Co-ordinate of centroid of a triangle. A(x1, y1)
Let ‘G’ be the centroid of a triangle having G(x, y)

vwehretirceesAPA(xis1, ay1m), eBd(ixa2n, ya2n)dan‘Gd’ cCu(xts3, tyh3e) P
median AP in the ratio 2:1.

Now, using mid-point formula for BC,
y1 + y2
(x’, y’) = ( x1 + x2 , 2 )
2

or p(x’, y’) = ( x2 + x3 , y2 + y3 ) B(x2, y2) C(x3, y3)
2 2

Again, using section formula for AP

(x, y) = ( m1 x2 + m2 x1 , m1 y2 + m2 y1 )
m1 + m2 m1 + m2

>2×` x2 + x1 j + 1 × x1 2×a y2 + y1 k + 1 × y1 H
2 , 2
G(x, y) =
2+1 2+1

\ G(x, y) = [ x1 + x2 + x3 , y1 + y2 + y3 ]
3 3

It is the co-ordinates of centroid of DABC.

Worked out Examples

1. Find the co-ordinate of a point which divides the line joining the points A(3, –2) and

B(–7, 4) in the ratio 3:2.

(i) internally (ii) Externally

Solution :

A(3, –2) 3 2 B(–7, 4)
PRIME Opt. Maths Book - IX
The given points are A(3, –2) & B(–7, 4)
Ratio of the straight line = 3:2.

i) Using section formula internally,
+ m1 y2 + m2 y1
(x, y) = ( m1 x2 + m2 x1 , m1 + m2 )
m1 m2

3 × (–7) + 2 (3) 3 × 4 + 2x (–2)
=( 3+2 , 3+2 )

= a– 15 , 8 k
5 5

= a–3, 8 k
5

ii) Using section formula externally,
m1 x2 – m2 x1 m1 y2 – m2 y1
(x, y) = ( m1 – m2 , m1 – m2 )

3 (–7) –2 (3) 3 (4) –2 (–2)
=( 3–2 , 3–2 )

= ( –21 – 6 , 12 + 4 )
1 1

= (–27, 16)

126

2. In what ratio does x-axis cuts the line joining the points Y
(1, 5) and (5, –3)? Also find the point of intersection. A

Solution : O
Y’
Let, A point p(x, 0) on x-axis cuts the line joining the

points A(1, 5) and B(5, 3) in the ration k:1.

Now, Using section formula, X’ P(x, 0) X
m1 y2 + m2 y1 B
y = ( m1 + m2 )

or, 0= kx (–3) + 1 (5)
k +1

or, –3K + 5 = 0

or, k= 5
3
\ The required ratio is 5:3.

Again,
m1 x2 + m2 x1
x= m1 + m2

= 5×5+3×1
5+3
28
= 8

= 7
2

\ The point of intersection is ( 7 , 0)
2

3. If (2, 4) is the mid-point of the joining the points (–2, a) and (b, 5), find the value of
‘a’ and ‘b’.

Solution :

A(–2, a) (2, 4) B(b, 5)

Here, (2, 4) is the mid-point of line joining the points (–2, a) & (b, 5)

Using mid-point formula, y1 + y2
2
x= x1 + x2 and y =
2

or, 2= –2 + b and 4 = a+5
2 2

or, –2 + b = 4 and a + 5 = 8

\ b = 6 and a = 3

\ a = 3, b = 6

4. If AC = 15 units where A(3, 4), B(7, 7) and C(a, b) are the collinear points, find the
co-ordiante of ‘c’.

Solution :

A(3, 4) B(7, 7) C(a, b)

Here,
A(3, 4), B(7, 7) and C(a, b) are the collinear points where,
AC = 15 units

PRIME Opt. Maths Book - IX 127

Using distance formula,

d = (x2 – x1)² + (y2 – y1)²

d(AB) = (7 – 3)² + (7 – 4)²

= 16 + 9
= 5 units.
Then, BC = AC – AB = 15 – 5 = 10 units.
ie. B cuts AC in the ratio 5:10 = 1:2

Now, Using section formula, m1 y2 + m2 y1
m1 + m2
x= m1 x2 + m2 x1 and y=
m1 + m2

or, 7= 1× a + 2×3 and 7= 1×b+2×4
1+2 1+2

or, a = 6 = 21 and b + 8 = 21

\ a = 15 and b = 13

\ Co-ordinate of point C is (15, 13)

5. If (1, –2) is the centroid of a triangle having vertices (p, 2), (–2, –4) and (2, q), find the
value of p and q.

Solution :
Centroid of DABC having vertices A(p, 2), B(–2, –4) and C(2, q) is (1, –2)

Now, Using centroid formula, y1 + y2 + y3
3
x= x1 + x2 + x3 and y=
3

or, 1= P –2 + 2 and 2–4+q
3 and –2 = 3
or, p = 3 q=–4

\ p = 3 q = –4

6. Find the points of trisection of the line joining the points (1, 2) and (4, 5).
Solution :

A(1, 2) P Q B(4, 5)

Let, P and Q be the points of trisection of the line joining the points A(1, 2) and B(4,

5)

Where
P cuts AB in the ratio 1:2
Q cuts AB in the ratio 2:1
Then,

Using section formula, m1 y2 + m2 y1
m1 + m2
(x, y) = ( m1 x2 + m2 x1 , )
m1 + m2

P(x, y) = ( 1 × 4 + 2 × 1 , 1×5+2×2 )
1 + 2 1+2

= (2, 3)

128 PRIME Opt. Maths Book - IX

Again

Q(x, y) = ( 2 × 4 +1× 1 , 2×5+1×2 )
2 +1 2+1

= (3, 4)
\ P(2, 3) and Q(3, 4) are the points of trisection.

7. If three vertices of a parallelogram are A(4, 3), A(4, 3) D(a, b)
B(1, 2) and C(2, –5), find the co-ordinate of C(2, –5)
fourth vertex D.

Solution : P
The three vertices of a parallelogram are A(4,

3), B(1, 2) and C(2, –5) B(1, 2)
Let, fourth vertex be D(a, b)
Then,

Using mid-point formula for diagonal AC.
y1 + y2
(x, y) = ( x1 + x2 , 2 )
2

P(x, y) =( 4+2 , 3–5 )
2 2

= (3, –1)

It is the mid-point of diagonal BD also in the parallelogram,

So, x= x1 + x2 and y = y1 + y2
2 2

or 3= 1+ a and –1 = 2+b
2 2

or, 1 + a = 6 and 2 + b = – 2

\ a = 5 and b = – 4

\ Fourth vertiex is (5, –4)

PRIME Opt. Maths Book - IX 129

Exercise 4.3

1. i) What do you mean by median of a triangle?
ii) What is centroid? Write down its co-ordinate.
iii) Write down section formula to show the section point externally.
iv) In what ratios does the points of trisection cuts a line segment.
v) In what condition the quadrilateral will be the parallelogram using mid-point?

2. Find the co-ordinate of a point under the following conditions.
i) Which divides the line joining the points A(–3, 9) and B(2, –1) in the ratio 2:3

internally.
ii) Which divides the line joining the points (1, 2) and (3, 4) in the ratio 4:5

externally.
iii) Which divides the line joining the points (3, 5) and (1, 7) at mid-point.
iv) Which divides the line joining the points (–2, –3) and (5, –3) in the ratio 3:4

internally.
v) Which divides the line joining the points (2 + 3 , 2 – 3 ) and (2 – 3 , 2 + 3 )

at mid-point.

3. Find the followings:
i) Find the centroid of a triangle having vertices A(1, 1), B(3, 7) and C(2, 1).
ii) If a point (–2, 3) is the centroid of a triangle having vertices (–4, m), (n, 4) and

(–1, –3). Find the value of ‘m’ and ‘n’.
iii) If (2, 3) is the mid-point of the line joining the points (3, b) and (a, –5), find the

value of ‘a’ and ‘b’.
iv) If (2, 1) is the mid-point of line joining the points (2 + 3 , a), and (b, 2 + 3 ),

find the value of ‘a’ and ‘b’.
v) Prove that the points (–2, –1), (1, 0), (4, 3) and (1, 2) are the vertices of a

parallelogram.

4. Find the followings
i) In what ratio does (2, P) cuts the line joining the points (–2, 6) and (4, 3)? Also

find the value of ‘p’.
ii) In what ratio does y-axis cuts the line joining the points (–3, 2) and (4, –3)? Also

find the point of intersection.
iii) In what ratio does x-axis cuts the line joining the points (0, –6) and (5, 4)? Also

find the point of intersection.
iv) X-intercept of the line joining the points A(–1, 4) and B(4, 4) is 1, find the y-

intercept of the point.
v) In what ratio does the point (2, 6) cut the line joining the points (1, 8) and (4, 2)?

5. Find the followings.
i) In the collinear points A(1, –2), B(4, 2) and C, length of AC = 10 units, find the

130 PRIME Opt. Maths Book - IX

co-ordiante of the point C.
ii) If PR = 20 units in the collinear points P(3, 2), Q(–1, –1) and R(a, b), find the co-

ordinate of ‘R’.
iii) Find the points of trisection of the line joining the points (1, 8) and (4, 2).
iv) Find the co-ordinate of the points which divides the line joining the points

(–5, –5) and (25, 10) in three equal parts.
v) Find the co-ordinate of fourth vertex of a parallelogram having three of the

vertices are (8, 5), (–7, –5) and (–5, 5).

6. PRIME more creative questions.
i) Find the length of the median drawn from first vertex of a triangle having

vertices A(4, 2), B(–3, –4) and C(–1, –8)
ii) Find the co-ordinate of the points which divides the line joining the points

(–6, –5) and (8, 7) in four equal parts.
iii) The mid-point of the sides of DABC are (2, 1), (3, –1) and (7, 3)respectively find

the co-ordinate of the veritecs of DABC.
iv) The three vertices of a parallelogram are (–3, 2), (–5, –3) and (5, –6) find the co-

ordinate fourth vertex which is opposite to (–5, –3).
v) Find the co-ordinate of centroid of triangle having first vertex (3, 7) and mid-

point of its opposite side is (3, –2)
vi) Find the length of the medians of a triangle having vertices (2, 5), (–4, 1) and

(–8, –3)

Answer

1. Show to your teacher.

2. i) (–1, 5) ii) (–7, –6) iii) (2, 6)
iv) (1, –3) v) (2, 2)

3. i) (2, 3) ii) m = 8, n = –1 iii) a = 1, b = 11
iv) a = – 3 , b = 2 – 3 iii) 3:2, (3, 0)
ii) 3:4 (0, – 1 )
4. i) 2:1, P = 4 7
iv) 2:3, 4 v) 1:2

5. i) (7, 6) ii) (–13, –10) iii) (2, 6), (3, 4)
iv) (5, 0), (15, 5) v) (10, 15)

6. i) 10 units ii) (– 5 , –2), (1, 1), ( 9 , 4)
2 2
iii) (6, 5), (–2, –3) and (8, 1)

iv) (7, –1)

v) (3, 1) vi) 10 units, 1 unit, 85 units

PRIME Opt. Maths Book - IX 131

4.4 Equation of straight lines:

Let us consider a linear equation 3x – 2y – 4 = 0 where x and y are the variable and can

solve it as below to plot the points in graph.

3x – 2y – 4 = 0

or 3x – 4 = 2y

or, y= 3x – 4 Y
2

x024

y –2 1 4

From the graph we conclude that the linear equation
represents a straight line and the equation 3x – 2y – 4 = 0
represent a straight line.

There are three types of straight lines which have different X’ O X
forms of equations according to the conditions of their Y’
position. But every equation have the degree one. The
common types are the straight line parallel to x-axis, parallel

to y-axis and making an angle with x-axis. We can see them

in the following examples.

Parallel to x-axis → Horizontal line Parallel to y-axis → Vertical line
Y Y

A B PR
X’ X X’
X
C D
QS
Y’ Y’

Making an angle with x-axis
YB

X’ q X
A

Y’

Here,

we discuss about the equation of such straight lines.

132 PRIME Opt. Maths Book - IX

4.4.1 Inclination of a straight line

The angle made by a straight line with x-axis in positive direction (Anticlockwise) is

called inclination of the straight line. YB
BY

q

X’ A X X’ q X
A
Y’ Y’
In the given diagrams ‘q’ is the inclination of the straight line AB.

4.4.2 Slope (gradient) of a straight line Y
The steepness of a straight line with the horizontal line which

is measured as the tangent of the inclination of the straight q rise

line, called the slope (gradient) of the straight line. run
In the adjoning figure,
X’ qX

Slope (m) = Tanq = rise Y’
run

Example : YB

X’ 30° X
A

Y’
Here,

Inclination of the straight line AB is (q) = 30°
\ Slope (m) = Tanq
= Tan30°
= 1

3

4.4.3 Slope of straight line joining the two points
Y
)
(x , y B(x2, y2)
1 R
1
A
q

Q

q M NX
PO

Let, ‘q’ be the inclination made by a straight line joining the points A(x1, y1) and B(x2, y2)

Slope(m) = Tanq = rise
run

PRIME Opt. Maths Book - IX 133

Draw, AM^OX,
BN^OX
AR^BN
Then, AR = MN = ON – OM B=Nx2––AxM1 ==ryu2n–
BR = BN – RN = y1 = rise
In right angled DBRA,
rise
Tanq = run

or, m= y2 – y1
x2 – x1

\ Slope (m) = y2 – y1
x2 – x1

Notes :

• For parallel straight lines, slopes are equal (m1 = m2).
• For perpendicular straight lines, product of slopes equal to –1 (m1 × m2 = –1).
• For x – intercept (a) and y – intercept (b) of a straight line are given, slope = – b .
a

4.4.4 Equation of straight line parallel to Y

x-axis A B

Let, AB is a straight line parallel to x-axis as well as X’ b b X
A’B’. It means AB is equidistance from each and very –b B’
point on x-axis equal to b (say). So ‘b’ is the ordinate O
(y-co-ordinate of any point lies on AB.) –b
A’
\ y = b is the equation of straight line parallel to
x-axis.

For the straight line AB, ‘b’ is positive and for the Y’
straight line A’B’, b is negative.
\ y = b [b may be +ve as well as -ve]

4.4.5 Equation of straight line parallel to Y
y-axis P P’

Let, PQ be the straight line parallel to y-axis which is X’ a X
at a distance of ‘a’ from y-axis. In this situation any a
point on PQ must have its abscissa as x = a.
a
\ x = a is the equation of straight line parallel to
y-axis. –a Q’
Q
\ x = a [a my be +ve as well as -ve]
Y’

134 PRIME Opt. Maths Book - IX

Worked out Examples

1. Find the slope of straight line joining the points (1, –2) and (–5, 4). Also find the

inclination of the line with x-axis in positive direction.

Solution :

The given point are A(1, –2) & B(–5, 4)
y2 – y1
Now, Slope of AB(m1) = x2 – x1

= 4+2
–5 – 1

= –1

Again,
Slope (m) = Tanq
or, –1 = Tanq
or, Tanq = Tan135°

\ q = 135°

\ Inclination (q) = 135°

2. Prove that the points (2, –3), (–1, 2) and (–4, 7) are the collinear points.

Solution :
The given points are A(2, –3), B(–1, 2) and C(–4, 7)

Now, = y2 – y1 = 2+3 = 5
Slope of AB x2 – x1 –1 –2 –3

Slope of BC = y2 – y1 = 7–2 = 5
x2 – x1 –4+1 –3

Here,

Slope of AB = Slope of BC = 5
–3

Hence, A, B and C are collinear points.

3. Find the equation of straight line passes through a point (3, 4) which is parallel to
x-axis.

Solution :
The equation of straight line parallel to x-axis is,
y = b .................... (i)
It passes through a point (3, 4)
i.e. 4 = b
\ b = 4
From equation (i)
y = b= 4
\ y = 4 is the required equation

PRIME Opt. Maths Book - IX 135

4. Find the equation of straight line parallel to y-axis which Y
A
is 5 units left from y-axis.

Solution :

Let, AB be a straight line parallel to y-axis.

It is 5 units left from y-axis,

i.e. a = –5 X’ X
Then,
5 units
Equation of straight line parallel to y-axis is, B

x=a Y’

or, x = –5

\ x + 5 = 0 is the required equation.

5. Find the slope of sides of DABC where BC is parallel to OX.

Solution : Y
A
In DABC,
120°
AB = AC
BC||OX O
Y’
\ A = 120°

\B= \C=x=? B C
Then, X’ X

\ A + \ B + \ C = 180°

or, 120° + x + x = 180°

or, x = 30°

\ \ B = \ C = 30°

Now,

Angle made by side BC with x-axis (q) = 0°

Angle made by side AB with x-axis (q) = 30°

Angle made by side AC with x-axis (q) = 180° – 30°

= 150°
\ Slope of BC (m) = Tanq = Tan0 = 0
Slope of AB (m) = Tanq = Tan30° = 1
3

Slope of AC (m) = Tanq = Tan150° = – 1
3

136 PRIME Opt. Maths Book - IX

Exercise 4.4

1. i) What do you mean by slop of straight line?
ii) Write down the slope of a straight line joining the points (a, b) and (b, a).
iii) Write down the formula of equation of straight lines parallel to x-axis and also

parallel to y-axis.
iv) Write down the slope of a straight line which makes an angle 60° with x-axis in

negative direction.
v) Find the inclination of a straight line with x-axis which has slope 3 .

2. Find the slope of the straight line under the following conditions.

i) Which makes an angle 30° with x-axis in positive direction

ii) Which makes an angle 120° with x-axis in positive direction.

iii) Which makes an angle 45° in negative direction.

iv) Y B v) B Y

X’ 30° X X’ 60° X
A A

Y’ Y’

3. Answer the followings:

i) Find the slope of line joining the points (2, 5) and (6, 7).

ii) Find the gradient of the line joining the points (a + b, a – b) and (a – b, a + b)

iii) Find the slope of line joining the points ( 3 , 3) and (3 3 , 1).
iv) Prove that the points (3, 2), (5, 6) and (8, 12) are collinear.
x y
v) If the points (a, 0), (x, y) and (0, b) are collinear, prove that a + b =1

4. Find the inclination of a straight line under the following conditions.

i) Line joining the points (2, 3) and (3, 4)

ii) Line joining the points (4, 3 3 ) and (2, 3 )

iii) Line joining the points (4 3 , 5) and (2 3 , 7)

iv) Y B(5, 7) v) Y
B(–3, 8)

X’ A A(1, 3) X X’ A A(2, 3) X
q q

Y’ Y’

PRIME Opt. Maths Book - IX 137

5. Find the equation of straight line parallel to x-axis under the followings.
i) Passes through a point (3, 5)
ii) Passes through a point (4, –2)
iii) Which is 3 units distance from the x-axis.

iv) Which is –4 units distance from the x-axis.

v) Which is 3 units down from the x-axis.

6. Find the equation of straight lien parallel to y-axis under the followings:
i) Passes through a point (4, –2)
ii) Passes through a point (–5, –2)
iii) Which is 3 units right from y-axis.
iv) Which is 4 units left from y-axis.
v) Which is 2 units distance from the y-axis.

7. Prime more creative questions:

i) Find the slope of sides of equilateral DABC where BC is parallel to x-axis in the

given diagram. Y

A

B C
X
X’ O

Y’
ii) Find the slop of sides of isoscles right angled triangle PQR from the given

diagram where QR||OX.
Y
P

X’ O Q R
X

Y’
iii) Find the slope of diagonals of a square ABCD as shown in diagram where BC||OX.

Y

AD

B C
X’ O X

Y’

138 PRIME Opt. Maths Book - IX

iv) If slope of line joining the points A(3, b) and B(1, –3) is 2, find the value of ‘b’.
v) If angle made by a straight line joining the points (3 3, 4) and ( 3, P) is 30° in

positive direction, find the value of ‘p’.

1. Show to your teacher. Answer

2. i) 1 ii) – 3 iii) –1
3
v) – 3 iii) – 1
iv) 1 3
3 ii) –1
iii) 150°
3. i) 1 ii) 60° iii) y – 3 = 0
2 v) 135° iii) x – 3 = 0
iii) 1, –1
4. i) 45° ii) y + 2 = 0
iv) 45° v) y + 3 = 0

5. i) y – 5 = 0 ii) x + 5 = 0
iv) y + 4 = 0 v) x – 2 = 0

6. i) x – 4 = 0 ii) 1, 0, –1
iv) x + 4 = 0 v) p = 2

7. i) 3, 0, – 3
iv) b = 1

PRIME Opt. Maths Book - IX 139

4.5 Equation of straight line in standard form:

4.5.1. Slope-intercept form: Y P(x, y)

When slope (m) and y - intercept (c) of a straight Bq rise
line are given to find the equation of the line. run
Let, AB be a straight line which makes an angle ‘q’ Q

with x-axis and y-intercept ‘c’ where, X’ A q X
\ BAX = q OM
\ Slope of AB(m) = Tanq.
y -intercept (OB) = C Y’
P(x, y) be any point in the line AB,
Draw PM^OX and BQ^PM
Then,
\ PBQ = \ BAX = q
BQ = OM = x
PQ = PM – QM = PM – OB = y – C

Now,
In right angled DPQB
rise
Tanq = run

or, Tanq = PQ
BQ

or, m= y–C
x
\ y – c = mx

\ y = mx + c

It is the equation of st. line in slope-intercept form.

4.5.2. Double intercepts from:
When x and y intercepts of a straight line are given, to find the equation of the
straight line.

Intercept: The part of a straight line separated Y

by any two points taken in the line is called B
intercept.

Here, A straight lien AB marks, b AX
x - intercept = OA = a Oa
y - intercept = OB = b

140 PRIME Opt. Maths Book - IX

Proof: Y

Let,

A straight line AB cuts x-axis at A and y-axis at B, B(0, b)
P(x, y)
where,
x-intercept (OA) = a
y-intercept (OB) = b

Also, the co-ordiantes of A and B are (a, 0) and (0, b) X’ X
respectively.
P(x, y) be any point in AB. Here, The points A(a, 0), O A(a, 0)
P(x, y) and b(0, b) are the collinear points.
Y’

Then y2 – y1 y –0 y
x2 – x1 x –a x–a
Slope of AP = = =

Slope of PB = y2 – y1 = b – y = b–y
x2 – x1 0 – x –x

As we known for the collinear points,
Slope of AP = Slope of PB

or, x y a = b–y
– –x

or, – xy = bx – ab – xy + ay

or, ab = bx + ay

or, bx + ay = ab

Dividing both sides by ‘ab’,
bx ay ab
or, ab + ab = ab

or, x + y =1
a b

\ It is the required equation of double intercept form.

4.5.3. Perpendicular form (Normal)

When length of perpendicular drawn from origin to the straight line and inclination

of perpendicular over given, to find the equation of the straight line.

Let,

A straight line AB is given as in diagram where
OC is perpendicular to AB and \ COX = a
& perpendicular distance (OC) = p. Y
Draw PM^OX B
\ OAC = 90° – a and
\ MPA = 90° – (90° – a) = a
Then, C

In right angled DPMA, p P(x, y)

Also, Tana = MA a a (90°– a)
PM O
X’ X
MA MA
or, Tana = y

or, MA = yTana. Y’

PRIME Opt. Maths Book - IX 141

Again, In right angled DOCA,
OC
Cosa = OA

or, Cosa = p
OM + MA
p
or, Cosa = x + yTana

or, xCosa + Cosa × yTana = p

or, xCosa + Cosa. Sina .y =p
Cosa

\ xCosa + ySina = p

It is the required equation.

Worked out Examples

1. Find the equation of straight line having x-intercept 6 and B

y-intercept 8. 8
O
Solution : For a straight line AB,

x-intercept (a) = 6

y-intercept (b) = 8
Then, the equation of straight lien is,
x y
a + b =1 A

or, x + y =1 6
6 8

\ 4x + 3y = 24 is the required equation.

2. Find the equation of straight line which makes Y
y-intercept 3 and angle with x-axis in positive direction

is 30°. B
Solution : Let AB be a straight line,

Where, 6
30°
y-intercept (c) = 3 AO

\ BAX = 30° X
\ Slope of AB(m) = Tanq
= Tan30°
=1
3

Then, Equation of straight line AB is,

y = mx + c

or, y = 1 x + 3
3

or, y = x + 3 3
3

or, x + 3 3 = 3y

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

142 PRIME Opt. Maths Book - IX

3. Find the equation of straight line AB from the given Y

diagram. BC
2
Solution : 60°

For the straight line AB, X’ O X
\ COA (a) = 60° Y’ X
Perpendicular distance (p) = OC = 2 units

Then, Equation of straight line AB is, A
(1, 3)
xcosa + ysina = p

or, xcos60° + ysin60° = 2

or, x× 1 +y× 3 =2
2 2

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

4. Find the equation of straight line passes through a point Y

(1, 3) where the straight line makes equal but opposite OB
A
intercepts on the axes.
Y’
Solution : X’
Let, AB be a straight line which makes equal but opposite

intercepts on the axes.

i.e. x - intercept (a) = k (say)

y - intercept (b) = –k

Then, the equation of straight line is,
x y
a + b =1

or, x + y =1
k –k

or, x – y = k ................................... (i)

It passes throught a point (1, 3)

or, 1 + 3 =1
k –k

or, – 2 =1
k

\ k = –2

Then, the equation of straight line is, from equation (i) is
x y
–2 – –2 =1

or, x – y = – 2

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

PRIME Opt. Maths Book - IX 143

5. Find the equation of straight line passes through a point Y
(–2, 3) which bisects the line intercepted between the axes.

Solution : B(0, b)

Let, AB be a straight line which makes, (–2, 3) b
x - intercept (OA) = a = A(a, 0)
y - intercept (OB) = b = B(o, b) X’ a O X
AB is bisected by the point (–2, 3) A(a, 0)

Then, using mid-point formula, Y’

x= x1 + x2 , y = y1 + y2
2 2

or, –2 = a + 0 , 3 = 0+b
2 2

\ a = – 4 b=6

Then, the equation of straight line is,

x + y =1
a b

or, x + y =1
–4 6

or, –3x + 2y = 12

\ 3x – 2y + 12 = 0 is the required equation.

6. Find the equation of straight line AB from the given diagram.

Solution : Y A (3, 0)
X
Here, AB is a straight line passes through a point (3, 0) B
It makes equal angles on the axes.
\ OA = OB = 3 3
\ a = b = 3 X’ O

Then, the equation of straight line is, Y’ 3

x + y =1
a b

or, x + y =1
3 3

or, x + y = 3

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

144 PRIME Opt. Maths Book - IX

Exercise 4.5

1. i) Write down the formula to find equation of st. line in slope intercept from with

usual meaning.
ii) Write down the formula to find the equation of st. line in perpendicular form

with usual meaning.
iii) Write down the equation of straight line making equal intercepts ‘a’ on the axes.

iv) Write down the equation of st. right line in slope-intercept form which passes

through the origin. x y
2 4
v) Find the intercepts of the st. line having equation + = 1.

2. Find the equation of straight line under the following condition.
i) Having intercepts 3 and 4 on the axes.
ii) having x-intercept –6 and y-intercept 8 on the axes.
iii) Making equal intercepts on the axes and passes through a point (2, 3).
iv) making equal but opposite intercepts on the axes and passed through (3, –4).
v) Making x-intercept 3 and equally inclined on the axes.

3. Find the equation of straight line AB under the followings.
Y Y
i) ii)

B

4 X’ O 150° X
O2 X A
X’ Y’ A
iii) iv) 3
Y
B
B Y’

Y

B(0, 4)

X’ A(–4, 0) O X X’ 60° A X
O

Y’ Y’

PRIME Opt. Maths Book - IX 145

v) Y X
X’
3
AO

B

Y’

4. Find the equation of straight line under the following conditions.
i) Having y-intercept 3 and slope = 2.
ii) Having y-intercept –4 and making an angle 30° with x-axis in positive direction.
iii) Having y-intercept 2 and making an angle 120° with x-axis in negative direction.
iv) Making y-intercept –3 and equally inclined on the axes.
v) Making equal angles on the axes and passes through a point (0, 4).

5. Find the equation of straight line under the following conditions.
i) P = 3 and a = 30°
ii) Perpendicular distance from the origin = 4 and a = 150°

iii) Y iv)

B B
C
C
X’ A 2 units X 60° A
O
30°
O OC = 2 units

Y’

v) Y

B
C

X’ A O X
OC = 2 units
\ BAO = 30° Y’

6. Find the equation of straight line under the following conditions.
i) Find the equation of straight line having perpendicular distance from the origin

is 4 units and angle made by the perpendicular to the x-axis in positive direction
at the origin is 150°. Also prove that it passes through a point (–3 3, –1)

146 PRIME Opt. Maths Book - IX