Optional Math 9

Equating the corresponding x-component.
7 – a = 3
or, 7 – 3 = a
or, 4 = a
\ a=4

Exercise 6.1

1. i) Differentiate between vector and scalar in two points.

ii) Which of the following quantities are the scalar quantities?

Length, work, mass, density, force, acceleration, power, time

iii) Which of the quantities given above are the vector quantities?

iv) Which type of line segments given below ? For what purpose are they used?

A B P Q

v) Write down the components of vector AB from the given diagram. Also write

down the column vector AB .
B

4

A 3


2. Study the given graph and write down the components and vector for the followings.
i) AB ii) CD iii) PQ
iv) RS v) MN

BP

A CS

Q
M

DR
N

3. Show the following vectors in directed line segment.

i) PQ = d3 n ii) AB = d–42n iii) CD = d––34n
5

iv) RS = d3 n v) a = d–86n
–5

PRIME Opt. Maths Book - IX 247

4. i) If A(3, 2) and B(5, 6) are the any two points, find the components and vector of
AB .

ii) If A(1, 4) is a point, find the components and vector of OA .
iii) If P(3, 2) is a point find the position vector of P.
iv) If P(–1, 4) and Q(3, 1) are any two points, find PQ .
v) If M(3, –2) and N(–1, – 5) are any two points, find MN .

5. PRIME more creative questions:
i) A(4, –5) displaced to B(3, –6) in the vector AB and G(2, 5) displaced to H(1, 4)

in the vector . Prove that AB = GH .
ii) D(–3, 2) displaced to E(1, –3) in vector DE and K(1, –1) displaced to L(–3, 4) in

the vector KL . Prove that KL = – DE
iii) If A(3, 2), B(4, 5), C(1, 7) and D(2, 10) are the four points. Prove that AB = CD .

Also prove that AB || CD .
iv) If P(3, 7), Q(1, 4), R(5, –1) and S(3, –4) are the four points, prove that PQ = RS .

Also prove that they are parallel to each other.
v) If A(2, 4), B(7, 8), P(3, –1) and Q(–2, –5) are the four points, prove that AB = –PQ .

Also prove that they are parallel to each other.

6. i) If A(3, x), B(1, 4), C(5, –1) and D(3, –4) and AB = CD , find the value of ‘x’.
ii) If P(–2, –5), Q(3, –1), R(7, 8), S(m, 4) and PQ = –RS , find the value of ‘m’.
iii) If A(2, 4), B(6, 3), C(–3, 5) and D(a, b) are any four points where AB = CD , find

the co-ordinate of D.
iv) If P(–3, 2), Q(–2, 4), R(0, –2), S and PQ = RS , find the co-ordiante of S.
v) If CD = –RS in the points C(P, 2), D(1, –3), R(1, –1) and S(–3, q), find the value of

p and q.

Answer

1. Show to your teacher.

2. Show to your teacher.

3. Show to your teacher.

4. i) 2, 4, d2n ii) 1, 4, d1 n iii) d32 n iv) d–43n v) d––43n
4 4

6. i) x = 7 ii) m = 2 iii) (1, 4) iv) (1, 7) v) (–3, 4)

248 PRIME Opt. Maths Book - IX

6.2 Vector operations

6.2.1 Magnitude of a vector Y
B(4, 6)
Let us consider the two points A(1, 2) and B(4, 6).

AB = e x - comp. o = d AR n
y - comp. RB

= d x2 – x1 n A(1, 2) R
y2 – y1

d4 – 1n OM N X
6 –2
=

= d3 n
4

Where,
Length of AB = d(AB)

= AR2 + RB2

= 32 + 42

= 5 units

\ Magnitude of AB = 5 units.

It is written as modulus of vector AB

= | AB |
= 5 units.

The modulus of a vector which denotes the length of the

vector is called magnitude of the vector.
ex - comp. o
i.e. If AB = y - comp.

| AB | = (x - comp.)2 + (y - comp.)2
It is called the absolute value of vector too.

6.2.2 Direction of a vector

The angle made by the directed line segment for a vector with positive direction of x -

axis (taking the ratio of y - component and x-component as the slope of a line segment)

is called direction of the vector.

i.e. for direction q of ayv-eccotomr pw.e have
slope = Tanq = x - comp

PRIME Opt. Maths Book - IX 249

Let us consider: Y B (2, 2 3 )
x
Here, AB = d y - comp n R
- comp QX

= AR A (1, 3)
q
dn
BR

= d PQ n q
BQ – RQ OP

= dOQ – OPn
BQ – AP

= dx2 – x1 n = d 2 – 1 n = d 1 n
y2 – y1 2 3 – 3 3

Then,

y - comp.
Slope of AB = x - comp

or, Tanq = 3
1

or, Tanq = Tan60°

\ q = 60°

\ Direction of AB is 60°

The angle made by a vector with x-axis in positive direction

which is calculated by taking the tangent ratio is called

di.eir.eTcatnioqn=ofyxa vector. where ‘q’ is the direction

- comp.
- comp

Important points for direction of vector:
1. If x-component and y- component both are positive, direction= q.

2. If x-component is –ve and y- component Y
is positive (+ve), direction = 180° – q.
• If x-component is -ve and (–, +) (+, +)
y-component -ve, direction = 180° direction = 180° – q direction = q
+q
• If x-component is +ve and X’ O X
y-component -ve, direction = 360°
–q (–, –) (+, –)

Note : As domain for Tan is –90° < q < 90° so direction = 180° + q direction = 360° – q
we find most of the angles written in both
negative and positive sign. Y’

250 PRIME Opt. Maths Book - IX

6.2.3 Unit vector

The vector having magnitude one is called unit vector.

i.e. AB = 1

Where,

If AB = 1
dn
Here, 0

AB = x2 + y2 = 12 + 02 = 1

\ AB is the unit vector.

Taking another vector PQ ,
KKKKKKKLJ ONPOOOOOO
If PQ = 3 5
4 5

Here, PQ = x2 + y2

=

= 9 + 16
25

=1

\ PQ is a unit vector.

Formula calculating of unit vector.

If a is a vector.

Unit vector along a (a) = 1 ^a h
a

= vector
its mod ulus

6 = 10 units
If a = d n

8

Then, |a | = x2 + y2 = 62 + 82

Then,

Unit vector along a = 1 ^a h
a

\ a = 16
10 d n
8
KJKKKKKLK PONOOOOOO
a = 3 5
4 5

PRIME Opt. Maths Book - IX 251

6.2.4 Addition of vectors
Let us consider AB & BC are any two vectors having same direction, they can be
combined as AB + BC which will be the new vector AC .

A BC

A C

Let us consider,
13

a = d n and b = c m
24

13 4
Then, a + b = d n + c m = c m

246

The sum of any two vectors is the resultant vector obtained by

adding their corresponding components.
If, a = dx1 n & b = dx2 n
y1 y2

Resultant vector of their sum = a + b
= dx1 n + dx2 n

y1 y2

= dx1 + x2 n
y1 + y2

Note : Difference of any two vectors is also same as the addition.

i and j vector in the form of column vector
The unit vectors are the vectors having magnitude one where such unit vectors taken
along x-axis and y-axis are denoted by i and j vectors.

i.e. i = unit vector along x-axis Y

= d1 n B(0, 1) X
O A(1, 0)
0

j = unit vector along y-axis

= d0 n

1

In the given diagram,

252 PRIME Opt. Maths Book - IX

i = OA = d1 n

0

j = OB = d0 n

1

By using such unit vectors any kind of column vector AB , OA , OP , PQ , d can be

expressed in i and j vectors where,

a = dx n = dxn + d0n = xd1n + yd0n = xi + yj
y 0y 01

6.2.5 Properties of vector addition

i) Commutative property:

a + b = b + a

ii) Associative property:
a + (b + c) = (a + b) + c

iii) Addition with negative vector:
a + (–a) = (–a) + a = 0

iv) Addition identity:
a + o = o + a = a

6.2.6 Multiplication of a vector by scalar.

Let us consider,

a = 4 = 2 × 2 = 2 2 n
cm d n d
6 2×3 3

Here, 2 is the common constant value of both components, which is the scalar value.

Multiplication of a vector by a scalar results the new vector
obtained by multiplying each component with the given
scalar.
It multiples the magnitudes without changing the direction
of the vector if scalar is positive.

Example:

x
If a = d n is a vector.

y

k is a scalar. x kx
Then, ka = k d n = d n

y ky

PRIME Opt. Maths Book - IX 253

Example : If a = c2m & b = c0 m , find the value of 2 a + b .
1 1
Solution:

Here, a = 2 , b = 0
cm cm
11

Then, = 2 2 m + 0 m
2a + b c c
11

= 4 + 0 m
cm c
21

= 4
cm
3

Worked out Examples

1. If a = ( 3, 1), find its magnitude.
Solution,
a = ( 3, 1)
Magnitude of a is,

|a | = x2 + y2

= ^ 3h2 + ^1h2
= 4
= 2 units

2. Find the direction of a = e 3 o
1
Solution :

a = e 3o

1

For the direction of a q,

Tanq = y - component
x - component

Tanq = 1
3

or, Tanq = Tan30°
\ q = 30°

254 PRIME Opt. Maths Book - IX

3. Find the magnitude and direction of AB where A(1, –4) and B(5, 0) are any two

points.

Solution :

Taking the points A(1, 4) and B(5, 0)

AB = dx2 – x1 n = d5 – 1 n = d4n
Then, y2 – y1 0 + 4 4

Magnitude of AB is,

AB = x2 + y2

= 42 + 42

= 32

= 4 2 units

Again,

For the direction of AB ‘q’,

Tanq = y - component
x - component
4
= 4

=1

or, Tanq = Tan45°

\ q = 45°

4. If a = d6 n , find the unit vector of a .
8
Solution : a = 6
dn
8
|a | = x2 + y2

= 62 + 82

= 100

= 10 unit

Then, unit vector of a is,

a= 1 ^a h
Ua

= 1 6
10 dn
8
KLKKJKKKJLKKKKKKKK 115500ONOOOOPOOOOOPOONOO
= 6
= 8
3
4

PRIME Opt. Maths Book - IX 255

5. If direction of a = 2 3 i – pj is 330°, find the value of ‘p’. Also find the magnitude

of a .
Solution :

a =2 3 i – pj = e2 3 o
–p

For direction q = 330°

y - component
Tanq = x - component

or, Tan 330° = –p
23

or, – 1 = –p
3 23

\ p = 2

Again, magnitude of a is,

|a | = (x - comp.)2 + (y - comp.)2

= (2 3 )2 + (–p)2

= 12 + 4
= 4 units.

6. If a = d42n , b = d13n , find the magnitude of a + 2b .
Solution :

a = d4n b = d1 n
2 3

Then,

a + 2b = d4n + 2d13 n
2

= e4 + 2o
2 + 6

= d6n
8

\ Magnitude of a + 2b is,
| a + 2b | = x2 + y2
= 62 + 82
= 100
= 10 units.

256 PRIME Opt. Maths Book - IX

Exercise 6.2

1. i) What do you men by magnitude of a vector?
ii) Define the term direction of a vector.
iii) What do you mean by position vector?

iv) What do you men by unit vector?

v) Is a = d1 n a unit vector?

0

2. Find the magnitude of the following vectors.

i) a = d3 n ii) b = ( 8, 17)
4

iii) AB = d6n iv) AB for A(1, 1) and B(–3, –2)
8

v) PQ for P(3, 7) and Q(–1, 7)

3. Find the direction of the following vectors.

i) a = (2 3, 2) ii) b = d– 327 n

iii) AB = d4n iv) PQ for P(–2, –1) and Q(–6, –5)
0

iii) CD for C(–2, 5) and D(–2, 8)

4. Find the unit vector of the following vectors.

i) a = d2 n ii) b = d–86n
1

iii) AB = ^2 2, 2h iv) PQ for P(3, –2) & Q(–1, 1)

v) RS for R(–1, –3) & S(7, 3)

5. Write down the followings in i and j vectors.

i) a = d2 n ii) b = d––43n
3

iii) 3a where a = d2 n iv) PQ where p(3, – 2) and Q(1, –5)
–3

v) 2AB where A(–3, 1) and B(1, 4)

6. If a = d3n and b = d1 n , find the followings.
2 2

i) a + b ii) a – b

iii) 2a + b iv) 3a – 2b

v) 3a + 2b

PRIME Opt. Maths Book - IX 257

7. PRIME more creative questions:

a. i) If a 1 = 2 n , find the magnitude of a +b .
= c m and b d
23

ii) If a = 2 and b = 4 m , find the magnitude of 2a +b .
dn c
30

iii) If a = 2 m and b = 3 m, find the magnitude of a + 2b .
c c
44

iv) If A(–3, –1) and B(2, 4), find the magnitude and direction of AB .

v) If magnitude of a = dxn is 5 units, find the value of ‘x’.
4

b. i) If a = 3 i – 3 j , find the magnitude and direction of a .
ii) If magnitude of a = m i – 8 j is 10 units, find the value of ‘m’.
iii) If direction of a = p i – 27 j is 60°, find the value of ‘p’.

iv) If a = d 0 n , b = d k n and a + b = 6 units, find the positive value of ‘k’.

– 3 –2 3
Also find the direction of a + b .
v) If a = 5 i – j , b = –m i – 3j and magnitude of a – 2b is 5 2 units, find the
negative value of ‘m’. Also find the direction of a – 2b .

Answer

1. Show to your teacher.

2. i) 5 units ii) 5 units iii) 10 units iv) 5 units v) 4 units

3. i) 30° OOOOOOONO ii) 300° iii) 0° iv) 225° v) 90°
KKKKKKJKK P ii) KKKKKJLKK–5453 POONOOOOO KKJKKKKKL–5354 OOOOOONO v) KKJKKKKLK 5534 OPNOOOOOO
4. i) L 2 iii) c 2 , 1 m iv) P
5 5 5
1
5

5. i) 2 i + 3j ii) –3 i –4j iii) 6 i – 9j

iv) –2 i – 3j v) 8 i + 6j

6. i. 4 ii. d20n iii. d76n iv. d72n v. d1101n
4

7. a. i. 34 units ii. 10 units iii. 4 13
iv. 5 2 units, 45° v. +3

b. i. 2 3 units 330° ii. +6 iii. –9 3
iv. +3, 300° or 240° v. –5, 135°

258 PRIME Opt. Maths Book - IX

6.3 Concepts of vector geometry: C
B
6.3.1 Triangle law of vector addition

If the point A is displaced to B by AB and the point B is
displaced by C by BC , then the sum of AB and BC (which
is denoted by AB + BC ) is given by the vector AC where the
point A is displaced to C. The vector AC is called the resultant A
vector of the vectors AB and BC .
i.e. AB + BC = AC .
Also, AB + BC = AC .
or, AB + BC – AC = 0
or, AB + BC + CA = 0

Note : As the application of triangle law resultant vector can be calculated in different
polygons like quadrilateral pentagon, hexagon etc.

6.3.2 Quadrilateral law of vector addition C

In a quadrilateral ABCD, AC is a diagonal. By using D

triangle law of vector addition,
AB + BC = AC ^In TABCh ............................ (i)
AC + CD = AD ^In TACDh

Then, A B
AC + CD = AD
or, AB + BC + CD = AD (from equation ‘i’)

[It is quadrilateral law of vector addition]
or, AB + BC + CD – AD = 0
or, AB + BC + CD + DA = 0

6.3.3 Parallelogram law of vector addition:

DC

AB

In a parallelogram ABCD, AC is a diagonal where,

AB + BC = AC [ a using triangle law]

or, AB + AD = AC [ a AD = BC

\ AB + AD = AC is the parallelogram law of vector addition.

i.e.

PRIME Opt. Maths Book - IX 259

The sum of any two vectors having same initial point is equal
to the diagonal vector represented by the parallelogram
completed in the given vectors is called parallelogram law
of vector addition.

B C
b
c

Oa A

OA and OB are any two vectors having same initial point ‘0’ where,

OA = a and

OB = b

OC = c is a diagonal vector completed in the given vectors OA and OB .

\ a + b = c.

6.3.4 Addition of column vectors:

Let OA = a x1 k and OB = a x2 k the Y
y1 y2

position vector of points A(x1, y1)

and B(x2, y2) respectively construct C

a parallelogram OACB as shown in Q
B(x2, y2)
diagram to presents the given vectors. A(x1, y1) NP
M
Draw the perpendiculars,

AM^OX, BN^OX, CP^OX and BQ^CP.
Then, DAOM , DCQB.

\ OM = BQ = NP = x1 O X
CQ = AM = y1
Where,

OP = ON + NP = x2 + x1
PC = PQ + QC = NB + QC = y2 + y1.
\ Co-ordinate of c is (OP, CP) = (x1 + x2, y1 + y2)
According to parallelogram law of vector addition,

OA + OB = OC

i.e. a x1 k + a x2 k = a x1 + x2 k.
y1 y2 y1 + y2

260 PRIME Opt. Maths Book - IX

6.3.5 Theorems related to position vector a O
Theorem 1: Position vector of a point which divides a
line segment at mid-point. b
?
Let, a point M cuts a line segment AB at mid-point where,
the position vectors are,

OA = a M B
B
OB = b A

OM = ?
Now, taking
AM = MB

or, AM = MB [\Being in same direction]

or, AO + OM = MO + OB [\Using triangle law]

or, – OA + OM – MO = b

or, –a + OM + OM = b

or, 2OM = a + b

or, OM = a +b
2

\ OM = OA + OB
2

Theorem 2: Position vector of a point which cuts the O
b
line segment in the ratio m:n internally.

Solution :

Let, a point ‘P’ cuts a line segment AB in the ratio m:n a ? n
internally, where position vectors of A and B are mP

OA = a

OB = b A
OP = ?

Here,

AP = m
PB n

or, n AP = m PB

or, n AP = mPB

or, n (AO + OP) = m (PQ + PB) [\ Using triangle law ]

or, n (–a + OP) = m (–OP + b)

or, –n a + nOP = m OP + mb

or, nOP + mOP = na + mb

or, (m + n) OP = na + mb

\ OP = mb + na
m+n

i.e. OP = mOB + nOA
m+n

PRIME Opt. Maths Book - IX 261

Theorem 3 : Position vector of a point which cuts a line O

segment in the ratio m:n externally. ?
Solution :

Let, a point P cuts the line segment AB in the ratio m:n a b P

externally, where the position vectors are, n
OA = a B

OB = b Am

OP = ?

Here, from the concept of section point,

AP = m
BP n

or, n AP = mBP

or, n (AO + OP) = m (BO + OP)

or, n (–a + OP) = m (–b + OP)

or, –n –na + nOP = m –mb + mOP)

or, mb – na = OP(m – n)

\ OP = mb – na
m–n

i.e. OP = mOB – nOA
m–n

Worked out Examples

1. Express the sides of regular hexagon ABCDEF, in terms of a and b where AB = a

and BC = b . ED
Solution:

In the regular hexagon ABCDEF,

AB = a and BC = b FO C

Join CF & AD which are intersected At ‘0’. b
Then,

CD = BO = BA + BC = –a + b A aB
DE = OF = BA = – a

EF = DO = CB = – b

FA = OB = OC + OA = AB + CB = a – b .

262 PRIME Opt. Maths Book - IX

2. In a parallelogram ABCD, OA = a , OB = a and OC = c for any point ‘0’. Find OD in

terms of a , b & c . DC
Solution:

In a parallelogram ABCD,

OA = a

OB = b ? c
OC = c A B
OD = ?

Then, using triangle law,

OD = OA + AD ab

= OA + BC O
= OA + BO + OC

= a – b + c

\ OD = a – b + c

3. In the parallelogram ABCD. AB = a . Find AC + DB in terms of a .

Solution:

ABCD is parallelogram, AB = a D C
B
Now,

Using triangle law,

AC + DB = ^AB + BCh + ^DA + ABh

= AB + BC + CB + AB

= 2AB + BC – BC A
= 2AB a
= 2a

\ AC + DB = 2a

4. Find the position vector of a point p under a & b from the given diagram where

2AP = 3PB.

Solution : O

OA = a b
OB = b

2AP = 3PB a B
Now, using triangle law in the given condition,

2 AP = 3 PB P

or, 2 ^AO + OPh = 3 ^–OP + bh A

or, ^–a + OPh = 3^–OP + bh

or, 2OP + 3OP = 2a + 3b

or, 5 OP = 2a + 3b

\ OP = 1 ( 2a + 3b )
5

PRIME Opt. Maths Book - IX 263

Exercise 6.3

1. i) Explain triangle law of vector addition.
ii) What do you mean by i and j vectors?
iii) Write down the formula of position vector of mid-point of a line segment AB.
iv) Write down the formula of position vector of section point of a pine segment PQ

in the ratio.
v) State parallelogram law of vector addition.

2. Find the following:

i) If a = ` 2 j and b = ` 1 j , find a +b and write down in i and j vector.
3 –2

ii) If OA = 4 i + 3 j and OB = 5 i – 2 j , find AB .

iii) If position vectors of P and Q are 2 i – j and 5 i – 3 j , find PQ .

iv) If 2a + 3b = (0, –7) and b = (2, –3), find a .

v) If a + b = ` 5 j, a = ` 0 j, find b in i and j form.
1 4

3. Prove that the followings:
i) AB + BC + CA = 0 in DABC.
ii) PQ + QR + RS + SP = 0 in quadrilateral PQRS.
iii) AB + BC + CD + DE = AE in a pentagon ABCDE.
iv) DE + EF + FG + GH + HI + ID = 0 in the given diagram.

HG

IF

D E

v) AC + BD = 2BC in a parallelogram ABCD. O
?


4. i) Express OP in terms of a and b where P is the b

mid- point of AB in the adjoining diagram.

a B

P
A

264 PRIME Opt. Maths Book - IX

ii) Prove that 2AR = AP + AQ in the given diagram where R is the mid-point of
PQ.
A

Q

R
P

iii) Find OC in terms of p and q where AC = 2CB in the adjoining diagram.
O

q

p B

C
A

iv) Find OP in terms of m and n where 3AP = 2PB .

O

n

m B

P
A

v) Prove that AB + AC + AD + EA + FA = 4AB in a regular hexagon ABCDEF.
AF

BE

CD

PRIME Opt. Maths Book - IX 265

5. PRIME more creative questions:
a. i) If A and B are the mid points of sides QR and RS of a parallelogram PQRS

respectively, prove that 2^PA + PBh = 3PR .

ii) In a parallelogram ABCD, AB = a & AD = b . Find the vectors CB , DB & AP
in terms of a & b where P is the intersecting point of diagonals.

iii) In the given diagram ABCD is a parallelogram and P is the intersecting point of
diagonals. Prove that OA + OB + OC + OD = 4 OP for any point ‘O’.
AB

P

DC

O

iv) Prove that AD + BE + CF = 0 from the given diagram where D, E, & f are the
mid-point of sides of DABC.
A

E
F

C

BD

v) Find the followings from the given diagram in the single vector.

AB + BC DO + OC

DC + CA DO + AD
AB

O

D

C

vi) Prove that the position vector of centroid of a triangle is g = 1 ^a + b + c h .
3

b. i) If 4a – 3b and 2a + 5b are the position vectors of p and Q, find PQ .

ii) If a = ` –2 j , b = ` 7 j & m = 2, prove that m^a + bh = ma + mb .
1 –3

266 PRIME Opt. Maths Book - IX

iii) If AB = –5 i – 7 j and BC = 7 i + j , find AC .

iv) Where and how vector operations are used in our daily life? Prepare report and
present in your class.

v) Present a + b , a – b , b – a and 2a + b in arrow diagrams.

6. Project work
Collects the formula and operation system of vectors in a chart paper and present

into the classroom.

Answer

1. Show to your teacher

2. i. 3 i + j ii. i – 5j iii. 3 i – 2j

iv. (–3, 2) v. 5 i – 3j

4. i. OP = 1 (a + b) iii. OC = 1 (p + 2q)
2 3

iv. OP = 1 (3m + 2n)
5

5.a. ii. b, a –b , 1 (a + b) v. AC, DC, DA, AC
2

b. i. –2a + 3b iii. 2 i – 6j

v. ,D C, D C
b
a+b b b–a
a–b
b

aA BA B

aa

AC

a 2a + b

O B

a +b

PRIME Opt. Maths Book - IX 267

Vector

Unit Test Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Define the term position vector of a point.

2. a. If a = c –3 m , find the direction of a .
3

b. Find the unit vector of 3i – 4j .

c. If magnitude of a vector a = ` m j is 10 units, find the value of ‘m’.
–6

3. a. If a = a 2 k and b = a 3 k, find the magnitude and direction of a + 2b .
–1 2

b. Prove that AB + BC + CD + DA = 0 in a quadrilateral ABCD.

4. If M is the mid - point of a line segment AB where OA = a and OB = b , find the
position vector of M in terms of a and b.

268 PRIME Opt. Maths Book - IX

Unit 7 Transformation

Specification Grid Table

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

No. of Questions 1 – 1 1 3 10 18
Weight 1 – 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 know the concept of transformation.
• Students are able to identify the types of isometric and non- isometric

transformations.
• Students are able to know the formula to find image of a point.
• Students are able to find the image of an object for different transformations

and can plot them in graph.

Materials Required:
• Model of transformation.
• Geo-board.
• Flash card.
• Graph paper.
• Chart paper.
• Formula chart of transformations.

PRIME Opt. Maths Book - IX 269

7.1 Transformation

Let us see the shape of fishes of same size as well as the printed clothes as shown in the
diagram. Here one fish is transferred to the another fish continuously and size of fishes
can be seen which is the concept of transformation in the first example. One flower is
transferred to the another continuously and shape of flowers are printed in the clothes as
shown in diagram in second example which give the concept of transformation.

The process of changing the position, shape and size
of objects under any geometrical conditions is called
transformation.

7.1.1 Types of transformation :
Isometric transformations :

The transformation in which objects and its image are always congruent is called
isometric transformation. Example : Reflection, Rotation and translation.
Non-Isometric transformations :
The transformation in which object and image are not congruent but only similar is called
non-isometric transformation. Example: Enlargement

7.1.2 Reflection on mirror line :

Image of an object is shifted from one place to another place by the mirror line as the
reflection axis where image and object are always congruent in reflection.

270 PRIME Opt. Maths Book - IX

Here, B M
M is the mirror line. C P
DABC is an object.
B’

Draw, BP^M, BP = PB’

AQ ^M, AQ = QA’ A Q A’
R C’
CR^M, CR = RC’

Then, DA’B’C’ so formed is called image of

DABC.

7.1.3 Reflection using co-ordinate: Y
i) Reflection on x-axis. P(x, y)

Theoretical proof (x, 0)
Let us consider a point p(x , y) is reflected to

p’(x’ , y’) by the reflection axis as x-axis (y=0).

Draw PM^OX, X’ X

\ PM = MP’ where M is (x , 0) which is the mid- OM

point of PP’.

Then, Using mid-point formula,
y1 + y2
x= x1 + x2 , y= 2 P’(x’, y’)
2 Y’

or, x = x + x1 , 0= y + y1
2 2

or, 2x = x + x’ , y + y’ = 0

or, x’ = x , y’ = -y
\ p(x , y) " p’(x’ , y’) = p’(x , -y)

ii) Reflection on y-axis:

Theoretical proof P(x, y) Y
Let us consider a point p(x, y) is reflected to P’(x’, y’)

p’(x’, y’) under the reflection M(0, y)
OX
axis y - axis (x = 0)
Draw PM = OY, Y’

Then, PM = MP’ where M (0, y) is the X’

mid - point of PP’.

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

or, x= x + x' , y + y'
2 y= 2

or, x + x’ = 0 y + y’ = 2y

` x’ = – x y’ = y
` p(x, y) $ p’(x’, y’) = p’(–x, y)

PRIME Opt. Maths Book - IX 271

iii) Reflection on y = x P(x, y)
Y
Theoretical proof :
Let us consider a point p(x, y) is reflected A
M
to p’(x’, y’) by the a reflection axis as y = x

Draw PM^AB x=y P’(x’, y’)
O X
\ PM = MP’ where M (x, y) is the mid point X’
of pp’

Then, Using mid - point formula,
y1 + y2
x= x1 + x2 , y= 2 B
2
Here,
Y’
x=y
y1 + y2
or, x1 + x2 = 2
2
Here,

x=y

or x + x' = y + y' [ x1 + x2 = y1 + y2 ]
2 2 2 2

or, 2x + 2x’ = 2y + 2y’

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

Again, Line pp’ and y = x are perpendicular to each other,

so, m1 × m2 = –1

or, y2 – y1 × c– coeff. of x m = –1
x2 – x1 coeff. of y

or, y' – y × 1 = –1
x' – x

or, y’ – y = –x’ + x

or, x + y = x’ + y’

Adding equation (i) and (ii)
2x = 2y’
` y’ = x
Subtracting equation (i) and (ii),
or, 2y = 2x’
or, y = x’
` x’ = y
` p(x, y) $ p’(x’, y’) = p’(y, x)

iv) Reflection on y = –x
Theoretical proof :
Let us consider a point p(x. y) is reflected to p’(x’, y’) under a reflection axis
y= –x
Draw PM^AB
DPM = DMP’
` M is the mid - point of pp’.

272 PRIME Opt. Maths Book - IX

Then, Using mid - point formula, y1 + y2 P(x, y)
2 Y
x= x1 + x2 , y= A
2 A
M
or, x= x + x' , y= y + y'
2 2

Taking y = – x P’(x’, y’)
X’
or, y + y' = x + x' X
2 2
or, y + x = – x’ – y’ ........................... (i) O y = –x

Again, B B

The lines pp’ and y = – x are perpendicular lines, Y’

Soor,, myx122 × ym12 = –1
– x1 ×
c– coeff. of x m = –1
– coeff. of y

y' – y
or, x' – x × (–1) = – 1

or, x’ – x = y’ – y

or, x – y = x’ – y’ ............... (ii)

Now, adding equation (i) and (ii),
2x = –2y’
\ y’ = – x

Subtracting equation (i) and (ii),
2y = – 2x’
\ x’ = – y
\ p(x, y) $ p’(x’, y’) = p’(–y, –x)

v) Reflection on y = k (parallel to x - axis)

Theoretical proof : Y
Let us consider a point p(x, y) is reflected P(x, y)
to p’(x’, y’) under reflection axis y=k.
Draw PM = AB MB
y=k P’(x’, y’)
\ PM=MP’ where, A OX

M(x, k) is the mid-point of PP’. Y’

Then, Using mid-point formula, X’
y1 + y2
x= x1 + x2 , y= 2
2

or, x= x + x' , y + y'
2 y= 2

or, x + x’ = 2x y + y’ = 2k

\ x’ = x y’ = 2k - y
\ p(x, y) " p (x’, y’) = p’ (x, 2k - y)

PRIME Opt. Maths Book - IX 273

vi) Reflection on x = h (parallel to y-axis) YA
Theoretical proof:
P(x, y) P’(x’, y’)
Let us consider a point p(x, y) is translated to M X
p’(x’, y’) under a reflection axis x = h.
O
Draw PM = AB X’
\ PM = MP’

ie. M( x, y) is the mid-point of pp’.

Then, Using mid-point formula, x=h
y1 + y2
x= x1 + x2 , y= 2 Y’
2 B

or, x= x + x' , y + y'
2 y= 2

or, x + x’ = 2h y + y’ = 2y

\ x’ = 2h – x y’ = y
\ p(x, y) $ p’(x’, y’) = p’ (2h – x, y)

S.No Reflection axis Object Image
(x. y) (x, – y)
1. About x - axis (y = 0) (x. y) (–x, y)
(x. y) (y, x)
2. About y-axis (x = 0) (x. y) (– y, – x)
(x. y) (2h – x, y)
3. About y = x (x = y) or x – y = 0 (x. y) (x, 2k – y)

4. About y = – x (x = – y) or x + y = 0

5. About x = h (parallel to y - axis)

6. About y = k (parallel to x - axis)

7.1.4 Properties of reflection :

i) The object and image are congruent.
ii) The object and image are equidistance from the axis of reflection.
iii) The line joining the object point A and image point A’ is perpendicular to the

mirror line.

iv) For the objects point A, B and C, AA’ ' BB’ ' CC’.

274 PRIME Opt. Maths Book - IX

Worked out Examples

1. Find the image of the plane figure given below under the given mirror line M.
A BM

C

Solution:
Here, DABC is the given object mirror line M is the reflection axis.

Draw, BP^M, BP = PB’

AQ^M, AQ = QA’

CR^M, CR = RC’ BM
A

P
C’ Q B’

R

C A’
Then, Join A’, B’ & C’ so DA’B’C’ is the image of DABC after reflection under a mirror

line M.

2. Find the image of a point A(–2, 3) under reflection about x-axis.
Solution:
Under reflection about x-axis.
P(x, y) → P’(x, –y)

A(–2, 3) → A’(–2, –3)

3. Find the image of DABC having vertices A(2, 3), B(4, 6) & C(7, 1) under reflection
about x = 0. Also plot them in graph.

Solution :
Under reflection about x = 0 (y-axis)
P(x, y) → P’(–x, y)

A(2, 3) → A’(–2, 3)
B(4, 6) → B’(–4, 6)

C(7, 1) → C’(–7, 1)

PRIME Opt. Maths Book - IX 275

Y
B’ B

A’ A

X’ C’ O C
X

Y’

Here,
DABC is the given object.
Y = axis (x = 0) is the mirror line.
DA’B’C’ is the image.

4. A reflection axis transform A(3, 2) to A’(1, 2). Find the axis. Also find the image of B(5,

5) and C(6, –3) with it and plot the DABC and DA’B’C’ in graph.
Solution : The point A(3, 2) → A’(1, 2) with an axis of reflection x = h.

\ P(x, y) → P’(2h – x, y)

A(3, 2) → A’(2h – 3, 2)

By equating A’, we get

2h – 3 = 1

or 2h = 4

\ h = 2
\ The reflection axis is x = 2.

Again,

B(5, 5) → B’(–1, 5) C(6, –3) → C’(–2, –3)
Y

B’ B

A’ X
A C

X’ O PRIME Opt. Maths Book - IX

C’
Y’

Here, DABC is an object x = 2is axis. DA’B’C’ is the
276

5. If A(2a – 1, – 10) → A’(3b + 1, – 3) under a reflection axis y = – x. Find the value of a

and b.
Solution : Under reflection about y = –x.

P(x, y) P’(–y, –x)

\ A(2a – 1, –10) A’(10, 1 – 2a)

By the question

A(2a –1, –10) A’(3b + 1, – 3)

By equating the corresponding elements of A’.

3b + 1 = 10 and 1 – 2a = – 3

or, 3b = 9 and 4 = 2a

\ – b = 3 and a=2

\ a = 2

b= 3

6. Transform a triangle having vertices A(1, 2), B(–2, 4) and C(0, – 4) under reflection
about x = 2 followed by reflection about y = –3. Also plot the object and image in

graph. Y B’
Solution: B
Under reflection about x = 2,

P(x, y) →p(2h – x, y) = p(4 – x, y) A A’
\ A(1, 2) → A’(3, 2)

B(–2, 4) → B’ (6, 4) X’ O X
C(0, – 4) → C’ (4, –4)

Again, under reflection about y = –3 C’’
C C’
P(x, y) → P’(x, 2k – y) =P(x, –6 – y)

\ A1(3, 2) → A”(3, – 8)

B1(6,4) → B” (6, – 10)

C1(4, – 4) → C” (4, –2)

Here, A’’ B’’
DABC is an object Y’
DA’B’C’ is first image
DA”B”C” is final image.

PRIME Opt. Maths Book - IX 277

Exercise 7.1

1. i) What is transformation? Write down the name of transformations.
ii) What is reflection in transformation.
iii) Write down the formula of a point P(a, b) under reflection about the axis x = h.
iv) Write down the formula of a point A(p, q) and reflection about the axis y = k.
v) What will be the image of a point P(x, y) under reflection about x-axis followed

by reflection about y - axis.

2. Find the image of the following plane figures under the given reflection axis ‘M’.

i) A ii) B M

B

C A
M
iii) M
R iv)
Q

Q

S PR
P

v) M vi)
T S R

PQ M

M

3. Find the image of point A(3, –2) under the following reflection axis.

i) about x-axis ii) about x = 2 iii) about y = x

iv) about y – 3 = 0 v) about y = – x

4. Find the image of object p(–3, 5) under the following reflection axis.

i) under x + y = 0 ii) under x + 2 = 0 iii) under y = 0

iv) under x = 0 v) under y = – 2

278 PRIME Opt. Maths Book - IX

5. Find the co-ordinate of image for the followings.
i) Find the image of the points A(2, 2) and B(6, 7) under reflection about x-axis.

Also plot them in graph. Also join AB and A’B’.
ii) Find the image of DPQR having vertices P(–2, 1), Q(–5, 4) & R(–6, –2) under

reflection about y-axis. Also plot them in graph.
iii) Find the image of quadrilateral having vertices A(1, 2), B(3, 5), C(6, 6) and D(7,

1) under reflection about y = –2. Also plot them in graph.
iv) Find the image of quadrilateral having vertices P(3, 1), Q(4, 5), R(7, 4) and S(8,

–2) under reflection about X = –1. Also plot them in graph.
v) Find the coordinate of image of triangle having vertices A(2, 1), B(4, 5) & C(5,

–3) under y = x to DA’B’C’. Also plot DABC and DA’B’C’ in graph.

6. Find the co-ordinate of image for the followings.
i) If a point A(3, 7) is transferred to A’(–7, –3) with a reflection axis, find the axis.
ii) A(5, –2) is transferred to A’(–1, –2) under a reflection axis, find the axis.
iii) P’(–3, –4) is the image of P(–3, –2) under a reflection axis. Find the axis.
iv) A’(–1, 5) is the image of A(1, 5) under a reflection axis, find the axis.
v) If a point A(2x + 1, y – 3) is reflected to A’(1, 5) under reflection about y = x. Find

the value of ‘x’ and ‘y’.

7. PRIME more creative questions:
i) A reflection axis gives image A’(–3, 2) of an object of A(–1, 2). Find the reflection

axis. Also find the image of points B(2, 5) and C(–2, –4) with it and plot the
object DABC and its image in graph.
ii) A’(2a + 1, 5 – b) is the image of A(1 – 3b, – 7) under reflection about x + y = 0.
Find the value of ‘a’ and ‘b’.
iii) Find the image of DABC having vertices A(1, 4), B(–3, –1) and C(0, –5) under
reflection about x + 2 = 0 to DA’B’C’. Again transform DA’B’C’ to DA’’B’’C’’ under
reflection about x = – y. Also plot them in graph.
iv) Transform a triangle having vertices A(–2, 1), B(2, 5) and C(4, 0) under reflection
about x-axis followed by reflection about x – y = 0. Also plot the object and image
in graph.
v) Transform a triangle having vertices P(2, –1), Q(4, 4) and R(6, 1) under
reflections about x = 0 following by y = 0. Also plot the object and image in
graph.

PRIME Opt. Maths Book - IX 279

Answer

1. Show to your teacher.
2. Show to your teacher.

3. i. A’ (3, 2) ii. A’ (1, –2) iii. A’(–2, 3)
iv. A’ (3, 8) v. A’ (2, –3)

4. i. P’(–5, 3) ii. P’(–1, 5) iii. P’(–3, –5)
iv. P’ (3, 5) v. P’ (–3, –9)

5. i. A’ (2, –2) B’ (6, –7) ; graph
ii. P’(2, 1), Q’ (5, 4), R’ (6, –2) ; graph.
iii. A’ (1, –6), B’(3, –9), C’(6, –10), D’(7, –5) ; graph.
iv. P’ (–5, 1) Q’ (–6, 5), R’ (–9, 4), S’ (–10, –2) ; graph.
v. A’ (1, 2), B’ (5, 4), C’(–3, 5) ; graph.

6. i. Reflection about y = –x.
ii. Reflection about x = 2.
iii. Reflection about y = –3.
iv. Reflection about y – axis.
v. x = 2, y = 4

7. i. Reflection about x = –2 ; B’ (–6, 5) and C’ (–2, –4) ; graph.

ii. a = –4, b = –2.
iii. A’ (–5, 4), B’ (–1, –1), C’(–4, –5) ; A” (–4, 5), B” (1, 1), C” (5, 4) ; graph.
iv. A’ (–2, –1), B’(2, –5), C’(4, 0) ; A” (–1, –2), B”(–5, 2), C” (0, 4) ; graph.
v. P’ (–2, –1), Q’ (–4, 4), R’ (–6, 1) ; P” (–2, 1), Q” (–4, –4), R”(–6, –1) ; graph.

280 PRIME Opt. Maths Book - IX

7.2 Rotation: 90° P’
O
Let us consider a point P is transferred to P’ with an angle P
90° in clockwise direction about a fixed point called centre of
rotation, where OP = OP’ and \POP’ = 90°

The transformation of an object from one
position to another place depands on centre,
direction and angle is called rotation.
Congruent image of an object is formed in
rotation.

w Rotation takes place about a fixed point, called the centre of rotation.
– Clockwise (negative) direction.
– Anti clockwise (positive) direction.

w Angle of rotation
– Quarter turn (90°)
– Half turn (180°)
– Full turn (360°)
w Full turn gives invariant image as an object.

w Let us consider a DABC is rotated to DA’B’C’ under [O, – 90°).
B

A
C
A’

O B’
C’

Here, DAOA’ = \BOB’ = COC’ = 90°
Anti clockwise direction is used to rotate the DABC to DA’B’C’.
Centre of rotation O is taken for the rotation.
Then,
Image of DABC is drawn under the above conditions.
Where DA’B’C’ is the image.

PRIME Opt. Maths Book - IX 281

7.2.1 Rotation using co-ordinates:

Taking an object point p(x, y) which transferred to p’(x’, y’) under rotation according to

the following conditions. Then the image p’(x’, y’) can be expressed in formula as follows.

i) Quarter turn about the centre as origin: ( + 90°) Y

Theoretical proof :

Here, P’(x’, y’) P(x, y)
\ POP’ = 90° (+ve and –ve)

i.e. m1 × m2 = – 1 +ve 90°
y–0 y'–0

or, x–0 × x'–0 = – 1

or, yy’= – xx’ X’ O –ve 90° X

or, y’ = – xx' ...........................(i)
y

Again, we have,

OP2 = OP’2 = r2 P’(x’, y’)
Y’
or, (x – o)2 + (y –0)2 = (x’ – 0)2 + (y’ – 0)2

or, x2 + y2 = x2 + x2 x'2
y2

or, y2(x2 + y2) = x’2(x2 + y2)

or, y2 = x’2

` x’ = ±y

Again,

From equation (i),

y’ = – x × y or – x × –y
y y

= – x or + x

\ P(x, y) $ P’(x’, y’) = P’ (–y, x) or (y, –x)

i.e. For anticlockwise (+ 90°) Y
P(x, y)
P(x, y) $ P’(–y, x)

For clockwise (–90°)

P(x, y) $ P’(y, –x)

ii) Half turn (+ 180°) about centre origin.

Theoritical proof : +180°
O
Here, X’ X

] POP’ = 180° (+ve and –ve)

Where O is the mid - point of line PP’. –180°

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

or, x= x + x' , y= y + y' Y’
2 2 PRIME Opt. Maths Book - IX

282

or, x + x’ = 0 y + y’ = 0

\ x’ = – x, y’ = – y
\ P(x, y) $ P’(x’, y’) = P’(–x, –y).

iii) Quarter turn about centre (a, b). P(x, y)
Let us consider A(a, b) is the centre of
+90°
rotation and P(x, y) be the object point. P’(x’, y’)
The image P(x’, y’) can be calculated by
using the following steps :

Step - I –90°
A(a, b)
Firstly a and b which are the components
P’(x’, y’)
of the centre of rotation should be to the

x - components and y - component of the

object point P(x, y).

i.e. P(x, y) centre (x – a, y – b)
(a, b)

Step - II

The new point (x – a, y –b) have to be rotated as for the centre origin.
(x, y) +90° (–y, x)

` (x – a, y –b) $ (–y + b, x –a)

(x, y) –90° (y, – x)

` (x – a, y – b) $ (y – b, –x + a)

Step - III
The components of the centre (a, b) should be added to the corresponding elements

of newly occurred points which will be the formula of rotation for the point p(x, y)
about the centre of rotation (a, b) as
(–y + b, x – a) +90° (–y + b + a, x – a + b)
(y – b, –x + a) –90° (y – b + a, –x + a + b)

\ P(x, y) +90° P’(x’, y’) = P’(–y + a + b, x – a + b) = p’{–(y – b) + a, (x – a) + b}
P(x, y) +90° P’(x’, y’) = P’(y + a – b, –x + a + b) = p’{(y – b) + a, – (x – a) + b}

Examples :

Image of P (2, –3) under rotation through +90° with centre (1, 2).

We have

P(x, y) +90° P’(–y + a + b, x – a + b)
centre(a, b) P’(–y + 1 + 2, x – 1 + 2)



P’(– y + 3, x + 1)

\ P(x, y) P’(3 + 3, 2 + 1)

P’(6, 3)

PRIME Opt. Maths Book - IX 283

Y P’(6, 3)
A(1, 2)

X’ O X

P(2, –3)

Y’

iv) Half turn about centre (a, b)

Here , P(x, y)

\ PAP’ = 180° (+ve or – ve) +180°
A(a, b)
i.e. A(a, b)e is the mid - point of PP’.
–180°
Then,
P’(x’, y’)
Using mid - point P’(x’, y’) formula,
y1 + y2
x= x1 + x2 , y= 2
2

or, x= x + x' , y + y'
2 y= 2

or, x + x’ = 2a, y + y’ = 2b

\ x’ = –x + 2a y’ = –y + 2b

\ P(x, y) $ P’(x’, y’) = P’(–x + 2a, –y + 2b) = P’{–(x – a) + a, – (y – b) + b}

S.No. Centre Direction Angle Object Image

1. (0, 0) +ve 90° (x, y) (–y, x)

2. (0, 0) +ve 90° (x, y) (y, –x)

3. (0, 0) ––––– 180° (x, y) (–x, –y)

4. (a, b) +ve 90° (x, y) (–y + a + b, x – a + b)

5. (a, b) –ve 90° (x, y) (y + a – b, –x + a + b)

6. (a, b) ––––– 180° (x, y) (–x + 2a, – y + 2b)
[(2a – x), (2b – y)]

7.2.2 Properties of rotation:

i) Three informations are given to rotate the objects which are direction, angle and centre.
ii) Object and image are congruent in it.
iii) Centre of rotation is the intersecting point of perpendicular bisectors of line

joining the object and corresponding images.
iv) Anticlockwise direction is called positive direction and clockwise is called negative direction.
v) If centre of rotation is inside the object, the object and image will be invariant.

284 PRIME Opt. Maths Book - IX

Worked out Examples

1. Rotate the given triangle under positive quarter turn with center ‘O’.
A

B
O

Solution: C
B’

A’
C’

A
B

O

C

Here, DABC is an object.
O is center of rotion.
\ AOA’ – \ BOB’ = \ COC’ = 90° in positive direction (Anticlockwise).
DA’B’C’ is the image.

2. Find the image of a point A(–3, 2) under rotation about –90° with centre origin.
Solution :
Under rotation through –90° about (0, 0)
P(x, y) → P’(y, –x)
\ A(–3, 2) → A’(2, 3)

3. Find the image of DPQR under rotation about positive quarter turn. Also plot the
object and image in graph. Where P(1, 2) Q(3, –2) & R(5, 4) are the vertices.

Solution:
Under rotation through +90° about (0, 0)
P(x, y) → P’(–y, x)
\ P(1, 2) → P’(–2, 1)
Q(3, –2) → Q’(2, 3)
R(5, 4) → R’(–4, 5)

PRIME Opt. Maths Book - IX 285

Y

P’
Q’ P

R

X’ R’ X
O

Q
Y’

Here,
DPQR is the object.
under rotation about +90°.
DP’Q’R’ is the image.

4. Find the co-ordinate of image of a point P(3, –2) under rotation about half in anti

clockwise direction with centre origin and then about + 90° with centre (–1, 2).

Solution:

The given point is P(3, –2) under rotation through 180°, about (0, 0)

P(x, y) →p’(–x, –y)

\ P(3, –2) →p’(–3, 2)

Again, under +90° with centre (–1, 2)

P(x, y) → p’(–y + b + a, x – a+ b)

→ p’ (–y + 1, x + 3)

\ P(–3, 2) → p” (–2 + 1, x + 3)

→ p” (–1, 5).

5. Transform A(2, 4), B(5, 1) & C(–3, 2) under rotation about positive quarter turn with

centre origin followed by rotation about 180° with centre (3, –1). Also plot the object

and image in graph.

Solution:

The given vertices of DABC are A(2, 4), B(5, 1) and C(–3, 2)

Under rotation about +90° with centre (0, 0).

P(x, y) → p’(–y, x)

\ A(2, 4) → A’ (–4, 2)

B (5, 1) → B’ (–1, 5)

C (–3, 2) → C’ (–2, –3).

Again, Under rotation about 180° with centre (3, –1).

P(x, y) → p’ (–x + 2a, – y + 2b)

→ p’ (–x + 6, – y – 2)

\ A’(–4, 2) → A’’ (4 + 6, –2 – 2) = A’’ (10, – 4)

B’(–1, 5) → B’’ (1 + 6, – 5 – 2) = B’’ (7, – 7)

C’(–2, – 3) → C’’ (2 + 6, 3, 2) = C’’(8, 1)

286 PRIME Opt. Maths Book - IX

Y
B’

A

A’ C C’’
X’ O B

X

C’
A’’

Y’ B’’

Here, DABC is the object.
DA’B’C’ is first image.
DA’’B’’C’’ is final image.

Exercise 7.2

1. i) What is rotation in transformation?
ii) Write down the formula for rotation about 180° with centre (a, b).
iii) Write down the image of p(x, y) under rotation about –90° followed by rotation

about +90° with centre origin.
iv) Find the image of A(a, b) under reflection about y = –x followed by rotation

about –270° with centre origin.
v) Find the image of a point m(p, q) under rotation about 180° with centre origin

followed by reflection about y-axis.

2. Draw the image of the following diagrams under the following conditions given in

question. A ii) P
i)

B R Q
C
O (–90°)
O (+90°)
PRIME Opt. Maths Book - IX 287

iii) Q iv) E

R P AD
S BC

O (–90°) G O (–90°) A
v) E vi) W
T
A Q
V S
HF U
C
R
BD

O (+270°) O (–270°)

3. Find the image of the following points under rotation about negative quarter turn

with centre origin.

i) (–5, –3) ii) (5, 2) iii) (–2, –7)

iv) (–6, 3) v) (–7, –1) vi) (5, –3)

4. Find the image of the following points under rotation about positive quarter turn.

i) (–7, 2) ii) (–5, –3) iii) (8, –2)

iv) (3, 7) v) (–6, 2) vi) (–2, –8)

5. Find the image of the following points under rotation about half turn.

i) (–2, –6) ii) (5, –3) iii) (3, 6)

iv) (–5, 4) v) (–4, –5) vi) (2, –5)

6. i) Find the co-ordinate of image of DABC having vertices A(–2, 3), B(–3, –1) and
C(–5, 4) under rotation about –270°. Also plot the object and image in graph.

ii) Find the co-ordinate of the image of DPQR having vertices P(2, 1), Q(3, 5) & Q(5,
0) under rotation about +270°. Also plot the DPQR and DP’Q’R’ in graph.

iii) Find the co-ordinate of the image of quadrilateral having vertices A(–3, 2), B(–1,
5), C(2, 4) & D(3, –1) under rotation about positive quarter turn. Also plot them
in graph.

iv) Find the image of quadrilateral having vertices A(3, 4), B(4, 6), C(6, 2) & D(4, 1)
under rotation about 180° with center origin. Also plot the object and image in
graph.

v) Transform a triangle having vertices A(1, –2), B(4, –5), C(6, 0) under rotation
about +90° with centre origin to DA’B’C’ and then rotation about 180° with

288 PRIME Opt. Maths Book - IX

center origin to DA’’B’’C’’. Also plot the object and image in graph.

7. Find the image under the following.
i) P(3, –2) under +90° with center (1, 2).
ii) A(3, 4) under –90° with center (–1, 2).
iii) M(–3, 3) under 180° with center (2, –1).
iv) Triangle having A(1, 1), B(3, 5) and C(5, 0) under +ve quarter turn with center

(–2 –1). Also plot them in graph.

8. i) Which rotation gives the image of point A(2, –1) to A’(1, 2)? Also find the image
of the points B(3, 2) and C(5, –4). Also plot DABC and DA’B’C’ in graph.

ii) Find centre of rotation, angle of rotation and direction of rotation which gives
A(2, –3) to A’(–3, –2) and B(4, 1) to B’(1, –4) by plotting the points in graph.

iii) Find the image of P(a, b) under rotation about +90° with center origin and again
under 180° with center origin.

iv) Find the centre, direction and angle of rotation where DABC having A(3, 2),
B(5, 1) and C(4, –2) is rotated to A’(–2, 3) B’(–1, 5) and C’(2, 4) by plotting in
graph.

v) Find the center direction and angle of rotation which transformed DABC having
A(3, –4), B(5, –6) & C(7, –1) to A’(1, 4), B’(3, 6), C’(–2, 8) by plotting in graph.

9. PRIME more creative question:
i) Transform a quadrilateral having vertices A(2, 1), B(1, –2), C(–3, –2) and

D(–5, 1) under rotation about 180° with centre origin followed by rotation
about + 90° centre (3, –4). Also plot the object and image in graph.
ii) Find the image of DPQR having vertices P(–4, 6), Q(–1, –2) and R(3, –5) under
negative quarter turn with centre origin followed by rotation about 180° with
centre origin Also plot the object and image in graph.
iii) Transform a triangle having vertices A(2, 3), B(1,5) and C(–2, 4) under reflection
about y = x. Following by rotation about rotation about +90° with centre in
origin. Also plot the image and object in graph.
iv) Triangle having vertices P(3, 4), Q(–2, 6) and R(1, – 5) is reflected under x = –2
to DP’Q’R’ and rotated again to DP’’Q’’R’’ under +270° with centre origin. Also
plot them in graph.
v) If a point P(x, y) is reflected under x-axis and then under y-axis continuously is
there any difference it with rotation about 180° with centre origin? Discuss the
application of rotation in group and a report.

PRIME Opt. Maths Book - IX 289

1. Show to your teacher. Answer iii. (–7, 2)
2. Show to your teacher. vi. (–3, –5)
3. i. (–5, –3) ii. (2, –5)
iv. (3, 6) v. (–1, 7)

4. i. (–2, –7) ii. (3, – 5) iii. (2, 8)
iv. (–7, 3) v. (–2, –6) vi. (8, –2)

5. i. (2, 6) ii. (–5, 3) iii. (–3, –6)
iv. (5, –4) v. (4, 5) vi. (–2, 5)

6. i. A’(–3, –2), B’(1, –3), C’(–4, –5) ; graph.
ii. P’(1, –2), Q’(5, –3), R’(0, –5) ; graph.
iii. A’(–2, –3), B’(–5, –1), C’(–4, 2), D’(1, 3) ; graph.
iv. A’(–3, –4), B’(–4, –6), C’(–6, –2), D’(–4, –1) ; graph.
v. A’(2, 1), B’(5, 4), C’(0, 6) ; A”(–2, –1), B”(–5, –4), C”(0, –6) ; graph.

7. i. (5, 4) ii. (1, –2) iii. (7, –5)
iv. A’(–4, 2) , B’(–8, 4) , C’(–3, 6) ; graph.
v. P’(0, 5) , Q’(–4, 6) , R’(2, 8) ; graph.

8. i. Rotation about +90° with centre origin, B’(–2, 3) , C’(4, 5) ; graph.
ii. Centre (0, 0), angle 90°, negative direction.

iii. P(a, b) →P’(–b, a) → P”(b, –a)
iv. Centre (0, 0) angle 90°, positive direction
v. Show to your teacher.

9. i. A’ (–2, –1), B’(–1, 2), C’(3, 2) D’(5, – 1); A”(0, 5), B”(–3, 6), C”(–3, 10), F’(0, 12);
graph.

ii. P’(6, 4), Q’(–2, 1) R’(–5, –3) ; P”(–6, –4), Q”(2, –1), R”(5, 3); graph.
iii. A’(3, 2), B’(5, 1), C’(4, –2) ; A”(–2, 3), B”(–1, 5), C”(2, 4); Graph.
iv. P’(–7, 4), Q’(–2, 6), R’(3, –5) ; P”(–4, – 7), Q”(–6, –2) & R”(5, 3) ;
v. P(x, y) → P’ (x, – y) → P” (–x, – y). It is same as rotation about 180° with centre

290 PRIME Opt. Maths Book - IX

7.3 Translation P’

Here, vector AB is the magnitude & direction of

translation. P

DPQR is, an object. Q’ R’

PP’ = QQ’ = RR’ = AB A B
PP’ || QQ’ || RR’ ' AB R

DP’Q’R’ is the image of DPQR Q

The transformation of an object from one place to another place by
the magnitude and direction of a given vector is called translation.

7.3.1 Translation using co-ordinate
Let us consider P(x, y) is an object point and OA is the translation vector <aF .

b

According to properties of translation, P is Y
shifted to P’ where direction is according
P’(x’, y’)
to OA and PP’ ' OA.

Also PP’ = OA

i.e. P’ is defined as ‘a’ units more in P(x, y) A(a, b)
b
horizontal direction than of p and ‘b’ units
M
more in vertical direction than p.

\ x’ = x + a and y’ = y + b O a X

\ P(x, y) $ P’(x’, y’) = P’(x + a, y + b) X’

Also, If a point P is translated to P’ under a

displacement and direction of AB having Y’
tchoe-torarndsinlaattieosnAc(oxm1,pyo1n) eanntdcBan(xb2,eyd2e)ftinheend

as,

a = x2 – x1 P’(x + a, y + b) where a = x2 – x1 and b = y2 – y1
b = y2 – y1
\ P(x, y) AB

PRIME Opt. Maths Book - IX 291

i) Translation using co-ordinate:

Y

A’(7, 5)
A(4, 3)

X’ O B’(6, 1) C’(9, 1)
C(6, -1) X

B(3, -1)

Y’
Here,
A(4, 3) is translated to A’(7, 5)
B(3, –1) is translated to B’(6, 1)
C(6, –1) is translated to C’(9, 1)

i.e. All the points are translated with constant number 3 for x-component and 2 for
y - component.

i.e. A (4, 3) → A’(4 + 3, 3 + 2) = A’(7, 5)

i.e. Translation vector is T = <3F

2

i.e. under translation T = <aF

b
P(x, y) → P’(x + a, y + b)

iii) Translation using vector:

Let us consider a translation vector is AB where A(1, 2) and B(3, 5) are any two

points.

Then, according to the concept of column vector AB

AB = <x2 – x1F = <3 – 1F = <2F
y2 – y1 5–2 3

\ Translation vector T = AB = <2F
3

Properties of translation:
i) Object and image are congruent in it.
ii) For the object points A & B and A’ & B’ for vector T //AA’//BB’.
iii) Length of AA’ and BB’ is always equal to magnitude of T .
iv) Image is always in the same direction of vector T .

292 PRIME Opt. Maths Book - IX

Worked out Examples

1. Find the image of a point A(3, –2) under a translation vector of T = <1F .
3
Solution:

Under translation vector T = <1F
3
P(x, y) → P’(x + a, y + b)

→ P’(x + 1, y + 3)

A(3, –2) → A’(3 + 1, –2 + 3)

→ A’(4, 1)

T = <1F
3
\ A(3, –2) A’(4, 1)

2. If A (3, –1) and B(1, 2) are any two points, find the image of point A and B under AB .

Solution :

Here,

Translation vector AB from the given points A(3, –1) & B(1, 2) is,

AB = <x2 – x1F = <1 – 3F = <–2F =T
y2 – y1 2 + 1 3

Then, under translation vector T = <–2F
3

p (x, y) → P’(x + a, y + b) = P(x – 2, y + 3)

A(3, –1) → A’(3 – 2, –1 + 3) = A’(1, 2)

T = <–2F
3 A’(1, 2)
\ A(3, –1)

B(1, 2) → B’(1 – 2, 2 + 3) = B’(–1, 5)

3. Find the image of DABC having vertices A(3, 1), Y
B’
B(–1, 5) & C(1, –3) under a translation vector T
B
= <32F . Also draw the object and image in graph.
Solution: Under a translation vector T A
P(x, y) → P’(x + a, y + b) = <3F O
2
C’
→ P’(x + 3, y + 2) A’
C X
Then, Y’
293
A(3, 1) → A’(3 + 3, 1 + 2) = A’(6, 3) X’
B(–1, 5) → B’(–1 + 3, 5 + 2) = B’(2, 7)

C(1, –3) → C’(1 + 3, –3 + 2) = C’(4, –1)

Here,
DABC is the object.
DA’B’C’ is the image of DABC

PRIME Opt. Maths Book - IX

4. Draw the image of given triangle under the given vector AB .
Q

AB Q Q’
P
R A P’ B
Solution: P R R’
Here, Draw PP’, QQ’, RR’ || AB .
Then,
Taking PP’ = QQ’ - RR’ = AB
DPQR is the object.
DP’Q’R’ is the image.

5. Translate a triangle having vertices A(1, –1), B(–2, 2) and C(3, 3) under T1 = <32F and
<–13F
again translate under T = . Also plot the object and image in graph.
Solution :


We have,

Under translation about T1 = <3F
2

P(x, y) → P’(x + a, y + b) = P’(x + 3, y + 2)
\ A(1, –1) → A’(1 + 3, –1 + 2) = A’(4, 1)
B(–2, 2) → B’(–2 + 3, 2 + 2) = B’(1, 4)
C(3, 3) → C’(3 + 3, 3 + 2) = C’(6, 5)

<1 F Y
–3
Again, under translation about T2 =
P(x, y) → P’(x + 1, y – 3)

\ A’(4, 1) → A’’(5, –2) C’
C C’’
B’(1, 4) → B’’(2, 1) B’
A’ X
C’(6, 5) → C’’(7, 2) B’’ A’’
O
Here, B
DABC is the object. X’ A
DA’B’C’ is the first image.
DA’’B’’C’’ is the final image.

Y’

294 PRIME Opt. Maths Book - IX

Exercise 7.3

1. i) What do you mean by translation in transformation? m
n
ii) What is the formula of p(a, b) under translation about T = ?
iii)
Find the image of a point p(x, y) under translation about T1 = 1 followed by
2
<2F
translation about T2 = 3

iv) Find translation vector T1 Which gives A(2, 3) → A’(5, 2).
<2F
v) If A’(3, 5) is the image after translation T = 3 , find the point A.

2. Find the image of the following figures under the given vector.

i) P B ii) AM

B

A R N
Q

C

iii) S iv) B P
T
A R D
B C

Q AQ

v) A G

H

M F
BD

CE N

3. Find the image of the following points under the translation vector T = <2F .
1

i) A(3, 2) ii) P(–2, 5) iii) M(–3, –1)

iv) N(4, –6) v) O(–3, –4)

4. Find the image of the points P(3, –2) under the translation vector AB for the

following points A and B.

i) A(3, 2) and B(1, 0) ii) A(1, –2) & B(3, 1)

iii) A(2, 1) and B(4, 5) iv) A(–1, 2) & B(2, 1)

PRIME Opt. Maths Book - IX 295

v) A(4, 1) and B(1, –2)

5. Find the image of the following triangles under the given vector. Also plot the object

and image in graph.

i) A(1, –2), B(–3, 4) and C(3, –3) under a vector T= <2F
3

ii) P(2, 5), Q(–2, 1) and R(5, 2) under a vector T = <–2F
–3

iii) A(–2, –4), B(–4, –1) and C(3, 0) under a vector T = <1F
3

iv) K(2, 3), L(4, 6) & M(6, 1) under a vector T = <–2F
4

v) X(–2, 5), Y(3, 1) & Z(–4, –1) under a vector T = <3 F
–4

6. i) If a translation T gives image of an object A(1, 3) → A’(3, 4). Find the value of

translation vector T.
ii) If a point P(3, –1) is translated to P’(5, 2) under a vector T, find the value of

translation vector T.

iii) If a vector T = <1F translate a point P to P’(3, 7), find the co-ordiante of the point
P. 3

iv) Find the co-ordiante of image of DABC having vertices P(1, 2), Q(3, 5) and R(6,

–2) under a translation vector PQ . Also plot DP’Q’R’ and DP’Q’R’ on graph paper.

v) Find the co-ordinate of the image of DABC having A(3, 1), B(7, 3), C(5, –2) under

a translation vector AB .

7. PRIME more creative questions

i) Find the image of ABCD having vertices A(1, 8), B(–3, 9), C(0, 13) & D(4, 12)

under translation about T1 = AB . Also transform the image so obtained under
<–3F
T2 = –2 and plot the image and object in graph.

ii) Transform a triangle having vertices A(–4, 6), B(3, –2) and C(1, 2) under

translation about T1 = <1 F followed by translation about T2 = <–2F . Also plot
–2 3

the object and image in graph.

iii) Transform a triangle having vertices A(–3, 2), B(1, 5) and C(2, 3) under reflection

about x = 2 followed by translation about AB . Also plot the object and image in

graph.
iv) Reflect p(x, y) under x = 1 followed by x = 3 and compare its image with
4
translation about T 0 . Write down the conclusion of it.

v) What is translation? Write down its properties. Where is it applicable in our

daily life? Discuss in group and prepare a report.

296 PRIME Opt. Maths Book - IX