Case II. When the axes of reflections intersect at a point A
Q Q'
Let Q be a point on the plane and AB and CD are two
axes of reflections intersect each other at point O. M
If Q' be the reflection of Q about AB and Q" be the image N C
of Q' about CD. QQ' intersects AB at M and Q'Q" intersects CD
at N. Q" is the image of Q under the combined reflection.
Here, ®QOQ" = ®QOQ' + ®Q'OQ" Q''
=2MOQ' + 2®Q'ON D O
B
= 2(MOQ' + ®Q'ON)
=2MON
Hence, if the axes of two reflections intersect at a point, then the reflection followed by reflection is
a rotation. The centre of rotation is the point of intersection of two axes and the angle of rotation is twice
the angle between two axes and the direction is towards the second axes.
Example 3: If the image of a point (2, a) under the reflection in Y-axis followed by reflection in the
line y = 4 is (– 2, 3), find the value of a.
Solution:
The image of point (2, a) under the reflection in Y-axis is (– 2, a).
The final image of point (– 2, a) under the reflection on the line y = 4 is (– 2, 2 × 4 – a) i.e. (– 2, 8 – a).
From question, the final image is (– 2, 3). So,
(– 2, 8 – a) = (– 2, 3)
or, 8 – a = 3
or, a = 8 – 3
\ a = 5
Example 4: If R1 is the reflection about y = 4 and R2, the reflection about y = – x, find R2R1 (– 8, 2). Show
that R2R1 is a rotation.
Solution:
y = –x Y
Let A(– 8, 2) is the given point. A''(-6,8)
Reflecting about y = 4,
P(x, y) → P'(x, 2 × 4 – y) A'(-8,6)
So, for reflection R1, M(-4,4) y=4 Transformation
A(– 8, 2) → A'(– 8, 2 × 4 – 2)
i.e. A'(– 8, 6) A(-8,2) OX
Reflecting about y = – x, X'
P(x, y) → P'(– y, – x)
So, for reflection R2, Y'
A'(– 8, 6) → A"(– 6, 8)
Hence, image of point A(– 8, 2) under the combined transformation R2R1 is A"(– 6, 8).
As the lines of reflection R1 and R2 intersect each other at point (–4, 4), R2R1 is a rotation through – 90°
about centre M(– 4, 4).
Combination of Reflections / 247
As P(x, y) (a, b), – 90° P'(y – b + a, – x + a + b)
A(– 8, 2) (–4, 4), – 90° A1{2 – 4 + (– 4), – (– 8) + (– 4) + 4} i.e. A1(– 6, 8)
Hence, from above result A" and A1 are same and R2R1 is a rotation.
Example 5: Find the co-ordinates of the vertices of the image of DABC having the vertices A(3, 1),
B(4, – 1) and C(6, 2) under reflection in the line x – y = 0 followed by another reflection in Y-axis.
Show the above combined transformation on a graph.
Solution:
Here, the vertices of DABC are A(3, 1), B(4, – 1) and C(6, 2). First reflect DABC on the line x – y =
0 i.e. y = x. Y C'
As, P(x, y) y = x P'(y, x), we have C"
A(3, 1) → A'(1, 3)
B(4, – 1) → B'(– 1, 4) B' B" C
A" A'
C(6, 2) → C'(2, 6)
A
Again reflect DA'B'C' on Y-axis.
As, P(x, y) Y-axis P'(– x, y), we have X' O X
B
A'(1, 3) → A"(– 1, 3) Y'
B'(– 1, 4) → B"(1, 4)
C'(2, 6) → C"(–2, 6)
Showing the above transformation on a graph.
Alterantely,
We have, the axes of reflection x – y = 0 and Y-axis Y
intersect each other at origin O making an angle of 45°. So, the C1
combined reflection in the line x – y = 0 followed by reflection
in Y-axis is a single rotation through 2 × 45° = 90° about
centre O. A1 B1 B
The image of combined reflection is directly obtained as follows: B X
B
As, P(x, y) [(0,0),90°] P'(– y, x), we have X' O
A(3, 1) → A"(– 1, 3)
B(4, – 1) → B"(1, 4) Y'
C(6, 2) → C"(–2, 6)
The graph of transformation is shown in the figure.
248 / Optional Mathematics Class 10
Exercise 7.2A
4. Find the image of the point A(3, – 2) under the
a. Reflection in Y–axis followed by reflection in X–axis.
b. Reflection in y = x followed by reflection in y = – x.
c. Reflection in Y–axis followed by reflection in x = 4.
d. Reflection in y = x followed by reflection in x = – 1.
2. a. Find the image of a point (3, 7) under the reflection in Y-axis followed by reflection in the line
x = 3.
b. Find the image of a point A(3, 8) under the reflection in X-axis followed by the reflection in the
line y = – 3.
c. Find the image of the point (2, – 3) under the reflection about in x = –1 followed by reflection
in y = – 2.
d. Find the image of a point P(8, – 4) under the reflection in y = – 1 followed by the reflection in
the line y = 4.
3. a. Find the image of the point (2, 7) under the reflection in X–axis followed by reflection in y = x.
b. If r1 is the reflection about the line x = y and r2 is the reflection about the Y-axis, find the image
of a point M(2, – 7) under the combined transformation of r1○r2.
c. R1 and R2 denote reflection in the line x = 2 and line x = – y respectively. Which point has the
image A'(–3, 2) under the combined transformation R1○R2.
4. If Rx be the reflection about X–axis, Ry be the reflection about Y–axis, R1 be the reflection in
x = 2 and R2 be the reflection in y = 3, find the images of the given points by the following
transformations.
a. RxR1(2, 4) b. RxRy(1, 5) c. R1R2(–3, 2)
d. R1Rx(1, –4) e. R2Ry(6, 3) f. Rx2(1, –2)
5. a. If the image of the point (3, b) under the reflection in the X-axis followed by the reflection in
the line y = – 3 is (3, 4), find the value of b.
b. If the image of the point (a, 6) under the reflection in the Y-axis followed by the reflection in Transformation
the line x = 2 is (9, 6), find the value of a.
c. If the image of the point (p + 3, – 3) under the reflection in the line x = 3 followed by the
reflection in the line x = – y is (3, – 2), find the value of p.
6. a. If the image of the point (1 – p, q) under the reflection in Y-axis followed by the reflection in
the line x = y is (2, 3), find the values of p and q.
b. If the image of the point (a + 2, 2) under the reflection in the line y = 3 followed by the
reflection in the line x + y = 0 is (b – 2, – 3), find the values of a and b.
c. If the image of the point (2, a) under the reflection in the line y = x followed by the reflection
in the line y = b is (5, 2), find the values of a and b.
Combination of Reflections / 249
Exercise 7.2B
1. a. If R1 is the reflection about Y-axis and R2, the reflection about x = – 3, find the image of DABC
under the combined reflection R2R1 where the vertices of DABC are A(4, 1), B(1, 3) and
C(3, 5). Show that R2R1 is a translation.
b. A triangle with vertices A(1, 2), B(4, – 1) and C(2, 5) is reflected successively in the lines x = 5
and x = – 2. Find the co-ordinates of the vertices of the images and represent them graphically.
Also state the single transformation given by the combination of these transformations.
c. A triangle with vertices N(2, 5), V(4, – 1) and M(1, 2) is reflected successively in the lines
y = – 1 and y = 2. Find the co-ordinates of the vertices of the images and represent them graphically.
Also state the single transformation given by the combinations of these transformations.
2. a. If R1 is the reflection about y = 3 and R2, the reflection about y = x, find R2R1(3, – 4), show that
R2R1 is a rotation.
b. If R1 is the reflection about x = – 4 and R2, the reflection about y = – x, find R1R2(2, 4), show
that R1R2 is a rotation.
c. If R1 is the reflection about x = 3 and R2, the reflection about Y-axis, find R1R2(2, 5), show that
R1R2 is a translation.
3. a. Let P(2, 4) and Q(5, 2) be two points on the plane. The line PQ is reflected about X-axis; find
the image P'Q'. The line P'Q' is again reflected about the line y = – x. Find the image P''Q'' of
P'Q' and show the lines PQ, P'Q' and P''Q'' on the same graph.
b. A triangle with vertices P(3, 1), Q(6, 2) and R(4, 5) is reflected successively in the lines y = x
and y = – 2. Find the co-ordinates of the vertices of the images and represent them graphically.
Also state the single transformation given by these transformations.
4. a. State the single transformation equivalent to the combination of reflections on the X-axis and
Y-axis respectively. Using this single transformation find the co-ordinates of the vertices of
the image of ∆ABC having vertices A(2, 3), B(3, – 4) and C(1, – 2). Also draw the object and
image on the same graph.
b. A triangle with vertices N(1, 2), V(4, – 1) and M(2, 5) is reflected successively in the lines
x = – 1 and y = 2. Find the images under these transformations by stating co-ordinates and
represent them graphically. Also state the single transformation, given by these transformations.
c. Find the co-ordinates of the vertices of the image of ∆ABC having vertices A(1, 1), B(2, 3)
and C(4, 2) under the reflection about X-axis followed by the reflection about x = – 1, show the
combined transformation in the graph.
d. If R1 is the reflection about the line x = – 3 and R2, the reflection about the line y = 4, find the
image of DPQR under the combined reflection R1R2 where the vertices of DPQR are P(3, – 2),
Q(– 1, – 2) and R(2, – 4). Show that R1R2 is a rotation.
250 / Optional Mathematics Class 10
7.3 Combination of Rotations
Review T T'
C
Rotation is a kind of transformation in which an object is rotated
through certain angle about the given centre. In the given figure the
triangle T is rotated through 90° about C. The object and the image in
rotation are congurent.
If P(x, y) be a point on the plane and P' be the image of P
under the various rotations, then
1. Rotation through 90° about O(0, 0), 4. Rotation through 90° about M(a, b),
P(x, y) + 90°, O(0, 0) P'(– y, x) P(x, y) + 90°, M(a, b) P'(– y + a + b, x – a + b)
– 270°, O(0, 0) – 270°, M(a, b)
2. Rotation through 180° about O(0, 0), 5. Rotation through 180° about M(a, b),
P(x, y) + 180°, O(0, 0) P'(– x, – y) P(x, y) + 180°, M(a, b) P'(2a – x, 2b – y)
– 180°, O(0, 0) – 180°, M(a, b)
3. Rotation through 270° about O(0, 0), 6. Rotation through 270° about M(a, b),
P(x, y) + 270°, O(0, 0) P'(y, – x) P(x, y) + 270°, M(a, b) P'(y + a – b, – x + a + b)
– 90°, O(0, 0) – 90°, M(a, b)
Two Rotations
A point or an object once rotated can further be rotated. There are two cases for the combination of
two rotations.
Case I : When the centres of two rotations are same.
Case II : When the centres of two rotations are different.
Case I. When the centres of two rotations are same Y P'(4,2) Transformation
X
Let P(2, – 4) be a point on the plane and O(0, 0) be the O
centre of rotation. P'(4, 2) be the image of P(2, – 4) under the rotation X' P(2,-4)
through 90º about O(0, 0). Again, P"(– 4, – 2) be the image of
P'(4, 2) under the rotation through 180º about the same origin P''(-4,-2)
O(0, 0). P"(– 4, – 2) is the image of P(2, – 4) under the combined Y'
rotation.
Here, reflex ®POP"
=®POP' + ®P'OP"
= 90º + 180º
= 270º
Hence, if the centres of two rotations are same, then the rotation followed by rotation is a rotation.
The centre of single rotation is the common centre of two rotations and the angle of the single rotation is
the sum of the angles of two rotations in the same direction.
Example 1: If R1 = rotation of +90° about (0, 0) and R2 = rotation of +90° about (0, 0), find the image
as indicated.
a. R1R2(3, 4) b. R2R1(– 4, 2)
Combination of Rotations / 251
Solution:
a. Here, R1R2(3, 4) = R2 followed by R 1 b. Here, R2R1(– 4, 2) = R1 followed by R2
=R1(– 4, 3) =R2(– 2, – 4)
= (– 3, – 4) = (4, – 2)
\ R1R2(3, 4) = (– 3, – 4) \ R2R1(– 4, 2) = (4, – 2)
Example 2: If R1 represents a positive quarter turn about origin and R2, a half turn about the origin.
Find R2R1(DABC), where A(6, 1), B(2, 2) and C(5, 4) are the vertices of DABC. Also represent the
transformation on a graph.
Solution:
Here, R1 = rotation of +90° about origin and R2 = rotation YC
of 180° about origin.
As R1 and R2 are the rotations about same centre (0, 0), X' B A
the centre of combined rotation R2R1 is (0, 0) and the angle of X
combined rotation is 90° + 180° = 270°. O
B'
Thus the image under this combined rotation is given by
P(x, y) 270°, O(0, 0) P'(y, – x) and C'
A(6, 1) A'(1, – 6) A'
Y'
B(2, 2) B'(2, – 2)
C(5, 4) C'(4, – 5)
Showing the above transformation on a graph.
Case II. When the centres of two rotations are different
Let A(3, – 2), B(5, – 2) and C(3, – 5) be the vertices of B'' A" Y C'
a ∆ABC. ∆ABC is rotated through 90º about O(0, 0) to get
the image ∆A'B'C'. ∆A'B'C' is further rotated through 180º 5
about the point M(0, 3) to get the image A"B"C".
∆A"B"C" is the image of ∆ABC under the combined 4 B'
rotation. Angle of rotation is 90º + 180º = 270º. C'' M(0,3) A'
For the centre of rotation, join two vertices A and B
(or C) with their corresponding images A" and B" (or C") 2
respectively and draw perpendicular bisectors of these lines
AA" and CC". The intersection point N(4, 4) is the centre of 1
the combined rotation.
X' -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 X
-1
-2 A C
-3
-4
-5 B
Y'
Hence, if the centre of two rotations are different, then the rotation followed by rotation is a rotation.
The angle of rotation of combined rotation is the sum of the angles of two rotations and centre of combined
rotation is obtained by drawing the perpendicular bisectors of the lines joining any two points of given
object with their corresponding final images.
252 / Optional Mathematics Class 10
Example 3: A(2, – 1), B(5, – 2) and C(3, 2) are the vertices of DABC. Find the image DA'B'C' of DABC
under the rotation through +90° about origin. Again find the image DA"B"C" of DA'B'C' under
the rotation through +90° about point (– 1, – 2). Draw the object and image on the same graph.
Solution:
Here, the given DABC has the vertices A(2, – 1), B(5, – 2) and C(3, 2).
For the image under the rotation through +90° about origin,
P(x, y) 90°, O(0, 0) P'(– y, x) and Y
B'
A(2, – 1) A'(1, 2)
B(5, – 2) B'(2, 5)
C(3, 2) C'(– 2, 3) C' C
A'
Again for the image under the rotation through B"
A"
+90° about the point (– 1, – 2) X' O X
A B
P(x, y) +90°, M(a, b) P'(– y + a + b, x – a + b) and
C"
A'(1, 2) A"(– 2 – 1 – 2, 1 + 1 – 2) Y'
i.e. A"(– 5, 0)
B'(2, 5) B"(– 5 – 1 – 2, 2 + 1 – 2)
i.e. B"(– 8, 1)
C'(– 2, 3) C"(– 3 – 1 – 2, – 2 + 1 – 2)
i.e. C"(– 6, – 3)
Exercise 7.3A
1. a. Find the image of the point (– 4, – 3) under the rotation through 180º about the origin followed
by the rotation through – 90º about the origin.
b. Find the image of the point (2, 3) under the rotation through – 90º about origin followed by the
rotation through 180º about (– 2, 0).
c. Find the image of the point (– 3, 2) under the rotation through – 90º about (– 1, – 1) followed by Transformation
the rotation through – 90º about (2, 2).
2. Let R1 be the rotation through 90º about origin, R2 be the rotation through 180º about origin and
R3 be the rotation through 90º about (2, 3), find the images of the given points by the following
transformations :
a. R2R1(1, 4) b. R1R2(– 2, 2) c. R1R3(– 4, 1)
d. R2R3(– 3, 2) e. R22(1, 3) f. R3R2(p, q)
3. a. If a point A(a, b) has its image A'(3, – 2) under the rotation through 90° about origin followed
by the rotation through 180° about origin, find the values of a and b.
b. If the image of the point (6, q – 1) under the rotation through 180° about origin followed by the
rotation through – 90° about origin is (5, p + 3), find the value of p and q.
Combination of Rotations / 253
4. a. If the image of a point A(u, 3) under the rotation through 180º about N(2, 2) followed the
rotation through 90º about origin is A'(v + 3, 1), find the values of u and v.
b. If the image of a point B(p + 2, q – 3) under the rotation through 270º about A(1, 2) followed
by the rotation through – 90º about B(– 3, – 1) is B'(– 6, – 5), find the values of p and q.
Exercise 7.3B
1. a. If R1 represents a postitive quarter turn about origin and R2, a positive half turn about origin.
Find R2R1(DABC) where A(7, – 1), B(2, – 2) and C(5, 3) are the vertices of DABC. Also
represent the transformation in a graph.
b. If R1 represents a half turn about origin and R2, a negative quarter turn about origin. Find
R1R2(DPQR) where P(3, 2), Q(7, 1) and R(5, 5) are the vertices of DPQR. Also represent the
transformation in a graph.
c. A triangle ABC has vertices A(– 5, 2), B(– 4, 4), C(– 2, 1), find the co-ordinates of the vertices
of image ∆A'B'C' after negative quarter turn about origin. Also find the co-ordinates of the
vertices of image ∆A''B''C'' of ∆A'B'C' after half turn about origin.
2. a. P(– 6, – 2), Q(– 1, – 1) and R(– 4, 3) are the vertices of DPQR. Find the image DP'Q'R' of DPQR
under the rotation through +90° about origin. Again find the image DP"Q"R" of DP'Q'R' under
the rotation through +90° about point (– 3, 0). Draw the object and image on the same graph.
b. D(2, 3), E(6, 4) and F(4, 0) are the vertices of DDEF. Rotate DDEF through +90° about origin
to get DD'E'F'. Again find the image DD"E"F" of DD'E'F' under the rotation through 180° about
point (– 3, – 1). Draw the object and image on the same graph.
3. a. If AB is a line segment with end points A(3, 5) and B(5, 2). Line AB is rotated through 90°
about M(1, – 1), find the image A'B'. Line A'B' is again rotated through – 90° about N(– 4, 0),
find the image A"B" and also show the lines AB, A'B' and A"B" on the same graph.
b. A triangle PQR with vertices P(3, 1), Q(4, – 1) and R(6, 1) is rotated through 90° about
M(1, 1), find the co-ordinates of image DP'Q'R'. DP'Q'R' is again rotated through 180° about
N(– 1, 2), find the co-ordinates of image of DP"Q"R".
4. a. DPQR has the vertices at P(1, 1), Q(5, 1) and R(4,3). DP'Q'R' is the image of DPQR under the
rotation through +90° about origin. DP'Q'R' s again rotated through 180° about M(– 1, 0) to get
image DP"Q"R". Find the co-ordinates of vertices of the images and also find the centre and
angle of the combined rotation.
b. Let a triangle ABC has the vertices A(3, 0), B(2, – 3) and C(4, – 2). If ∆A'B'C' be the image of
∆ABC under the rotation through 90° about origin. Find the co-ordinates of ∆A'B'C' and again
rotate ∆A'B'C' through 180° about M(– 1, 1). Find the co-ordinates of last image ∆A"B"C".
Also find the centre and angle of the combined rotation.
254 / Optional Mathematics Class 10
7.4 Combination of Enlargements
Review
An object is enlarged about any point with given scale factor. In the
given figure triangle T is enlarged about O with scale factor 2. The object
and image in enlargement are similar to each other.
If P(x, y) be a point on the plane and P' be the image of P under the T'
various enlargements, then
1. Enlargement with centre O(0, 0) and scale factor k, T
P(x, y) → P'(kx, ky)
2. Enlargement with centre M(a, b) and scale factor k, C
P(x, y) → P'{k(x – a) + a, k(y – b) + b}
Two Enlargements
A point or an object once enlarged (reduced) can further be enlarged (or reduced). The single
transformation equivalent to above two transformations can be studied in two different cases.
Case I : When centres of two enlargements are same.
Case II : When centres of two enlargements are different.
Case I. When centres of two enlargements are same Y B''
Let AB be a line segment having ends A(2, 1)
and B(1, 2). AB is enlarged with centre O(0,0) and scale
factor 2 i.e. E1[O, 2]. If A'B' be the image of AB under
this enlargement then A'B' = 2AB and the ends of A'B' are
A'(4, 2) and B'(2, 4). Again A'B' is enlarged about O with scale factor B' A' A''
B X
3 i.e. E2[O, 3]. If A"B" be the image of A'B' under this enlargement
then A"B" = 3A'B' and the ends of A"B" are A"(12, 6) and
B"(6, 12). X' O A
Now, A"B" = 3A'B' = 3(2AB) = 6AB.
Similarly, the image of the points A(2, 1) and B(1, 2) under Y'
the enlargement E[O, 6] are A(2, 1) → A1(6 × 2, 6 × 1) i.e. A1(12, 6) and B(1, 2) → B1(6 × 1, 6 × 2) Transformation
i.e. B1(6, 12) which are the points A" and B" respectively.
Hence, A"B" is the image of AB under the enlargement with centre O(0, 0) and scale factor 6 i.e. 2×3.
∴ AB E1[O, 2] A'B', A'B' E2[O, 3] A"B". Hence, E2[O, 3]°E1[O, 2] = E[O, 2×3] = E[O, 6].
Example 1: A point (a, 3) is first transformed by E1[(0, 0), 2] and then by E2 [(0, 0), 2 ] so that the final
3
image is (8, b – 3). Find the values of a and b.
Solution:
Here, E1[(0, 0), 2] and E2[(0, 0), 23] are given two enlargements. As these two enlargements have the
same centre (0, 0), the (0, 0) and scale factor 2
combination of these transformations has centre at × 2 = 34.
3
Combination of Enlargements / 255
i.e. E2[(0, 0), 32]⸰E1[(0, 0), 2] = E[(0, 0), 43]
Image under this combined transformation is
P(x, y) P' 4 x, 4 y
3 3
4 4 4a
i.e. (a, 3) 3 a, 3 ×3 i.e. 3 , 4
From question, image of (a, 3) is (8, b – 3), so 4a ,4 = (8, b – 3)
Hence, and 3 b– 3
4a
3 = 8 4=
or, 4a = 24 or, b = 4 + 3
\ a = 8 \ b = 7
Example 2: A triangle PQR has the vertices P(1, 3), Q(3, 1) and R(2, 4), find the image of DPQR
under the enlargement with centre O and scale factor – 1. Again enlarge DP'Q'R' with the same
centre and scale factor – 2 and find the co-ordinates of the vertices of the image DP"Q"R" of this
enlargement. Present DPQR and DP"Q"R" on the same graph.
Solution:
Vertices of DPQR are P(1, 3), Q(3, 1) and R(2, 4). When DPQR is enlarged with centre O(0, 0) and
scale factor – 1. Y R"
P(1, 3) → P'(– 1 × 1, – 1 × 3) i.e. P'(– 1, – 3)
Q(3, 1) → Q'(– 1 × 3, – 1 × 1) i.e. Q'(– 3, – 1) P"
R
R(2, 4) → R'(– 1 × 2, – 1 × 4) i.e. R'(– 2, – 4) P
Again, ∆A'B'C' is enlarged with centre O(0, 0) and scale factor – 2 OQ Q"
X
P'(– 1, – 3) → P"(– 2 × – 1, – 2 × – 3) i.e. P"(2, 6) X' Q'
Q'(– 3, – 1) → Q"(– 2 × – 3, – 2 × – 1) i.e. Q"(6, 2)
R'(– 2, – 4) → R"(– 2 × – 2, – 2 × – 4) i.e. R"(4, 8) P'
R' Y'
Now drawing DPQR, DP'Q'R' and DP"Q"R" on the same
graph paper, we get the figure as shown. Y R1
Alternately, X' Q' P1 Q1
Here, the first enlargement is E1[O, – 1] and the second R X
enlargement is E2[O, – 2]. P
The combined enlargement is OQ
E2E1 = E2[O, – 2]⸰E1[O, – 1] = E[O, – 2 × – 1] = E[O, 2]
Now, for the image of this combined enlargements, we get Y'
P(x, y) E[O, 2] P'(2x, 2y)
P(3, 1) → P1(2 × 3, 2 × 1) i.e. P1(6, 2)
Q(1, 3) → Q1(2 × 1, 2 × 3) i.e. Q1(2, 6)
R(2, 4) → R1(2 × 2, 2 × 4) i.e. R1(4, 8)
256 / Optional Mathematics Class 10
Case II. When the centres of the enlargements are different
Let AB be a line segment with ends at Y
A(4, 3) and B(0, 5). AB is enlarged with centre B''
B
M1(– 2, – 1) and scale factor 1 i.e. E1 M1(– 2, – 1), 1 .
2 2
A'B' is the image of AB under the enlargement E1
and co-ordinates of A' is (1, 1) and B' is (– 1, 2) and M B' A A''
A'B' = 1 AB. Again, A'B' is enlarged about M2(– 3, 0) A' X
2
with scale factor 3. X' M1 O
i.e. E2 [M2(– 3, 0), 3] M2
A"B" is the image of A'B' under the enlargement
E2 and co-ordinates of A" is (9, 3) and B"(3, 6). Hence Y'
1 3
A"B" = 3A'B' = 3× 2 AB = 2 AB.
From the figure join AA" and BB" and produce them to meet at M(– 6, 3). The intersection point M
is the centre of combined enlargement and scale factor of combined enlarement is 3× 1 = 32.
2
Example 3: If E1 represents the enlargement [(0, 0), 2], E2, the enlargement [(2, 4), 3 ], find the images
2
under the combined transformation as given below
Solutiona:. E2E1(2, – 5) b. E1E2(– 4, 6)
The given enlargements are E1 = [(0, 0), 2] and E2 = [(2, 4), 23].
a. E2E1(2, – 5) = E1 followed by E2 b. E1E2(– 4, 6) = E2 followed by E1
=E2[2 × 2, 2 × (– 5)] =E1 23 (– 4 – 2) + 2, 3 (6 – 4) + 4
2
=E2(4, – 10) =E1 32 × 3
(– 6) + 2, 2 × 2 + 4
=32 (4 – 2) + 2, 3 (– 10 – 4) + 4
2 =E2(– 9 + 2, 3 + 4)
=32 ×2 3
+ 2, 2 × (– 14) + 4 =E2(– 7, 7)
=(3 + 2, – 21 + 4) = [2 × (– 7), 2 × 7] Transformation
= (5, – 17) = (– 14, 14)
Example 4: Let ∆ABC has vertices A(2, 2), B( 1, – 1) and C(3, 0).
a. Find the image ∆A'B'C' of ∆ABC under the enlargement with centre (0, 2) and scale factor 2.
b. Again find the image ∆A"B"C" of ∆A'B'C' under the enlargement with centre (0, – 2) and
scale factor – 21.
c. Also find the centre and scale factor of this combined enlargement.
Solution:
a. The vertices of ∆ABC are A(2, 2), B(1, 1) and C(3, 0). The co-ordinates of the vertices of the image
∆A'B'C' under the enlargement about (0, 2) and scale factor 2, then
Combination of Enlargements / 257
A(2, 2) → A'(2 × 2, 2(2 – 2) + 2) i.e. A'(4, 2)
B(1, –1) → B'(2 × 1, 2(– 1 – 2) + 2) i.e. B'(2, – 4)
C(3, 0) → C'(2 × 3, 2(0 – 2) + 2) i.e. C'(6, – 2)
b. As ∆A'B'C' has vertices A'(4, 2), B'(2, – 4) and C'(6, – 2) the co-ordinates of the vertices of the image
∆A"B"C" under the enlargement with centre (0, – 2) and scale factor – 12.
A'(4, 2) → A''(– 21 × 4, – 12 (2 + 2) – 2) i.e. A"(– 2, – 4)
Y
B'(2, – 4) → B''(– 21 × 2, – 12 (– 4 + 2) – 2) i.e. B"(– 1, – 1)
C'(6, – 2) → C''(– 21 × 6, – 21 (– 2 + 2) – 2) i.e. C"(– 3, – 2) A A'
O
c. For the centre and the scale factor of the combined C X
enlargement draw ∆ABC, ∆A'B'C' and ∆A"B"C" on X' B
B'' C'
the graph paper as shown. Join AA" and BB" and they C''
intersect each other at (0, – 1).
Hence, the centre of combined enlargement is (0, – 1) A'' B'
and scale factor is – AA"BB" = – 1. Y'
Exercise 7.4A
1. a. Find the image of a point (5, 6) under an enlargement with its centre O and scale factor 2
followed by enlargement with centre O and scale factor – 3.
b. Find the image of point A(– 5, 4) under the enlargement E1((0, 0), 32] followed by enlargement
E2[(0, 0), 2].
c. If E1 represents the enlargement [O, 3] and E2 represents the enlargement [O, 2] then find
E2E1(3, – 5) and E1E2(– 3, 4).
2. If E1 represents the enlargement [(0, 0), 2], E2 represents [(2, 1), – 1] and E3 represents [(– 2, 2), 12],
find the images of the given points by the following transformations :
a. E2E1(2, 3) b. E1E3(4, – 2) c. E2E3(4, 6)
d. E3E1(– 1, 3) is first e. E3E2(a, b) 0), 3] and f. EE23[E(01(,p0,)q, )43] so that the final image
3. a. A point (a, 2) transformed by E1[(0, then by
is (4, b + 2). Find the values of a and b.
b. A point (2, u) is first transformed by E1[(0, 0), 2] and then by E2[(0, 0), 23] so that the final image
(v – 2, 9). Find the values of u and v.
c. A point (p, q + 2) is first transformed by E1[(0, 0), 4] and then by E2[(0, 2), – 32] so that the final
image is (q – 7, 11p + 1). Find the values of p and q.
d. A point (a, 2) is first enlarged by E1[(0, 0), 2] and then by E2[(2, – 3), – 2] so that the final image
is (– 2, 1 – b). Find the values of a and b.
258 / Optional Mathematics Class 10
Exercise 7.4B
1. a. A triangle ABC has the vertices at A(4, 1), B(1, 2) and C(3, 3). Find the image of DABC under
the enlargement with centre O and scalar factor – 1. Again enlarge DA'B'C' with the same
centre and scale factor – 2 and find the co-ordinates of the vertices of the image DA"B"C" of
this enlargement. Present DABC and DA"B"C" on the same graph.
b. A triangle PQR has the vertices at P(– 2, – 1), Q(– 1, – 3) and R(– 3, – 2), find the image of
DPQR under the enlargement with centre O and scalar factor – 1. And enlarge DP'Q'R' with
centre O and scale factor 2 and find the co-ordinates of the vertices of the image DP"Q"R".
Present DPQR and DP"Q"R" on the same graph.
c. If E1 represents the enlargement [(0, 0), 2] and E2 represents [(0, 0), – 12], find the image of
E2E1(DABC) where the vertices of DABC are A(3, 0), B(1, 2) and C(3, 3).
2. a. A(1, 2), B(3, 4) and C(3, 2) are the vertices of ∆ABC. Enlarge it by E1[O, 2] and E2[O, 32]:
i. Find the image ∆A'B'C' of ∆ABC under the enlargement E1.
ii. Find the image of ∆A"B"C" of ∆A'B'C' under the enlargement E2.
iii. Draw ∆ABC, ∆A'B'C' and ∆A"B"C" on the same graph.
b. If A(1, 1), B(4, 1) and C(2, 3) be the vertices of a ∆ABC :
i. Find the image ∆A'B'C' of ∆ABC under the enlargement with centre O(0, 0) and scale factor – 1.
ii. Find the image ∆A"B"C" of ∆A'B'C' under the enlargement with centre M(– 1, 1) and
scale factor – 2.
iii. Find the centre and scale factor of this combined transformation.
3. a. AB is a line segment with end A(3, 1) and B(2, 3). AB is enlarged by E1[(0, 0), 2] to A'B;
and then A'B' is enlarged by E2[(4, – 2), 12] to A"B". Find the co-ordinates of ends of A'B' and
A"B". Also draw them on the same graph.
b. DPQR has vertices at P(– 4, – 1), Q(– 1, – 4) and R(– 2, – 1). DPQR is enlarged under the
enlargement E1[(2, 1), – 1] and the image so obtained is enlarged by E2[(0, 0), 2]. Find the co-
ordinates of DP"Q"R" and also draw DPQR and DP"Q"R" on the same graph.
c. If E1 represents [(0, 0), 2] and E2 represents [(2, 3), – 1], find the image E2E1(DPQR) where the vertices Transformation
of DPQR and P(4, 1), Q(0, 3) and R(2, 4). Also draw DPQR and DP"Q"R" on the same graph.
4. a. ABCD is a rectangle with vertices at A(1, 1), B(4, 1), C(4, 3) and D(1, 3). Rectangle ABCD is
first enlarged by E1[(0, 0), 2] and then by E2[(0, 0), – 1] to get the final emage A'B'C'D'. Find
the co-ordinates of the vertices of the image and also show them on the same graph.
b. PQRS is a parallelogram with vertices P(1, 1), Q(5, 2), R(6, 4) and S(2, 3). PQRS is enlarged
first by E1[(3, – 1), – 1] and then by E2[(0, 0), – 2]. Find the co-ordinates of the vertices of the
image and also show them on the same graph.
Combination of Enlargements / 259
7.5 Combination of Different Transformations
Introduction
An object can be transformed successively by two different transformations instead of the same
transformations. Translation followed by rotation, reflection followed by rotation, enargement followed by
rotation etc. are some examples of combination of different transformations.
Consider P(x, y) be a point which is rotated through – 90° about origin, its image will be P'(y, – x).
Again the image P'(y, – x) is translated by the vector Y
a
b then its new image will be P"(y + a, – x + b). a
b P"(y+a,-x+b)
Here, if R = rotation through – 90° about origin an T
a P'(y,-x)
= translation by b , then P(x,y) 90°
T⸰R(x, y) = T(y, – x) = (y + a, – x + b)
\ T⸰R(x, y) = (y + a, – x + b) X' O X
Example 1: A' is the image obtained by rotating A(– 2, 3) through + 90° aboutYo'rigin. If A" is the image
of A' obtained by translation 5 , find the co-ordinates of A' and A".
Solution: –2
The point A(– 2, 3) is rotated through + 90° about origin. So, its image is A'(– 3, – 2).
5
Again the image A'(– 3, – 2) is translated by – 2 , so we obtain the co-ordinates of the final image is
A"(– 3 + 5, – 2 – 2) i.e. A"(2, – 4).
Example 2: If R = reflection on the line y – x = 0 and E = enlargement with scale factor 2 and centre at
origin, what is the image of (3, 10) under the combined transformation R⸰E and E⸰R?
Solution:
Here, R = reflection on the line y – x = 0 or, y = x. Hence, P(x, y) P'(y, x)
E = enlargement of scale factor 2 and centre at origin. Hence, P(x, y) P'(2x, 2y)
Now, R⸰E(3, 10) = R(2 × 3, 2 × 10) And, E⸰R(3, 10) = E(10, 3)
= R(6, 20) = (2 × 10, 2 × 3)
= (20, 6) = (20, 6)
Hence, R⸰E(3, 10) = (20, 6). Hence, E⸰R(3, 10) = (20, 6).
Therefore, in this example, R⸰E(3, 10) = E⸰R(3, 10).
Example 3: If R1 = reflection on the line x + y = 0 and R2 = rotation through + 90° about origin, what
point should have the image (3, – 6) under the combined transformation R1⸰R2?
Solution:
Here, R1 = reflection in the line x + y = 0 i.e y = – x and R2 = rotation through + 90° about origin.
Let A(x, y) be a point then
R1⸰R2(x, y) = R2 followed by R1
=R1(– y, x) = (– x, y)
By question the image under the combined transformation R1⸰R2 is (3, – 6).
260 / Optional Mathematics Class 10
i.e. (– x, y) = (3, – 6)
\ x = – 3 and y = – 6.
Hence, the required point is (– 3, – 6).
Example 4: If T denotes the translation a and E denotes the enlargement with centre O and scale
4
factor – 2, the point (– 2, 4) has image (7, 2b) under the combined transformation T⸰E. Find the
values of a and b.
Solution:
a
Here, T = 4 be the given translation and E be the enlargement with centre O and scale factor – 2.
Now, we have T⸰E(– 2, 4) = T(– 2 × – 2, – 2 × 4) = T(4, – 8)
= (4 + a, – 8 + 4)
= (4 + a, – 4)
By question T⸰E(– 2, 4) = (7, 2b), so we have
(4 + a, – 4) = (7, 2b)
So, 4 + a = 7 and – 4 = 2b
or, a = 7 – 4 \ b = – 2
\a=3
Hence, a = 3 and b = – 2.
Example 5: If a triangle ABC has vertices A(1, 3), B(1, 6) and C(3, 5), find the co–ordinates of the
image ∆A'B'C' after positive quarter turn about origin. Find the image ∆A"B"C" of the ∆A'B'C'
after the reflection in Y–axis.
Solution:
As, A(1, 3), B(1, 6) and C(3, 5) are the vertices of ∆ABC. The image under the positive quarter turn
(rotation through 90º) about origin is
P(x, y) → P'(– y, x)
Hence, A(1, 3) → A'(– 3, 1)
B(1, 6) → B'(– 6, 1)
C(3, 5) → C'(– 5, 3)
∆A'B'C' is the image of ∆ABC under the rotation of positive quarter turn about origin. Again, the Transformation
image under the reflection about Y–axis is P(x, y) → P'(– x, y)
Hence, A'(– 3, 1) → A"(3, 1)
B'(– 6, 1) → B"(6, 1)
C'(– 5, 3) → C"(5, 3)
∴ ∆A"B"C" is the image of ∆A'B'C' under the reflection about Y–axis.
Example 6: A triangle ABC having vertices at A(1, 1), B(2, 3) and C(4, 2) is transformed to ∆A'B'C' by
– 8
an enlargement with centre O(0, 0) and scale factor 2 followed by a translation with a vector – 7
. Find the co-ordinates of ∆A'B'C' and draw ∆ABC and ∆A'B'C' on the same graph.
Combination of Different Transformations / 261
Solution:
The vertices of ∆ABC are A(1, 1), B(2, 3) and C(4, 2). Let ∆A1B1C1 be the image of ∆ABC after an
enlargement with centre O(0, 0) and scale factor 2, then
Y
A(1, 1) → A1(2 × 1, 2 × 1) i.e. A1(2, 2)
B(2, 3) → B1(2 × 2, 2 × 3) i.e. B1(4, 6) B1
C(4, 2) → C1(2 × 4, 2 × 2) i.e. C1(8, 4) B1
Again, ∆A1B1C1 is translated with a vector – 8 to get B C X
image ∆A'B'C', then
– 7 A1
A
A1(2, 2) → A'(2 – 8, 2 – 7) i.e. A'(–6, –5) X' B' O
B1(4, 6) → B'(4 – 8, 6 – 7) i.e. B'(–4, –1) C'
C1(8, 4) → C'(8 – 8, 4 – 7) i.e. C'(0, –3) A
Drawing ∆ABC and ∆A'B'C' on the same graph.
Y'
Exercise 7.5A
1. a. Point (– 6, 11) is rotated about the origin through – 90° and the image so obtained is reflected
in the line y = – x. Find the coordinates of the image so formed.
b. Point (– 9, 15) is rotated about the origin through – 180° and the image so formed is reflected
in the line y = x. Find the co-ordinates of the image so obtained.
c. Find the co-ordinates of the image of a point (3, 2) when it is first rotated by +180º about origin
O and then reflected in the Y–axis.
d. The image formed by reflecting the point (3, 4) on the X–axis is rotated about origin O(0, 0)
through +90º, find the co-ordinates of the final image formed.
e. Determine the co-ordinates of the image of a point (3, 7) when it is first reflected in the
Y-axis and then rotated through an angle of +90º about the origin. Also write down a single
transformation which denotes this combined transformation.
2. a. Find the co-ordinates of the image of a point (– 4, 3) when it is first rotated about the origin O
– 5
through – 90° and then translated by 6 .
b. Find the co-ordinates of the image of a point (– 4, 3) when it is first rotated about the origin O
2
through 180° and then translated by – 3 .
c. A point (4, 3) is rotated through 270° about origin O and the image so obtained is translated by
3
a translation vector – 4 . Find the co-ordinates of this image.
d. Find the co-ordinates of the image of a point (3, 4) when it is first rotated through 90° about
(– 1, – 1) and then translated by 5 .
– 4
262 / Optional Mathematics Class 10
3. If R1= reflection in the line x + y = 0, E = enlargement with scale factor – 2 and centre at origin,
3
T = translation through – 2 and R2 = rotation through – 270° about origin. Find the image of the
following combined transformations.
a. R1⸰R2(2, 3) b. R1⸰E(– 2, 3) c. T⸰E(– 4, – 2)
d. T⸰R1(5, – 2) e. E⸰R2(6, 2) f. R1⸰T(– 3, – 3)
4. a. If E = enlargement with scale factor – 2 and centre at origin and R = reflection in the line x – y
= 0. What point should have the image (8, – 12) under the combined transformation E⸰R?
b. If R= rotation through 270° about origin and E = enlargement with scale factor 3 and centre at
origin, find a point whose image under the combined transformation E⸰R is (6, – 9).
3
c. A point M is first translated by – 5 and the image so obtained is then reflected on the line x =
3. If the final image is at (2, 1), find the co-ordinates of M.
d. A point P is reflected in the line y + 5 = 0 and the image so obtained is then rotated through
180° about (2, 3). If the final image is at (0, 8), find the co-ordinates of P.
a
5. a. If E = enlargement about origin with scale factor 2 and T = 3 denotes a translation, the point
(3, 7) has image (2, b + 12) under the combined transformation T⸰E. Find the value of a and b.
b. If R1 = rotation through + 90° about origin and R2 = reflection in the line x + y = 0, the point
(u + 3, 6) has image (5, v – 7) under the combined transformation R1⸰R2. Find the values of u
and v.
c. A point A(a, b + 2) is reflected in the line y = 3 and the image so obtained is rotated through +
90° about origin. If the final image is at (2 – 3a, b + 6), find the values of a and b.
Exercise 7.5B
1. a. If ∆ABC has vertices A(1, 3), B(1, 6) and C(3, 5), find the co-ordinates of the image ∆A'B'C'
after positive quarter turn about origin. Find the image ∆A"B"C" of the ∆A'B'C' after the
reflection in Y-axis.
b. The vertices of a triangle ABC are A(2, – 5), B(3, 0) and C(1, 2). ∆ABC is translated through
2 to get ∆A'B'C' and ∆A'B'C' is again rotated through 90° about origin to get ∆A"B"C". Find Transformation
3
the co-ordinates of ∆A'B'C' and ∆A"B"C".
c. Draw a triangle having A(–1, 1), B(4, 2) and C(2, 3) on a graph paper. ∆ABC is translated by
– 2 and plot the figure ∆A'B'C' on the same graph paper. ∆A'B'C' is again reflected on the line
0
y = x to form ∆A"B"C" on the same graph paper. Find the co-ordinates of A", B" and C".
2
2. a. A triangle XYZ with vertices X(1, 2), Y(– 1, – 2) and Z(5, 0) is translated by a vector 1 . The
image so formed is reflected in X-axis. Write the co-ordinates of the vertices of the image thus
obtained and represent DXYZ and its image on the same graph.
Combination of Different Transformations / 263
1
b. A triangle with vertices A(1, 0), B(2, 1) and C(3, – 1) is translated by 2 . Now, draw its image
on a graph. Again its image is reflected in the line x = 2. Draw its image on the same graph.
3. a. Triangle PQR having vertices P(2, 1), Q(5, 3) and R(7, 1) is reflected on the Y-axis. The image
so formed is enlarged by E[(0, 0), 2]. Write the co-ordinates of the vertices of the images thus
obtained and represent DPQR and its images in the same graph.
b. A(2, 0), B(3, 1) and C(1, 1) are the vertices of triangle ABC. Find the image of ∆ABC under
the reflection in the line x = y followed by the enlargemet E[(– 3, – 4), 2]. Present the triangle
ABC and its images on the same graph.
5. a. A triangle having vertices A(2, 5), B(– 1, 3) and C(4, 1) is rotated about origin through + 90°.
The image so formed is reflected in the line x = 2. Find the vertices of the image triangles and
show all the triangles in the same graph.
b. A triangle with vertices A(3, 5), B(7, 2) and C(5, 8) is reflected about the line x = – 3 and then
the image formed is rotated about the point (0, 0) through 90° in clockwise direction. Find the
vertices of the image so formed and also draw DABC and its images on the same graph.
4. a. K(2, 5), L(– 1, 3) and M(4, 1) are the vertices of a triangle PLM. Find the co-ordinates of the
vertices of the image of DKLM under the rotation of negative 90° about the origin followed by
the enlargement E[(0, 0), 2]. Present the object and its images on the same graph.
b. The vertices of a DABC are A(1, 1), B(3, 1) and C(2, 3). DA'B'C' is obtained by rotating DABC
about the origin through half turn. The image thus formed is enlarged about the same centre
with scale factor 2 to get DA"B"C". Find the co-ordinates of the final image and represent all
three triangles on the same graph.
6. a. Translate the quadrilateral OABC having vertices O(0, 0), A(2, 0), B(3, 2) and C(1, 2) by
0
translation vector 2 . Reflect the image so formed in the line x = 3. Rpresent the image and
object in the graph.
b. A square having vertices A(2, 2), B(6, 2), C(6, 6) and D(2, 6) is given. The square ABCD is
rotated through 180° about M(4, 0) to get A'B'C'D', again A'B'C'D' is reflected on the line y = x
to get A"B"C"D". Find the co-ordinates of A'B'C'D' and A"B"C"D" and draw them on a graph.
c. A quadrilateral ABCD has vertices A(2, 3), B(5, 4), C(7, 1) and D(3, 0). It is enlarged with
centre O(0, 0) and scale factor – 1 to form figure A'B'C'D'. Then the figure A'B'C'D' is rotated
through – 90° about O, determine the co-ordinates of the vertices of A'B'C'D' and A"B"C"D".
d. Draw a figure with vertices A(2, 2), B(4, 0), C(8, 4) and D(4, 6). The figure ABCD is reflected
in Y-axis to form a figure A'B'C'D' and figure A'B'C'D' is enlarged with centre O(0, 0) and
scale factor – 12. Find the co-ordinates of the vertices of A'B'C'D' and A"B"C"D".
264 / Optional Mathematics Class 10
7.6 Inversion Tranformation
The transformations that we have studied in earlier classes, all had the property that every straight
line has been mapped to a straight line. Transformations with this property are called linear. We will now
investigate a specific transformation which is not linear, that is, sometimes lines are mapped to point sets
which are not lines. A P'
Inversion Circle
Reflection is one of the tranformations, we have studied in previous P
classes. We have got the idea of the reflection of a point or a line or a shape on OB
a straight line as axis. The images are congruent to their corresponding objects.
Here, we will have the idea of the reflection of a point on a curve (concave or
convex). When we reflect a point on a concave or convex curve (an arc of a C
circle), we get its image as a point. When we observe our face in concave (or
convex) mirror, our image is not congruent to our face. Similarly, when we reflect a straight line segment
on the curve, its image will be a curve, not a straight line segment. So, this tranformation is an example of
inversion transformation. To get the image under the concave or convex curve (or mirror), we have to
comptele the given curve as a complete circle on its curvature. In the given figure, to find the image of point
P under the reflection in the given curve AB, we have to complete a circle with AB as an arc of the circle
as shown. In the figure, ABC is the circle of radius r which is centered at O. For any point P, P ≠ O , the
inverse (image) of P is the unique point P' on the ray starting from O and passing through P such that OP
× OP' = r2. The point O is called the centre of inversion and circle ABC is called the circle of inversion,
while r is called the radius of inversion. We refer to the mapping P→P' as inversion with respect to the
circle ABC. Note that if P is inside the circle then P' will be outside the circle and vice versa.
When a point P is given inside the circle with centre at O and radius r. A P'
The inversion point P' is on the ray OP outside the circle. To find the point P',
drop a perpendicular to OP through P which intersect the circumference of the OP
circle at A and B. Construct a tangent to the cirlce at A and B and both these B
tangents intersect each other a point P' on OP. Here, we can prove that OP × OP' = r2
by joining joining OA. From two similar triangles, DOAB and DOP'P,
OOPA' = OP = AP Transformation
OA PP'
i.e. OP × OP' = OA2 = r2.
Similarly, when a point Q is inside the given circle with centre at O and A
radius r. The inversion point Q' is on the ray OQ inside the circle. To find the O Q' M Q
point Q', find a point M which is mid-point of OQ. Draw a circle with centre B
at M and diameter OQ. The new circle intersects the given circle at A and B as
shwon in the figure. Join AB theat intersect OQ at Q' and we can prove that OQ
× OQ' = r2. For this proof, draw a tangent AP and join OA. The triangle DOAP
and DOPA are similar, then by the property of similar triangles, we have,
OA = OQ = AQ
OQ' OA OQ'
Inversion Tranformation / 265
i.e. OP × OP' = OA2 = r2.
If P' and Q' are the points of inversion of P and Q with respect to the circle P'
with centre at O and radius r. Join two trianlges OPQ and OP'Q' as shown in the
figure, then by the definition of inversion points P
O Q Q'
OP × OP' = OQ × OQ' = r2 i.e. OP = OOPQ'.
OQ' B
Again, POQ and Q'OP' are the common angles of both the triangles
DOPQ and DOQ'P'. Now, from the theorem of similar triangles, DOPQ and
DOQ'P' are similar.
Example 1: In the given figure, O is the centre of inversion circle with radius r A
and P' is the inversion point of P. If OP = 2cm and r = 4cm, find OP'. r
Solution:
Here, radius of inversion, OA = r = 4cm O P P'
We have, OP × OP' = r2
or, 2 2 × OP' = 42
or, OP' = 16 = 8 = 4 2
22 2
Hence, OP' = 4 2 cm
Example 2: Let O be the circle with centre at the origin and radius 1. Find inversion with respect to this
circle for the following points: (2, 0), (0, 2), (0, 1), (0, – 2), (– 3, 0), (0, – 1/2), (1/5, 0) and (– 1, 0).
Solution:
Here, the centre of inversion is O(0, 0) and radius r = 1. For the point P(2, 0), OP = 22 + 02 = 2 and
the image of P lies on the line OP i.e. X-axis. So, let P'(a, 0) be the image, then
OP × OP' = r2 = 1 Y
or, 2 × OP' = 1 2
or, OP' = 1
2
1
or, a2 + 02 = 2 1
or, a2 = 1
4
1
or, a = ± 2 X' -2 -1 O P' 1 P 2 X
As point P(2, 0) lies on the positive X-axis, the image of
P(2, 0) is P' 12, 0 . Here, the remaining given points are
-1
(0, 1), (0, 2), (0, – 2), (– 3, 0), (0, – 1/2), (1/5, 0) and (– 1, 0).
As the inversion or the image of each point P is a point on -2 Y'
the ray starting from centre O(0, 0) and passing through P
such that
OP × OP' = r2 = 12 = 1. As, OP = 2 and OP × OP' = 1 i.e. OP' = 1/OP = 1/2
P(0, 2) → P'(0, 1/2) As, OP = 1 and OP × OP' = 1 i.e. OP' = 1/OP = 1
P(1, 0) → P'(1, 0) As, OP = 2 and OP × OP' = 1 i.e. OP' = 1/OP = 1/2
P(0, – 2) → P'(0, – 1/2)
266 / Optional Mathematics Class 10
P(– 3, 0) → P'(– 1/3, 0) As, OP = 3 and OP × OP' = 1 i.e. OP' = 1/OP = 1/3
P(0, – 1/2) → P'(0, – 2) As, OP = 1/2 and OP × OP' = 1 i.e. OP' = 1/OP = 2
P(1/5, 0) → P'(5, 0) As, OP = 1/5 and OP × OP' = 1 i.e. OP' = 1/OP = 5
P(– 1, 0) → P'(– 1, 0) As, OP = 1 and OP × OP' = 1 i.e. OP' = 1/OP = 1
Example 3: Let O be the circle with centre at the origin and radius 2. Find inversion with respect to this
circle for the following points. (0, – 1), (– 4, 0), (– 1, 0), (0, – 2), (2, 2), (– 2, 2) and (3, 3).
Solution:
Here, the centre of inversion is O(0, 0) and radius r = 2. For the point P(0, – 1), OP = 02 + (–1)2 = 1
and the image of P lies on the line OP i.e. Y-axis. So, let P'(0, a) be the image, then
OP × OP' = r2 = 4
or, 1 × OP' = 4
or, OP' = 4 Y
or, 02 + a2 = 4
or, a2 = 16 4
or, a = ± 4
As point P(0, –1) lies on the negative Y-axis, the Q
image of P(0, – 1) is P'(0, – 4). 2
Q'
Now, for the point Q(– 2, 2), OQ = = (–2)2 + 22 = 2 2. The X' O-4 -2 2 4 X
image of Q lies on the line OQ which is y = – x. As, y = – x P
lies in the second and the fourth quadrant, let Q'(a, a) be the
image of Q, then -2
OQ × OQ' = r2 = 4
or, 2 2 × OQ' = 4 P'
or, OQ' = 2 -4 Y'
or, a2 + a2 = 2
or, 2a2 = 2
or, a = ± 1
As, Q lies in the second quadrant, its image Q' is at (– 1, 1).
And the remaining given points are (– 4, 0), (– 1, 0), (0, – 2), (2, 2), (– 2, 2) and (3, 3). As the inversion or the Transformation
image of each point P is a point on the ray starting from centre O(0, 0) and passing through P such that OP
× OP' = r2 = 22 = 4.
P(– 4, 0) → P'(– 1, 0) As, OP = 4 and OP × OP' = 4 i.e. OP' = 4/OP = 1
P(– 1, 0) → P'(– 4, 0) As, OP = 1 and OP × OP' = 4 i.e. OP' = 4/OP = 4
P(0, – 2) → P'(0, – 2) As, OP = 2 and OP × OP' = 4 i.e. OP' = 4/OP = 2
P(2, 2) → P'(1, 1) As, OP = 2√2 and OP × OP' = 4 i.e. OP' = 4/OP = √2
P(3, 3) → P'(2/3, 2/3) As, OP = 3√2 and OP × OP' = 4 i.e. OP' = 4/OP = 4/(3√2)
Inversion Tranformation / 267
Example 4: Find the image of point a point K(– 3, 1) under the reflection on the cirlce with center at O(0, 0)
and radius 1. Use other reference points and corresponding images to find the formula for finding
the image under inversion.
Solution:
Let O be the circle with center at the origin and radius 1. Following are some examples of inversion
with respect to this circle.
SN Points Length Condition Length Images Analysis for Images
1. A(0, 1) OA = 1 OA×OA' = 1 OA' = 1 A'(0, 1) A'(0/(02 + 12), 1/(02 + 12))
2. B(2, 0) OB = 2 OB×OB' = 1 OB' = 1/2 B'(1/2, 0) B'(2/(02 + 22), 0/(02 + 22))
3. C(0, –2) OC = 2 OC×OC' = 1 OC' = 1/2 C'(0, –1/2) C'(0/((–2)2 + 02), –2/((–2)2 + 02))
4. D(1, 1) OD = 2 OD×OD' = 1 OD' = 1/ 2 D'(1/2, 1/2) D'(1/(12 + 12), 1/(12 + 12))
5. E(2, 1) OE = 5 OE×OE' = 1 OE' = 1/ 5 E'(2/5, 1/5) E'(2/(22 + 12), 1/(22 + 12))
6. F(–1, –2) OF = 5 OF×OF' = 1 OF' = 1/ 5 F'(–1/5, –2/5) F'(–1/((–1)2 + (–2)2), –2/((–1)2 + (–2)2))
7. G(–2, –2) OG = 2 2 OG×OG' = 1 OG' = 1/2 2 G'(–1/4, –1/4) G'(–2/((–2)2 + (–2)2), –2/((–2)2 + (–2)2))
8. H(2, –1) OH = 5 OH×OH' = 1 OH' = 1/ 5 H'(2/5, –1/5) H'(2/(12 + (–2)2), –1/(12 + (–2)2))
9. J(–3, 0) OJ = 3 OJ×OJ' = 1 OJ' = 1/3 J'(–1/3, 0) J'(–3/((–3)2 + 02), 0/((–3)2 + 02))
10. K(–3, 1) OK = 10 OK×OK' = 1 OK' = 1/ 10 K'(–3/10, 1/10) K'(–3/((–3)2 + 12), 1/((–3)2 + 12))
11. P(x, y) P'(x/(x2 + y2), y/(x2 + y2))
Y For point H(2, – 1)
1.5
OH = 22 + (–1)2 = 5
A' A D E Image of H lies on the
1 line through O and H.
The line lies on the 2nd
D' and the 4th quadrant.
Every point on OH has
0.5
E' its co-ordinates of the
O B' B-2 -1.5 -1 -0.5 0.5 1 1.5 2 form (2a, – a).
Let H'(2a, –a) be the
H' image of of H.
F'
X OH × OH' = r2 = 12
X' -0.5 C' or, 5 × OH' = 1
or, (OH')2 = 1/5
-1 H or, (2a)2 + (–a)2 = 1/5
or, 5a2 = 1/5
or, a = ±(1/5)
-1.5 As H, H' lies in 4th
quadrant.
∴ The image is H'(52, – 15)
C
F -2 Y'
268 / Optional Mathematics Class 10
Example 5: Find the image of point a point P(3, – 4) under the reflection on the cirlce with center at
O(0, 0) and radius 2. Use other reference points and corresponding images to find the formula for
finding the image under inversion.
Solution:
Let O be the circle with centre at the origin and radius 1. Following are some examples of inversion
with respect to this circle.
SN Points Length Condition Length Images Analysis for Images
1. A(0, 1) OA = 1 OA×OA' = 4 OA' = 4 A'(0, 4) A'(4×0/(02 + 12),4×1/(02 + 12))
2. B(0, 4) OB = 4 OB×OB' = 4 OB' = 1 B'(0, 1) B'(4×0/(02 + 42),4×4/(02 + 42))
3. C(–4, 0) OC = 4 OC×OC' = 4 OC' = 1 C'(–1, 0) C'(4×–4/((–4)2 + 02),4×0/((–4)2 + 02)
4. D(–1, 0) OD = 1 OD×OD' = 4 OD' = 4 D'(–4, 0) D'(4×–1/((–1)2 + 02),4×0/((–1)2 + 02)
5. E(2, 0) OE = 2 OE×OE' = 4 OE' = 2 E'(2, 0) E'(4×2/(22 + 02),4×0/(22 + 02)
6. F(2, 2) OF = 2 2 OF×OF' = 4 OF' = 2 F'(1, 1) F'(4×2/(22 + 22),4×2/(22 + 22)
7. G(–2, 2) OG = 2 2 OG×OG' = 4 OG' = 2 G'(–1, 1) G'(4×–2/((–2)2 + 22),4×2/((–2)2 + 22)
8. H(2, 1) OH = 5 OH×OH' = 4 OH' = 4/ 5 H'(8/5, 4/5) H'(4×2/(22 + 12),4×1/(22 + 12)
10. I(3, –4) OI = 5 OI×OI' = 4 OI' = 4/5 I'(12/25, –16/25) I'(4×3/(32 + (–4)2),4×4/(32 + (–4)2)
11. P(x, y) P'(4x/(x2 + y2), 4y/(x2 + y2))
A' Y For point I(3, –4)
4B
OI = 32 + (–4)2 = 5
3 Image of I lies on the line
through O and I. The line
G F lies on the 4th and the 2nd
quadrants.
2
G' A B' F' H Every point on OI has its
co-ordinates of the form
1
H' (3a, – 4a).
C D' C' D A E E' A' Let I'(3a, –4a) be the
X' O-4 -3 -2 -1 1 2 3 4 X image of of I.
I' OI × OI' = 22 = 4 Transformation
-1 or, 5 × OI' = 4
or, (OI')2 = (4/5)2
-2 or, (3a)2 + (–4a)2 = 16/25
or, 25a2 = 16/25
-3 or, a = ±(4/25)
-4 Y' As I, I' lies in 4th quadrant.
I ∴ The image is I'(1225, – 2156)
Inversion Tranformation / 269
Finding Inversion Point
Let C(h, k) be the centre of inversion and r the radius of inversion. P(x, y) be any point and
P'(x', y') the inversion point of P with respect to the given circle. Y P'(x',y')
Let us draw CA⊥OX, PB⊥OX and P'D⊥OX and also
draw CN⊥P'D which intersects PB at M. C P (x,y)
Here, CM = AB = OB – OA = x – h, CN = AD (h,k) M N
= OD – OA = x' – h, PM = PB – MB = y – k, P'N = P'D – ND
= y' – k and PC2 = (x – h)2 + (y – k)2.
Now, DCP'N ~ DCPM and hence,
CN = P'N = P'C X' O AB DX
CM PM PC Y' P'
or, CN = P'N = P'C × PC
CM PM PC PC
or, CN = P'N = PrC2 2 [P'C × PC = r2]
CM PM
or, x' – h = y' – k = (x – h)2 r2 (y – k)2 P y'–k
x – h y – k +
y–k
Equating the first and the second ratio with the third one, we get
C x–h M C x'–h N
x' – h = (x – h)2 r2 (y – k)2 and y' – k = (x – h)2 r2 (y – k)2
x – h + y – k +
or, x' – h = (x – r2(x – h) k)2 or, y' – k = (x – r2(y – k) k)2
h)2 + (y – h)2 + (y –
r2(x – h) r2(y – k)
\ x' = (x – h)2 + (y – k)2 + h \ y' = (x – h)2 + (y – k)2 + k
Hence, the inversion point of P(x, y) with respect to the circle with centre at C(h, k) and radius r is
P'(x', y') = P' (x – r2(x – h) k)2 + h, (x – r2(y – k) k)2 + k .
h)2 + (y – h)2 + (y –
From this relation, if centre, object and image are given, then we can find the radius of inversion from the
following relations.
Here, x' = (x – r2(x – h) k)2 + h ⇒ x' – h = (x – r2(x – h) k)2 ⇒ r2 = (x' – h) {(x – h)2 + (y – k)2}
h)2 + (y – h)2 + (y – x – h
Similarly, y' = (x – r2(y – k) k)2 + k ⇒ y' – k = (x – r2(y – k) k)2 ⇒ r2 = (y' – k) {(x – h)2 + (y – k)2}
h)2 + (y – h)2 + (y – y – k
When the centre of inversion is origin, the image of point P(x, y) can be found as P'(x', y') = P' r2x , r2y .
x2 + y2 x2 + y2
Again, the centre can be found when object point and image point are given by the relation
Here, x' = r2x ⇒ r2 = x' (x2 + y2) and y' = r2y ⇒ r2 = y' (x2 + y2)
x2 + y2 x x2 + y2 y
Example 6: Find the inversion point of A(3, 4) with respect to the circle with centre at (– 1, 1) and
radius 2 units.
Solution:
Here, centre (h, k) = (– 1, 1) and radius (r) = 2 units.
Given point is A(3, 4) i.e. x = 3, y = 4 and inversion point A'(x', y') = ?
270 / Optional Mathematics Class 10
We have the inversion point,
(x', y') = (x – r2(x – h) k)2 + h, (x – r2(y – k) k)2 + k
h)2 + (y – h)2 + (y –
= {3 22{3 – (– 1)} 1)2 + (– 1), {3 – 22(4 – 1) – 1)2 + 1
– (– 1)}2 + (4 – (– 1)}2 + (4
= (3 4(3 + 1) – 1, (3 + 4× 3 (3)2 + 1
+ 1)2 + (3)2 1)2 +
= 16 – 1, 12 + 1 = 162–525, 12 + 25 = –259, 37
25 25 25 25
Hence, the required inverstion point is A' –259, 37 .
25
Example 7: Find the radius of inversion if point of A(3, 3) with respect to the circle with centre at
(2, 3) is A'(6, 3).
Solution:
Given, centre (h, k) = (2, 3) and A(3, 3) I[(2, 3), r] A'(6, 3).
We know that r2 = (x' – h) {(x – h)2 + (y – k)2}
x – h
Here, h = 2, k = 3, x = 2, y = 3, x' = 6 and y' = 3.
Hence, r2 = (6 – 2) {(3 – 2)2 + (3 – 3)2} = 4 (12 + 02} = 4
3 – 2 1
\ r = 2 units.
Exercise 7.6
1. Find the image of the following points given on the graph under the reflection on the cirlce with
centre at origin and the radius 1. Y
1.5
C
1
A
0.5
FE B X0.5 1 1.5 2 Transformation
X' O-2 -1.5 -1 -0.5
-0.5 D
-1
-1.5
Y'
Inversion Tranformation / 271
2. In the given figure, O is the centre of inversion circle with radius r and P' is B A
the inversion point of P.
a. If OA = 1cm and OA' = 4cm, find r. r A'
b. If OA' 4cm, and r = 2cm, find OA. O
c. If OA = 4 2 cm and OA' = 2cm, find r.
3. Let O be the circle with centre at the origin and radius 1. Find inversion with respect to this circle
for the following points. c. N 0, – 12
g. Q(1, 1)
a. C(1, 0) b. K(0, –1) d. E(2, 0)
e. M(0, 2) f. S – 12, 0 h. P(2, 2).
4. Let O be the circle with centre at the origin and radius 2. Find inversion with respect to this circle
for the following points.
a. M(2, 0) b. S(0, –2) c. R(0, –1) d. L(4, 0)
e. N(0, 4) f. Q(–2, 0) g. J(2, 2) h. G(1, 1)
5. Let O be the circle with centre at the origin and radius 1. Find inversion with respect to this circle
for the following points.
a. A(1, –1) b. L(2, –2) c. D(1, 2) d. F(2, 1)
e. J(1, 3) f. R(3, 1) g. P(3, –2) h. Q(3, 5).
6. Let O be the circle with centre at the origin and radius 2. Find inversion with respect to this circle
for the following points.
a. C(3, 0) b. D(0, –4) c. E(0, –5) d. F(0, 3)
e. G(0, –3) f. H(–2, 3) g. J(5, 4) h. M(3, 4)
7. Let C(2, 3) be the circle with centre at the origin and radius 2. Find inversion with respect to this
circle for the following points.
a. A(4, 3) b. B(5, 3) c. E(2, 7) d. F(3, 5)
e. G(3, 6) f. Q(3, 3) g. R(4, 4) h. H(6, 5)
8. a. If the inversion point of P(0, 1) under the reflection on the circle with centre at origin (0, 0) and
having radius 'r' is P'(0, 4), find the value of r.
b. Find the radius of inversion if the point of A(1, 0) with respect to the circle with centre at (0, 0)
is A'(9, 0).
c. Find the radius of inversion if the point of B(– 6, 6) with respect to the circle with centre at
– 34, 3
(0, 0) is B' 4 .
9 a. If the inversion point of M(3, 3) under the reflection on the circle with centre at origin (2, 3) and
having radius 'r' is M'(6, 3), find the value of r.
b. Find the radius of inversion if the point of N(2, 3) with respect to the circle with centre at (2, 1)
is N'(2, 11).
c. Find the radius of inversion if the point of Q(1, 1) with respect to the circle with centre at (2, 3)
is Q' – 53, – 151 .
10. a. Find the inversion point of A(6, 9) with respect to inversion circle 3x2 + 3y2 = 27.
b. Find the inversion point of P(1, 2) with respect to inversion circle x2 + y2 – 4x + 2y – 4 = 0.
272 / Optional Mathematics Class 10
7.7 Transformation using Matrices
We have already discussed different types of transformations like translation, reflection, rotation
and enlargement. The image of a point or an object under these transformations can be obtained by using
2 × 1 and 2 × 2 matrices as follows.
Translation using 2×1 Matrices Y
aa P'(x+a,y+b)
The column vector b can be written as 2 × 1 matrix b
a
a b
. Hence 2 × 1 matrix b describes a translation of a unit in the
a
P(x,y) b
horizontal direction and b units along vertical direction as shown
X' O X
in figure. Y'
Now, if we write the co-ordinates of any point P(x, y) in the
xa
form of a position vector y and is added with a translation vector T = b we get the position vector of
x a x+a x+a x
a new point y + b = y + b . Clearly the new point P' y + b is the image of point P y under the
a
translation through T = b .
Example 1: Find the image of point A(2, 3) under the translation through translation vector T = 4
and show them in the same graph. 2
Solution: Y
The given point is A(2, 3) and the translation vector A'(6,4)
T= 4 . The image of A under the given translation is 2 units
2
A(2,3) 4 units
2 + 4 = 2+4 = 6 .
3 2 3+2 5 OX
Transformation
X'
Hence, the image of A(2, 3) under the given translation is
A'(6, 5) as shown in figure. Y'
Transformations using 2×2 Matrices
a b
If the position vector of any point P(x, y) be pre-multiplied by a 2 × 2 matrix c d , the position
x' x' a b x ax + by
vector of P is transformed into a new point y' . i.e. y' = c d y = cx + dy .
Transformation using Matrices / 273
ax + by x a b
Hence, cx + dy be the image of y under the transformation using a 2 × 2 matrix c d .
1 2
For example, a 2 × 2 matrix 3 4 transformed a point P(3, 4) as follows. Writing point (3, 4) in
3 1 2 1 2 3 3 + 8 11
matrix form, 4 and pre-multiplying this point by 3 4 , we have 3 4 4 = 9 + 16 = 25 . Hence,
1 2
image of P(3, 4) under the transformation using matrix 3 4 is P'(11, 25).
Reflections using 2×2 Matrices
The image of P(x, y) under the reflection in X-axis is P'(x, – y).
i.e. P(x, y) X-axis P'(x, – y)
Reflection
If P'(x', y') be the image of P(x, y) under the reflection in X-axis, then
x' x 1 × x + 0 × y 1 0 x
y' = – y = 0 × x – 1 × y = 0 – 1 y
1 0
Hence, the matrix 0 – 1 represents the reflection in X-axis.
So, the image of the point (– 5, 6) under the reflection in X-axis is obtained as
1 0 – 5 1 × (– 5) + 0 × 6 – 5
0 – 1 6 = 0 × (– 5) – 1 × 6 = – 6
– 1 0 x – x x
Similarly, 0 1 y = y is the image of y under the reflection in Y-axis.
0 1 x y x
1 0 y = x is the image of y under the reflection in the line y = x.
0 – 1 x – y x
– 1 0 y = – x is the image of y under the reflection in the line y = – x.
Example 2: Find a 2×1 matrix which transforms a point (a, b) into the point (a + 4, b – 5). Transform the
point (– 4, 6) using the same 2 × 1 matrix.
Solution:
p
Let matrix q transforms a point (a, b) to the point (a + 4, b – 5), then
a+p a+4
b + q = b – 5
p4 [Equating the corresponding elements]
\ q = – 5
4 – 4 + 4 0
The image of (– 4, 6) under the transformation by – 5 is 6 – 5 i.e. 1 .
Hence, the required image of (– 4, 6) is (0, 1).
274 / Optional Mathematics Class 10
Example 3: To what transformation is the matrix 0 1 associated ? Use this matrix to transform the point
(1, 2). 1 0
Solution:
0 1
Given tranformation matrix is 1 0 .
0 1 x 0 × x + 1 × y y 0 1
As 1 0 y = 1 × x + 0 × y = x , the matrix 1 0 is associated with the reflection in y = x.
Now, 0 1 1 0 × 1 + 1 × 2 2
1 0 2 = 1 × 1 + 0 × 2 = 1
Hence, the image of the point (1, 2) under this transformation is (2, 1).
Example 4: Find the image of ∆ABC having vertices A(– 5, 2), B(– 3, 5) and C(– 1, 3) using transformation
matrix – 1 0
Solution: 1 . Plot ∆ABC and its image on the same graph paper.
0
The ∆ABC has vertices A(– 5, 2), B(– 3, 5) and Y
C(– 1, 3). These co-ordinates are represented by a matrix as
– 5 – 3 – 1 B(-3,5) B'(3,5)
M = 2 5 3 and such matrix is called a co-ordinate matrix.
To find the resulting image of the points A, B and C by C C'
– 1 0 (-1,3) (1,3)
transformation matrix 0 1 , we multiply the matrix M by
given transformation matrix as follows. A(-5,2) A'(5,2)
X' O X
A B C A' B' C'
– 1 0 – 5 – 3 – 1 = 5 3 1 Y'
0 1 2 5 3 2 5 3
Hence, the co-ordinates of image ∆A'B'C' of ∆ABC under the transformation using given matrix Transformation
are A'(5, 2), B'(3, 5) and C'(1, 3). Drawing ∆ABC and ∆A'B'C' on the same graph, we have the given
figure.
Rotations using 2×2 Matrices
The image of P(x, y) under the rotation through 90° about origin is P'(– y, x)
i.e. P(x, y) [(0, 0), 90°] P'(– y, x)
Rotation
If P'(x', y') be the image of P(x, y) under the rotation through 90° about origin then
x' – y 0 × x – 1 × y 0 – 1 x
y' = x = 1 × x – 0 × y = 1 0 y
0 – 1
Hence, the matrix 1 0 represents the rotation through 90° about origin.
So, the image of (– 3, 8) under the rotation through 90° about origin is obtained as
Transformation using Matrices / 275
0 – 1 – 3 0.(– 3) – 1.8 – 8
1 0 8 = 1.(– 3) – 0.8 = – 3 . The image of (– 3, 8) is (– 8, – 3).
0 1 x y x
Similarly, – 1 0 y = – x is the image of y under the rotation through – 90° about origin.
–1 0 x –x x
And 0 –1 y = –y is the image of y under the rotation through 180° about origin.
Example 5: –1 0
If A(a, b) and B(c, d) are transformed by 0 –1 into A'(– 2, 3) and B'(5, – 2), find the
values of a, b, c and d.
Solution:
a c
Here, the matrix form of A(a, b) and B(c, d) is b d .
–2 5
The matrix form of A'(– 2, 3) and B(5, – 2) is 3 –2 .
–1 0 a c –2 5 –a –c –2 5
Now, 0 –1 . b d = 3 –2 i.e. –b –d = 3 –2
Equating the corresponding elements, we get
a = 2, b = – 3, c = – 5 and d = 2.
Example 6: A(1, 0), B(4, 3) and C(1, 2) are the vertices of DABC. DABC is mapped onto the DA'B'C'
whose vertices are A'(0, – 1), B'(3, – 4) and C'(2, – 1). Find the 2×2 matrix which represents this
transformation. Which is the single transformation for this mapping?
Solution:
1 4 1 0 3 2
The matrix form of the vertices of DABC and DA'B'C' are 0 3 2 and –1 –4 –1
respectively.
a b
Let c d be the required 2×2 transformation matrix, then
A B C A' B' C'
a b 1 4 1 = 0 3 2
c d 0 3 2 – 1 –4 –1
or, ca 4a + 3b a + 2b = 0 3 2
4c + 3d c + 2d – 1 –4 –1
Equating the corresponding elements of two equal matrices, we get
a = 0 Similarly, c = – 1
and a + 2b = 2 and c + 2d = – 1
or, 0 + 2b = 2 or, – 1 + 2d = – 1
\ b = 1 \ d=0
0 1
Hence, the required 2×2 matirx is – 1 0 .
276 / Optional Mathematics Class 10
Enlargements using 2×2 Matrices
The image of P(x, y) under the enlargement with centre O and scale factor k is P'(kx, ky).
i.e. P(x, y) [(0, 0), k] P'(kx, ky)
Enlargement
If P'(x', y') be the image of P(x, y) under the enlargement with centre O and scale factor k then
x' kx k×x + 0×y k 0 x
y' = ky = 0×x + k×y = 0 k y
Hence, the matrix k 0 represents the enlargement with centre O and scale factor k.
0 k
The image of (2, – 3) under the enlargement with centre O and scale factor 3 can be obtained as
3 0 2 3 × 2 + 0 × – 3 6 Transformation
0 3 –3 = 0 × 2 + 3 × – 3 = – 9 . The image of (2, – 3) is (6, – 9).
Example 7: The vertices of a rectangle are A(– 3, 3), B(2, 3), C(2, 5) and D(– 3, 5). ABCD is transformed
by a 2×2 matrix so that the co-ordinates of its image are A'(– 6, 6), B'(4, 6), C'(4, 10) and D'(– 6, 10).
Find the 2×2 matrix.
Solution:
The vertices of rectangle ABCD are A(– 3, 3), B(2, 3), C(2, 5) and D(– 3, 5). The co-ordinate matrix
A B C D
– 3 2 2 –3
of the vertices of rectangle ABCD is 3 3 5 5 .
A' B' C' D'
– 6 4 4 – 6
The matrix form of the image of ABCD is 6 6 10 10
a b
Let c d be the required transformation matrix, then
A B C D A' B' C' D'
a b – 3 2 2 – 3 – 6 4 4 – 6
c d 3 3 5 5 = 6 6 10 10
or, – 3a + 3b 2a + 3b 2a + 5b – 3a + 5b = – 6 4 4 – 6
– 3c + 3d 2c + 3d 2c + 5d – 3c + 5d 6 6 10 10
Equating the corresponding elements of the above equal matrices, we get
– 3a + 3b = – 6 ... ... (1)
i.e. a – b = 2 ... ... (2)
2a + 3b = 4
– 3c + 3d = 6 ... ... (3)
i.e. c – d = – 2 ... ... (4)
2c + 3d = 6
Transformation using Matrices / 277
Solving equations (1) and (2), we get Again, solving equations (3) and (4), we get
3c – 3d = – 6 [3 × (3)]
3a – 3b = 6 [3 × (1)]
2a + 3b = 4 2c + 3d = 6
5a = 10 5c = 0
\ a = 2 \ c = 2
and b = a – 2 = 2 – 2 = 0 and d = 0 + 2 = 2
Hence, a = 2, b = 0, c = 0 and d = 2 and the required matrix is 2 0 . Y
0 2
Unit Square
In the figure, OABC is a square having area 1 square unit. C B X
The square is called a unit square. Hence the vertices of a square X' O A
in order is O(0, 0), A(1, 0), B(1, 1) and C(0, 1).
Y'
Example 8: Find the transformation matrix in which a unit square is transformed into the parallelogram
0 3 4 1 .
0 1 3 2
Solution:
Here, the matrix form of a unit square is 0 1 1 0 .
0 0 1 1
a b
Let c d be the transformation matrix.
Then we have, a b 0 1 1 0 = 0 3 4 1
c d 0 0 1 1 0 1 3 2
or, 00 a a + b b = 0 3 4 1
c c + d d 0 1 3 2
Comparing corresponding elements of equal matrices, a = 3, b = 1, c = 1 and d = 2
a b 3 1
∴ The required transformation matrix is c d = 1 2
Example 9: The a triangle ∆ABC with vertices A(– 2, 3), B(– 4, 1) and C(1, 1) is transformed to ∆A'B'C'
by the matrix 3 0
Solution: 2 , find the vertices of ∆A'B'C'.
0
As the co-ordinates of the vertices of ∆ABC are A(– 2, 3), B(– 4, 1) and C(1, 1), its co-ordinate
A B C
– 2 – 4 1 3 0
matrix is 3 1 1 . To find the image ∆A'B'C' of ∆ABC under the transformation matrix 0 2 , we
multiply the co-ordinate matrix by as follows:
278 / Optional Mathematics Class 10
A B C A' B' C' A' B' C'
3 0 – 2 – 4 1 = 3×(– 2) + 0×3 3×(–4) + 0×1 3×1 + 0×1 – 6 – 12 3
0 2 3 1 1 0×(–2) + 2×3 0×(–4) + 2×1 0×1 + 2×1 = 6 2 2
Hence, the co-ordinates of image ∆A'B'C' of ∆ABC are A'(– 6, 6), B'(– 12, 2) and C'(3, 2).
Example 10: A rectanlge ABCD with vertices A(– 1, 1), B(2, 1), C(2, 3) and D(–1, 3) is mapped onto a
parallelogram A'B'C'D' by a 2×2 matrix so that the vertices of A'B'C'D' are A'(1, – 7), B'(4, 2),
C'(8, – 6) and D'(5, – 15). Find the 2×2 matrix.
Solution:
The matrix of vertices of ABCD is – 1 2 2 – 1 and that of A'B'C'D' is 1 4 8 5 .
1 1 3 3 – 7 2 – 6 – 15
a b
Let c d be the required transformation matrix, then
A B C D A' B' C' D'
a b – 1 2 2 – 1 1 4 8 5
c d 1 1 3 3 = – 7 2 – 6 – 15
or, – a + b 2a + b 2a + 3b – a + 3b = 1 4 8 5
– c + d 2c + d 2c + 3d – c + 3d – 7 2 – 6 – 15
Equating the corresponding elements of above equal matrices, we get
– a + b = 1 ... ... (1)
2a + b = 4 ... ... (2)
– c + d = – 7 ... ... (3)
2c + d = 2 ... ... (4)
Solving equations (1) and (2), we get Again, solving equations (3) and (4), we get
– 2a + 2b = 2 [2 × (1)] – 2c + 2d = – 14 [2 × (3)]
2a + b = 4 2c + d = 2
3b = 6 3d = – 12
\ b = 2 \ d = – 4
and – a + 2 = 1 and – c – 4 = – 7
\ a=2–1=1 \ c=–4+7=3 Transformation
Hence, a = 1, b = 2, c = 3 and d = – 4 and the required matrix is 1 2 .
3 – 4
Example 11: The vertices of a square O(0, 0), A(2, 0), B(0, 2), C(2, 2) are transformed by the matrix M,
2 0
where M = 0 2 to O'A'B'C' and then transformed to O"A"B"C" under the translation 6.
2
Find the co-ordinates of O', A', B', C' and O", A", B", C".
Solution:
The vertices of square OABC are O(0, 0), A(2, 0), B(0, 2) and C(2, 2) and its co-ordinate matrix is
O A B C
0 2 0 2 2 0
0 0 2 2 . Image of square OABC under the transformation matrix 0 2 is
Transformation using Matrices / 279
O A B C O' A' B' C'
2 0 0 2 0 2
0 2 0 0 2 2 = 2×0 + 0×0 2×2 + 0×0 2×0 + 0×2 2×2 + 0×2
0×0 + 2×0 0×2 + 2×0 0×0 + 2×2 0×2 + 2×2
O' A' B' C'
0 4 0 4
= 0 0 4 4
Hence, the vertices of O'A'B'C' are O'(0, 0), A'(4, 0), B'(0, 4) and C'(4, 4) and O'A'B'C' is translated
O" A" B" C" O " A" B" C"
by 6 and its image is 0 + 6 4 + 6 0 + 6 4 + 6 6 10 6 10
2 0 + 2 0 + 2 4 + 2 4 + 2 = 2 2 6 6
And the vertices of O"A"B"C" are O"(6, 2), A"(10, 2), B"(6, 6) and C"(10, 6).
We can summarise the following table for the different types of 2×2 transformation matrices.
SN Transformation Object Image Equations Matrix
Reflection on X-axis (x, y) (x, – y) x' = 1.x + 0.y
(x, y) (– x, y) y' = 0.x – 1.y 1 0
1. (x, y) (y, x) x' = –1.x + 0.y 0 – 1
Reflection on Y-axis (– y, – x) y' = 0.x + 1.y
(– y, x) – 1 0
2. (– x, – y) x' = 0.x + 1.y 0 1
(y, – x) y' = 1.x + 0.y
Reflection on y = x (kx, ky) 0 1
3. (x, y) x' = 0.x – 1.y 1 0
y' = –1.x + 0.y
4. Reflection on y = – x (x, y) 0 – 1
(x, y) x' = 0.x – 1.y – 1 0
5. Rotation through 90° (x, y) y' = 1.x + 0.y
about origin (x, y) 0 – 1
(x, y) x' = –1.x + 0.y 1 0
6. Rotation through (x, y) y' = 0.x – 1.y
180° about origin – 1 0
x' = 0.x + 1.y 0 – 1
7. Rotation through y' = –1.x + 0.y
270° about origin 0 1
x' = k.x + 0.y – 1 0
Englargement with y' = 0.x + k.y
8. centre at origin and k 0
x' = 1.x + 0.y 0 k
scale factor k y' = 0.x + 1.y
1 0
9. Identity transformation 0 1
280 / Optional Mathematics Class 10
Exercise 7.7A Transformation
6
1. a. Find the image of a point A(– 4, – 2) under the translation through translation vector T = 4 .
3
b. Find the image of a point M(2, – 5) under the translation through translation vector T = – 4 .
2. a. If a point A(3, 5) is transformed to A'(7, 2) through a tanslation vector T, find T.
b. If a translation vector T1 transforms a point M(6, 6) to M'(– 2, 8), find the vector T1.
c. Find a 2×1 matrix which transforms a point (p, q) into the point (p – 8, q + 5).
1 0
3. a. If a point A(3, 6) is transformed to A' by a tranformation matrix 0 2 , find the co-ordinates of A'.
1 2
b. If a transformation matrix 0 3 transforms a point B(3, 7) to B', find the co-ordinates of B'.
0 1
4. a. What transformation does the matrix 1 0 represent?
0 1
b. To what transformation is the matrix – 1 0 associated?
– 1 0
c. Which types of transormation does 0 – 1 represent?
k 0
d. What transformation does the matrix 0 k represent?
5. a. Write a transformation matrix which represents the reflection in the line y = – x.
b. Write a transformation matrix which represents the reflection in Y-axis.
c. Write a transformation matrix which represents the rotation through +90° about origin.
d. Write a transformation matrix which represents the enlargment with scale factor 2 and centre
at origin.
6. a. Find a 2×1 matrix which transforms a point (a, b) into the point (a + 3, b – 6). Using the same
matrix transform the point (6, 10).
b. Find a 2×1 matrix which transforms a point (3, 8) into the point (0, 2). Using the same 2×1
matrix transform the point (10, 8).
0 – 1
7. a. If a point is transformed to (4, – 2) by a transformation matrix – 1 0 , find the given point.
1 0
b. A transformation matrix 0 2 transforms a point A(a + 1, b) onto A'(3, – 6). Find the values
of a and b.
2 1
c. If a point B(3, 4) is mapped onto B'(2p, 4q – 1) by a transformation matrix 1 3 , find the
values of p and q.
u 2
d. If a transformation matrix v – 1 1 maps a point C(2, – 3) onto C'(– 8, 3), find the values of u
and v.
Transformation using Matrices / 281
8. a. If a line AB with ends at A(2, 3) and B(5, 1) is transformed into A'B' by a translation matrix
1 2 , find the co-ordinates of A' and B'.
0 3
2 3
b. Find the image of the points A(1, 2) and B(3, 4) under the transformation by matrix 3 2 .
0 1
9. a. If ∆ABC with vertices A(3, 2), B(5, 5) and C(7, 3) is transformed by a matrix 1 0 , find the
co-ordinates of image of ∆ABC.
b. If A(2, 3), B(– 1, 2), C(0, 4) and D(3, 6) are the vertices of a quadrilateral, find its image given
1 0
by matrix 0 – 1 .
0 – 1
c. A transformation T has matrix – 1 0 , find the images A', B' and C' of the points A(1, 1), B(3,
1) and C(2, 4).
– 1 0
10. a. Which types of transformation does 0 1 represent ? Also find the image of point
(– 3, – 1) by using this matrix.
b. What transformation does the matrix 0 – 1 denote ? Using the given matrix, find the image
– 1 0
of the point (– 3, 2). 0 – 1
1 0
c. To what transformation is the matrix associated ? Use this matrix to transform the
point (a, b). 2 1 2
3 4 followed by translation 4
11. a. Find the image of B(3, 4) under the transformation by matrix .
b. Find the image of A(4, 7) under the transformation by a matrix 1 2 followed by – 2 .
2 1 – 3
12. a. Find the image of the points A(2, 3) and B(3, –2) under the transformation by matrix 2 1 .
–1 3
b. If P(–1, 3), Q(2, 1) and R(3, 4) are the vertices of DPQR, find the image given by matrix 1 2 .
3 1
13. a. If A(a, b) and B(c, d) are transformed by 0 1 into A'(2, – 3) and B'(– 2, 5), find the values
–1 0
of a, b, c and d.
b. If P(u + 3, v) and Q(5, – 3) are transformed by 2 0 into P'(8, – 4) and Q'(3w + 1, x – 5), find
0 2
the values of u, v, w and x.
c. If C(a + 5, – 2) and D(2b – 1, 6) are transformed by 3 0 into C'(9, – 4) and D'(15, 12), find
0 c
the values of a, b and c.
282 / Optional Mathematics Class 10
Exercise 7.7B Transformation
0 1
1. a. Pre-multiply the co-ordinates of the points A(1, 1), B(4, 2) and C(4, 5) by the matrix – 1 0
to obtain A', B' and C'. Plot the points A, B, C and A', B', C' on the grid and deduce the
transformation represented by this matrix.
b. Pre-multiply the co-ordinates of vertices P(2, – 1), Q(3, 3) and R(5, 2) of the triangle PQR by
– 1 0
the matrix 0 – 1 to obtain DP'Q'R'. Plot both triangles PQR and P'Q'R' on squared paper and
deduce the transformation represented by this matrix.
c. A triangle ABC with vertices A(– 2, 3), B(– 4, – 1) and C(– 3, 5) is transformed to ∆A'B'C' by
2 0
the matrix 0 3 , find the co-ordinates of ∆A'B'C' and draw them in the same graph.
d. ∆PQR is a triangle with vertices P(2, 3), Q(– 5, 4) and R(3, 2). Transform the given triangle by
1 3
a matrix 2 1 and find the co-ordinates of the vertices of its image.
2. a. Write a 2×2 transformation matrix which represents reflection on Y-axis and hence apply this
matrix to the DPQR whose vertices are P(1, 2), Q(3, – 1) and R(5, 3) and find the image of
DPQR.
b. Write a 2×2 transformation matrix which represents a rotation through +270° about origin and
hence apply this matrix to DABC having vertices at A(1, 1), B(– 3, 2) and C(– 2, 6) and find the
image of DABC.
c. An enlargement at the centre O has the scale factor 2. Write down the matrix to represent the
enlargement and hence apply this matrix to the ∆ABC whose vertices are A(1, 0), B(3, 2) and
C(2, 3) and find the image of ∆ABC.
3. a. Find the 2×2 transformation matrix which transforms square ABCD with vertices
A(2, 3), B(4, 3), C(4, 5) and D(2, 5) into a square A'B'C'D' with vertices A'(3, 2), B'(3, 4),
C'(5, 4) and D'(5, 2).
b. Find the 2×2 transformation matrix which transforms line segment AB with ends A(3, 2) and
B(5, 4) into a line segment A'B' with ends A'(– 3, 2) and B'(– 5, 4).
c. DPQR with the vertices P(– 3, – 4), Q(– 4, – 6) and R(– 1, – 8) is mapped onto the DP1Q1R1,
whose vertices are P1(4, 3), Q1(6, 4) and R1(8, 1). Which is the single transformation for this
mapping ? And also find the 2 × 2 matrix which represents this transformation.
4. a. DABC with vertices A(3, 6), B(4, 2) and C(3, 3) maps onto DA'B'C' with the vertices
A'(– 6, 3), B'(– 2, 4) and C'(– 3,3). Find 2×2 matrix that represents this transformation. Also
find which is the single transformation for this mapping.
b. DPQR with the vertices P(– 2, 4), Q(6, 2) and R(1, – 1) is mapped onto the DP1Q1R1, whose
vertices are P1(4, 2), Q1(2, – 6) and R1(– 1, – 1). Which is the single transformation for this
mapping ? Also find 2 × 2 matrix which represents this transformation.
Transformation using Matrices / 283
c. A triangle ABC having vertices A(3, 6), B(5, – 3) and C(– 4, 2) is transformed by a 2×2 matrix
so that the co-ordinates of the vertices of its image are A'(– 3, – 6), B'(– 5, 3) and C'(4, – 2). Find
the 2×2 matrix.
5. a. DABC with vertices A(5, 1), B(4, 4) and C(2, 1) is mapped onto a triangle A'B'C' by a 2×2
matrix so that the vertices of DA'B'C' are A'(10, 2), B'(8, 8) and C'(4, 2). Find the 2×2 matrix.
b. Two points A'(4, 2) and B(6, – 2) are mapped onto A'(2, 1) and B'(3, – 1) by a 2×2 matrix. Find
the 2×2 matrix.
6. a. A square ABCD with vertices A(2, 0), B(5, 1), C(4, 4) and D(1, 3) is mapped onto a
parallelogram A'B'C'D' by a 2×2 matrix so that the vertices of the parallelogram are A'(2, 2),
B'(7, 3), C'(12, – 4) and D'(7, – 5). Find the 2×2 transformation matrix.
b. Points P(0, 0), Q(2, 0), R(2, 1) and S(0, 1) are transformed by a 2×3 matrix to P'(0, 1), Q'(2, 4),
R'(2, 5) and S'(0, 1) respectively. Find the 2×2 matrix.
c. A square PQRS having the vertices P(0, 3), Q(1, 1), R(3, 2) and S(2, 4) is mapped onto a
parallelogram P'Q'R'S' with the vertices P'(6, – 6), Q'(3, – 1), R'(7, – 1) and S'(10, – 6) by a 2 × 2
matrix. Find 2 × 2 matrix.
d. A square ABCD whose vertices are A(0, 0), B(0, 1), C(1, 1), D(1, 0) is mapped onto the
square A'B'C'D' by a 2 × 2 matrix so that the vertices of the square A'B'C'D' are A'(0, 0),
B'(0, – 2), C'(2, – 2) and D'(2, 0). Find the 2 × 2 matrix.
7. a. Find a 2×2 transformation matrix in which a unit square 0 1 1 0 is transformed into
0 0 1 1
0 3 5 2
a parallelogram 0 1 2 1 .
b. Find a 2×2 transformation matrix in which a unit square 0 1 1 0 is transformed into
0 0 1 1
0 3 4 1
a parallelogram 0 1 3 2 .
8. a. Fina a 2×2 transformation matrix in which trasforms a unit square to the parallelogram
0 6 8 2 .
0 2 6 4
b. Fina a 2×2 transformation matrix in which trasforms a unit square to the parallelogram
0 4 6 2 .
0 1 3 2
284 / Optional Mathematics Class 10
9. a. a 2 transforms a unit square matrix to a parallelogram 0 4 c 2 , find
If the matrix b 2 0 2 3 d
the values of a, b, c and d.
b. If the matrix 2 c transforms a unit square matrix to a parallelogram 0 2 a + 1 2 ,
1 d 0 1 b – 2 4
find the values of a, b, c and d.
10. a. DABC with vertices A(3, 2), B(– 3,1), C(0, 4) maps onto DA'B'C' with vertices A'(8, 7), B'(c, – 1)
2 a
and C'(4, 2d). If a matrix 1 b represents above transformation, find the values of a, b, c and d.
3 4
b. A matrix c – 1 transorms DPQR to DP'Q'R'. If DPQR has vertices P(4, – 1), Q(1, b) and R(2, 2)
and DP'Q'R' has vertices P'(3a – 1, 9), Q'(– 9, d) and R'(14, 2), find the values of a, b, c and d.
11. a. The vertices of a square O(0, 0), A(2, 0), B(2, 2) and C(0, 2) are transformed by the matrix
M= 0 2 to O'A'B'C' and then transformed to O"A"B"C" under the translation – 3 . Find
1 0 2
the co-ordinates of O'A'B'C' and O"A"B"C".
b. Describe the transformation on the rectangle 0 2 2 0 by matrix 2 0
0 0 1 1 0 2 to O'A'B'C'
2
and then transform to O"A"B"C" under the translation 4 . Find the co-ordinates of both
images.
12. a. A square ABCD with vertices A(1, 1), B(3, 1), C(3, 3) and D(1, 3) is transformed by 1 2
2 – 1
3 4
to A'B'C'D'. Again A'B'C'D' is transformed by – 3 1 to A"B"C"D". Find the co-ordinates of
vertices of images A'B'C'D' and A"B"C"D".
b. – 1 2 Transformation
If a rectangle PQRS is transformed by 0 – 3 to P'Q'R'S' and P'Q'R'S' is agin transformed by
1 2
2 3 to P"Q"R"S". If the vertices of PQRS are P(– 1, 1), Q(3, 1), R(3, 3) and S(– 1, 3), find
the co-ordinates of vertices of P'Q'R'S' and P"Q"R"S".
13. a. Using matrix method show that a reflection in the line y = x followed by a reflection in the
Y-axis is equivalent to the rotation through 90° about origin.
b. Show by using matrix method that reflection about the line y = x followed by the rotation about
the origin through 90° positive is the reflection about Y-axis.
Transformation using Matrices / 285
8. STAsdTafsdIfSa TICS
Curriculum Contents Included Periods
17
Dispersion
– Quartile Deviation and its co-efficient (Continuous Series)
– Mean Deviation (from Mean and Median) and its co-efficient (Continuous Series)
– Standard Deviation and its co-efficient and analysis (Continuous Series)
– Co-efficient of Variation (Continuous Series) (Co-efficient of variance)
Objectives
On completion of this unit, students will be able to:
find median and quartile of continuous data.
find the mean deviation from mean or median and its co-efficient of given data.
find the standard deviation and variance and their co-efficients of given data.
Teaching Materials
Find the standard deviation from the follwing data.
Class Interval 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55
5
Frequency 7 10 16 9
fx2
Calculating the standard deviaion 700
4000
Class Frequency (f) Mid-value (x) fx 14400
14400
5 – 15 7 10 70 12500
46000
15 – 25 10 20 200
25 – 35 16 30 480
35 – 45 9 40 360
45 – 55 5 50 250
Total 47 1360
Here, N = 47, ∑fx = 1360 and ∑fx2 = 46000
Standard deviation (σ) = ∑fx2 – ∑fx 2 = 46000 – 1360 2 = 978.723 – 837.302 = 11.892
N N 47 47
Specification Grid
Contents Knowledge Understanding Application Higher ability Total Total
Statistics Each of 1 mark Each of 5 marks Quesitons Marks
Each of 2 marks Each of 4 marks
– – 3 10
12
286 / Optional Mathematics Class 10
8 STATISTICS
8.1 Partition Values
Median of Grouped Data
In continuous frequency distribution, the class corresponding to the cumulative frequency just greater
N
than or equal to 2 is called the median class.
Arrange the data in the ascending (or descending) order if necessary. The median is calculated by
using the following formula.
N
2 – cf
f
Median (Md) = + × h
where, = the lower limit of median class
N = the total of frequencies
cf = cumulative frequency preceeding the median class
f = the frequency of the median class
h = size of the median class
Example 1: Calculate median from the following dada:
Marks 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100
No. of Students 10 20 10 25 10 50 40 30
Solution:
Computation of Median
Marks Frequencies (f) Cumulative Frequencies (cf)
20 – 30 10 10
30 – 40 20 30
40 – 50 10 40
50 – 60 25 65
60 – 70 10 75
70 – 80 50 125
80 – 90 40 165
90 – 100 30 195
Position of median = N = 195 = 97.5
2 2
Cumulative frequency just greater than N (i.e. 97.5) is 125. Statistics
2
Median class is 70 – 80.
Here, = 70, cf = 75, f = 50, h = 80 – 70 = 10
N – cf 97.5 – 75
2 f 50
Median (Md) = + × h = 70 + × 10 = 70 + 4.5 = 74.5
Partition Values / 287
Example 2: If the median of frequency distribution given below is 46, find the missing frequencies.
Wages (Rs.) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Total
No. of Workers 12 30 ? 65 ? 25 19 230
Solution:
Let a and b be the two missing frequencies. The cumulative frequency table is as follows:
Class Frequency Cumulative frequency
10 – 20 12 12
20 – 30 30 42
30 – 40 a 42 + a
40 – 50 65 107 + a
50 – 60 b 107 + a + b
60 – 70 25 132 + a + b
70 – 80 19 151 + a + b
Total 230
Here, N = 151 + a + b = 230
a + b = 79 ... ... (1)
Median = 46
The median class is 40 – 50.
N
2 – cf
f
Median (Md) = + × h
Where, = 40, f = 65, c.f. = 42 + a and h = 10
or, 46 = 40 + 115 – (42 + a) × 10
65
or, 6 = 115 – 42 – a ×2
13
or, 39 = 73 – a
∴ a = 73 – 39 = 34
and a + b = 79
or, 34 + b = 79
b = 79 – 34 = 45
Hence, the missing frequencies are 34 and 45.
Quartiles of Grouped Data
Quartiles of a set of data are the items which divide the total number of items into four equal parts.
Hence, there are three quartiles, the first quartile (Q1), the second quartile (Q2) and the third quartile (Q3).
In the continuous frequency distribution, quartiles are calculated by using the formula.
iN – cf
4
Qi = + f × h, where, i = 1, 2, 3
288 / Optional Mathematics Class 10
N – cf
4
Thus, the first quartile (Q1) = + f × h
N
2 – cf
The second quartile (Q2) = + f × h
3N
4 – cf
The third quartile (Q3) = + f × h
Example 3: Find the first quartile (lower quartile) and the third quartile (upper quartile) of the follow-
ing data:
Wages (Rs.) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of Workers 15 20 23 9 5 4
Solution:
COMPUTATION OF QUARTILES
Wages (Rs.) No. of Workers (f) Cumulative Frequencies (cf)
20 – 30 15 15
30 – 40 20 35
40 – 50 23 58
50 – 60 9 67
60 – 70 5 72
70 – 80 4 76
N th 76 th
4 4
Here, N = 76, and the first quartile = item = = 19th item.
Cumulative frequency just greater than 19 is 35.
Therefore, the first quartile lies in the class interval 30 – 40.
Now, = 30, N = 76 = 19, cf = 15, f = 20, h = 10
4 4
N – cf 19 – 15
4 f 20
Q1 = + × h = 30 + × 10 = 30 + 2 = 32
Hence, the first quartile (Q1) is 32.
th th
3N 3 × 76
The third quartile = 4 item = 4 = 57th item.
Cumulative frequency just greater than 57 is 58.
Therefore, the first quartile lies in the class interval 40 – 50.
Now, = 40, 3N = 3 × 72 = 57, cf = 35, f = 23, h = 10 Statistics
4 4
3N – cf 57 – 35
4 f 23
Q1 = + × h = 40 + × 10 = 30 + 9.57 = 39.57
Hence, the third quartile (Q3) is 39.57.
Partition Values / 289
Deciles of Grouped Data
Deciles of the given data are the items which divides given set of data into ten equal parts after
arranging them in ascending (or descending) orders. Therefore, there are nine deciles, D1, D2, D3, ... , D9.
In the continuous frequency distribution, deciles are calculated by using the formula.
iN – cf
10
Di = + f × h, where, i = 1, 2, 3, ... ..., 9
N
10 – cf
Thus, the first decile (D1) = + f × h
2N – cf
10
The second decile (D2) = + f × h
3N
10 – cf
The third decile (D3) = + f × h
... ... ... ... .... .... .... .... .... .... .... .... ....
9N
10 – cf
The ninth decile (D9) = + f × h
Percentiles of Grouped Data
Percentiles are the values of given data which divide the given set data into 100 equal parts after
arranging them in ascending (or descending) order. Here, there are 99 pertiles P1, P2, P3, ... ..., P99.
In the continuous frequency distribution, percentiles are calculated by using the formula.
iN – cf
100
Pi = + × h, where, i = 1, 2, 3, ... ..., 99
f
N – cf
100 ×
Thus, the first percentile (P1) = + h
f
2N – cf
100
The second percentile (P2) = + × h
f
3N – cf
100
The third percentile (P3) = + × h
f
... ... ... ... .... .... .... .... .... .... .... .... .... 99N
100
– cf
The ninety ninth percentile (P99) = + f × h
290 / Optional Mathematics Class 10
Example 4: Calculate the 3rd decile and the 67th percentile of the following data.
Age (yrs) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
13
No. of workers 11 15 25 26 23
Solution:
Calculation of the Deciles
Ages (yrs) Number of workers (f) Cumulative frequencies (cf)
10 – 20 11 11
20 – 30 15 26
30 – 40 25 51
40 – 50 26 77
50 – 60 23 100
60 – 70 13 113
Here, N = 113
3N th 3 × 113 th
10 10
The 3rd decile = item = item = 33.9th item
Cumulative frequency just greater than 33.9 is 51.
Therefore, the 3rd decile lies in the class interval 30 – 40.
3N
So, l = 30, 10 = 33.9, cf = 26, f = 25, h = 10
3N – cf
10 f
Hence, D3 = + × h
= 30 + 33.9 – 26 × 10 = 30 + 7.9 × 2
25 5
= 30 + 3.16 = 33.16
Hence, the 3rd decile (D3) is 33.16.
67N th 67 × 113 th
100 4
The 67th percentile = item = item = 75.71th item
Cumulative frequency just greater than 75.71 is 77.
Therefore, the 67th percentile lies in the class interval 40 – 50.
So, l = 40, 67N = 75.71, cf = 51, f = 26, h = 10
100
67N
100 – cf
Hence, P67 = + f × h
67 × 113 – 51 75.71 – 51 Statistics
100 26
= 40 + 26 × 10 = 40 + × 10
= 40 + 9.50
= 49.50
Hence, the 67th percentlile (P67) is 49.50.
Partition Values / 291
Exercise 8.1
1. Find the median of the following data.
a. Marks obtained 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35
b. No. of students 11 16 15 6 854
Interval 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
c. Frequency 32456
Class 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180
d. Frequency 5 4 22 25 14 4
x 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65
f 5 8 10 9 11 1
40 – 50
2. a. The median of the given data is 32, find the value of p.
Class Interval 0 – 10 10 – 20 20 – 30 30 – 40
Frequency 2 3 4 p 6
50 60
b. If the medain of the given data is 34, calculate the value of a.
Age (yrs) 10 – 20 20 – 30 30 – 40 40 – 50
No. of workers 4 12 a 9 5
c. In the data given below the median (Md) = 45, find the value of k.
x
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
f 4 7 9 18 k 7 3
d. The median (Md) of the data given below is 27.5 years. Find the value of b.
Age in years 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of people 2 3 4 b 3
50 – 60
3. Find the lower or the first quartile (Q1) from the data given below.
a.
Class interval 10 – 20 20 – 30 30 – 40 40 – 50
b. Frequency 2 5655
Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
c. Frequency 4 6 10 20 10 6
Weight (kg) 8 – 10 10 – 12 12 – 14 14 – 16 16 – 18 18 – 20
No. of children 6 4 3 5 7 5
d. Wages (Rs.) 100 – 125 125 – 150 150 – 175 175 – 200 200 – 225
No. of workers 3 4 6 5 2
292 / Optional Mathematics Class 10
e. Marks 90 – 100 80 – 90 70 – 80 60 – 70 50 – 60 40 – 50 30 – 40
No. of students 4 9 12 50 10 3 2
4. Calculate the upper or the third quartile (Q3) from the table given below.
a.
Marks obtained 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of students 5 6 8 10 7 4
b.
Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students 5 4 5 10 7 6
c. Class 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 100 – 110 110 – 120
Frequency 8 10 16 14 10 5 2
5. a. According to the given data, find the maximum marks obtained by lower 75% students.
Class 0–5 5 – 10 10 – 15 15 – 20 20 – 25
Frequency 6 4 7 5 8
b. According to the given data, find the maximum marks obtained by lower 25% students.
Class 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30
Frequency 4 6 8 8 4
6. Calculate the fourth and seventh deciles from the table given below.
a.
Marks obtained 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of students 5 6 12 13 8 6
b.
Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students 5 4 5 10 7 6
c. Class 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 100 – 110 110 – 120
Frequency 8 10 16 14 10 5 2
25 – 30
7. a. According to the given data, find the 23rd and 59th percentiles.
Class 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25
Frequency 6 20 27 15 8 5
b. According to the given data, find the 47th and 85th percentiles.
Class 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40
10 6
Frequency 14 36 28 12 14 Statistics
Partition Values / 293
8.2 Quartile Deviation and Mean Deviation
Introduction
Dispersion is the measure of variations in the values of the variable. It measures the degree of
scatterness (or spread out) of the observations in the given data. Quartile deviation, Mean deviation and
Standard deviation are some different tools for calculating the dispersion.
Quartile Deviation
The dispersion depending on the lower quartile (Q1) and upper quartile (Q3) is the quartile deviation.
For the given data, the quatile deviation or semi interquartile range is given by
Q.D. = Q3 – Q1 , where Q1 and Q3 are the first and third quartiles respectively.
2
The co-efficient of the quartile deviation is calculated by the formula,
Co-efficient of Q.D. = Q3 – Q1
Q3 + Q1
Mean Deviation
The absolute value of the deviation of the items from the average is used in the caluclation of the
mean deviation. The mean deviation is calculated from any one average of mean or median. The mean
deviation is denoted by M.D. Similarly, the another average deviation is the quartile deviation.
For continuous series, the mean deviation is given by
M.D. = ∑f|x – A|2 , where x is the mid-value of the class intervals and A is any average mean ( ‒x )
N
or median (Md) or mode (Mo).
The co-efficient of the mean deviation is calculated by the formula,
Co-efficient of M.D. = M.D.
A
Example 1: Find the quartile deviation and its co-efficient from the following data.
Marks 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
No. of students 12 20 25 18 5
294 / Optional Mathematics Class 10
Solution: No. of students (f) Cumulative frequency (c.f.)
12 `12
Marks 20 32
0 – 20 25 57
20 – 40 18 75
40 – 60 5 80
60 – 80
80 – 100 N = 80
Total
Here, total number of observations (N) = ∑f = 80
N th 80 th 20th
4 4
First quartile (Q1) = item = item = item.
Cumulative frequency equal to or just greater than 20 is 32, so the Q1 lies in the corresponding class i.e.
20 – 40 and
N – cf
4 f
Q1 = + × h
= 20 + 20 – 12 × 20 = 20 + 8 = 28
20
3N th 3 × 80 th 60th
4 4
Third quartile (Q3) = item = item = item.
Cumulative frequency equal to or just greater than 60 is 75. So, Q3 lies in the corresponding class i.e.
60 – 80 and
3N – cf
4 f
Q3 = + × h
= 60 + 60 – 57 × 20 = 60 + 3.33 = 63.33
18
The quartile deviation (Q.D.) = Q3 – Q1 = 63.33 – 28 = 35.33 = 17.67
2 2 2
Now, the co-efficient of Q.D. is Q3 – Q1 = 63.33 – 28 = 35.33 = 0.39
Q3 + Q1 63.33 + 28 91.33
Hence, the quartile deviation is 17.67 and its co-efficient is 0.39.
Example 2: Find the mean deviation from mean and its co-efficient. Statistics
Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of students 4 7 9 18 12 7 3
Partition Values / 295
Solution: Frequency (f) Mid-value (x) fx |x – x‒ | f |x – ‒x |
Marks 4 15 60 30 120
7 25 175 20 140
10 – 20 9 35 315 10 90
20 – 30 18 45 810 0 0
30 – 40 12 55 660 10 120
40 – 50 7 65 455 20 140
50 – 60 3 75 225 30 90
60 – 70 60 2700 700
70 – 80
Total
Here, total no. of observations (N) = ∑f = 60, ∑fx = 2700
Arithmetic mean ( ‒x ) = ∑fx = 2700 = 45 x‒ |
Now, N 60
∑f |x – 700
mean deviation from mean M.D. = N = 60 = 11.667
Co-efficient of M.D. = M.D. = 11.66 = 0.259
‒x 45
Hence, mean deviation is 11.667 and its co-efficient is 0.259.
Example 3: Compute the mean deviation from median of the following frequency table. Also find its
co-efficient.
Class Interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 8 10 12 20 12 6
Solution:
Computation of Mean Deviation
Class Mid value (x) f c.f. |x –Md| f |x –Md|
0 – 10 5 8 8 27 216
10 – 20 15 10 18 17 170
20 – 30 25 12 30 7 84
30 – 40 35 20 50 3 60
40 – 50 45 12 62 13 156
50 – 60 55 6 68 23 138
Total 68 824
N th
2
Median (Md) = Value of item
296 / Optional Mathematics Class 10
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