Final PDF file of OPT Class 9 2077

we can conclude for the image of a point under the reflection about the line y = – x.

i.e. P(x, y) = y = – x P'(– y, – x)
Reflection

Example 6:

Find the image of ∆PQR with vertices P(– 2, 1), Q(2, 3) and R(– 1, 5) under the reflection about the

line y = – x. Also draw ∆PQR and its image on the same graph paper.

 The vertices of given triangle PQR are P(– 2, 1), Q(2, 3) and Y

R(– 1, 5). The co-ordinates of the vertices of the image ∆P'Q'R' R
M
under the reflection about the line y = – x can be obtained by
P'
using formula as follows: Q
y = – x
We have, P(x, y) Reflection P'(– y, – x)

Hence, P(– 2, 1) → P'(– 1, 2) R'

Q(2, 3) → Q'(– 3, – 2) X' P X
O

R(– 1, 5) → R'(– 5, 1) Q'
Y'
Drawing ∆PQR and its image ∆P'Q'R' on the same graph, we have
the figure as shown.

Reflection About the Line x = a Y

Let M(1, 3) be a point and the line x = 4 be the axis of reflection. M(1, 3) M'(7, 3)
The line x = 4 is a line parallel to Y-axis and at 4 units right from X
Y-axis. From point M, draw MN perpendicular on the line x = 4 and
produce it upto M' so that MN = NM', then the co-ordinates of M'
is (7, 3) as shown in the figure.

Hence, M'(7, 3) is the image of the point M(1, 3) under the reflection X' X

about line x = 4. O
Y'
x=4 x = 4
Reflection
i.e. M(1, 3) M'(7, 3)

Let us see the following table of some points on the plane and their corresponding images under the Transformation
reflection about the line x = 4.

SN Points in the plane Corresponding images

1. P(2, 4) P'(6, 4)

2. Q(– 1, 3) Q'(9, 3)

3. R(5, – 3) R'(3, – 3)

4. S(3, – 5) S'(5, – 5)

5. L(a, b) L'(2 × 4 – a,b)

From above table we can see that the image of any point under the reflection about the line x = a, is obtained
by changing the x co-ordinate of the given point by 2a – x. So we can conclude for the image of a point

Reflection ~ 247

under the reflection about the line x = a.

i.e. P(x, y) = x=a P'(2a – x, y)
Reflection

Reflection About the Line y = b Y

Let Q(2, – 5) be a point and the line y = – 3 be the axis of reflection. X' O X

From point Q, draw QM perpendicular on the line Q'(2, – 1)

y = – 3 and produce it upto Q' so that QM = MQ', the coordinates of M
Q(2, – 5)
point Q' will be (2, – 1) as shown in the figure. y = – 3

i.e. Q(2, – 5) y = – 3 Q'(2, – 1)
Reflection

Y'
Let us see the following table of some points on the plane and their corresponding images under the
reflection about the line y = – 3.

SN Points in the plane Corresponding images

1. A(– 2, 1) A'(– 2, – 7)

2. B(3, 4) B'(3, – 10)

3. C(1, – 2) C'(1, – 4)

4. D(– 4, – 3) D'(– 4, – 3)

5. R(a, b) R'(a, 2 × – 3 – b)

From the above table we can see that the image of any point is obtained by changing the y-coordinate of
the object by 2b – y where b is the distance of the axis of reflection with direction from X-axis. So we can
conclude for the image of a point under the reflection about the line y = b.

i.e. P(x, y) = y=b P'(x, 2b – y)
Reflection

Example 7:

A(1, 1), B(4, 3) and C(– 1, 5) are the vertices of a triangle ABC, find the co-ordinates of the vertices

of the image of ∆ABC under the reflection about the line y = 2.

 Here A(1, 1), B(4, 3) and C(– 1, 5) are the vertices of given triangle. We can find the co-ordinates of

the vertices of the image of ∆ABC under the reflection about the line y = 2 by using formula.

We have, P(x, y) y=b P'(x, 2b – y)
Reflection

Hence, A(1, 1) → A'(1, 2 × 2 – 1) i.e. A'(1, 3)

B(4, 3) → B'(4, 2 × 2 – 3) i.e. B'(4, 1)

C(– 1, 5) → C'(– 1, 2 × 2 – 5) i.e. C'(– 1, – 1)

∴ The coordinates of the vertices of image ∆A'B'C' are A'(1, 3), B'(4, 1) and C'(– 1, – 1) respectively.

248 ~ Perfect Optional Mathematics Class 9

Example 8:

If A'(2, 3), B'(4, 1) and C'(6, 4) are the images of the point A, B and C of DABC undert he reflection

on the line y – 3 = 0, determine the co-ordinates of the vertices of DABC. Aslo draw DABC and

DA'B'C' on the same graph.

 The image P'(x', y') of point P(x, y) under the reflection on the line y = b is P'(x, 2b – y).

Hence, x' = x and y' = 2b – y. Y
i.e. x = x' and y = 2b – y'.

Hence, the object for the image P'(x', y') is P(x, y) or P(x', 2b – y'). N
C'
Here, the line of reflection is y – 3 = 0 i.e. y = 3. A'
A C
The object of image A'(2, 3) is A(2, 2×3 – 3) i.e. A(2, 3). X' B' X
The object of image B'(4, 1) is B(4, 2×3 – 1) i.e. B(4, 5). O

And the object of image C'(6, 4) is C(6, 2×3 – 4) i.e. C(6, 2). Y'

The object DABC and image DA'B'C' is shown on the graph.

Note: We can find these co-ordinates of image ∆A'B'C' by drawing in graph paper.

Conclusion About Reflection Using Co-ordinates

SN Object Axis of Image point Short Symbol
Point Reflection

1. P(x, y) X-axis P'(x, – y) P(x, y) X-axis P'(x, – y)
Reflection

2. P(x, y) Y-axis P'(– x, y) P(x, y) Y-axis P'(– x, y)
Reflection

3. P(x, y) line y = x P'(y, x) P(x, y) y=x P'(y, x)
Reflection

4. P(x, y) line y = – x P'(– y, – x) P(x, y) y = – x P'(– y, – x)
Reflection
Transformation
5. P(x, y) line x = a P'(2a – x, y) P(x, y) x=a P'(2a – x, y)
Reflection

6. P(x, y) line y = b P'(x, 2b – y) P(x, y) y=b P'(x, 2b – y)
Reflection

Reflection ~ 249

EXERCISE : 7.1A

1. Find the image of the following figures under the reflection about the line.

a. A b. M D c. M
N

CE

B C N
M
AB
N

d. P e. M f. B
B DA
O
R CM C
Q
M S A N
N N

2. Reflect DABC given in the graph of the following figures.

a. Y b. Y c. Y
A A A

B CB C B C
X' O X O X X' O X
X'

Y' Y' Y'
X-axis Y-axis line y = 1

d. Y e. Y f. Y
A A

B C X' OA X X' BC
X' O X OX

B C Y'
line y = – x
Y' Y'
line x = 1 line y = x

250 ~ Perfect Optional Mathematics Class 9

3. Draw the axis of reflection with the help of following pair of object and its image in each case.

a. A A' b. Q' c. Q R
P'

B B' P
PR P'

R' R'

C C' Q Q'

4. a. A point (3, 2) is reflected on the Y-axis, find co-ordinates of the image so obtained.

b. A(1, 2) and B(– 3, 5) are the ends of a line segment AB, find the image of line segment AB under

the reflection about X-axis and the co-ordinates of A' and B'.

c. A point M(7, 2) is reflected on the line x = 3, find the co-ordinates of the image of M.

5. If P(2, 3) and Q(– 4, – 3) are the points, find the co-ordinates of their images under the reflection about
the following lines.

a. Y-axis b. X- axis c. line x = y

d. line y = – x e. line x = – 3 f. line y = 2

6. Find the axis of reflection with the help of the following pair of point and corresponding image.

a. A(2, 5) and A'(4, 5) b. P(– 3, 5) and P'(3, 5)

c. Q(– 2, – 3) and Q'(– 2, 3) d. M(3, – 4) and M'(– 4, 3)

e. B(2, – 4) and B'(2, 8) f. C(4, – 3) and C'(3, – 4)

[Hint: Draw perpendicular bisector of AA' for question (a) and so on]

EXERCISE : 7.1B

1. a. If A(1, 1), B(0, 5) and C(4, 3) are the vertices of ∆ABC, find the co-ordinates of image ∆A'B'C'
under the reflection about Y-axis. Draw ∆ABC and its image on the same graph paper.

b. Draw quadrilateral PQRS in your square copy and find the image of PQRS under the reflection Transformation
about X-axis where the vertices of PQRS are at P(1, 3), Q(4, 3), R(5, 1) and S(3, 1).

2. a. Determine the vertices of the image ∆A'B'C' formed when the DABC with vertices A(– 2, 8),
B(8, 6) and C(4, 2) is reflected on the line x – 2 = 0. Also draw both triangles on the same graph
paper.

b. Find the image of ∆ABC with vertices A(1, 6), B(4, 5) and C(0, 2) under the reflection about the
line x = 4. Also draw ∆ABC and its image on the same graph paper.

c. If the hexagon ABCDEF with vertices A(– 1, – 2), B(1, – 2), C(3, – 4), D(1, – 6), E(– 1, – 6) and
F(– 2, – 4) is reflected on the line y = – 1. Find the co-ordinates of the vertices of image of
ABCDEF and draw the given hexagon and its image on the same graph paper.

Reflection ~ 251

3. a. Find the co-ordinates of the vertices of the image of pentagon ABCDE with vertices A(– 2, 1),
B(1, 1), C(3, 3), D(1, 5), E(– 2, 4) under the reflection about the line y = – x. Also draw the object
and image on the same graph paper.

b. Find the image of ∆PQR under the reflection about the line x + y = 2 where the vertices of ∆PQR
are at P(– 1, 1), Q(– 2, – 1) and R(1, – 2).

[Hint : Draw the line of axis x + y = 2 on graph paper and the find the image of ∆PQR by
drawing]

c. Find the image of ∆DEF under the reflection about the line x – y = 3, where the vertices of ∆DEF
are at D(– 4, 1), E(0, 4) and F(– 2, 5).

4. a. If A(1, 3), B(3, 4) and C(4, 1) are the vertices of ∆ABC, find the co-ordinates of the image
∆A'B'C' under the reflection about X-axis and again reflect ∆A'B'C' on the Y-axis to get ∆A"B"C"
as image and find the co-ordinates of the vertices of ∆A"B"C" also.

b. If P(– 4, 1), Q(– 2, 1) and R (– 2, 3) are the vertices of ∆PQR. Reflect ∆PQR on the line x = – 1 to
get ∆P'Q'R' and again reflect ∆P'Q'R' on the line y = – x to get ∆P"Q"R". Write the co-ordinates
of the vertices of ∆P'Q'R' and DP"Q"R".

5. If A(2, 4), B(– 1, 2) and C(3, 2) are the vertices of ∆ABC, find the vertices of the image of ∆ABC
under the reflection about the following axis of reflection

a. Y-axis b. line x = y

c. line x = – 2 d. line y = 3

6. a. Under the reflection, the image of A(1, 2) is A'(1, – 2), then find the image of DABC in which co-
ordinaes of points B(4, 4) and C(3, 7) under the same reflection. Also show DABC and DA'B'C'
on the same graph.

b. A triangle formed by the points A(–3, 3), B(– 6, 3) and C(– 3, 9) is reflected under the reflection
line as A(–3, 3) to A'(– 3, 3). Then find the image of B and C under the same reflection. Also
show DABC and DA'B'C' on the same graph.

c. A triangle formed by the points P(2, 3), Q(4, 5) and R(6, 2) is reflected under the reflection line
as P(2, 3) to P'(– 2, 3). Then find the images of Q and R under the same reflection. Also draw
DPQR and DP'Q'R' on the same graph.

7. a. If A'(2, – 5), B'(3, – 2) and C'(1, – 3) are the images of the points A, B and C of DABC under the
reflection on line y = 0. Determine the co-ordinates of the vertices of DABC and show DABC
and DA'B'C' on the same graph.

b. If A'(2, 3), B'(4, 1) and C'(6, 4) are the images of point A, B and C of DABC under the reflection
on the line x = 0. Determine the co-ordinates of the vertices of DABC. Also draw DABC and
DA'B'C' on the same graph.

8. a. If P'(6, 3), Q'(2, 2) and R'(4, 5) are the image of the points P, Q and R of DPQR under the line
x = 4. Determine the co-ordinates of vertices of DPQR. Also draw DPQR and DP'Q'R' on the
same graph.

b. If A'(2, 0), B'(8, 2) and C'(4, 6) are the images of the vertices A, B and C of DABC under the
reflection in the line y = 2. Determine the co-ordinates of vertices of DPQR. Also draw DPQR
and DP'Q'R' on the same graph.



252 ~ Perfect Optional Mathematics Class 9

7.2 Rotation

Introduction

Rotation is a transformation in which each point on the object is rotated through an angle about a fixed

point in the given direction. The fixed point is called the centre of rotation and the angle is called the angle

of rotation.

Image P' R' B' C' D'
P DA
TransformationImageImage

Q' A'

90° 60° Object Object O
Centre
O Q C

Object R B

The following are the properties of rotation.
a. The object and the image under the rotation are congruent.

b. Every point on the object is rotated through same angle in the same direction about centre of rotation
to reach its image.

Example 1:

Find the image of line PQ under the rotation through + 90° P
about O.

OQ

 To find the image of line PQ under the rotation through + 90° P' P
Q'
about O. Join OP and OQ by dotted line and draw an arc taking
O as centre and OP as radius as shown in figure. At O make OQ
POP' = 90° with compass or protractor. Join OP' by dotted
line to meet the arc at P'.

Similarly, draw an arc taking O as centre and OQ as radius and
make QOQ' = 90° to get Q'. P' and Q' are the images of P and Q
respectively under the rotation through 90° about O. Hence, P'Q' be
the image of PQ.

Rotation ~ 253

Example 2: R

Find the image of ∆PQR under the rotation through – 70° P
(clockwise) about O. Q

O

 For the image of ∆PQR under the given rotation. Join OP, R

OQ and OR by dotted line as shown. Draw an arc taking O as P
centre and OP as radius. Make POP' = 70° and OP' meet the
Q P'
arc at P'. Here, P' is the image of point P. Similarly, find the
70°
images of the remaining vertices Q and R. Join P'Q', Q'R' and
R'P' to get the image P'Q'R' under the rotation through 70° O Q'

about O.

R'

Example 3: Q'
P'
If DP'Q'R' be the image of the ∆PQR under the rotation, find
the centre and the angle of rotation.

P

Q R R'

 Trace ∆PQR and ∆P'Q'R' in the exercise book. Join PP' and P Q'
P'
RR' (or QQ'). Draw the perpendicular bisectors of PP' and RR'.

If these two perpendicular bisectors meet at O, then O is the
centre of rotation. Join OP and OP', then measure POP' (70°)

which is the angle of rotation. The angle of rotation can also be
measured as QOQ' or ROR'.

Q R R'

80°

O

Rotation Using Co-ordinates

The image of the geometrical figures under the rotation through certain angles about given points as centre
can be obtained with the help of co-ordinates.

254 ~ Perfect Optional Mathematics Class 9

Rotation Through 90° About Origin (Positive Quarter Turn)

Let P(3, 1) be a point and O(0, 0) be the centre of rotation and 90° Y P(3, 1)
P'(– 1, 3) X
be the angle of rotation. Join OP and draw an arc taking O as centre
and OP as radius shown in figure. Make POP' = 90° to get the O

image P' of point P, then the co-ordinates of P' will be (– 1, 3) as

shown in figure.

[90°, O(0, 0)] X'

i.e. P(3, 1) Reflection P'(– 1, 3)

Without drawing arc, we can find the image P' when the angles of Y'
Y

rotations are 90°, 180°, 270°. Draw PM⊥OX and join MO to get L P'(– 1, 3)
O
shaped line as shown in figure. Then rotate this L shaped line OMP

through 90° about origin O in positive direction to get the image P(3, 1)
M
P'(– 1, 3). X' X

Y'

Let us see the following table of some points on the plane and their corresponding images under the rotation
through 90° about origin.

SN Points in the plane Corresponding images

1. A(2, 3) A'(– 3, 2)

2. B(– 3, 1) B'(– 1, – 3)

3. C(– 4, – 2) C'(2, – 4)

4. D(4, – 3) D'(3, 4)

5. M(a, b) M'(– b, a) Transformation

From the above table, we can see that the image of point P(x, y) under the rotation through 90° about origin

O is [90°, O(0, 0)]
Reflection
i.e. P(x, y) = P'(– y, x)

Example 4:

If A(1, 1), B(4, 1) and C(2, 3) are the vertices of a triangle ABC, find the co-ordinates of the vertices

of the image of ∆ABC under the rotation through 90° about origin.

 As A(1, 1), B(4, 1) and C(2, 3) are the vertices of ∆ABC. The vertices of the image ∆A'B'C' of ∆ABC

under the rotation through 90° about origin can be obtained as follows:

Rotation ~ 255

[90°, O(0, 0)] Y

We have, P(x, y) Reflection P'(– y, x)

Hence, A(1, 1) → A'(– 1, 1) B'

B(4, 1) → B'(– 1, 4) C
C'
C(2, 3) → C'(– 3, 2)
A' A BX
Drawing ∆ABC and its image ∆A'B'C' on the same graph paper, we X' O
have the figure as shown.

Rotation Through 180° About Origin (Half turn) Y'

Let us see the following table of some points on the plane and their corresponding images under the rotation
through 180° about origin.

SN Points in the plane Corresponding images

1. A(2, 4) A'(– 2, – 4)

2. B(– 3, 5) B'(3, – 5)

3. C(– 2, – 3) C'(2, 3)

4. D(4, – 1) D'(– 4, 1)

5. M(a, b) M'(– a, – b)

From the above table, we can see that the image of point P(x, y) under the rotation through 180° about origin is

[180°, O(0, 0)]

i.e. P(x, y) = Reflection P'(– x, – y)

Rotation Through 270° (or – 90°) About Origin (Negative Quarter Turn)

Let us see the following table of some points on the plane and their corresponding images under the rotation
through 270° about origin.

SN Points in the plane Corresponding images

1. A(– 4,– 3) A'(– 3, 4)

2. B(– 3, 1) B'(5, 2)

3. C(2, – 1) C'(– 1, – 2)

4. D(4, 3) D'(3, – 4)

5. Q(a, b) Q'(b, – a)

From the above table, we see that the image of a point P(x, y) under the rotation through 270° about origin
O(0, 0) is

[270°, O(0, 0)]

i.e. P(x, y) = Reflection P'(y, – x)

256 ~ Perfect Optional Mathematics Class 9

Example 5:

If P(1, 1), Q(2, – 2), R(5, 0) and S(3, 3) are the vertices of a quadrilateral PQRS, find the co-ordinates

of the vertices of its image under the rotation through 180° about origin. Also draw object and

image on the same graph paper.

 As P(1, 1), Q(2, – 2), R(5, 0) and S(3, 3) are the vertices of given quadrilateral PQRS. The co-ordinates

of the vertices of its image under the rotation through 180° about origin can be obtained as follows.

We have, Y

[180°, O(0, 0)]
P(x, y) Rotation P'(– x, – y)
Q' S
P
Hence, P(1, 1) → P'(– 1, – 1) X' R' RX
Q(2, – 2) → Q'(– 2, 2) O
R(5, 0) → R'(– 5, 0) P'

S(3, 3) → S'(– 3, – 3) Q

Draw quadrilateral PQRS and its image quadrilateral P'Q'R'S' on S'

the same graph paper as shown in figure. Y'

Rotation Through 90° About M(a, b)

Let P(5, 6) be a point and M(2, 1) be the centre of rotation and angle Y
P'(– 3,4)
of rotation be 90°. Draw a L shaped line from P and M as shown in P(5,6)

figure and then rotate PNM through 90° about M as shown in the

figure. The co-ordinates of image P' will be at (– 3, 4).
[90°, M(2, 1)]
i.e. P(5, 6) Rotation P'(– 3, 4)

X' O M(2,1) N X

Y'
Let us see the following table of some points on the plane and their corresponding images under the rotation
through 90° about M(2, 1).

SN Points in the plane Corresponding images Transformation

1. A(4, 5) A'(– 2, 3)

2. B(– 1,3) B'(0, – 2)

3. C(– 2, – 4) C'(7, – 3)

4. D(4, – 1) D'(4, 3)

5. Q(a, b) Q'(– b + 2 + 1, a – 2 + 1)

From the above table, we can see that the image of a point P(x, y) under the rotation through 90° about

M(a, b) is P(x, y) = [90°, M(a, b)] P'(– y + a + b, x – a + b)
Rotation

Rotation ~ 257

Rotation Through 180° About M(a, b)

Let us see the following table of some points on the plane and their corresponding images under the rotation
through 180° about M(– 3, 2).

SN Points in the plane Corresponding images

1. A(1, 4) A'(– 2, 3)

2. Q(– 2, 5) Q'(– 4, – 1)

3. R(– 4, – 2) R'(– 2, 6)

4. S(5, – 1) S'(– 11, 5)

5. T(a, b) T'(– a + 2 × – 3, – b + 2 × 2)

From the above table we can see that the image of a point P(x, y) under the rotation through 180° about

M(a, b) is P(x, y) = [180°, M(a, b)] P'(–  x + 2a, – y + 2b)
Rotation

Rotation Through 270° About M(a, b)

The image of point P(x, y) under the rotation through 270° (or 90°, 180°) about any point M(a, b) can be

derived by following procedure also.

Procedure: Y Y1

Step 1: Draw new co-ordinate axes X1MX1' and Y1MY1' P(x – a,y – b)
having origin at M(a, b) and parallel to original P(x,y)
axes.

Step 2: Find the co-ordinates of P(x, y) according to new 270°

co-ordinate axes X1' M(a,b) X1
i.e. P(x – a, y – b). X' O X

Step 3: Find the image of P(x – a, y – b) under the rotation P'(y – b,– x + a)
through 270° about origin M. P'(y – b + a,– x + a + b)

Y' Y1'

i.e. P'((y – b), – (x – a)) = P'(y – b, – x + a)

As, P(x, y) [270°, O(0, 0)] P'(y, – x)
Step 4: Rotation

Change the co-ordinates of image P' to the original co-ordinate axes i.e. P'(y – b + a, – x + a + b)

Hence, P(x, y) = [270°, M(a, b)] P'(y + a – b, – x + a + b)
Rotation

Conclusion About Rotation Using Co-ordinates

SN Object Centre Angle of Image point Short Symbol
Point of Point Rotation

1. P(x, y) O(0, 0) + 90° P'(x, – y) P(x, y) 9R0°o,tOat(i0o,n0)  P'(– y, x)

2. P(x, y) O(0, 0) 180° P'(– x, – y) P(x, y) 18R0o°,taOt(i0o,n0)  P'(– x, – y)

258 ~ Perfect Optional Mathematics Class 9

3. P(x, y) O(0, 0) 270° P'(y, – x) P(x, y) 27R0o°,taOt(i0o,n0) P'(y, – x)
4. P(x, y) M(a, b) + 90°
5. P(x, y) M(a, b) 180° P'(– y + a + b, P(x, y)  90°, M(a, b)  P'(– y + a + b, x – a + b)
6. P(x, y) M(a, b) – 90° x – a + b) Rotation
Example 6:
P'(– x + 2a, P(x, y)  18R0o°,taMti(oan, b)  P'(– x + 2a, – y + 2b)
– y + 2b)

P'(y + a – b, P(x, y) – 9R0o°,taMti(oan, b) P'(y + a – b, – x + a + b)
– x + a + b)

If the vertices of a quadrilateral ABCD are A(1, 2), B(3, 3), C(2, 5) and D(– 1, 5). Find the

co-ordinates of the vertices of the image of ABCD under the rotation through – 90° about M(– 2, 2).

 As A(1, 2), B(3, 3), C(2, 5) and D(– 1, 5) are the vertices of a Y

quadrilateral ABCD, the image of this quadrilateral can be easily DC

found by drawing it on the graph paper as shown where M(– 2, 2) is M(– 2,2) B
the centre of rotation and – 90° is the angle of rotation.
A' A
The co-ordinates of the vertices of image A'B'C'D' are A'(– 2, – 1), X' B' D' X
O
B'(– 1, – 3), C'(1, – 2) and D'(1, 1) from above figure. C'

The co-ordinates of the vertices of image A'B'C'D' can also be found Y'

by using formula

P(x, y) [– 90°, M(a, b)] P'(y + a – b, – x + a + b)
Rotation

Example 7:

A triangle with vertices A(3, – 2), B(5, – 2) and C(3, – 5) is rotated through an angle x° about point

(a, b). The images of A, B and C are A'(2, 3), B'(2, 5) and C'(5, 3) respectively. Find the values of

x, a and b using graph. Y Transformation
B'
 Given triangle ABC and its image A'B'C' are drawn in

same graph as shown. For the centre and the angle of

rotation,

1. Join A with A' and C with C' (or, B with B'). A' C'
2. Draw the perpendicular bisectors of AA' and CC'.

3. The perpendicular bisectors meet each other at origin, X' OX
hence the centre of roation is the origin. AB

4. Join OC and OC', COC' represents the angle of rotation.
and the angle of rotation is 90° in anticlockwise direction.

C
Y'

Rotation ~ 259

EXERCISE : 7.2A

1. Find the images of the following geometrical figures under the rotation with O as centre of rotation
and the given angle of rotation.

a. Angle of rotation = 60° b. Angle of rotation = – 70° c. Angle of rotation
= 90° clockwise direction
BA
O P

O

O S

A B C R
Q

d. Angle of rotation e. Angle of rotation = f. Angle of rotation
= half turn negative quarter turn = 80° in negative direction

O X AD P

O Q S
C
Y Z B O
R

2. If the following figures are the pair of object and image under the rotation, find the centre and the angle
of rotation in each case.

a. A b. B'
A' C Image C'
Object
Image Object A'
B
B
A

B'

3. Rotate the following figure through 90°about the origin in anticlockwise direction and write down the

co-oardinates of the image.

a. Y b. YC

CD

B

X' AB X' A X
OX O

Y' Y'
260 ~ Perfect Optional Mathematics Class 9

4. Rotate the following figure through 90°about any point (1, 2) in clockwise and anticlockwise derection

and write down the co-ordinates of the images.

a. Y b. YS

C R
M M

P

X' OA X X' O QX

B

Y' Y'
5. a. A point A(– 4, – 3) is rotated through – 90° (clockwise) about origin, find its image.

b. P(– 2, 7) and Q(– 4, 3) are the ends of a line segment PQ, find its image under the rotation through
270° about origin.

6. a. A point M(– 4, 2) is rotated through 180° about the point N(2, 1), find the co-ordinates of the
image of point of M by drawing in graph paper.

b. A point A(4, – 2) is rotated through 270° about M(2, 1), find the co-ordinates of the image of A
by drawing in graph paper.

7. Find the image of the point (2, – 8) under the following rotation.

a. Origin as centre of rotation and angle of rotation = 180°

b. M(2, 4) as centre of rotation and angle of rotation = 90°

c. N(– 1, 2) as centre of rotation and angle of rotation = 270°

8. Obtain the co-ordinates of the images of the following points due to the rotation about origin through
90° clockwise direction.

a. A(2, 4) b. (2, 8) c. M(– 1, 4)

d. N(– 3, – 4) e. P(– 4, 6) f. Q(a, b)

9. Obtain the co-ordinates of the images of the following points under the rotation through positive
quarter turn about C(– 2, 3).

a. P(3, – 4) b. Q(4, – 8) c. M(– 2, – 4)

d. N(4, 3) e. S(– 5, 5) f. A(p, q) Transformation

EXERCISE : 7.2B

1. a. If A(2, 5), B(4, 7) and C(7, 4) be the vertices of triangle ABC. Find the co-ordinates of the
vertices of image of ∆ABC under the rotation through 90° about origin. Draw ∆ABC and its
image on the same graph paper.

b. On the graph paper draw the rhombus ABCD with the vertices A(4, 6), B(2, 3), C(4, 0) and
D(6, 3). If the rhombus ABCD maps to A'B'C'D' by a rotation about origin through 180°, find the
co-ordinates A', B', C' and D'. Draw this image A'B'C'D' on the same graph paper.

Rotation ~ 261

2. If P(1, 2), Q(3, 4) and R(6, 1) are the vertices of triangle PQR, find the co-ordinates of the image of

∆PQR under the following rotation. Also draw ∆PQR and its image on the same graph paper in each case:

a. Positive quarter turn about origin. b. Negative quarter turn about origin.

c. Half turn about origin.

3. Quadrilateral ABCD with vertices A(2, 2), B(3, 0), C(7, 1) and D(5, 4) is rotated through following
angles about origin. Find the image A'B'C'D' by drawing on graph paper and find the coordinates of
the image in each case:

a. 90° b. 180° c. 270°

4. a. ∆MNS is rotated through 90° clockwise direction about A(2, 3), find the co-ordinates of the
vertices of its image under this rotation and draw DMNS and its image on the same graph paper.
Where coordinates of M, N and S are (– 3, 0), (– 5, 2) and (– 4, 5) respectively.

b. A(3, 1), B(3, – 4) and C(6, – 4) are the vertices of DABC. Let R be the rotation about origin

through a certain angle. If the image of A(3, 1) under R is A'(– 3, – 1), find the images of B and

C under same rotation.

5. A(0, 1), B(2, 4), C(5, 2) and D(6, 5) are the vertices of a parallelogram ABCD. Find the images of the
vertices of this parallelogram under the rotation about origin and the angle of rotation and direction as
mentioned below.

a. Quarter turn in clockwise direction, b. Quarter turn in anticlockwise direction.

c. Half turn.

6. A(3, 4), B(– 2, 1) and C(4, – 2) are the vertices of ∆ABC. When the vertices are rotated through certain
angle in a certain direction about origin, the co-ordinates of the image of A becomes A'(– 4, 3).

a. Find the angle of rotation and its direction

b. Using the same angle of rotation and direction, find the images of the points B and C.

7. a. If P(– 3, 2) and Q(– 1, 4) are the ends of a line segment PQ and the image of P and Q under the
rotation through certain angle and certain centre are P'(– 1, 2) and Q'(1, 0) respectively. Find the
centre of rotation and the angle of rotation with direction.

b. If ∆ABC and ∆A'B'C' are object and image under the rotation where A(2, 1), B(3, 3), C(5, 1) and
A'(1, 0), B'(– 1, 1) and C'(1, 3). Find the centre of rotation and angle of rotation.

8. If A(1, 0), B(1, 3) and C(– 1, – 1) are the vertices of ∆ABC, find

a. Image ∆A'B'C' and its vertices under the rotation of – 90° about M(2, 0).

b. Image ∆A"B"C" of ∆A'B'C' under the rotation of 180° about origin and also find the vertices of

∆A"B"C"

9. If P(2, – 1), Q(3, – 3) and R(5, – 2) are the vertices of ∆PQR, find

a. image ∆P'Q'R' and its vertices under the reflection on the line y = x.

b. image ∆P"Q"R" of DP'Q'R' and its vertices under the rotation through 270° about M(2, 3).

10. a. A triangle withe vertices P(2, 4), Q(6, 4) and R(4, 2) is rotated through angle of x° about point

(a, b) then the images of P, Q and R are P'(4, – 2), Q'(4, – 6) and R'(2, – 4) respectively. Find the

values of x, a and b using graph.

b. A triangle with vertices A(1, 2), B(3, 3) and C(2, 5) is rotated by an angle of y° about (c, d) then
the images of A, B and C are A'(– 2, – 1), B'(– 1, – 3) and C'(1, 2) respectively. Find the values of
y, c and d using graph.



262 ~ Perfect Optional Mathematics Class 9

7.3 Translation

Introduction

Translation is a transformation in which each point of the given object is displaced through definite distance

and direction. The displacement is defined by a vector h .
k

P A

Object R B D A'
R' D'
Q C
B' C'
Image P'

Object Image

Q'

The following are the properties of translation:
1. The object and the image under the translation are congruent.
2. The lines joining any point of object with its corresponding image are parallel and equal.

Example 1: Q
Find the image of ∆ABC under the translation through
translation vector →PQ . A

P
BC

 To get the image of ∆ABC under the translation vector →PQ A A' C' Transformation

draw the lines through A, B and C which are parallel and equal Q
to PQ in the same direction to get the images of A, B and C and B'
join them to get ∆A'B'C'.

P
BC

Translation Using Co– ordinates

The image of the geometrical figures under the translation through certain translation vector can be
obtained with the help of co-ordinates.

Translation ~ 263

Translation Through Translation Vector a Y
b
4 A(– 2, 3) 4 units
Let A(– 2, 3) be a point and – 5 be the translation vector. To
5 units
translate A(– 2, 3) through 4 , it has to move 4 units right and 5 X' X
– 5 O
A'(2, – 2)
units down. That is A has translated to A' through translation vector

AA' = 4 .
– 5

Let us see the following table of some points on the plane and their

corresponding images under the translation through vector 4 . Y'
– 5

SN Points in the plane Corresponding images

1. A(2, 4) A'(6, – 1)

2. B(– 4, 3) B'(0, – 2)

3. C(– 3, – 2) C'(1, – 7)

4. D(3, – 5) D'(7, – 10)

5. M(a, b) M'(a + 4, b – 5)

From above table we can see that the image of any point P(x, y) under the translation through translation

vector a is obtained by adding a to x co-ordinate and b to y co-ordinate of the given point.
b

i.e. P(x, y) = (a, b) P'(x + a, y +b)
Translation

x a x+a
y b y+b
i.e. P = Translation P'

Example 2:

If A(– 4, – 4), B(– 2, – 1) and C(– 1, – 5) are the vertices of a triangle ABC. Find the co-ordinates of

the image of ∆ABC under the translation T = 6 . Draw DABC and its image on the same graph paper.
5

 As A(– 4, – 4), B(– 2, – 1) and C(– 1, – 5) are the vertices of Y

DABC. The co-ordinates of the vertices of image of ∆ABC can B(4, 4)

be obtained by using formula

P(x, y) (a, b) P'(x + a, y + b) A
Translation
(2, 1)
Hence, X' X
B(– 2, – 1) O C(5, 0)

A(– 4, – 4) → A'(– 4 + 6, – 4 + 5) i.e. A'(2, 1)

B(– 2, – 1) → B'(– 2 + 6, – 1 + 5) i.e. B'(4, 4) A
C(– 1, – 5) → C'(– 1 + 6, – 5 + 5) i.e. C'(5, 0)
(– 4, – 4)

C(– 1, – 5) Y'

Drawing ∆ABC and ∆A'B'C' on the same graph paper we have figure as shown.

264 ~ Perfect Optional Mathematics Class 9

Example 3:

A(– 4, 2), B(0, 1) and C(– 2, – 3) are the vertices of ∆ABC. If A'(– 1, 6) be the image under the

translation of vertex A. Find the images B' and C' of B and C under the same translation.

 Here, as the image of A(– 4, 2) under the translation is A'(– 1, 6),

The translation vector T = A→A'

= O→A' – →OA

= – 1 –  – 4
6 2

= – 1 + 4 = 3
6 – 2 4

Now, under the same translation, the image of B(0, 1) is B'(0 + 3, 1 + 4).

i.e. B'(3, 5)

and the image of C(– 2, – 3) is C'(– 2 + 3, – 3 + 4) i.e. C'(1, 1).

Example 4:

Translate a point A(2, 7) under translation T1 = 4 to the point A'. Translate A' to A" under
– 3

another translation T2 = 2 . Find the translation vector which translates A to A".
4

 Here, the image of A(2, 7) under the translation T1 = 4 is A'(2 + 4, 7 – 3).
– 3
i.e. A'(6, 4)

Again the image of A'(6, 4) under the translation T2 = 2 is A"(6 + 2, 4 + 4).
4
i.e. A"(8, 8).

Let T be the translation vector which translate A(2, 7) to A"(8, 8).
Hence, T = A→A" = O→A" – →OA

= 8 –  2 = 8 – 2 = 6 .
8 7 8 – 7 1

Example 5: Transformation

Translate the point M(4, – 5) by the translation vector T = – 3 and find the image point M' and
4

write down the translation vector which maps M' back to M.

 Let M'(x1, y1) be the image point of the M(4, – 5) under the translation T = – 3 .
4
Therefore, M'(x1, y1) = M'(4 – 3, – 5 + 4) = M'(1, – 1)

The image point M' is M'(1, – 1)

Again, let T' = a be the translation vector which maps M' back to M.
b

Translation ~ 265

Now, (1 + a, – 1 + b) = (4, – 5)

i.e. 1 + a = 4 ∴ a=3

and – 1 + b = – 5 ∴ b = – 4

∴ Required translation vector is 3 which is called inverse of T and written as T– 1.
– 4

i.e. If T = a then T-1 = – a
b – b

EXERCISE : 7.3A

1. Translate the following geometrical figures in magnitude and direction of a given vector →a. Obtain the
translated images.

a. B C b. A D

→a

A D B →a
c. B A d. A
C

F

→a

CD BE

→a

EF CD

2. If the translation vector is defined by T = 2 , find the images of the following points under the
translation. – 4

a. A(2, 5) b. P(– 3, – 2) c. M(– 2, 6)

d. B(– 4, 3) e. C(4, – 7) f. F(0, 0)

3. a. What is the translation vector if point A(3, – 5) is translated to A'(– 1, 2).

b. If a translation vector maps A(– 4, 8) to A'(2, 3), find the translation vector.

c. Find the translation vector which maps P(x, y) to P'(x + a, y + b).

4. Translation – 4 maps triangle 1 onto triangle 2. Give the translation vector
– 1



4Triangle

3Triangle 1Triangle

2Triangle

a. Triangle 1 to triangle 3 b. Triangle 2 to triangle 4
c. Triangle 1 to triangle 4 d. Triangle 3 to triangle 2

266 ~ Perfect Optional Mathematics Class 9

5. a. Translate a point A(3, 5) under T1 = 2 to the point A'. Translate A' to A" under another
– 6

translation T2 = – 3 . Find the single translation which translates A to A".
– 2

b. Translate a point A(2, – 4) under the translation T = 4 to the point A'. Reflect A' on the line x = y
3

to the point A". Find the co-ordinates of A' and A".

EXERCISE : 7.3B

1. a. P(– 1, – 3) and Q(2, 1) are two points. With translation vector (4, – 2), the points P and Q are
translated to the points P' and Q'. Find the co-ordinates of P' and Q'. Draw the line PQ and P'Q'
on the same graph paper.

b. X(– 2, 2), Y(0, 5) and Z(4, 3) are the vertices of a triangle XYZ. Find the co– ordinates of the

vertices of image of ∆XYZ under the translation through the translation vector 2 . Draw
– 3

∆XYZ and its image on the same graph paper.

c. W(2, 2), X(4, 6), Y(7, 5) and Z(5, 1) are four points. Join WX, XY, YZ and ZW. Translate the
points W, X, Y and Z with translation vector (– 4, – 5). Find the corresponding images due to
translation. Show WXYZ and its image on the same graph.

2. a. Which is the translation vector if a point A(2, 6) is translated to the point A'(5, 9)? With the same
translation vector where will be the points B(4, 4) and C(– 2, – 5) translated?

b. Under the translation, ∆ABC with vertices A(1, – 1), B(2, – 3) and C(4, – 2) is translated to
∆A'B'C'. If after translation the co-ordinates of C' be (3, 3), which is the translation vector and
with the same translation vector, find the co-ordinates of A' and B'.

c. If a point P(4, 6) is translated to P'(0, – 3). If a triangle ABC with vertices A(0, 2), B(4, 1) and
C(2, 5) is translated through the translation vector →PP' , find the co-ordinates of the vertices of

∆A'B'C'.

3. If A(1, 0), B(7, 6) and C(3, – 2) are the vertices of ∆ABC, find the co-ordinates of the image after

translating it in the following magnitude and direction.

a. 12 →AB b. 2 →AC c. 41 →BC

4. If P(1, 1), Q(3, 1), R(5, 3), S(2, 3) are the vertices of a quadrilateral.

a. Find the co-ordinates of the vertices of the image P'Q'R'S' under the translation vector (1, – 4). Transformation

b. Find the co-ordinates of the vertices of the image P"Q"R"S" of P'Q'R'S' under the translation
vector (– 5, 3).

5. If A(1, – 1), B(3, – 2) and C(5, 1) are the vertices of ∆ABC

a. Find the vertices of image ∆A'B'C' under the translation with translation vector 2 .
– 3

b. Find the vertices of image ∆A"B"C" of ∆A'B'C' under the reflection on the line of x = – y.

c. Find the vertices of image ∆A'"B"'C"' of ∆A"B"C" under the rotation through – 90° about origin.



Translation ~ 267

7.4 Enlargement

Introduction

Enlargement (or reduction) is a transformation in which the size of an object is changed without changing
its shape if the size of object increase, we call it an enlargement and if the size of an object decrease, we
call it a reduction.
The enlargement is made with the help of a fixed point called centre of enlargement and the scale factor
i.e. the ratio of the corresponding sides of image and object. As enlargement changes the size of the
object, it will have no real sense in case of a point.

O

P D'
A'
Q Object R
AD
P'

Object

Image BC

Image

Centre Object Image Q' R' B' C'

The following are the properties of enlargement:

1. The object and the image under the enlargement are similar.

2. Scale factor (k) = Length Length of the side of image figure figure
of the corresponding side of object

3. If the scale factor k > 1, then the transformation is called enlargement.

4. If the scale factor 0 < k < 1, then the transformation is called reduction.

5. If the scale factor k = 1, then the transformation is identity.

6. If the scale factor k < 0, then the image will be on the opposite side of the object from centre of enlargement.

Example 1:

Find the image of ∆ABC under the enlargement with A C
centre of enlargement O and scale factor 2. [It is denoted by C'
E(O, 2)] B
O
 For the image ∆ABC under the given enlargement, perform
A'
the following steps.
Step 1: Join OA, OB and OC. A B'
Step 2: Produce OA upto A' such that OA' = 2OA. B C
Step 3: Produce OB upto B' such that OB' = 2OB. O
Step 4: Produce OC upto C' such that OC' = 2OC.
Step 5: Join A'B', B'C' and C'A'.

Hence, ∆A'B'C' is the required image of ∆ABC under the enlargement E(O, 2).

268 ~ Perfect Optional Mathematics Class 9

Example 2: P
Find the image of ∆PQR under the enlargement with centre
of enlargement O and scale factor – 21 .

QR

 As the scale factor negative i.e. – 12 , the image of ∆ABC will be O
P
on the opposite side of centre O and size will be half of given
Q R
figure. i.e. image of each side will be half of its corresponding R' O

object side. Q'

Step 1: Join OP, OQ and OR. P'

Step 2: Produce PO upto P' such that OP' = 21 OP.

Step 3: Produce QO and RO upto Q' and R' in the same direction
so that OQ' = 21 OQ and OR' = 12 OR.

Step 4: Join P'Q', Q'R' and R'P'.
Hence, ∆P'Q'R' is the image of ∆PQR under the given enlargement.

Enlargement Using Co– ordinates

The image under the enlargement with given centre of enlargement and scale factor can be obtained with

the help of co– ordinates.

Enlargement With Origin as Centre Y Q'(4, 8)

Let P(3, 1) and Q(2, 4) be two points and k = 2 be the scale factor

and O(0, 0) be the centre of enlargement. Join OP and OQ and Q(2, 4) Transformation
produce OP upto P' such that OP' = 2OP and produce OQ upto Q'

such that OQ' = 2OQ and join P'Q' which is the image of PQ. The P'(6, 2)

co-ordinates of P' and Q' are (6, 2) and (4, 8) respectively. X' P(3, 1) X

[O(0, 0), 2] O
Y'
i.e. P(3, 1) Enlargement P'(6, 2)

[O(0, 0), 2]

i.e. Q(2, 4) Enlargement Q'(4, 8)

Enlargement ~ 269

Lets us see the following table of some points on the plane and their corresponding images under the
enlargement with centre O(0, 0) and scale factor 2.

SN Points in the plane Corresponding images

1. P(2, 4) P'(4, 8)

2. Q(– 1, 3) Q'(– 2, 6)

3. R(– 2, – 3) R'(– 4, – 6)

4. S(2, – 5) S'(4, – 10)

5. M(a, b) M'(2a, 2b)

From the above table we can see that the image of any point P(x, y) under the enlargement with centre
O(0, 0) and scale factor 2 is

[O(0, 0), 2]

i.e. P(x, y) = Enlargement P'(2x, 2y)

Let us see the following table of some point in the plane and their corresponding images under the
enlargement with centre O(0, 0) and scale factor k.

SN Scale factor Points in the plane Corresponding images

1. 3 A(2, 4) A'(6, 12)

2. – 1 B(2, 3) B'(– 2, – 3)

3. – 2 C(– 4,1) C'(8, – 2)

4. – 1/2 D(2, 6) D'(– 1, – 3)

5. 3/2 E(– 2,4) E'(– 3, 6)

6. 2 F(2, 0) F'(4, 0)

7. –1/3 G(6, 3) G'(–2, – 1)

8. k M(a, b) M'(ka, kb)

From the above table we can see that the image of any point P(x, y) under the enlargement with centre
O(0, 0) and scale factor k is

[O(0, 0), k]

P(x, y) = Enlargement P'(kx, ky)

Example 3:

If ∆ABC with vertices A(2, 2), B(6, 2) and C(4, 4) is enlarged with the centre of enlargement O(0, 0)
3
and scale factor 2  . Find the co-ordinates on the vertices of the image of ∆ABC and draw ∆ABC

and its image of the same graph paper.

 As the ∆ABC has vertices at A(2, 2), B(6, 2) and C(4, 4) the image of these vertices under the
O(0, 0),32 
enlargement with centre O(0, 0) and scale factor 3 by using formula P(x, y) P'  3 x, 3 y
2 Enlargement 2 2

270 ~ Perfect Optional Mathematics Class 9

Hence, A(2, 2) → A' 3 ×2, 3 ×2 Y
2 2

i.e. A'(3, 3)

B(6, 2) → B' 3 × 6, 3 ×2 C'
2 2
C
i.e. B'(9, 3) A' B'
X
C(4, 4) → ' 3 × 4, 3 ×4 A B
2 2
X' O
i.e. C'(6, 6) Y'

Hence, co-ordinates of image ∆A'B'C' are A'(3, 3), B'(9, 3) and C'(6, 6). Drawing ∆ABC and ∆A'B'C' on the

same graph paper we get the figure as shown:

Enlargement With Centre Any Point

The image of point P(x, y) under the enlargement with centre M(a, b) and scale factor k can be derived by

the following procedure.
Procedure:

Step 1: Draw a new co– ordinate axes X1MX1' and Y1MY1' having origin at M(a, b) and parallel to
original axis.

Step 2: Find the co– ordinates of P(x, y) according to new Y Y1 P'(y – b + a,– x + a + b)
Step 3: co-ordinate axes i.e. P1'(x – a, y – b). P1'(y – b,– x + a)

Find the image of P(x – a, y – b) under the P1(x – a,y – b)
enlargement with centre M(a, b) (origin of new P(x,y)
axes) and scale factor k,

i.e. P1'[k(x – a), k(y – b)]. [O(0, 0), k] X1' M(a,b) X1
Step 4: O X
As P(x, y) Enlargement P'(kx, ky) X'
Y' Y1'
Change the co-ordinates of image P' to the original

co-ordinates axes

i.e. P'[k(x – a) + a, k(y – b) + b]

i.e. C'(6, 6) Transformation

Hence, the image of P(x, y) under the enlargement with centre M(a, b) and scale factor k is given by

P(x, y) = [M(a, b) k] P'[k(x – a) + a, k(y – b) + b]
Enlargement

Example 4:

If ∆PQR with vertices P(– 1, 2), Q(– 1, – 1) and R(1, 0) is enlarged with centre of enlargement
M(1, 1) and scale factor – 2. Find the co-ordinates of the vertices of the image of ∆PQR and draw
∆PQR and its image on the same graph paper.

Enlargement ~ 271

 As P(– 1, 2), Q(– 1, – 1) and R(1, 0) are the vertices of ∆PQR, Y

the image of these vertices under the enlargement with centre

M(1, 1) and scale factor – 2 by using formula Q'

[M(1, 1) – 2] R'
P
P(x, y) Enlargement P'{– 2(x – 1) + 1, – 2(y – 1) + 1}

Hence, X' OM X
P(– 1, 2) → P'{– 2(– 1 – 1) + 1, – 2(2 – 1) + 1} R
i.e. P'(5, –  1) Q P'
Q(– 1, – 1) → Q'{– 2(– 1 – 1) + 1, – 2(– 1 – 1) + 1}
i.e. Q'(5, 5) Y'

R(1, 0) → R'{– 2(1 – 1) + 1, – 2(0 – 1) + 1}

i.e. R'(1, 3)

The vertices of image ∆P'Q'R' under the given enlargement are P'(5, – 1), Q'(5, 5) and R'(1, 3).

Drawing ∆PQR and ∆P'Q'R' on the same graph paper, we have the above figure.

Example 5:

If A(– 1, – 1), B(0, – 3) and C(2, 0) are the vertices of ∆ABC and ∆A'B'C' is the image of ∆ABC

under the enlargement where A'(0, 2), B'(2, – 2) and C'(6, 4). Find the centre and scale factor of

this enlargement.

 As we have the co– ordinates of ∆ABC and its image ∆A'B'C' under the enlargement, we can find the

centre and the scale factor of enlargement by drawing.

Step 1: Draw ∆ABC and ∆A'B'C' with the help of the given Y C'
co-ordinates. A'

Step 2: Join AA' and BB' (or CC') and produce them to meet each
other at M.

Step 3: Co-ordinates of centre M is (– 2, – 4) and scale factor, X' C X
AO
A'B' MB'
k = AB = MA = 2. B'

B
M(– 2,– 4)

Hence, the required centre of enlargement is (– 2, – 4) and scale factor (k) = 2. Y'
Alternative Method:

Let (a, b) be the centre and k be the scale factor of enlargement E. Under the enlargement,

P(x, y) → P'{k(x – a) + a, k(y – b) + b}

Hence, A(– 1, – 1) → A'{k(– 1 – a) + a, k(– 1 – b) + b}

B(0, – 3) → B'{k(0 – a) + a, k(– 3 – b) + b}

C(2, 0) → C'{k(2 – a) + a, k(0 – b) + b}

But the vertices of image A'B'C' are A'(0, 2), B'(2, 2) and C'(6, 4)

So, k(– 1 – a) + a = 0 … … … (1)

272 ~ Perfect Optional Mathematics Class 9

k(0 – a) + a = 2 … … … (2)

k(2 – a) + a = 6 … … … (3)

and k(– 1– b) + b = 2 … … … (4)

k(– 3 – b) + b = 2 … … … (5)

k(0 – b) + b = 4 … … … (6)

Subtracting (1) from (2), we get

k=2

Putting the value of k in (1)

2 (– 1 – a) + a = 0

or, – 2 – 2a + a = 0

or, a = – 2.

Putting the value of k in (4)

2 (– 1 – b) + b = 2

or, – 2 – 2b + b = 2

or, b = – 4

Hence, the centre of enlargement is (– 2, – 4) and scale factor 2.

EXERCISE : 7.4A

1. Find the image of the following geometrical figures under the enlargement with centre O and scale
factor given in each case by drawing.

a. Scale factor = 2 b. Scale factor = 3
A 2
D
A

O B C BC
O

c. Scale factor = 1 d. Scale factor = – 12 Transformation
2 P

A S

B OE O
R
Q
Enlargement ~ 273
C D

2. Find the centre of enlargement and scale factor from the following pair of geometrical figures and their
respective images.

a. P' b. A D

A' D'

Q' P R' B' C'
Q B C

R

3. If A(4, 2) be a given point, find its image under the enlargement represented by E(O, k) in the following
cases.

a. E [O, 2] b. E  O, 1 c. E O, – 21 d. E O, 3
2 2

4. If P(3, – 4) is a given point. Find the image of point P under the following enlargements.

a. E M(3, 4), 1 b. E[M(1, 1), 2]
2

5. a. If A(2, 3) and B(– 1, 4) are two ends of a line segment AB. Find the images of A and B under the
enlargement with centre as origin and scale factor 2.

b. An enlargement E[C, 2] maps the point (3, – 1) to the point (5, 0). Find the co-ordinates of the
centre C.

c. What is the value of a and b when the point M(1, a) is enlarged with centre as origin and scale
factor – 2 to form an image M'(b, – 8)?

EXERCISE : 7.4B

1. a. If P(2, 5) and Q(– 3, 4) are the ends of line segment PQ, find the co-ordinates of the image P' and
Q' under the enlargement with centre at origin and scale factor – 2.

b. A(1, 3), B(4, 1) and C(3, 5) are the vertices of DABC, find the co-ordinates of the vertices of
image of DABC under the enlargement with centre O(0, 0) and scale factor 2.

2. P(– 4, 6), Q(– 2, 2) and R(0, 4) are the vertices of DPQR. Find the image of DPQR under the enlargement

with centre O(0, 0) and the given scale factor.

a. 2 b. – 1 c. 12 d. – 2

Draw DPQR and its image on the same graph paper in each case.

3. A(6, 3), B(6, 6), C(3, 3) and D(3, 0) are the vertices of a parallelogram ABCD. Under the enlargement

with centre as origin and the following scale factor, find the image of parallelogram ABCD.

a. 2 b. – 2 c. 31 d. – 13

Draw the graph of ABCD and its image on the same graph in each case.

274 ~ Perfect Optional Mathematics Class 9

4. A(1, 1), B(4, 1) and C(3, 3) are the vertices of ∆ABC. Find the co-ordinates of vertices of the image
of ∆ABC under the following enlargements.

a. Centre of enlargement (1, – 1) and scale factor 2.

b. Centre of enlargement (– 1, 2) and scale factor – 1.

Draw the object and respective image on the same graph in each case.

5. a. A(2, 5), B(– 1, 6), C(5, 4) and D(6, 3) are four points. Find the image of the quadrilateral ABCD
under the enlargement with centre at (1, 2) and scale factor 2.

b. P(2, 1), Q(3, 3), R(6, 4) and S(5, 2) are the vertices of a parallelogram PQRS. Find the co-ordinates
of the vertices of image of parallelogram PQRS under the enlargement with centre (1, 2) and
scale factor – 2 and draw PQRS and its image on the same graph paper.

6. a. With the centre (0, 0) and scale factor 2, if the vertices A, B and C of the triangle ABC have
mapped into the points A'(6, 2), B'(– 4, 4) and C'(4, 0) respectively. Find the vertices of the
triangle ABC. Present ∆ABC and ∆A'B'C' on the same graph.

b. With centre at (1, 0) and scale factor 2, if the vertices X, Y and Z of a triangle XYZ have mapped
into the point X'(5, 4), Y'(17, 4) and Z'(9, 12) respectively, find the vertices of the triangle XYZ.
Present ∆XYZ an ∆X'Y'Z' on the same graph.

7. a. If ∆A'B'C' is the image under the enlargement of ∆ABC where A'(2, 1), B'(5, 1) and C'(3, 3) are
the vertices of ∆A'B'C' and A(1, 3), B(7, 3) and C(3, 7) are the vertices of ∆ABC. Find the scale
factor and the centre of enlargement.

b. An enlargement maps the ∆ABC with the vertices A(2, 1), B(2, 4) and C(5, 3) onto the DA'B'C'
with the vertices A'(– 2, 1), B'(– 2, 10) and C'(7, 7). Find the centre and the scale factor of
enlargement.

8. If A(2, 1), B(2, 4) and C(5, 2) are the vertices of DABC.

a. Enlarge ∆ABC with centre at (3, 5) and scale factor 2 to get ∆A'B'C' and find the co-ordinates of

A', B' and C'. 1
2
b. Enlarge ∆A'B'C' with centre at (– 3, 1) and scale factor to get ∆A"B"C" and find the co-ordinates

of A", B" and C".

Also draw DABC, DA'B'C' and DA"B"C" on the same graph.

9. If P(– 5, – 2), Q(– 1, – 2) and R(– 3, – 4) be the vertices of ∆PQR 1
2
a. Find the image ∆P'Q'R' under the enlargement with centre at (– 1, 0) and scale factor and find Transformation

the co-ordinates of P', Q' and R'.

b. Translate ∆P'Q'R' with translation vector 5 to get ∆P"Q"R" and find the co-ordinates of P", Q"
4

and R".

c. Rotate ∆P"Q"R" through 90° about M(– 1, 2) to get ∆P1Q1R1 and find the co-ordinates of P1, Q1
and R1.

d. Reflect ∆P1Q1R1 on the line x = y to get ∆P2Q2R2 and find the co-ordinates at P2, Q2 and R2.
Also draw ∆PQR, ∆P'Q'R', ∆P"Q"R", ∆P1Q1R1 and ∆P2Q2R2 on the same graph paper.



Enlargement ~ 275

8 STATISTICS

Curriculum Contents Included Periods
12
a. Partition Values

- Quartiles, Deciles, Percentiles (Individual and discrete series only)
b. Dispersion
- Quartile deviation and its co-efficient
- Mean deviation (from mean and median) and its co-efficient (individual

and doscrete series only)
- Standard deviation and its co-efficient (individual and discrete series

only)

Objectives

On completion of this unit, students will be able to
find the quartiles, deciles and percentiles of individual as well as discrete series

define dispersion and find the quartile deviation and its co-efficient of the given data.

find the mean deviation from mean and median and its co-efficient (individual and discrete
series)

find the standard deviation and its co-efficient (individual and discrete series)

Materials

2x + 2, 3x – 1, 2x + 11, 4x – 7, 34, 36 and 49 are in ascending order. If the first quartile (Q1) is 29,
what will be the value of x?

The given data in ascending order are 2x + 2, 3x – 1, 2x + 11, 4x – 7, 34, 36 and 49

Here, number items (n) = 7
th th th
n + 1 7 + 1 8
Now, Q1 = 4 = 4 = 4 = 2nd item

Hence, Q1 = 2nd item

29 = 3x – 1

or, 3x = 29 + 1

or, 3x = 30

∴ x = 10.

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Statistics Each of 1 mark Each of 5 marks Questions Marks
Each of 2 marks Each of 4 marks
– – 3 10
12

276 ~ Perfect Optional Mathematics Class 9

8.1 Partition Values

Partition Values (Measures of Positions)

The values that divide the given statistical data into number of equal parts are called the partition values.
The partition values are (i) median (ii) quartiles (iii) deciles and (iv) percentiles.

Median

Median of a distribution is the value of the variable which divides the distribution into two equal parts after

arranging them into ascending or descending order. It is denoted by Md.

Median of Individual Series

n + 1 th
2
If the number of observation is odd, the median is item.

n th n th
2 2
If the number of observations is even, the median is the average of and + 1 items.

n th n th
2 2
item + + 1 item

i.e. Median (Md) = 2 Median (Md)

Example 1:

Find the median of 2, 7, 14, 6, 8, 10, 9, 8, 7.

 Arranging the given observations in ascending order:

2, 6, 7, 7, 8, 8, 9, 10, 14
The number of observations n = 7.

n + 1 th
2
Median = item i.e. 5th item.

∴ The median (Md) = 8.

Example 2:

Find the median of 10, 12, 13, 8, 9, 15.

 Arranging the given observations in ascending order, we have,

8, 9, 10, 12, 13, 15.
The number of observations n = 6.

n th n th
2 2
Median is the average of and + 1 items. Statistics

n th n th 6 th 6 th
2 2 2 2
item + 2 + 1 item item + 2 + 1 item 3rd item + (3 + 1)th item
2
Median (Md) = = =

= 3rd item + 4th item = 10 + 12 = 22 = 11.
2 2 2

Partition Values ~ 277

Alternative Method

n + 1 th 6 + 1 th 7 th
2 2 2
Median = item = item = = 3.5th item (Lies inside 3rd item and 4th item)

= 3rd item + 0.5 (4th item – 3rd item)

= 10 + 0.5(12 – 10) [here, 4th item = 12 and 3rd item = 10]

= 10 + 0.5 × 2

= 10 + 1 = 11

Example 3:

If 7, x + 3, 2x – 5, 2x , 2x + 3, 2x +7, 3x and 40 are in the ascending order with median 21.5, find

the value of x.

 Here, given data is 7, x + 3, 2x – 5, 2x, 2x + 3, 2x + 7, 3x, 40 and the number of items (n) = 8

Median = n + 1 th = 8 + 1 th = 9 th = 4 21 th = 4.5th item
2 2 2
item item item item

(lies exactly midway between 4th item and 5th item)

Hence, 21.5 = 4th item + 0.5 (5th item – 4th item)

or, 21.5 = 2x + 0.5 (2x + 3 – 2x) [here, 5th item = 2x + 3 and 4th item = 2x])
or, 21.5 = 2x + 0.5 × 3
or, 21.5 = 2x + 1.5
or, 2x = 21.5 – 1.5 = 20

∴ x = 10.

Median of a Discrete Series

In case of discrete frequency distribution, the median is the variable corresponding to the cumulative

frequency equal to or just greater than N + 1.
2

Example 4:

Find the median of the following data:

Marks (x) 10 15 20 25 30

No. of students (f) 6 11 17 13 8

 From the given infformation, we can calculate the median as folloiw.

Computation of Median

Ages No. of Students (f) Cumulative frequency (c. f.)

10 6 6

15 11 17

20 17 34

25 13 47

30 8 55

Total N = 55

278 ~ Perfect Optional Mathematics Class 9

Here, N + 1 = 55 + 1 = 56 = 28.
2 2 2

Cumulative frequency equal to or just greater than 28 is 34. Hence, the median is corresponding value of

c.f. 34.

∴ Md = 20.

Quartiles

Median divides the total number of observations into two halves. In the same way, quartiles divide the total
number of observations into 4 equal parts.

Lower Half Upper Half

Q1 Md Q3
8, 10, 11, 14, 15, 17, 19, 21, 23, 24, 25, 25, 26, 27, 27, 28, 30, 35, 38

25%
50%
75%

The single value dividing the lower half into two equal parts is said to be lower quartile or the first quartile
and is denoted by Q1. The single value dividing the whole data into two equal parts is said to be the second
quartle and is denoted by Q2. Similarly, the single value dividing the upper half into two equal parts is
known as upper quartile or the third quartile and is denoted by Q3.

1st quarter 2nd quarter 3rd quarter 4th quarter

Q1 Q2 Q3

Thus, there are three quartiles Q1, Q2 (= Md) and Q3.
For example :
Lower half Upper half

9, 12, 15, 16, 20, 21, 23

Q1 Q2 = Md Q3

Example 5:

The marks obtained by 11 students of class IX in mathematics test of full marks 50 are : 42, 33, 18,

25, 26, 37, 20, 17, 19, 35, 36. Find the quartile marks. Statistics

 Arranging the given data in ascending order, we get,

17, 18, 19 , 20, 25, 26 , 33, 35, 36 , 37, 42
Q1 Q2 Q3

Here n = 11, therefore,

n + 1 th 11 + 1 th
4 4
∴ Q1 = item = item = 3rd item = 19

Partition Values ~ 279

2(n + 1) th n + 1 th 11 + 1 th
4 2 2
∴ Q2 = item = item = item = 6th item = 26

3(n + 1) th 3( 11 + 1) th
4 4
∴ Q3 = item = item = 9th item = 36

Example 6:

Find Q1 , Q2 and Q3 of the following data :
22, 14, 25, 20, 18, 17, 11, 23, 26, 16

 Arranging the given data in ascending order, we get,

11, 14, 16, 17, 18, 20, 22, 23, 25, 26

Here, number of items (n) = 10

Q1 = 1(n + 1) th Q2 (Md) = n + 1 th
4 2
item item

1(10 + 1) th 11 th
4 4
= item = item 10 + 1 th 11 th
2 2
= item = item
= 2.75th item (lies inside 2nd and 3rd item)

=2th item + 0.75(3rd item – 2th item) = 5.5th item [lies inside 5th and 6th item]

= 14 + 0.75(16 – 14) =5th item + 0.5(6th item – 5th item)

[here, 3rd item = 16 and 2nd item = 1 4] = 18 + 0.5(20 – 18)

= 14 + 0.75 × 2 [here, 6th item = 20 and 5th item = 18]
= 14 + 1.5 = 18 + 0.5 × 2

= 15.5 = 18 + 1

Q3 = 3(n + 1) th = 3(10 + 1) th = 19
4 4
item item

= 8.25th item [lies inside 8th and 9th item]
= 8th item + 0.25(9th item – 8th item)
= 23 + 0.25(25 – 23) [here, 9th item = 25 and 8th item = 23]
= 23 + 0.25 × 2
= 23 + 0.5
= 23.5

Example 7:

2x + 7, 3x + 5, 2x + 14, 5x – 4, 39, 41 and 54 are in ascending order. If the first quartile (Q1)

is 29, what will be the value of x?

 The given data is 2x + 7, 3x + 5, 2x + 14, 5x – 4, 39, 41 and 54

Here, number items (n) = 7

n + 1 th 7 + 1 th 8 th
4 4 4
Now, Q1 = = = = 2nd item

280 ~ Perfect Optional Mathematics Class 9

Hence, Q1 = 2nd item
29 = 3x + 5
or, 3x = 29 – 5
or, 3x = 24
∴ x = 8.

Example 8:

The marks obtained by 59 pupils in a test of full marks 100 are given below. Calculate the quartiles.

Marks (x) 0 10 20 30 40 50 60 70 80 90 100

No. of students (f) 3 5 7 9 12 6 5 4 4 3 1

 To calculate the quartiles, we construct the cumulative frequency table as follows.

Marks Frequency (f) Cumulative frequencies

0 3 3
10 5 8
20 7 15
30 9 24
40 12 36
50 6 42
60 5 47
70 4 51
80 4 55
90 3 58
100 1 59

Total 59

Here, N = 59.

Therefore, the first quartile Q1 = N+1 th
4
value

59 + 1 th
4
= value

= 15th value

From above table, cumulative freqency equal to or just greater than 15 is 15.

i.e. The row with c.f. 15 contains 15th value.

Hence, Q1 = 20

2(N + 1) th Statistics
4
Q2 = value

N + 1 th 59 + 1 th
2 2
= value = value = 30th value

From above table, cumulative frequency equal to or just greater than 30 is 36.

Partition Values ~ 281

i.e. The row with c.f. 36 contains the 30th value. Q2 = 40.

3(N + 1) th 3 × 60 th
4 4
Q3 = value = value = 45th value.

From above table, cumulative frequency equal to or just greater than 45 is 47.

i.e. The row with c.f. 47 contains the 45th value.

∴ Q3 = 60.
∴ Q1 = 20, Q2 = 40 and Q3 = 60.

Deciles

Deciles of a set of data are the items which divide the total number of items into ten equal parts. Hence,
there are nine deciles, the first decile (D1), the second decile (D2), the third decile (D3) and so on.

D1 D2 D3 D4 D5 D6 D7 D8 D9

In case of individual series and discrete series, the values of the deciles can be calculated as following after
arranging them in ascending or descending order.

i(n + 1) th
10
Di = the value of item, i = 1, 2, 3, 4, 5, 6, 7, 8, 9 and n = number of items.

Thus, D1 = the value of 1(n + 1 th item
10
2(n + 1 th
D2 = the value of 10 item

3(n + 1) th
10
D3 = the value of item

... ... ... ...

9(n + 1) th
10
D9 = the value of item

Example 9:

Calculate the values of D3 and D7 from the following data.

20, 22, 24, 17, 14, 29, 31, 37, 45, 21, 29, 42, 44, 19, 22

 Arranging the given data in ascending order,

14, 17, 19, 20, 21, 22, 22, 24, 29, 29, 31, 37, 42, 44, 45

Here, n = 15 and the third decile is

D3 = 3(n + 1) th = 3(15 + 1) th = 48 th = 4.8th item
10 10 10
item item item

D3 = 4.8th item [lies inside 4th item and 5th item]

=4th item plus 0.8 times difference between 5th item and 4th item.

=4th item + 0.8 (5th – 4th item)

= 20 + 0.8 (21 – 20) [Here, 5th item = 21 and 4th item = 20]

= 20 + 0.8 × 1

282 ~ Perfect Optional Mathematics Class 9

= 20 + 0.8

= 20.8

D7 = 7(n + 1) th = 7(15 + 1) th = 112 th = 11.2th item
10 10 10
item item item

D7 = 11.2th item [lies inside 11th item and 12th item]

=11th item + 0.2 times difference between 12th item and 11th item.

=11th item + 0.2 (12th item – 11th item)
= 31 + 0.2 (37 – 31) [Here, 12th item = 37 and 11th item = 31]
= 31 + 0.2 × 6
= 31 + 1.2
= 32.2

Example 10:

Calculate the values of D1, and D7 from the following data.

Marks (x) 5 10 15 20 25 30

No. of students (f) 268762

 From the given information, we can calculate the decile as follows.

Calculation of the Deciles

Marks Number of students (f) Cumulative Frequencies (cf)

52 2
10 6 8
15 8 16
20 7 23
25 6 29
30 2 31

Total N = 31

Here, N = 31,

D1 = 1(N + 1) th = 31 + 1 th = 32 th = 3.2th item
10 10 10
item item item

From above table, cumulative frequency equal to or just greater than 3.2 is 8.

i.e. The row with c.f. 8 contains 3.2th item.

TNhoetef:irst decile is D1 = value of variate corresponding to c.f. 8 = 10. Statistics

Arranging the above data (marks) in ascending order, we have
5, 5, 10, 10, 10, 10, 10, 10, ... ...
D1 = 3.2th item = 3th item + 0.2 (4th item – 3th item)
= 10 + 0.2 (10 – 10)

= 10

7(N + 1) th 7(31 + 1) th 224 th
10 10 10
D7 = item = item = item = 22.4th item

Partition Values ~ 283

From above table, cumulative frequency equal to or just greater than 22.4 is 23.

i.e. The row with c.f. 23 contains 22.4th value.

TNhoetes:eventh decile is D7 = value of variate corresponding to c.f. 23 = 20.

Arranging the above data (marks) in ascending order, we have
5, 5, 10, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 15, 15, 15, 20, 20, 20, 20, 20, 20, 20, 25, 25, 25, 25, 25, 25, 30, 30.
D7 = 22.4th item = 22th item + 0.4 (23th item – 22th item)
= 20 + 0.4 (20 – 20)

= 20

Percentiles

Percentiles of a set of data are the items which divide the total number of items into hundred equal parts.

Hence, there are ninety nine percentiles, the first percentile (P1), the second percentile (P2), the third
percentile (P3) and so on.

P1 P8 P16

P10 P20 P30 P40 P50 P60 P70 P80 P90

In case of individual series and discrete series, the values of the percentiles can be calculated as following

after arranging them in ascending or descending order.

i(n + 1) th
100
Pi = the value of item, i = 1, 2, 3, ... ... , 99.

1(n + 1) th
100
Thus, P1 = the value of item

2(n + 1) th
100
P2 = the value of item

... ... ... ....

10(n + 1) th
100
P10 = the value of item

... ... ... ....

99(n + 1) th
100
P99 = the value of item

Example 11:

Calculate the values of P13 and P75 from the following data.

16, 18, 20, 13, 10, 25, 27, 33, 17, 25, 38, 40, 15, 18

 Arranging the given data in ascending order,

10, 13, 15, 16, 17, 18, 18, 20, 25, 25, 27, 33, 38, 40

Here, n = 14 and the thirteen percentile is

P13 = 13(n + 1) th item
100

13(14 + 1) th 13 × 15 th 195 th
100 100 100
= item = item item = 1.95th item

284 ~ Perfect Optional Mathematics Class 9

=1st item + 0.95 (2nd item – 1st item)
= 10 + 0.95(13 – 10) [Here, 2nd item = 13 and 1st item = 10]
= 10 + 0.95 × 3
= 10 + 2.85
= 12.85

75(n + 1) th
100
P75 = item

75(14 + 1) th 75 × 15 th 1125 th
100 100 100
= item = item = item = 11.25th item

= 11th item + 0.25 (12th item – 11th item)

= 27 + 0.25(33 – 27) [Here, 12th item = 33 and 11th item = 27]
= 27 + 0.25 × 6
= 27 + 1.5
= 28.5

Example 12:

Calculate the values of P37 and P83 from the following data.

Marks (x) 5 10 15 20 25 3 35 40 45 50

No. of students (f) 10 15 18 22 30 35 26 18 15 8

 From the given information, we can calculate the percentile as follows.

Calculation of the Percentile

Marks Number of students (f) Cumulative Frequencies (cf)

5 10 10
10 15 25
15 18 43
20 22 65
25 30 95
30 35 130
35 26 156
40 18 147
45 15 189
50 8 197

Total N = 197

Here, N = 197, Statistics

37(N + 1) th
100
p37 = item

37(197 + 1) th 37 × 198 th 7326 th
100 100 100
= item = item = item = 73.26th item

Partition Values ~ 285

From above table, cumulative frequency equal to or just greater than 73.26 is 95.

The 37th percentile is P37 = value of variate corresponding to c.f. 95 = 25.

83(N + 1) th
100
P83 = item

83(197 + 1) th 83 × 198 th 16434 th
100 100 100
= item = item = item = 164.34th item

From above table, cumulative frequency equal to or just greater than 164.34 is 174.

The row with c.f. 174 contain 164.34th item.

The 83rd percentile is P83 = value of variate corresponding to c.f. 174 = 40.

EXERCISE : 8.1A

1. a. If the numbers of items is 11, then find the postition of Q1, Q2 and Q3.
b. If the numbers of iterms is 9, then find the postition of Q1, Q2.
2. a. If the numbers of items is 11, then find the postition of D2.
b. If the numbers of items is 35, then find the postition of D2.
3. a. Out of a total of 51 observations arranged in ascending order, if 13th observation is 30 then find Q1.
b. Out of a total of 44 observations arranged in ascending order, if 9th observation is 66 then find P20.
4. Find the median of the following data.

a. 7, 6, 8, 10, 22, 7, 6, 2, 4, 8, 3 b. 50, 67, 80, 29, 70, 13, 44, 65, 71, 66, 50, 34, 82

5. Find the median of the following data.

a. 7, 4, 10, 16, 13, 16, 22, 14, 8, 9 b. 40, 52, 62, 29, 72, 39, 67, 85, 92, 71, 89, 40

6. Calculate the first quartile (Q1) and the third quartile (Q3) from the following data.

a. 2, 5, 7, 10, 8, 7, 12, 14, 9, 4, 11 b. 13, 26, 24, 19, 17, 15, 12, 17, 19, 18, 23, 10, 18, 17

7. Find the deciles D3, D5 and D9 of the following data.
a. 34, 37, 52, 25, 62, 50, 44, 48, 67, 53 b. 14, 12, 18, 23, 13, 33, 39, 40, 29, 23, 28, 30, 32

8. Calculate the percentiles P17, P23, P43 and P78 of the following data.
a. 5, 10, 25, 14, 18, 15, 17, 26, 16, 19, 27 b. 43, 23, 40, 20, 34, 33, 48, 22, 26, 39, 42, 46, 39, 26

9. Calculate the third quartle and sixth decile from the following data.

a. 4, 8, 4, 10, 12, 9, 9, 14, 11, 10, 12, 18 b. 13, 19, 19, 24, 30, 28, 33, 40, 18, 17, 23, 29, 10

10. Calculate the seventh decile, fifty sixth percentile of the following data.

a. 25, 14, 16, 20, 13, 29, 23, 16, 17, 28, 33, 26

b. 9, 7, 6, 4, 12, 16, 11, 10, 8, 5, 13, 17, 20, 25

11. a. 7, 12, x + 5, 2x – 3, 2x, 3x – 3 and 47 are in ascending order. If its median is 21, find the value of x.

b. x + 10, 2x + 6, 3x, 2x + 13, 3x + 10 and 50 are in ascending order. If its median is 34, find the
value of x.

12. a. 2x + 3, 3x + 1, 3x + 4, 5x – 1, 52, 54 and 62 are in ascending order. If the first quartile (Q1) is 22,
what will be the value of x?

286 ~ Perfect Optional Mathematics Class 9

b. 8, 3x – 1, 4x – 2, 5x – 4, 5x, 7x – 2 and 28 are in ascending order. If the third quartile is 26, what
will be the value of x?

c. 14, 19, 2x + 4, 2x + 5, 3x – 4, 28, 3x + 2 and 34 are in ascending order. If the third quartile is 31,
what will be the value of x?

EXERCISE : 8.1B

1. Calculate the median form the following data.

a. Wages (Rs.) 200 225 250 260 300 320
7 20 8 3
No. of workers 8 5 29
20 2
b. Marks obtained 12 13 15 10 22 25 27
22
No. of students 3 8 9 12 8 3 2
1220
2. Calculate the the first quartile (Q1) and the third quartile (Q3) from the following data. 2
a.
37
x 10 12 14 16 18 20 8
75
f 8 12 16 25 12 8 5

b. 930 950 1000 1050 1100 1200 155
Area (sq. ft.) 2
80
No. of houses 8 10 12 7 6 4 8

3. Calculate the the third decile (D3) and the eighth decile (D8) from the following data.
a. Marks
25 27 30 32 35

No. of students 3 6 15 12 10

b. Marks 55 60 62 67 70 73

No. of persond 13 15 17 20 18 10

4. Calculate the percentiles P21, P43 and P89 from the data given below.

a. Height (cm) 135 140 142 145 150 152

No. of students 3 8 10 12 7 3

b. Marks obtained 10 20 30 40 50 60 70

No. of students 13 28 29 40 32 28 11

5. Calculate the values of D7, D9, D27 and D71 from the following data.
a. Marks
12 15 20 25 28 30 35
16 9 5
No. of student 8 12 50 55 Statistics
15 16 18
b. Age (months) 7 8 9 10 12 30 12 3

No. of cocks 5 10 12 15 18



Partition Values ~ 287

8.2 Dispersion

Quartile Deviation

For quartile deviation, let us have an idea about quartiles. We know that the median divides the given
observations, arranged in the ascending or descending order into two equal parts. Similarly, the quartiles
divide the given observations into four equal parts after arranging them in the ascending or

descending order. So there are three quartiles denoted by Q1, Q2 and Q3 known as the first, the second and
the third quartile respectively.

The difference between the third and the first quartiles is called inter quartile range.

So, Inter Quartile Range = Q3 – Q1

The average of deviation of the first quartile and the third quartile taken from the median is called quartile
deviation or semi-interquartile range.

So, Quartile Deviation (Q.D.) = (Q3 – Md) + (Md – Q1) = Q3 – Q1
2 2

∴ Q.D.= Q3 – Q1
2

As the units of quartile deviations is same as that of the given observation, the two distributions with

different units cannot be compared by quartile deviation. To facilitate the comparison of the quartiles of two

or more than two series, a relative measure, the co-efficient of quartile deviation is used where,

∴ Co-efficient of Quartile Deviation = Q3 – Q1 = Q3 – Q1
2 Q3 + Q1

Q3 + Q1

2

∴ Co-efficient Q. D. = Q3 – Q1
Q3 + Q1

Example 1:

Calculate the quartile deviation and its co-efficient of the the following data:

24, 32, 46, 48, 39, 42, 28, 25, 26, 24, 38.

 Arranging the given data in the ascending order, 24, 24, 25, 26, 28, 32, 38, 39, 42, 46, 48.

Number of observations (n) = 11

n + 1 th
4
First quartile (Q1) = item

11 + 1 th
4
or, Q1 = item

or, Q1 = 3rd item
∴ Q1 = 25

288 ~ Perfect Optional Mathematics Class 9

3(n + 1 th
4
Third quartile (Q3) = item

or, Q3 = 3 × 12 th
4
item

or, Q3 = 9th item

∴ Q3 = 42.

Quartile deviation (Q.D.) = Q3 – Q1 = 42 – 25 = 17 = 8.5
2 2 2

Co-efficient of Q.D. = Q3 – Q1
Q3 + Q1

= 42 – 25 = 17 = 0.25
42 + 25 67

Hence, Q.D. = 8.5 and co-efficient of Q.D. = 0.25

Example 2:

Calculate the quartile deviation and its co-efficient for the following data:

12, 4, 8, 5, 14, 7

 Arranging the given data in the ascending order, 4, 5, 7, 8, 12, 14

Number of observations (n) = 6

n + 1 th
4
First quartile (Q1) = item

6 + 1 th
4
or, Q1 = item

or, Q1 = 1.75th item
or, Q1 = 1st item + 0.75 (2nd item – 1st item)
= 4 + 0.75 × 1

= 4.75

3(n + 1 th
4
Third quartile (Q3) = item

or, Q3 = 3 × 7 th
4
item

or, Q3 = 5.25th item

or, Q3 = 5th item + 0.25(6th item – 5th item)

= 12 + 0.25 × 2 = 12.5 Statistics

Quartile deviation (Q.D.) = Q3 – Q1 = 12.5 – 4.75 = 3.87
2 2

Co-efficient of Q.D. = Q3 – Q1 = 12.5 – 4.75 = 7.75 = 0.45
Q3 + Q1 12.5 + 4.75 17.25

Hence, quartile deviation = 3.87 and its co-efficient = 0.45

Dispersion ~ 289

Example 3:

Calculate the quartile deviation and its co-efficient from the following data: Algebra

Ages 12 13 14 15 16 17 18

No. of Students 12 21 15 20 17 10 5

 From the above information, we can find the quartile deviation as follows.

Computation of Quartile Deviation

Ages(in yrs) (x) No. of Students (f) Cumulative frequency (c. f.)
12
12 21 12
13 15 33
14 20 48
15 17 68
16 10 85
17 5 95
18 100
N = 100

n + 1 th 100 + 1 th
4 4
First quartile = item = item = 25.25th item

Cumulative frequency just greater than 25.25 is 33. So the corresponding value i.e. 13 is the first quartile.

∴ Q1 = 13

Third quartile = 3(N + 1 th 3 × 101 th = 75.75th item
4 4
item = item

Cumulative frequency just greater than 75.75 is 85. So the corresponding value i.e. 16 is the third quartile

∴ Q3 = 16

Quartile deviation (Q.D.) = Q3 – Q1 = 16 – 13 = 1.5
2 2

Co-efficient of Q.D. = Q3 – Q1 = 16 – 13 = 3 = 0.103
Q3 + Q1 16 + 13 29

Hence, quartile deviation = 1.5 and its co-efficient = 0.103

Mean Deviation

Range and quartile deviation are based on two points of observations only. There are other measures of
dispersion such as mean deviation, standard deviation, which take into account of all data of distribution.

Mean Deviation or Average Deviation

Mean deviation or average deviation is defined as the arithmetic mean of the deviations of the items from
mean or median or mode. Since median is the central point of any distribution, mean deviation from median
generally gives the best result among the average deviations. Since mode is ill defined, mean deviation from
mode will generally not be used.

290 ~ Perfect Optional Mathematics Class 9

Individual Series

In case of individual series, mean deviation is calculated as follows:

Mean deviation (M.D.) = ∑|x – A| , where A = x–  or Md or Mo and |x – A| is read as modulus of x – A or
n

absolute value of x – A. That is magnitude of (x – A) or the value of deviation taken from average A ignoring

the negative sign.

Discrete Series

In case of discrete and continuous series, mean deviation can be calculated by the formula.

Mean Deviation (M.D.) = ∑f |x – A | where A = x– or Md or Mo and N = ∑f is total number of observations.
N ,

The relative measure of mean deviation or the co-efficient of mean deviation is defined as:

Co-efficient M.D. = M. D. = M. D.
Average A

Where A is the average from which mean deviation is calculated.

Example 4:

Calculate the mean deviation and its co-efficient from mean and median of the following series.
24, 28, 29, 33, 36, 35, 25.

 Arranging the given data in the ascending order, 24, 25, 28, 29, 33, 35, 36

Here, the numbers of observations (n) = 7 and sum of all observations ∑x = 210

Hence, arithmetic mean (x– ) = ∑x = 210 = 30
n 7

n + 1 th
2
Median Md = item

7+ 1 th
2
= item

=4th item = 29

Computation of Mean Deviation

M.D. from the mean M.D. from the median
x –x  x – x–  |x – –x |
x Md x – Md |x – Md|

24 – 6 6 24 – 5 5

25 – 5 5 25 – 4 4

28 – 2 2 28 –1 1

29 30 – 1 1 29 29 0 0 Statistics

33 3 3 33 4 4

35 5 5 35 6 6

36 66 36 7 7
∑x = 210 ∑|x – –x |
∑|x – Md|
= 28 = 27

Dispersion ~ 291

Now, mean deviation from mean M.D. = ∑|x – x– | = 28 = 4
n 7

Co-efficient of M.D. = M–x.D  . = 4 = 0.12 and mean deviation from median = 27 = 3.86
30 7

Co-efficient of M.D. = M.D. = 3.86 = 0.13
Md 29

Example 5:

Find the mean deviation from mean and its co-efficent of the following data.

Age in yrs 15 18 20 22 23 25
9
No. of People 5 8 10 12 15

 From the above information, we can find the mean deviation as follows.

Computation of Mean Deviation

Ages (yrs) No. of fx x–  x – –x  |x – –x | f|x – x– |
(x) people (f) 75
31.20
15 5 – 6.24 6.24 25.92
12.40
18 8 144 – 3.24 3.24 9.12
26.40
20 10 200 21.24 – 1.24 1.24 33.84
22 12 264 0.76 0.76
∑f|x – –x | 
23 15 345 1.76 1.76
= 138.88
25 9 225 3.76 3.76

N = 59 ∑fx =
1253

Here, N = 59, ∑fx = 1253, Mean (–x ) = ∑f x = 1253 = 21.24, ∑f|x – –x | = 138.88
N 59

Hence, mean deviation from mean (MD) = ∑f |x – –x | = 138.88 = 2.35
N 59

And the co-efficient of mean deviation from mean = M–xD  = 2.35 = 0.11
21.24

Example 6:

Find the mean deviation from median and its co-efficient of the following data.

Weight in kg 35 37 38 40 42 45
No. of students 6 8 12 10 9 5

292 ~ Perfect Optional Mathematics Class 9

 From the above information, we can find the mean deviation as follows.

Computation of Mean Deviation

Weight in No. of c.f Md x – Md |x – Md| f|x – Md|
kg (x) students (f)

35 6 6 35 – 38 3 18

37 8 14 37 – 38 1 8

38 12 26 38 38 – 38 0 0
40 10 36 40 – 38 2 20

42 9 45 42 – 38 4 36

45 5 50 45 – 38 7 35

N = 50 ∑f|x – Md|

=  117

Here, N = 50 and N + 1 = 50 + 1 = 51 = 25.5. The c.f. just greater than 25.5 is 26 and the median is
2 2 2

the corresponding value of c.f. 26. i.e. Median (Md) = 38. ∑f |x – Md|
N
Standard DeviationAgain, ∑f|x – Md|= 117 and mean deviation from median (MD) = = 117 = 2.34
50
Now, co-efficient MD 2.34
of mean deviation is = Md = 38 = 0.062

In the calculation of mean deviation, we use the absolute value of the deviation of the items from average.
To remove this drawback, we use another measure of dispersion named “Standard Deviation". It is denoted
by s and defined as the positive square root of the mean of the square of the deviations of the items taken
from their arithmetic mean.

Individual Series  ∑(xn– –x )2

The standard deviation of the set of n observations is given by σ =

The formula of calculating the standard deviation can be used in another form also.

σ=  ∑(xn– x– )2 =  ∑(x2 – 2.x.x– + x–  2)
n

=   ∑x2 – 2.–x ∑x+ –x  2∑1)
n

= ∑x2 – 2 x–   ∑nx  + x–  2 ∑n1
 n

= ∑x2 – 2 ∑nx ∑nx + ∑x 2 n Statistics
=  n nn

 ∑nx2 ∑x 2 
n
– 

 ∑(xn– –x )2 or  ∑nx2 ∑x 2 
n
Hence, σ = – 

Dispersion ~ 293

Where –x is the arithmetic mean of the given observations.
Co-efficient of Standard deviation = σ–x 

Variance of the given set of data is the square of their standard deviation. It is denoted by σ2 and σ2 =

∑(x – –x )2 or σ2 = ∑x2 – ∑x 2 
n n n

Co-efficient of variance = σx–  × 100%.
Discrete Series

For a discrete and continuous frequency distribution, the standard deviation is given by

 ∑f(xN– –x )2  ∑Nfx2 ∑fx 2 
N
σ= or – 

Where –x be the arithmetic mean and N be the total number of observations of the given data.

Co-efficient of standard deviation = σ–x 

Variance of the given set of data is the square of their standard deviation. It is denoted by and σ2

σ2 = ∑fx2 – ∑fx 2  σ2 = ∑fx2 – ∑fx 2 
N N N N
or

Co-efficient of variance = σ–x  × 100%

Example 7:

Find the standard deviation and its co-efficient of the following data.

10, 12, 18, 13, 15, 26, 22.

 Here, n = 7, and mean (–x ) = ∑x = 116 = 16.57
n 7
–x  x – –x  |x – –x |2
The standard deviation (σ) =  ∑(xn– –x )2 x 10 – 16.57 = – 6.57 43.1649
=  199.77143 10
12 – 16.57 = – 4.57 20.8849
12
18 – 16.57 = 1.43 2.0449
18
12.7449
= 28.5306 = 5.34 13 16.57 13 – 16.57 = – 3.57
15 15 – 16.57 = – 1.57 2.4619

26 26 – 16.57 = 9.43 88.9249

22 22 – 16.57 = 5.43 29.4849
∑x ∑|x – x– |2
= 116 = 2122

294 ~ Perfect Optional Mathematics Class 9

Alternative Method

Here, n = 7, and mean (–x ) = ∑x = 116 = 16.57 x x2
n 7 10 100
12 144
∑x2 ∑x 2  18 324
n n 13 169
The standard deviation (σ) =   –  15 225
26 676
∑x2 ∑x 2   21722 116 2  484
n n 7
=   –  –  ∑x2 = 2122

= 303.14 – 274.61 22

= 28.53 ∑x

= 5.34

The co-efficient of standard deviation = σ = 5.34 = 0.322
x–  16.57

Example 8:

Find the variance and its co-efficient from the following data.

Age in yrs 10 15 20 25 30 35
No. of People 5 4 15 25 10 8

 From the above information, we can find the standard deviation and variance as follows.

Computation of Standard Deviation and Variance

Age in yrs (x) No. of people (f) fx fx²

10 5 50 500
15 7 105 1575
20 15 300 6000
25 25 625 15625
30 10 300 9000
35 8 280 9800

N = 70 Sfx = 1660 Sfx² = 42500

Here, N = 70, ∑fx = 1660, ∑fx2 = 42500

Arithmetic mean (–x ) = ∑fx =  167060 = 23.71
N

 ∑Nfx2 ∑fx 2 
N
The standard deviation = –  Statistics

=  4275000 –  1660 2  607.14 – 562.37 = 44.77 = 6.69
70
=

Co-efficent of S.D. = σx–  = 6.69 = 0.28
23.71

Dispersion ~ 295

∑fx2 ∑fx 2  42500 1660 2
N N 70 70
Variance (σ2) = –  = – 

= 607.14 – 562.37 = 44.77

Hence, the deviation (σ2) = 44.77

Co-efficient of variance = σ × 100% = 6.69 × 100% = 0.2822 × 100% = 28.22%
–x  23.71

EXERCISE : 8.2

1. Calculate the quartile deviation and its co-efficient from the following data.

a. 25, 18, 28, 30, 35, 15, 10. b. 80, 60, 20, 45, 30, 115, 110, 120, 118, 90, 130.

2. a. If Q1 = 8.5 and Q3 = 12, then find the quartile deviation (Q.D.) and its co-efficient of Q.D.
b. If Q1 = 17.5 and Q.D. = 20, then find the third quartile (Q3)
c. If Q3 = 15 and Q.D. = 1, then find the first quartile deviation (Q3) and co-efficient of Q.D.
d. If the sum and difference of Q1 and Q3 are 18 and 2 respectively, then find the co-efficient of Q.D.
3. Calculate the quartile deviation and its co-efficient from the following data.

a. Wages (Rs.) 500 550 650 700 800 900

No. of workers 10 15 17 20 8 5

b. Ages (months) 22 24 25 26 28 32 35

No. of cattle 7 12 15 20 12 8 3

4. Calculate the mean deviation from mean and its co-efficient from the following data.

a. 22, 36, 38, 29, 34, 25, 26, 30. b. 7, 8, 9, 20, 6, 13, 8, 18, 5, 10, 16, 15.

5. Calculate the mean deviation from median and its co-efficient from the following data.

a. 22, 25, 26, 30, 36, 38, 29, 34. b. 64, 70, 72, 75, 70, 65, 66, 69, 76, 71, 78.

6. Calculate the mean deviation from mean and median and their co-efficients from the following data.

a. 8, 9, 12, 17, 25, 6, 10, 11. b. 77, 68, 40, 68, 80, 29, 112, 30, 40, 88, 90, 110.

7. a. From an individual series, if ∑x = 540, n = 5 and ∑|x – x– | = 52, then find the mean deviation from
mean and its co-efficient.

b. From an individual series, if Md = 125, n = 5 and ∑|x – Md| = 65, then find the mean deviation
from median and its co-efficient.

8. Calculate the mean deviation from mean and its co-efficient of the following data.

a. Wages (Rs.) 600 650 750 850 900 920

No. of workers 8 12 15 20 12 6

b. Marks 12 15 18 20 25

No. of students 5 7 8 6 4

296 ~ Perfect Optional Mathematics Class 9