Shubharambha OPT Mathematics 10 Final for CTP 2077

angle is found to be 30º. Find the height of the house and breadth of river.

(m) A flagstaff stands on the top of a pole. The angles of elevation of the top of
the pole and the top of the flagstaff at a point 40m away from the foot of
the pole are 45º and 60º respectively. Find the height of the flagstaff.

(n) A statue of height 7m stands on the top of a tower. The angle subtended
by the tower and the statue at a point on the ground are 45º and 15º
respectively. Find the height of the tower.

(o) A column and a flagstaff on its top subtend angles of 45º and 15º respectively
at a point on the ground. If the length of this flagstaff is 4.6m, find the
height of the column and the distance of the point on the ground from the
base of the column.

(p) A man, 2m high, observes that the top of a statue standing on a tower has
an elevation of 60º and the bottom has an elevation 45º at his eye. The
distance of the bottom of the tower from the man is 60m. Find the height
of the tower and the statue.

(q) The angle of elevation of the top of an incomplete house from a distance of
100m from its base is found to be 45º. What height should it be raised so
that the elevation may changes to 60º ?

4. (a) From the top of 200m high cliff the angles of depression of the top and
bottom of a tower are observed to be 45º and 60º respectively. Find the
height of the tower.

(b) From the top of a 100m high tower, the angles of depression of the top and
the bottom of a tree are 30º and 45º. How high is the tree ?

(c) From the top of a tower the angle of depression of the foot of a 20m high
column on the ground is 60º and the angle of depression of the foot of the
tower observed from the top of the column is 30º. What is the height of the
tower ?

(d) The angles of elevation of the top of a rock from the top and bottom of a
200m high tower are 30º and 45º respectively. Find the height of the rock.

(e) From the top of a 15m high clock-tower an observer finds the angle of
elevation of the top of a tree to be 45º and the angle of depression of the foot
of the tree to be 30º. Find the height of the tree.

(f) From the top of a 200m high island, the angle of elevation of the top of a
mountain is found to be 45º and the angle of depression of its foot is 30º.
Find the height of the mountain.

(g) A pole is 50m high. From the foot and the top of the pole the angles of
elevation of a clock-tower situated in front of the pole are found to be 60º
and 45º respectively. Find the height of the clock-tower and its distance
from the pole.

(h) A house is 25m high. The angle of elevation of the top of a tower which is
in front of the house is 30º from its top and 45º from its foot.

(i) What is the height of the tower ?

Infinity Optional Mathematics Book - 10 251

(ii) What is the distance between the house and the tower ?

(i) From the roof of a house 30 meters high, the angle of elevation of the top of
a tower is 45º and the angle of depression of its foot is 30º. Find the height
of the tower.

(j) When the top of the building is seen from the top and bottom of a tower,
the angles of depression and elevation are 60º and 30º respectively. If the
height of the tower is 160m, find the height of the building.

(k) There are two columns of heights 9m and 3m respectively. From the foot of
the second the elevation of the top of the first is found to be 60º. Find the
angle of elevation of the top of the second from the foot of the first.

(l) The height of a tree is twice that of a house, both of which lie on the same
horizontal ground. The angles of elevation of the top of the tree from
the bottom and the top of the house are complementary. If the distance
between the tree and the house is 20m, find the height of the tree.

5. (a) Two poles stand on the either side of the road. At the point mid-way
between the them, the angles of elevation of their tops are 45º and 60º.
Find the length of the shorter pole if the longer pole is 15m.

(b) Two houses of equal height stand on either side of road which is 50m. wide.
At a point on the road-way between the house the elevation of the tops of
the houses are 60º and 30º. Find their height and the position of the point.

(c) A ladder 25m long stands against a house at one side of a road at an angle
of 30º with the house. When it is turned so that is rests against another
house on the other side of the road, it makes an angles of 45º with the
house. Find width of the road.

(d) Two pillars are of equal height. A boy standing midway between them
observes the elevation of either pillar to be 45º. After walking 20m towards
one of them he observes its elevation to be 60º. Find the height of pillar and
the distance between them.

(e) A rope dancer was walking on a loose rope tied to the top of two equal posts
of height 20m. when he was 5m above the ground, it was found that the
stretched pieces of the rope made angles of 30º and 60º with the horizontal
line parallel to the ground. Find the length of rope.

6. (a) A man 1.8m high stands at a distance of 2.4m from the lamp post and it is
observed that his shadow is 3.6m long. Find the height of the lamp post.

(b) A ladder 10m long reaches a point 10m below the top of a vertical flagstaff.
From the foot of the ladder, the angle of elevation of the top of the flagstaff
is 60°. Find the height of the flagstaff.

(c) The angles of elevation of the top a tower as observed from the distance of
36m and 16m from the foot of the tower on the same side are found to be
complementary. Find the height of the tower.

252 Infinity Optional Mathematics Book - 10

(d) The angles of elevation of the top of a tree from two points at distance
of 16m and 9m from its base on the ground on the opposite sides are
complementary. Find the height of the tree.

(e) A vertical pole is divided at a point in the ratio of 1:9 from the base. If the
two parts of the pole subtend equal angles at a point 20m from the foot of
the pole, find the height of the pole.

(f) AB is a vertical pole with its foot B on a level ground. C is a point on AB
such that AC:CB = 3:2. If the parts AC and CB subtend equal angles at a
point on the ground, which is at a distance of 18m from the foot of the pole,
find the height of the pole.

(g) A man of 1.68m high observed the angles of elevation of the top of a house
and its window from a place and found to be 45º and 30º respectively. If the
height of the window from the ground is 11.68m, calculate the height of
the house.

(h) A poster hanging on a wall has a vertical height 3.66 m. From a point
5m away from the wall on the same plane, the angle of elevation of the
bottom edge of the poster was found to be 45º. What will be the angle of
elevation of the top of the poster if it is observed from the same point on
the horizontal plane ?

(i) The angle of elevation of the top of a tower is 45º from a point 10m above
the water level of a lake. The angle of depression of its image in the lake is
60º. Find the height of the tower above water level.

(j) A tower subtends an angle of 60º at a point on the same level of the foot of
the tower. At a second point 20m above the first, the angle of depression of
the foot of the tower is 45º. Find the height of the tower.

(k) An aeroplane flying horizontally at a height of 750m above the ground is
observed at an elevation of 60º. If after 5 seconds, the elevation is observed
to be 30º, find the speed of the aeroplane in km per hour.

(l) From the top of a cliff at the seashore a man finds the angle of depression of
a ship coming towards it to be 30º. After 20 minutes, the angle of depression
is observed to be 45º. If the height of the cliff is 2km, find th speed of the
ship.

(m) When the kite is at a height of 500m from the ground its string makes an
angle of 60º with the ground. When the speed of the wind decreases, the kite
falls down to the height of 500 m from the ground. Find the angle made by
3
the string of the kite with the ground.

(n) The shadow of a tower is 50 m long in the sun. At the same time, the
shadow of a pole 5 m in the length is 5 3 m long. Find the altitude of the
sun and the height of tower.

Infinity Optional Mathematics Book - 10 253

UNIT VECTORS

6

Review

Discuss the following questions in a group.

 Define magnitude of a vector with formula.

 Define direction of a vector with formula.

 Find the magnitude and direction of the following vectors.

(i) a = (3, 3) (ii) b = (–3 3, 3)

(iii) c = (–5, –5 3) (iv) d = (1, – 3)

 A vector AB displaces A(3, 5) to B(6, –3). Find the magnitude and direction of
AB.

 Define following vectors with one example of each.

(i) Row vector (ii) Column vector

(iii) Position vector (iv) Null or zero vector

(v) Unit vector (vi) Negative vector

(vii) Parallel vectors

 Find the unit vector of the following vectors:

(i) a = (3, 4) (b) b = (3 3, 3)

 Show that A 1, 1 and B 25, 1 are the initial point and terminal point of a
5

unit vector AB.
 Define triangle law of vector addition with figure.
 Define parallelogram law of vector addition with figure.

6.1 Product of the vectors

There are two types of product of vectors. B
(i) Scalar product or dot product
(ii) Vector product or cross product

Scalar Product or Dot Product

The scalar product or dot product of two vectors a and b is defined b A
as the product of their magnitudes multiplied by the cosine of the θ

Oa

254 Infinity Optional Mathematics Book - 10

angle made by the two vectors.

Symbolically, we write scalar product as a.b = |a||b|cosθ = abcosθ.

Where θ is the angle between the two vector a and b.

Case I

When the two vector a and b are perpendicular to each other, then the angle

between them θ = 90°. B
Now, a.b = |a|.|b| cosθ = abcos90° = 0

∴ a.b = 0

b

Thus, when the two vectors are perpendicular to each other,

their scalar product is zero. O A

Conversely, when a.b = 0, then a

|a||b|cosθ = 0

or, cosθ = 0

cosθ = cos90°

∴ θ = 90°

Hence, when the dot product of two vectors is zero, then they are perpendicular

(orthogonal) to each other.

Case II bB
When the two vectors a and b are like parallel, then angle O

between them θ = 0°. Now, OA
a.b = |a|b| cosθ = abcos0° = ab.1 = ab
∴ a.b = ab. a

Case III

When the two vectors a and b are unlike parallel, then angle between them θ

= 180°, Now, A 180° B
a.b =|a|b| cosθ = abcos180° = ab × (–1) = –ab.
∴ a.b = –ab. ab

O

Properties of scalar product

For two vectors a and b, the following properties hold in the scalar product.

(i) a.b is a scalar

(ii) a.b = b .a (Commutative property) Y
(iii) a2 = a.a = |a|2
(iv) k(a. b) = ka.b

(v) a.(b + c ) = a.b + a.c (Distributive property.

Mutually perpendicular unit vectors i and j: X' j X
Oi
The unit vector along x-axis is denoted by i where i = (1, 0)

The unit vector along y-axis is denoted by j where j = (0, 1)

Y'

Infinity Optional Mathematics Book - 10 255

Here, |i| = (1)2 + (0)2 = 1 Y

|j | = (0)2 + (1)2 = 1 Oi X X
Y
X' Y'
X' j
(a) i.i = |i| |i| cosθ = 1 × 1 × cos0° = 1 × 1 × 1 = 1
O

(b) j .j = |j | |j |cosθ = 1 × 1 × cos0° = 1 × 1 × 1 = 1

Y'

Y

j

(c) i . j = |i||j | cosθ = 1 × 1 × cos90° = 1 × 0 = 0 X' Oi X


Y'
Y

(d) j .i = |j | |i | cosθ = 1 × 1 × cos90° = 1 × 1 × 0 = 0 j
X' O i X

Scalar product in terms of components: Y'

If a = (x1, y1) and b = (x2, y2) are two vectors, then the two vectors in terms of
i and j can be written as

a = (x1, y1) = x1i + y1j

b = (x2, y2) = x2i + y2j

Now, dot product of two vectors a.b = (x1i + y1j ). (x2i + y2j )

= x1x2 (i.i ) + x1x2(i.j ) + y1x2 (j .i ) + y1y2(j .j )

= x1x2 × 1 + x1y2 × 0 + y1x2 × 0 + y1y2 × 1

= x1x2 + 0 + 0 + y1y2

∴ a.b = x1x2 + y1y2

Hence, when a = (x1, y1) and b = (x2, y2)

Then a.b = x1x2 + y1y2

When two vectors are perpendiculars, then

a.b = 0

i.e. x1x2 + y1y2 = 0
Also, if a = (x1, y1) then, a2 = a.a = x12 + y12

256 Infinity Optional Mathematics Book - 10

Angle between two vectors

Let a = (x1, y1) and b = (x2, y2) be the two vectors, then the angle between two
vectors a and b is given by

cosθ = a.b
|a||b|

Where, |a| = x12 + y12 and |b| = x22 + y22

WORKED OUT EXAMPLES

1. Find the scalar product of the given pair of vectors:

(i) p = 3i + 4j and q= 6i + j (ii) p = –3 and q = –7
Solution: 4 –2

(i) Here, p = 3i + 4j = (3, 4) = (x1, y1)

q= 6i + j = (6, 1) = (x2, y2)

Now, p.q = x1x2 + y1y2

= (3) × (6) + (4) × (1)

= 18 + 4

= 22

(ii) p = –3 = x1 and q = –7 = x2
4 y1 –2 y2

Now, p.q = x1x2 + y1y2

= (–3) (–7) + (4) (–2)= 21 – 8 = 13

2. Show that the two vectors, c = 7i – 4j and d = 4i + 7j are perpendicular

(orthogonal) to each other.

Solution:

Here, c = 7i – 4j = (7, –4) = (x1, y1)

d = 4i + 7j = (4, 7) = (x2, y2)

Now, c .d = x1x2 + y1y2

= (7) (4) + (–4) (7)

= 28 – 28 = 0

Since, c .d = 0, the vectors are perpendicular to each other.

Infinity Optional Mathematics Book - 10 257

3. If p = 5i – 4j , q= 7i + k j and p and q are perpendicular, then find the
value of k.

Solution: Here,

p = 5i – 4j = (5, –4) = (x1, y1)
q = 7i + kj = (7, k) = (x2, y2)
Since, two vectors p and q are perpendicular to each other, so

p.q = 0

or, x1x2 + y1y2 = 0
or, (5) (7) + (–4) (k) = 0

or, 35 – 4k = 0

or, 4k = 35

∴ k = 35
4

4. Find the angle between the following two vectors:
(i) p and q when |p| = 6, |q| = 4 and p.q = 12 2

(ii) a and b when a = 3 i + 3 j and b = 3 i + 3 3 j

Solution:
(i) Here, |p| = 6, |q| = 4 and p.q = 12 2

Now, angle between p and q is given by

cosθ = p.q = 12 2 = 1
|p||q| (6) (4) 2

or, cosθ = cos45°
∴ θ = 45°

(ii) Here, a = – 3 i + 3 j = (– 3, 3) = (x1, y1)

b = 3 i + 3 3 j = ( 3, 3 3) = (x2, y2)

Now, a.b= x1x2 + y1y2

= (– 3) ( 3) + 3 (3 3)

=–3+9=6

Again, |a||b| = x12 + y12 x22 + y22

= (– 3)2 + ( 3)2 ( 3)2 + (3 3)2 = 3 + 3 3 + 27
= ( 6 × 30)
= 180
258 Infinity Optional Mathematics Book - 10

= 6 5
Now, angle between a and b is given by

cosθ = a. b = 6 = 1
|a||b| 65 5

∴ θ = cos-1 1
5

5. Show that m = –5 and n = 10 are parallel vector.
Solution: –2 4

Here, m = –5 = x1 Alternative method:
–2 y1
m= –5
–2

n = 10 = x2 = – 1 10
4 y2 2 4

Now, m.n = x1x2 + y1y2 = –21 n

= (–5) (10) + (–2) × 4 i.e. m = – 1 n
= –50 – 8 2
= – 58
Here, 1 is a scalar quantity.
2

Here, |m||n| = ( x12 + y12) ( x22 + y22) ∴m and n are parallel.

= ( (–5)2 + (–2)2) ( (10)2 + (4)2)

= ( 25 + 4)( 100 + 16)

= 29 × 29 × 4

= 292 × 22

= 58

Here, m.n = – |m||n|
∴ The two vectors are unlike and parallel.

6. If a + b + c = 0, |a| = 6, |b| = 7 and |c |= 127, find the angle between a
and b.

Solution: Here,

a+b+c =0

or, a + b = – c [ Squaring on both sides, we get)
or, (a + b)2 = (–c )2
or, a2 + 2a.b + b2 = c 2
or, |a|2 + 2a.b + |b|2 = |c |2

Infinity Optional Mathematics Book - 10 259

or, 62 + 2 |a| |b|cosθ + 72 = ( 127 )2 [∴ a.b = |a|.|b|cosθ]

or, 2 × 6 × 7 cosθ = 127 – 36 – 49

or, cosθ = 4824 = 1
2

or, cosθ = cos60°

∴ θ = 60°

Hence, angle between a and b is 60°.

7. Given that a + 2b and 5a – 4b are orthogonal. If a and b are unit
vectors, then find the angle between a and b.

Solution:

Here, (a + 2b) and (5a – 4b) are orthogonal, so

(a + 2b) .(5 a – 4b) = 0

or, 5a2 – 4a.b + 10 a.b – 8b2 = 0

or, 5 × 12 + 6 a. b – 8 × 12 = 0 [since a and b are unit vectors]

or, 6|a| |b| cosθ = 3

or, 6 × 1 × 1 cosθ = 3

or, cosθ = 1
2

∴ θ = 60°
Hence, angle between a and b is 60°.

Exercise 6.1

Section 'A'
1. (a) Define scalar product.
(b) If a.b = 0 then what is the relation between a and b.
(c) If |p| = 6 3, |q| = 5 and θ = 30°, find the value of p.q.
(d) If |a| = 6 2, |b|= 6 and θ = 45°, find the value of a.b.
(e) If a.b = 30,|a| = 6 and angle between a and b is 60°, find the magnitude

of vector b.

260 Infinity Optional Mathematics Book - 10

Section 'B'
2. Find the scalar product of each of the following pair of vectors:

(a) a = (3, 4) and b = (2, 1) (b) a= 4 and b = 3
5 2

(c) m = 5i – 6j and n = 3i + 2j (d) p = – 6i + 4j and q = 3i – 2j

3. (a) If m = 3 and n = 1 , then find m.n.
4 1


(b) If a = x1i + y1j and b = x2i + y2j then show that a.b = x1x2 + y1y2.
4. (a) If |OA| = 6, |OB| = 7 and OA.OB = 21, find the value of ∠AOB.

(b) If |a|= 16, |b| = 3 and a.b = 24 3, find the angle between the vectors a
and b.

(c) If a = 3i + 4j and b = 8i – 6j , find the angle between a and b.

(d) If |OA| = 7i + 5j and |OB| = 5i – 7j , find the value of ∠AOB.

(e) There are two vectors OA and OB such that the position vector of A is (1,
2) and that of B is (3, 4). Find the angle between the given vectors.

(f) Find the angle between the vectors a = 2 and b = 2 .
1 –3

5. Prove that the following pairs of vectors are perpendicular to each
other.

(a) a = 3 and b = 5 (b) a = ––84 and b = 3
5 –3 –6

(c) a = 3i + 4j and b = 4i – 3j (d) p = 8i + 6j and q = 3i – 4j

(e) m = (4, 5) and n = (5, –4) (f) a = (–2, 6) and b = (6, 2)

6. Prove that the following pairs of vectors are parallel

(a) a = 2 and b = 8 (b) p = –34 and q = 9
4 16 –12

(c) a = 2i + 3j and b = 6i + 9j (d) c = i + j and d = –4i – 4j

7. (a) If a = mi + 3j , b = –5i – j and a.b = 7 then find the value of m.

(b) If a = pi + 6j , b = –i + j and a.b = 4, find the value of p.

Infinity Optional Mathematics Book - 10 261

8. (a) If a = 3i + 4j and b = 4i + xj are perpendicular to each other, find the
value of x.

(b) If a = –5 and b = p are orthogonal to each other, find the value
of p. 3 p+2

(c) For what value of x the vector a = x and b = 3 are perpendicular?
3 –4

9. (a) If a = 2 and b = 3 , find the value of a2 and b2.
3 –1

(b) If i and j are the unit vectors along x-axis and y-axis respectively, then
show that i .i = 1 and i .j = 0.

10. (a) If a = 2 and b = 4 are parallel vectors, find the value of m.
(b) If a = 3 m

–1 and b = 5 are parallel vectors, find the value of p.
2 p

11. (a) If the vectors a and b are mutually perpendicular, prove that):

(a + b)2 = (a – b)2

(b) If (x + y)2 = (x – y)2, prove x and y are perpendicular to each other.

12. (a) If a + 2b and 5a – 4b are perpendicular to each other and a and b are
unit vectors, find the angle between a and b.

(b) If |a + b| = |a – b|, prove a and b are perpendicular to each other.

13. (a) If p + q + r = 0, |p| = 6, |q|= 7 and |r | = 127, find the angle between
p and q.

(b) If a + b + c = 0, |a| = 4, |b| = 3 and |c | = 37, find the angle between
a and b.

14. If a and b are two vectors of unit length and θ is the angle between them,

show that: 1 |a – b| = sin θ .
2 2

15. In ∆ABC, if AB = 3i + 2j and BC = 8i – 12j then prove that ∆ABC is a right
angled triangle. In ∆ABC, which angle is right angle?

262 Infinity Optional Mathematics Book - 10

6.2 Vector Geometry

1. Mid Point Formula
If the position vector of A, B and M are a, b and m respectively, where M is

the mid-point of line segment AB, then prove that m = a + b.
2
Solution: A

Given : OA = a, OB = b, OM = m where M is the mid point

of AB. am M
B
To prove : m = a + b
2

Proof: M is the mid point of AB. So, AM = MB ....... (i) O b
OM – OA = OB – OM (Using law of ∆)

or, m – a = b – m

or, m + m = a + b

or, 2m = a + b

∴ m = a + b . Hence proved.
2

2. Section Formula for Internal Division

If a, b and p are the position vectors of A, B and P respectively where P
divides the line segment AB in the ratio of m:n internally, then prove

p = mb + na . A
m+ n
Solution: m

Given: P

OA = a, OB = b, OP = p where P divides AB internally a n

in the ratio m:n i.e. AP : PB = m : n. p

To prove : p = mb + na O B
m+n
b

Proof: Since P divides AB internally in the ratio of m:n, then

AP : PB = m:n

or, PABP = m
n

or, nAP = mPB ........ (i)

or, n(OP – OA) = m(OB – OP) [Using ∆ law]

or, n(p – a) = m(b – p)
or, np – na = mb – mp)
or, mp + np = mb + na

Infinity Optional Mathematics Book - 10 263

∴ p = mb + na . Hence proved.
m + n

3. Section formula for External Division

If a, b and p are the position vectors of A, B and P where P divides the line

segment AB in the ratio m : n externally, then prove that p = mb – na
m –n
Solution:
A
Given:

OA = a, OB = b, OP = p where P divides AB externally a Bm
in the ratio of m:n bn
i.e. AP :BP = m : n

To prove: p = mb – na OP
m –n
p

Proof : Since P divides AB externally in the ratio of m: n, then AP : BP = m:n, then

AP : BP = m : n

or, BAPP = m
n

or, n|AP | = m|BP |

or, n(OP – OA) = m(OP – OB)
or, n(p – a) = m(p – b)
or, np – na = mp – mb
or, mb – na = mp – np

∴ p = mb – na Hence proved.
m –n

4. If a, b, c and g are the position vectors of A, B, C and G of ∆ABC respectively

where G is the centroid of ∆ABC. Prove that g =a + b +c . A
3

Solution: G

Given: In ∆ABC, DC

B

OA = a, OB = b, OC = c and OG = g where G is the
centroid of ∆ABC.

To prove : g = a + b + c O
3

264 Infinity Optional Mathematics Book - 10

Proof:

D is the mid point of BC. So, by using mid point formula

OD = OB +OC
2
b + c
or, OD = 2 ......... (i) A

Since, centroid divides the median in the ratio 2:1 from 2
G
the vertex to the midpoint. 1
D
i.e. AG : GD = 2:1 (m:n = 2:1) O

Now, using section formula for internal division,

OG = 2 × OD + 1 × OA  p = na + mb
2+1 n + m

2 b+c +a
2
= 3

∴ g = a + b + c . Hence proved.
3
5. Prove by vector method that the diagonals of a parallelogram bisect

each other.

Solution: Given: ABCD is a parallelogram. AC and BD are two diagonals.

To prove : Diagonals AC and BD bisect each other.

Proof: Let the mid point of AC be M and mid point of BD be N.

Now, In ∆ABC, using ∆ law,

AB + BC = AC D C

or, 2AM = AB + BC N
M
∴ AM = AB + BC ........ (ii)
2 A
B
In ∆BDA, N is the midpoint of BD. So by using midpoint theorem

AN = AB + AD
2
AB + BC
AN = 2 [ BC = AD]

Here, AM = AN i.e. M and N represents the same point.

Hence, it is proved that diagonals of a parallelogram bisect each other.
Infinity Optional Mathematics Book - 10 265

6. In a parallelogram ABCD, M and N are the mid points of BC and CD
3
respectively. Show that AM + AN = 2 AC .

Given: In parallelogram ABCD, M and N are the mid points of BC and CD

respectively.

To prove : AM + AN = 3 AC DN C
2 M

Proof:

In ∆ ABC, using mid point formula, AB

AM = AB + AC ........ (i)
2

In ∆ACD, using mid point formula

AN = AD + AC ......... (ii)
2

Now, adding (i) and (ii) we get

AM + AN = AB + AC + AD + AC
2
or, AM + AN
= (AB + BC) + 2AC [ AD = BC ]
2

or, AM + AN = AC + 2AC [ By triangle law AB + BC = AC ]
2
3
∴ AM + AN = 2 AC . Hence proved.

7. If P and Q are the mid points of AB and AC of ∆ABC respectively,

prove that: PQ = 1 BC and PQ//BC. A
2

Solution: P Q
Given : In ∆ABC, P and Q are the mid point of AB and

AC respectively.

To prove : PQ = 1 BC and PQ//BC. BC
2

Proof: In ∆APQ, using ∆ law,

AP + PQ = AQ

or, PQ = AQ – AP ......... (i)

In ∆ABC, using ∆ law,

266 Infinity Optional Mathematics Book - 10

AB + BC = AC

or, BC = AC – AB
or, BC = 2AQ – 2AP [ Q and P are the mid points of AC and AB respectively]

or, BC = 2(AQ – AP )

or, BC = 2 × PQ [ From equation (i)]

or, PQ = 1 BC
2
1
By the definition of parallel vectors PQ = K. BC where K = 2

∴ PQ//BC

Hence, PQ = 1 BC and PQ//BC. Proved.
2

8. Prove by vector method that the lines joining the midpoints of the

adjacent sides of a quadrilateral taken in order is a parallelogram.

Solution:

Given : In quadrilateral ABCD. P,Q,R and S are the midpoint of sides AB,

BC, CD and AD respectively. D R
To prove: PQRS is a parallelogram. S
C

Construction: Join B and D.

A Q

Proof: In ∆ABD, P and S are the mid points of AB

P

and AD, respectively. So

1 B
2
PS = BD and PS //BD .......... (i)

Again, in ∆BCD, Q and R are the mid points of BC and CD respectively.

So, QR = 1 BD and QR//BD ......... (ii)
2

From equation (i) and (ii)

PS = QR and PS //QR ............. (iii)

When two lines are equal and parallel then the lines joining the same sides of

equal and parallel are also equal and parallel. So,

Infinity Optional Mathematics Book - 10 267

PQ = SR and PQ// SR ... (iv)
From equation (iii) and (iv), it is prove that PQRS is a parallelogram.
Hence the lines joining the midpoints of the adjacent sides of a quadrilateral

taken in order is a parallelogram.

9. Prove Pythagoras theorem vectorically.

Solution:
Given: ABC is a right angled triangle, where ∠ABC = 90°.
To prove: AC2 = AB2 + BC2
Proof: In ∆ABC, using ∆ law,

AB + BC = AC A

Now, squaring on both sides, we get

or, (AB + BC)2 = (AC)2

or, AB2 + 2AB.BC + BC2 = AC2

or, AB2 + 2 × 0 + BC2 = AC2 [ ∠ABC = 90°] B C

or, AB2 + BC2 = AC2

∴ AC2 = AB2 + BC2 Proved.

Hence, Pythagoras theorem is proved.

10. Prove by vector method that the length of diagonals of a rectangle
are equal.

Solution:

Given: ABCD is a rectangle where BD and AC are diagonals

To prove : AC = BD D C

Proof : In ∆ABC, using ∆ law, AB
AB + BC = AC
Squaring on both sides, we get
AC2 = (AB + BC ) 2

268 Infinity Optional Mathematics Book - 10

or, AC2 = AB2 + 2AB .BC + BC2

or, AC2 = AB2 + 2 × 0 + BC2 [ ∠ABC = 90°]

or, AC2 = AB2 + BC2 ........ (i)

Similarly, In ∆ABD,

BD = BA + AD

or, BD = –AB + BC [ AD = BC]

or, BD = BC – AB

Squaring on both sides, we get

BD2 = (BC – AB)2

or, BD2 = BC2 – 2BC. AB + AB2

or, BD2 = BC2 – 2 × 0 + AB2 [ ∠ABC = 90°]

or, BD2 = AB2 + BC2 ........... (ii)

Now, from equation (i) and (ii), we get

AC2 = BD2

∴ AC = BD. Hence, it is proved the diagonals of a rectangle are equal.

11. Prove by vector method that the angle at the semi circle is 90°.
Solution:

Given: O is the centre of circle. AB is a diameter. ∠APB is the inscribed angle
standing on the diameter AB.

To prove: ∠APB = 90° P
Proof: In ∆APO, using ∆ law,

PO + OA = PA ......... (i) AOB

In ∆POB, using ∆ law PO + OB = PB ......... (ii)

Now, taking dot product of (i) and (ii), we get

(PO + OA ). (PO + OB ) = PA .PB

or, (PO – AO ). (PO + AO ) = PA .PB [ AO = OB ]

or, PO 2 – AO 2 = PA .PB

Infinity Optional Mathematics Book - 10 269

or, |PO |2 – |AO |2 = PA .PB

or, |AO |2 – |AO |2 = PA .PB [ |PO | = |AO |]

or, O = PA .PB

∴ PA .PB = 0

Since dot product of two vectors PA and PB is zero, so they are at 90°.
∴ ∠APB = 90° Hence it is proved that the angles at the semi circle is 90°.

12. Prove by vector method that the diagonals of a rhombus bisect each

other at 90°.

Solution: D C
Given : ABCD is a rhombus where AC and BD are the

diagonals.

To prove: Diagonals AC and BD bisect each other at 90°. B
Proof: Since rhombus is a parallelogram and diagonals of a

parallelogram bisect each other, so the diagonals of a A
rhombus bisect each other.
Again, In ∆ABC, using ∆ law,

AB + BC = AC ....... (i)

In ∆DAB, using ∆ law,

DA + AB = DB
or, CB + AB = DB [ DA = CB ]

or, AB – BC = DB ......... (ii)

Now, taking the dot product of (i) and (ii), we get

(AB + BC ).(AB – BC ) = AC . DB
or, (AB )2 – (BC )2 = AC .DB
or, |AB |2 – |BC |2 = AC .DB
or, |AB |2 – |AB |2 = AC .DB [ |AB | = |BC |]

or, O = AC .DB

∴ AC .DB = 0
When dot product of two vectors is zero, they are perpendicular.

Hence, it is proved that diagonals AC and BD bisect each other at 90°.
270 Infinity Optional Mathematics Book - 10

13. Prove by vector method that the median of an isosceles triangle is

perpendicular to the base. A

Given: In ∆ABC, AB = AC and BD = DC.

To prove: AD⊥BC

Proof: In ∆ABC, using ∆ law BDC
AB + BC = AC

or, BC = AC – AB ........ (i)

In ∆ ABC, D is the mid point of BC, so, using midpoint formula,

AD = AB +2prAoCduc.t..o..f..(.i()iai)nd (ii), we get
Now, taking dot

BC .AD = (AC – AB ). AB + AC
2

or, AD . BC = (AC – AB ). AC + AB
2

oorr,, AADD.B. BCC==AACC2 22– AB2 [ |AC| = |AB|]
0
or, AD.BC = 2 – AC2
2

or, AD.BC = 0

Since dot product of two vectors is zero, they are perpendicular.

∴ AD⊥BC

Hence, it is proved that the median of an isosceles triangle is perpendicular

to the base.

14. In a right angled triangle, if a median is drawn from right angle to its
hypoteneous then the point of median is at equal distance from the
three vertices. Prove it by vector method.

Solution: A

Given: In ∆ABC, ∠ABC = 90°. BD is a median.

To prove: AD = BD = CD D

Proof: In ∆ABD, using ∆ law, BC
BA + AD = BD

Infinity Optional Mathematics Book - 10 271

or, BA = BD – AD........ (i)

In ∆BDC, using ∆ law,

BD + DC = BC

or, BC = BD + AD ........ (ii) [AD = DC]

Now, taking dot product of (i) and (ii), we get

BA.BC = (BD – AD). (BD + AD)

or, 0 = BD2 – AD2 [ ∠ABC = 90°]

or, AD2 = BD2

∴ AD = BD

Hence, it is proved that the mid point of the hypoteneus of a right angled

triangle is equidistant from its vertices.

AD = DC A
∴ AD = BD = DC Proved.

15. In the figure, AD = BD = DC. Prove that ∠ABC = 90°.

Solution: D

Given: In ∆ABC, AD = DC = BD

To prove: ∠ABC = 90°

Proof: In ∆ABD, using ∆ law BC

AD + DB = AB ............ (i)

In ∆BDC, using ∆ law

BD + DC = BC [ AD = DC]
or, –DB + AD = BC

or, AD – DB = BC ......... (ii)

Now, taking the dot product of (i) and (ii) we get

(AD + DB). (AD – DB) = AB.BC

or, AD2 – DB2 = AB.BC

or, AD2 – AD2 = AB.BC [AD = DB]

or, AB.BC = 0.

Since, dot product of two product is zero. They are perpendicular.

So, ∠ABC = 90°. Hence it is proved.

272 Infinity Optional Mathematics Book - 10

WORKED OUT EXAMPLES

1. The position vectors of A and B are 3i + 2j and 5i – 4j respectively.

Find the position vector of the midpoint M of the line AB. B
Solution:

Here, position vector of A = OA = 3i + 2j M
Position vector of B = OB = 5i – 4j O

Now, position vector of midpoint M of AB is given by A

OM = OA + OB – 4j
2
=
3i + 2j + 5i
2

= 8i – 2j
2

OM = 4i – j

∴ The position vector of the midpoint M of line AB is 4i – j .

2. M and N are two points with coordinates (4, 6) and (–1, 3) respectively.
Find the position vector of P which divides MN internally in the ratio
1:2. N(–1, 3)

Solution: Here, n

The position vector of M = OM = (4, 6) = 4i + 6j P

Position vector of N = ON = (–1, 3) = –i + 3j Om
m : n = 1:2 M(4, 6)

By using section formula for internal division, position vector of the point P is

given by

OP = mON + nOM
m+n

= 1(–i + 3j ) + 2 (4i + 6j )
=
1+2
–i + 3j + 8i + 12j

3

Infinity Optional Mathematics Book - 10 273

or, OP = 7i + 15j
3
7
or, OP = 3 i + 5j

Hence, the position vector of the point P is 7 i + 5j.
3

3. A(3, 4), B(2, –5) and C(4, – 5) are the three vertices of ∆ABC. Find the

position vector of the centroid of ∆ABC. A(3, 4)

Solution:

Here, position vector of A = OA = (3, 4) = 3i + 4j F E
Position vector of B = OB = (2, –5) = 2i – 5j G

Position vector of C = OC = (4, –5) = 4i – 5j B(2, –5) D C(4, –5)

Now, position vector of the centroid of ∆ABC is given by

OG = OA + OB + OC
3

= 3i + 4j + 2i – 5j + 4i – 5j
3

= 9i – 6j
3

∴ OG = 3i – 2j = (3, –2)

Exercise 6.2

Section 'A'

1. (a) In ∆ABC, M and N are the midpoints of AB and AC respectively then
write the relation between MN and BC in vector form.

(b) In a rhombus ABCD, AC and BD are diagonals then T E
write the value of AC .BD . F

(c) In the given ∆EFG, GT = ET = FT then write the

relation between EF and GF . G

Section 'B'

2. (a) If the position vectors of A and B are 3i – 5j and i – 5j respectively, find
the position vector of the mid-point M of the line segment AB.

(b) If the position vectors of A and B are 3i + 4j and i + 8j respectively, find
the position vector of the mid-point of the line segment AB.

274 Infinity Optional Mathematics Book - 10

(c) Find the position vectors of mid-point of the line segment AB if the co-
ordinate of the points A and B are (3, 5) and (9, –11) respectively.

(d) If the position vector of the mid-point of the line segment AB is 3i + j
where the position vector of A is 5i + 2j , find the position vector of the
point B.

3. (a) The position vector of the point A and B are 2i – 3j and 3i + 2j

respectively. Find the position vector of the point M which divides the

line segment AB internally in the ratio 2:3.

(b) Find the position vector of the point M which divides the line segment

joining A (5, 2) and B(3, 6) internally in the ratio 2:3.

(c) If OA = a, OB = b, OM = m and M divide BA in the ratio 3:2, then prove

that : m = 1 (3a + 2b ).
5

(d) The point C divides the line AB externally in the ratio 2:5. If the position

vectors of A and B are OA = –3 and OB = 8 respectively, find the
position vector of the point C. 4 7

4. (a) If the position vectors of the points A, B and C are 5 i + 4 j , – i + 3 j and

6 i – 2 j respectively, find the position vector of the centroid of ∆ABC.

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

positions vector of the centroid of ∆ABC. A
(c) In the given figure, AD is a median, where the position G

vector of the points A and D are 3 i – 2 j and 3 i – 4 j DC
respectively. Find the position vector of the centroid G. B

Infinity Optional Mathematics Book - 10 275

5. (a) In ∆OAB, if OA = a , OB = b and M is the middle point of AB, then show

1 A
2
that: OM = (a + b ).

(b) In the given figure, AM is the median of ∆ABC, prove

that : AM= 1 (AB + AC ). B MC
2 O

6. (a) In the figure, OA = a , OB = b . If AC = 3AB ,

find OC . a
b

AB C

D C
(b) In the given figure, ABCD is a parallelogram. If OA = a

, OB = b and OC = c , find OD in terms of a . b and c . A B

O
O

(c) In the given figure, OA = a , OB = b and AB:AP = 3:1,

express OP in term of a and b . ab

AP B
A

(d) In the given triangle, ABC, M is the mid-point of M N
C
AB and BC = 2MN, prove that N is the mid-point of
AC. B
A

(e) In the given figure, ∠ABC = 90°,
prove that: AC 2 = AB 2 + BC 2.



B C

276 Infinity Optional Mathematics Book - 10

Section 'C' A
7. In the figure, AD, BE and CF are the medians of ∆ABC.
FE
Prove that: AD + BE + CF = 0. G

B DC

A

8. In the given ∆ABC, M and N are the midpoints of sides AB

and AC respectively. Prove by vector method that M N
BC = 2MN and MN //BC. C



B

P

9. In the given ∆PEN, EP = EN and PF = FN then prove by F
vector method that EF⊥PN.



EN

10. In a given right angled ∆MAN, ∠AMN = 90°, AB = BN then prove by vector

method that AB = MB = BN. M

AB N

11. Prove that the perpendicular drawn from the vertex of an isosceles triangle to

the base bisects the base. DC

12. In a given rectangle ABCD, prove by vector method that
AC = BD.

AB

13. Prove by vector method that the diagonals of a parallelogram POST bisect

each other. E NT

14. In a given quadrilateral VOTE B, I, N and A are the A I

midpoints of VO, OT, TE and EV respectively. Prove by

vector method that, BINA is a parallelogram. V BO

Infinity Optional Mathematics Book - 10 277

T S
15. In a given rhombus REST, prove by vector method that RS
and TE bisect to each other at right angle.


RE

E

16. In a given figure, M is the centre of circle. Prove by vector

method that PE⊥TE. PMT


17. Prove that the parallelogram having equal diagonals is a rectangle.

P Q

18. In the given figure, PQRS is a parallelogram. M and N N
are two points on the diagonals SQ. If SM = NQ, prove M
by vector method that PMRN is parallelogram.
R
S

19. Prove that the straight line joining the middle points of non-parallel sides of
a trapezium is parallel to the parallel sides and half of their sum.



For Important Notes:
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278 Infinity Optional Mathematics Book - 10

UNIT

7 TRANSFORMATION

Review

Answer the following questions.
(a) Define one to one and onto function. Give an example of it in arrow diagram.
(b) Is the transformation one to one and onto function? Discuss.
(c) Define isometric transformation. Give examples of it.
(d) Define non isometric transformation. Give examples of it.

(e) If A(5, 6) and B(–2, 4) are two points then find AB.

7.1 Introduction of Transformation

A transformation is one to one and onto function from a plane to the same plane. So

the transformation is a function which maps each point of a plane with a point in

the same plane. There are four fundamental transformations. They are

(i) Reflection (ii) Rotation

(iii) Translation (iv) Enlargement

Reflection

Reflection in x-axis i.e. y = 0 line. Y
Let P(x, y) be an object point. Then the image of the point P(x, y)

P(x, y) under the reflection in x -axis is P'(x, –y) X' O X
i.e. P(x, y) x-axis P'(x, –y).
P'(x, –y)

Reflection in y-axis i.e. x = 0 line. Y'

Y

Let P(x, y) be an object point. Then the image of the point P'(–x, y) P(x, y)
X
P(x, y) under the reflection in y-axis is P'(–x y) X' O
i.e. P(x, y) y-axis P'(–x, y) Y'



Reflection in the line y = x i.e. x – y = 0. Y P(x, y)Py'(y=,xx)
Let P(x, y) be an object point. Then the image of the point X' OX
P(x, y) under the reflection in the line y = x is P'(y, x)
Y'
i.e. P(x, y) y = x P'(y, x)

Infinity Optional Mathematics Book - 10 279

Y
P'(-yP,(-xx,) y)
Reflection in the line y = –x i.e. x + y =0.
Y'
Let P(x, y) be an object point. Then the image of the point X

P(x, y) under the reflection in the line y = –x is P'(–y, –x) X'
i.e. P(x, y) y = –x P'(–y, –x)
y = –x


Reflection in the line parallel to y -axis i.e. x = h line. Y

Let P(x, y) be an object point. Then the image of the P(x, y) P'(2h–x, y)
X
point P(x, y) under the reflection in the line x = h is X'
P' (2h – x, y) i.e. P(x, y) x = h P'(2h – x, y) Y' x = h

Y P(x, y)
y=k

P'(x, 2k – y)
X

Reflection in the line parallel to x-axis i.e. y = k line.

Let P(x, y) be an object point. Then the image of the point

P(x, y) under the reflection in the line y = k is P' (x, 2k – y)
i.e. P(x, y) y = k P'(x, 2k – y)
X'

Rotation Y'

Rotation through positive quarter turn i.e. + 90° Y

Let P(x, y) be a given point. Then the image of P(x, y) under P'(-y, x) P(x, y)

the rotation through +90° about origin is P'(–y, x) i.e. X' +90°
P(x, y) R[O, +90°] P'(–y, x). The rotation through +90°
O X
–270°
about

origin is equivalent to the rotation through –270° about the Y'
origin.

Y
Rotation through negative quarter turn i.e. – 90°. P(x, y)

Let P(x, y) be a given point. Then the image of X' +270°O-90° X
P(x, y) under the rotation through –90° about origin is
P'(y, –x) i.e. P(x, y) R[O, –90°] P'(y, –x) P'(y, -x)

The rotation through –90° about origin is equivalent to the Y'

rotation through +270°about the origin.

280 Infinity Optional Mathematics Book - 10

Y

Rotation through half turn. P(x, y)

Let P(x, y) be a point. Then the image of P(x, y) under the X' +180° X
rotation through 180° about origin is P'(-x, -y) i.e. O-180°

P(x, y) R[O, ±180°] P'(–x, –y) P'(-x,-y)

The rotation through half turn in anticlockwise direction Y'

about origin is equivalent to the rotation through half turn in clockwise direction

about origin.

Translation Y
P'(x + a, y + b)
Translation by a vector T = a .
b

Let P(x, y) be a point. Then the image of the point P(x, y)
a P(x, y)
under the translation with vector T = b is X' O X

a Y'
b
T= P'(x + a, y + b)
P' (x + a, y + b) i.e. P(x, y)

Enlargement Y P'(kx, ky)

Enlargement with centre origin and scale factor k O P(x, y) X
i.e. E[(0, 0), k] Y'
Let P(x, y) be a given point. Then the image of the point X'
P(x, y) under E [(0, 0), k] is P'(kx, ky) i.e.

P (x, y) E[(0, 0), k] P'(kx, ky)

Y P'(kx–ka+a, ky–kb+b)

Enlargement with centre (a, b) and scale factor
k i.e. E[(a, b), k]

Let P(x, y) be a given point. The image of the point X' P(x, y) X
(a,b)
P(x, y) under E[(a, b), k] is P'(kx–ka+a, ky–kb+ b). O
i.e. P(x, y) E[(a, b), k] P'(kx – ka + a, ky – kb + b) Y'

Infinity Optional Mathematics Book - 10 281

7.2 Combined Transformations A''(–2, 4)Y y=x
X' O
Consider a point A(4, 2). The image of the point A(4, 2)
A(4, 2) after reflecting in x-axis is A'(4, –2). Again Y' X
the image of the point A'(4, –2) after reflecting in
the line y = x is A''(–2, 4). Hence the final image of A'(4, –2)
A(4, 2) after two successive reflections is A''(–2, 4) which
can be described by a single transformation rotation
through +90° about the origin. It is called combined
transformation of reflection in x-axis followed by
reflection in y = x line.

A''(–2, 4)
i.e. A(4, 2) x-axis A'(4, –2) y = x

R[O, +90°]

Let r1 and r2 be transformations which map a point P to the point P' and the point
P' to the point P'' respectively. Then the transformation which maps the point
P to the point P'' is said to be combined transformation of the transformations r1
and r2. The combined transformation is denoted by r2 or1 or r2r1. It is also called
as the transformation r1 followed by r2.



Combination of two reflections (Reflection followed by another reflection)

There are two cases of combination of two reflections.

When the axes of reflection are parallel. Y
Consider a point A(3, 3). The image of A(3, 3) under A(3,3)

the reflection in the line y = 2 is A'(3, 1) and the image y=2

of A'(3, 1) under the reflection in the line y = –1 is X' O A'(3,1)
X

A''(3, –3). So the final image of A(3, 3) is A''(3, –3). y = –1

i.e. A(3, 3) → A''(3, –3). This is a translation with the Y' A''(3,–3)

translation vector 2 0 = 0
–1 – 2 –6

If r1 is the reflection in the line y = k1 and r2 is the reflection in the line y = k2 then
the combined transformation of these two reflections where the lines of reflection

282 Infinity Optional Mathematics Book - 10

are parallel is a translation with the following translation vectors.

The translation vector of r1or2 = 2 0 – k2
k1

The translation vector of r2or1 = 2 0 – k1
k2

Similarly, if r1 is the reflection in the line x = h1 and r2 is the reflection in the line
x = h2 then the combined transformation of these two reflections where the lines of
reflections are parallel is a translation with the following translation vectors.

The translation vector of r1or2 = 2 h1 – h2
0

The translation vector of r2or1 = 2 h2 – h1
0

If the axes of reflections are parallel, the combination of two reflections (i.e. a
reflection followed by another reflection) is equivalent to the translation.

When the axes of reflection intersect at a point.

Consider two lines of reflection y = x and y = –x which intersect each other and

angle between them is 90°: y=–x X
The image of a point A(3, 1) under the reflection in A'(1,3) y = x
the line y = x is A' (1, 3) and the image of A'(1, 3)

under the reflection in the line y = –x is A'' (–3, –1). X' O A(3,1) X

So, the final image of A(3, 1) is A''(–3, –1) A"(-3, -1)

i.e. A(3, 1) → A''(–3, –1).

It is a rotation through 180° about origin. Here the Y'
angle of rotation is twice the angle between the two

lines of reflection and the centre of rotation is the point of intersection of the two

lines of reflection.

If the axes of reflection intersect each other at a point, the combination of
reflections is equivalent to a rotation in which
(i) The angle of rotation is twice the angle between the two axes of reflection.
(ii) The centre of rotation is the point where the axes of the reflection intersect

each other.
(iii) The direction of rotation is the direction from the first axis of reflection to

the second axis of reflection.

Infinity Optional Mathematics Book - 10 283

WORKED OUT EXAMPLES

1. If r1 is the reflection about a line x = y and r2 is reflection about the
y-axis, find the image point of the point A(2, –5) under the combined

transformation r1or2.
Solution: Here,

r1 is the reflection about a line x = y.
r2 is the reflection about the y-axis.
The given point = A(2, –5)

The combined transformation = r1or2.

We know that,
P(x, y) y-axis P'(–x, y)

A(2, –5) A'(–2, –5)

Again,
P(x, y) x = y P'(y, x)

A'(–2, –5) A''(–5, –2)

∴ The required image of A(2, –5) under r1or2 is A''(–5, –2).
Alternatively,

Since the axes of reflections intersect each other at origin making an angle

of 45°, the combined transformation r1or2 is equivalent to a rotation with
center (0, 0), angle of rotation = 2 × 45° = 90° and clockwise direction i.e.

R[(0, 0), –90°] = r1or2

We know that, P'(y, –x)
P(x, y) r1or2

A(2, –5) A'(–5, –2).

∴ The image of the point A(2, –5) under the combined transformation is

A''(–5, –2).

2. Find the co-ordinates of the point M whose image after reflection
about the line x = –y followed by the reflection about the line y = 2 is
M''(6, – 5).

Solution: Here,
First transformation is reflection about the line x = –y

284 Infinity Optional Mathematics Book - 10

Second transformation is reflection about the line y = 2.

Final image of the point M is M''(6, –5)

The co-ordinates of the point M = ?

Let the co-ordinates of the point M be (a, b)

Now,
P(x, y) x = –y P'(–y, –x)

M(a, b) M'(–b, –a)

Again, P(x, y) y = k P'(x, 2k – y)

M'(–b, –a) y = 2 M''(–b, 2 × 2 – (–a)) = M''(–b, 4 + a)

Comparing M''(–b, 4 + a) with M'' (6, –5)



So, –b = 6 4 + a = –5
∴b=–6 ∴a=–9

Hence, the co-ordinates of M = (–9, –6)

3. A triangle with vertices P(1, 2), Q(4, –1) and R(2, 5) is reflected

successively in the lines x = –1 and y = 2. Find the stating co-ordinates

and represent the images graphically under these transformations.

State also the single transformation given by the combination of

these transformations.

Solution: Here,

The given points are P(1, 2), Q(4, –1) and R(2, 5)

The reflection in the line x = –1
P(x, y) x = h P'(2h – x, y)
P(1, 2) x = –1 P'(2 × (–1) – 1, 2) = P'(–3, 2)
Q(4, – 1) x = –1 Q'(2 × (–1) – 4, –1) = Q'(–6, –1)
R(2, 5) x = –1 R' (2 × (–1) – 2, 5) = R'(–4, 5)

The reflection in the line y = 2,
P(x, y) y = k P'(x, 2k – y)
P'(–3, 2) y = 2 P''(–3, 2 × 2 – 2) = P''(–3, 2)
Q'(–6, –1) y = 2 Q''(–6, 2 × 2 – (–1)) = Q''(–6, 5)
R'(–4, 5) y = 2 R''(–4, 2 × 2 – 5) = R''(–4, –1)

Since, the axes of the reflection x = –1 and y = 2 intersect each other at

Infinity Optional Mathematics Book - 10 285

(–1, 2) making an angle of 90° between them, the combined transformation of
the reflection in the line x = –1 followed by the reflection in the line y =2 is the
rotation about (–1, 2) through 180° i.e. R[(–1, 2), 180°].
The ∆PQR and its images are shown in the graph given below.

Y

Q'' R' R

X' P' P X
Q' P'' O
R''
Q

Y'


Exercise 7.1

Section 'A'
1. (a) What is the image of P(x, y) under the reflection in the line y = x?

(b) What is the image of a point A(a, b) under the reflection in the line y = 0?

(c) What is the combined transformation of a reflection in the line y = 1
followed by another reflection in the line y = –2?

(d) What is the combined transformation of a reflection in the line y = x
followed by another reflection in y-axis?

(e) If r1 and r2 be the reflections about the lines x = 2 and y = 1 then write
the combined transformation under r1or2.
Section 'B'

2. (a) Find the image of a point A(2, 7) under the reflection in x -axis followed
by another reflection in y -axis.

(b) Find the image of a point B(–3, 5) if it is reflected in the line y = x and
then reflected in the line x = 0.

286 Infinity Optional Mathematics Book - 10

(c) A point P(1, –4) is reflected successively in the lines x = 2 and y = –1.
Find the final image of the point P under these transformations.

(d) Point (–4, –2) is reflected in x-axis and further reflected in the line y = –x.
Find the final image of the point.

3. (a) If R1 is the reflection about x -axis and R2 is the reflection about the
line y = x then find the image of the point (–3, 5) under the combined

transformation R1oR2.
(b) Let r1 be the reflection in the line x = –2 and r2 be the reflection in

the line y = –x. Find the image of the point A(4, 1) under the combined

transformation r2 or1.
(c) If F and G denote the reflections on y-axis and y = 3 respectively, find the

image of the point P(2, –3) under the combined transformation FoG.

(d) Let R1 be the reflection about x-axis and R2 be the reflection about
y-axis then find the image of the point (–8, –5) under the combined

transformation R2oR1.

4. Let r1 be the reflection about x-axis, r2 be the reflection in the line
y = x and r3 be the reflection in the line y = –1. Find the following.

(a) r1or2 (2, 5) (b) r1or3(–3, 1) (c) r2or3(5, –2)

(d) r3 or1 (–2, –3) (e) r2or1(–4, 6) (f) r2or2 (6,3)

5. (a) If R1 be the reflection on the line y = x and R2 be the reflection on the line
x = 0 then state what does R2oR1 represent? If a point A is transformed
by the above single transformation to A'(–2, 3), find the co-ordinates of

the point 'A'.

(b) F and G denote the reflections on x = –y and x = 4 respectively. What
point would have the image (2, –5) under the combined transformation
FoG?

(c) Let r1 be the reflection in the line y = 2 and r2 be the reflection in the
line x = –y. What point has the image X'(–3, 2) under the combined
transformation r1or2? Find it.

(d) Point (4, –3) is reflected in the line x = 0 at first and then the image so
formed is reflected in the line y = m so that the final image (–4, 9) is
obtained. Find the value of m.

Infinity Optional Mathematics Book - 10 287

Section 'C'

6. (a) Find the co-ordinates of a image of a triangle ABC with vertices A(1, 2),
B(4, –1) and C(2, 5) under the combined transformation of the reflection
on x-axis and then on y-axis.

(b) A triangle with vertices M(2, 3), N(–1, –2) and O(4, –1) is reflected
successively in the lines x = –1 and y = –2. Write the coordinates of the
images and show on the same graph.

(c) A quadrilateral A(6, 2), B(10, 2), C(11, 4) and D(5, 4) is reflected in the
lines y = x and x-axis. Find the stating co-ordinates and graphically
represent the images under these transformations.

7. (a) A triangle with vertices A(2, 3), B(4, 5) and C(1, 5) is reflected successively
in the lines x = –1 and y = 2. Find the stating co-ordinates and represent
the images graphically under these transformations. State also the
single transformation given by the combination of these transformation.

(b) Reflect the ∆XYZ with vertices X(–4, 0), Y(–6, 2) and Z(–4, 3) on the line
x = –3 and then on the line x = 1. Draw the graph of ∆XYZ and its images
after combined transformation. Also find the single transformation
which is equivalent to the given combined transformation.

(c) P(0, 1), Q(1, 2) and R(3, –4) are the vertices of a triangle PQR. Find
the co-ordinates of images of the vertices of ∆PQR under the reflection
on y-axis followed by the reflection on line y = –x. Also state the single
transformation given by the combination of these transformations.

8. (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 ∆PQR
having vertices P(4, 3) , Q(1, 1) and R(5, –1). Also draw the object and
image on the same graph.

(b) State the single transformation equivalent to the combination of
reflection on the line y = x followed by another reflection about y-axis.
Using this single transformation find the co-ordinates of the image of
quadrilateral ABCD with vertices A(2, 2), B(6, 2), C(7, 4) and D(3, 4).
Represent the object and image graphically.

(c) Find the single transformation equivalent to two successive reflections
in the lines y = 1 and y = 2. Using this single transformation, find the
image of ∆EFG having vertices E(2, –3), F(–1, 4) and G(3, 5). Represent
the object and image on the same graph.

288 Infinity Optional Mathematics Book - 10

Combination of Two Rotations

There are two cases in the combination of two rotations.

When two rotations are about the same Y
A(1, 3)
centres.
O
Consider a point A(1, 3). The image of the A''(3, -1)

point A(1, 3) after rotation through +90° A'(-3, 1) Y'

about origin is A'(–3, 1) and the image X' X
A'(–3, 1) after rotation through +180° about

origin is A''(3, –1).

The final image of A(1, 3) is A''(3, –1) which

is same as P(x, y) → P''(y, –x). This is the

rotation through –90° about origin.

A rotation through x° about a centre followed by another rotation through y°
about the same centre is equivalent to the rotation through (x + y)° about the
same centre.

When two rotations are about the different centres.

Rotate ∆XYZ through x° about a centre. Rotate the image ∆X'Y'Z' through y° about
the different centre to the final image ∆X''Y''Z''. The single transformation equivalent
to above two rotations is the rotation through (x + y)° about the third centre which
is the point of intersection of perpendicular bisectors of AA'', BB'' and CC''.

WORKED OUT EXAMPLES

1. If R1 is the rotation through +90° about (0, 0) and R2 is the rotation
through 180° about (0, 0) then find R2oR1 (–3, 5).

Solution: Here,

R1 = rotation through +90° about (0, 0)
R2 = rotation through 180° about (0, 0)
R2oR1 (–3, 5) = ?
Now,

R2oR1 (–3, 5) = R2[R1(–3, 5)]

= R2(5, 3) [ R1(x, y) = (y, –x)]

= (–5, –3) [ R2(x, y) = (–x, –y)]

Infinity Optional Mathematics Book - 10 289

2. ∆ABC has the vertices A(5, 4), B(6, 2) and C(2, 2).

(i) Find the image of ∆ABC after rotation through –90° about origin
followed by the rotation through 180° about origin.

(ii) Find the single transformation equivalent to these rotations.

(iii) Show ∆ABC and its images on the same graph paper.
Solution: Here,

(i) The given vertices are A(5, 4), B(6, 2) and C(2, 2).

Rotation through –90° about (0, 0).

P(x, y) P'(y, –x) A(5, 4) A'(4, – 5)

B(6, 2) B'(2, –6) C(2, 2) C'(2, –2)

Rotation through 180° about (0, 0)

P(x, y) P'(–x, –y) A'(4, –5) A''(–4, 5)

B'(2, –6) B''(–2, 6) C'(2, –2) C''(–2, 2)

(ii) The single transformation which is equivalent to the rotation through –90°

about (0, 0) followed by another rotation through 180° about (0, 0) is the

rotation about (0, 0) through (–90° + 180°) = +90° i.e. R[(0, 0), + 90°].

(iii) ∆ABC and its images ∆A'B'C' and ∆A''B''C'' are shown in the following graph.

Y

B'' A
A''
B
C X
C''
X' O

C'

A'
B'

Y'
290 Infinity Optional Mathematics Book - 10

Exercise 7.2

Section 'A'
1. (a) What is the image of P(x, y) under rotation through + 90° about origin?

(b) What point would have the image (–2, 5) under the rotation through
–180° about origin?

2. (a) What is the single transformation equivalent to rotation through + 80°
about (0, 0) followed by another rotation through +100° about (0, 0)?

(b) If r1 = rotation through 90° about origin and r2 = rotation through 180°
about (0, 0) then what does the combined transformation r1or2 represent?
Section 'B'

3. (a) Find the image of a point A(–2, 5) after rotation through 90° in
anticlockwise direction about origin followed by another rotation through
90° in clockwise direction about origin.

(b) Find the image of a point (6, 8) after rotation through 180° about origin
followed by another rotation through –270° about origin.

4. (a) Let R1 be the rotation through positive quarter turn about origin and
R2 be the rotation through half turn about origin. Find the image of
A(3, –6) under the combined transformation R2oR1.

(b) If F is the rotation through half turn about (0, 0) and G is the rotation

through quarter turn in clockwise direction about (0, 0) then find the

image of B(–3, –5) under the combined transformation FoG.
5. Let r1, r2 and r3 be the rotation through –90°, 180° and +90° respectively about

origin. Find the following.

(a) r1or2 (1, 3) (b) r1or3(–5, 8) (c) r2or3 (4, –9)

(d) r3or2(–3, –2) (e) r1or1 (–1, 4) (f) r32 (2, 6)

6. (a) What point would have the image (5, –2) under the combined

transformation of rotation through –90° about origin followed by another

rotation through 180° about origin.

Infinity Optional Mathematics Book - 10 291

(b) Let R1 be the rotation through 270° anticlockwise direction about origin
and R2 be the rotation through 90° in clockwise direction about origin.
Which point has the image A'(–3, 4) under the combined transformation

R1oR2? Find it.

Section 'C'

7. (a) A(2, 3), B(5, 1) and C(6, 4) are the vertices of ∆ABC.

(i) Find the image ∆A'B 'C' of ∆ABC after rotation through 90° in
clockwise direction about origin.

(ii) Find the image ∆A''B''C'' of ∆A'B'C' after rotation through 180° in
anticlockwise direction about origin.

(iii) Present ∆ABC, ∆A'B'C' and ∆A''B''C'' on the same graph paper.

(iv) Find the single transformation equivalent to above rotations.

(b) Find the images of ∆PQR with vertices P(–6, –3), Q(–2, –3) and R(–3, 0)
after rotation through 180 ° about origin followed by another rotation
through –270° about origin. Present ∆PQR and its images on the same
graph paper. Also find the single transformation equivalent to the
combination of above transformations.

8. (a) Find the single transformation equivalent to the rotation through –90°
about (0, 0) followed by the rotation through 180° about (0, 0). Using this
single transformation find the image of quadrilateral ABCD with vertices
A(2, 5), B(–6, 2), C(–4, 0) and D(–4, 3). Show the object and image on the
same graph paper.

(b) If R1 is the rotation through –20° about origin and R2 is the rotation
through –70° about origin then find the combined transformation to

represent R1oR2. Using the combined transformation R1oR2, find the
images of the points W(5, 1), X(–4, –3), Y(–2, 3) and Z(3, 2). Present the

object and the image on the same graph paper.

292 Infinity Optional Mathematics Book - 10

Combination of two translations

Consider a point A(2, 3). The image of the point

1 Y
4
A(2, 3) after translation by the vector T1 = is

A' (2 + 1, 3 + 4) = A'(3, 7)

A' 2
-3
The point A'(3, 7) is again translated by the 1 4 T2 =

=

translation vector T2 = 2 to the point T A'' 3
–3 T1oT2 = 1
1

A

A''(3 +2, 7 – 3) = A''(5, 4). So the final image of O X

A(2, 3) is A''(5,4).

Again, T2oT1 = 2 + 1 = 2+1 = 3 .
–3 4 –3 + 4 1

The image of the point A(2, 3) under combined transformation T2oT1 is

A'(2 + 3, 1 + 3) = A'(5, 4). Hence translation by T2oT1 = 3 is the single transformation
1

equivalent to the translation by T1 = 1 followed by another translation by
4

T2 = 2 .
–3

If the translation T1 = a is followed by the another translation T2 = c then the
b d

combination of two translations T1 and T2 is denoted by T1oT2 or T1T2 defined by

T1oT2 = T1 + T2 = a + c = a+c .
b d b+d

WORKED OUT EXAMPLES

1. If T1 is the translation by a vector 3 and T2 is the translation by the
2

vector –2 , find the image of the point A(2, –5) under the combined
1

translation T1oT2 and T2oT1.

Infinity Optional Mathematics Book - 10 293

Solution : Here,

The given point is A(2, –5)

T1 = 3 and T2 = –2
2 1

T1oT2 = 3 + –2 = 3–2 = 1
2 1 2+1 3

T2oT1 = –2 + 3 = –2 + 3 = 1
1 2 1+2 3

Under the translation by T1oT2 = 1
3

T1oT2 = a P'(x + a, y + b)
b
P(x, y)


T1oT2 = 1 A'(2 + 1, –5 + 3) = A'(3, –2)
3
A(2, –5)

Under the translation by T2oT1 = 1
3

P(x, y) T2oT1 = a P'(x + a, y + b)
b

A(2, –5) T2oT1 = 1 A'(2 + 1, –5 + 3) = A'(3, – 2)
3

2. M(5, 3), N(2, 1) and O(1, 4) are the vertices of ∆MNO. Find the image

of ∆MNO after translation with the vector –1 followed by another
–4
–3
translation with the vector –2 . Also show ∆MNO and its images on

the same graph.

Solution:

The given vertices are M(5, 3), N(2, 1) and O(1, 4)

Under the translation with the vector –1 ,
–4
T= a
b P'(x + a, y + b)
P(x, y)
–1

M(5, 3) –4 M' (5 – 1, 3 – 4) = M'(4, –1)

294 Infinity Optional Mathematics Book - 10

–1

N(2, 1) –4 N'(2 – 1, 1 – 4) = N'(1, –3)

–1 Y

O(1, 4) –4 O'(1 – 1, 4 – 4) = O'(0, 0) O
M
Under the translation with the vector –3 ,
–2

T= a
b P'(x + a, y + b)
P(x, y) X' O' N M' X
O

–3 O''
M''
M'(4, – 1) –2 M'' (4 – 3, –1 – 2) = M''(1, –3)
N'

–3 N''

N'(1, –3) –2 N'' (1 – 3, –3 – 2) = N''(–2, –5)

–3 Y'

O'(0, 0) –2 O'' (0 – 3, 0 – 2) = O''(–3, – 2)

∆MNO and its images are shown in the following graph.


Exercise 7.3

Section 'A'
1. (a) If A(x1, y1) and B(x2, y2) are the points, find the translation vector AB.

(b) If T1 = a and T2 = c then find T1oT2 and T2oT1.
b d

Section 'B'

2. (a) Find the co-ordinates of images of a point A(5, 2) under the translation

with vector 1 followed by another translation with vector –3 .
2 4

(b) Find the image of B(–3, –7) after translation with the vector T1 = 3
3

followed by another translation with the vector T2 = –4 .
2

3. (a) Let T1 = 1 and T2 = 2 be the two translations. Find the image of the
3 2

point (–4, 5) under the combined translation T1oT2 and T2oT1.
Infinity Optional Mathematics Book - 10 295

(b) If T1 is the translation with vector –4 and T2 is the translation with
–2

vector 1 then find the image of the point P(6, 4) under the combined
–3

translation T1oT2 and T2oT1.

4. (a) The image of a point P(3, 4) is the point P''(6, 10) under the translation

a followed by another translation –3 . Find the values of a and b.
b 2

(b) If T1 = –1 and T2 = 4 then find the co-ordinates of the point whose
–3 –3

image is (3, 4) under the combined translation T1oT2.

Section 'C'

5. (a) Find the images of ∆PQR with vertices P(2, 0), Q(–1, 3) and R(3, 4) under

the translation with vector 1 followed by another translation with vector
2
0
–3 . Present ∆PQR and its images on the same graph paper.

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

image of ∆ABC after translation with the vector AB followed by the

translation –2 . Show ∆ABC and its images on the same graph.
–1

6. (a) The vertices of ∆EFG are E(–2, 5), F(–2, 3) and G(2, 3). If T1 = 0 and
2
–2
T2 = 1 are two translations, find the image of ∆ABC under T1oT2 and

hence show the image and object on the same graph.

(b) If T1 = 1 and T2 = –3 are two translations then find the image of
2 1

rectangle PQRS having vertices P(2, 1), Q(5, 1), R(5, 2) and S(2, 2) under

T2.T1 and hence show the image and object on the same graph.

296 Infinity Optional Mathematics Book - 10

Y

Combination of two Enlargements

There are two cases in the enlargement followed A''(6,6)

by another enlargement. They are

When the centres of enlargement are same. A(1,1) A'(2,2) X
Consider a point A(1, 1). The image of A after O
enlargement with the centre (0, 0) and scale factor
2 i.e. E[O, 2] is A'(1 × 2, 1 × 2) = A'(2, 2). The image
of A'(2, 2) after enlargement with the centre (0, 0)

and scale factor 3 i.e. E[O, 3] is A''(2 × 3, 2 × 3) =

A''(6, 6). Hence, the final image of A(1, 1) is A''(6, 6). i.e. A(1, 1) → A''(6, 6) = A''(6 ×

1, 6 × 1)

In A''(6 × 1, 6 × 1), the scale factor is 6 which is the product of 2 and 3.



An enlargement with centre O and scale factor k1 i.e. E[O, k1] followed by another
enlargement with centre O and scale factor k2 i.e. E[O, k2] is equivalent to an
enlargement with same centre O and scale factor k1.k2

When the centres of enlargements are different

Consider a point A is transformed to A' under

enlargement with centre O and scale factor k1 A''
i.e. E[O, k1]. Again the point A' is transformed A'
to A'' under the enlargement with centre O'
O'
and scale factor k2 i.e. E[O', k2]. The combined
transformation of above two enlargements is

equivalent to the enlargement with the centre A
O''
O'' and scale factor k1.k2 i.e. E[O'', k1.k2]

O

Infinity Optional Mathematics Book - 10 297

WORKED OUT EXAMPLES

1. If E1 [(0, 0), 3] and E2 [(0, 0), –2] are the two enlargements then find the
image of A(2, 5) under E1oE2.

Solution: Here,

E1[(0, 0), 3] and E2 [(0, 0), – 2]are two enlargements. The combination of two
enlargements is also the enlargement with centre = (0, 0) and scale factor =

3 × (–2) = – 6 i.e. E[(0, 0), –6]
Now, P(x, y) E[(0, 0), k] P'(kx, ky)
A(2, 5) E[(0, 0), –6] A'(– 6 × 2, –6 × 5) = A''(–12, –30)

∴ The image of A(2, 5) under E1oE2 is A''(–12, –30).

2. If A(2, 0), B(3, 1) and C(1, 2) are vertices of ∆ABC, find the image of
∆ABC under an enlargement with centre (0, 0) and scale factor 2
followed by another enlargement with centre (1, 2) and scale factor
–3. Show the ∆ABC and its images on the same graph paper.

Solution:   Here, the vertices of ∆ABC are A(2, 0), B(3, 1), C(1, 2).

Under enlargement with centre (0, 0) and scale factor 2.
P(x, y) E[(0, 0), k] P'(kx, ky)
A(2, 0) E[(0, 0), 2] A' (2.2, 2.0) = A'(4, 0)
B(3, 1) E[(0, 0), 2] B'(2.3, 2.1) = B'(6, 2)
C(1, 2) E[(0, 0), 2] C'(2.1, 2.2) = C'(2, 4)

Under the enlargement with center (1, 2) and scale factor –3.
P(x, y) E[(a, b), k] P'(kx – ka + a, ky – kb + b)
A' (4, 0) E[(1, 2), –3] A''[(–3) × 4 – (–3) × 1 + 1, (–3) × 0 – (–3) × 2 + 2]

= A''(–8, 8)
B'(6, 2) E[(1, 2), –3] B''[(–3) × 6 – (–3) × 1 + 1, (–3) × 2 – (–3) × 2 + 2)

= B''(–14, 2)

298 Infinity Optional Mathematics Book - 10

C'(2, 4) E[(1, 2), –3] C''[(–3) × 2 – ( –3) × 1 + 1, (–3) × 4 – (–3) × 2 + 2)

= C''(–2, –4)

The ∆ABC and its images are shown in the graph.

Y

A''

C'

B'' C B'
X'
B X
O A A'

C''
Y'

Exercise 7.4

Section 'A'

1. (a) If E1[(0, 0), k1] and E2 = [(0, 0), k2] are two enlargements then find the
combined transformation E1oE2.

(b) If E1[(2, 3), –2] and E2[(2, 3), 3/2] then find E2oE1.
Section 'B'

2. (a) If E1[(0, 0), 3] and E2[(0, 0), – 1] be two enlargements then find the

image of A(–2, 3) under the combined transformation E1oE2 and E2oE1.

(b) If E1 is an enlargement with centre origin and scale factor 1 and E2 is an
2

enlargement with centre origin and scale factor 3 then find the image of

B(6, –12) under E1oE2 and E2oE1.

3. (a) E1[(1, 2), – 2 ] and E2[(1, 2), 3) are two enlargements. Find the image of
3

M(3, 4) under E2oE1 and E1oE2.

(b) If E1 is the enlargement with centre (0, 5) and scale factor –2 and E2 is

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the enlargement with centre (0, 5) and scale factor 3 then find the image

of N(–4, 2) under E2oE1 and E1oE2.
4. (a) Find the image of A(1, 5) after enlargement with centre origin and scale

factor 3 followed by another enlargement with centre (1, 2) and scale
factor –3.
(b) Find the image of B(–3, –5) after enlargement with centre (2, 3) and
scale factor 2 followed by another enlargement with centre (0, 0) and
scale factor –4.
5. (a) Let E1 [O, 2] and E2 [O, –3] be the two enlargements. What point would
have the image (–6, 18) under the combined transformation E1oE2.
(b) If the image of A under the enlargement E[(2, 3), – 1] followed by another

enlargement E[(0, 0), 3] is A''(9, 12), find the co-ordinates of A.

Section 'C'

6. (a) The vertices of ∆ ABC are A(6, 2), B(2, 3) and C(4, 5). If E1 = [(0, 0), 2]
and E2 = [(0, 0), – 1], then find the co-ordinates of image of ∆ABC under
enlargement E2oE1. Draw both triangles on the same graph.

(b) The vertices of ∆PQR are P(1, 2), Q(2, –1) and R(2, 0). If E1 [(0, 0), 2]
and E2 [(0, 0), 3], then find the co-ordinates of image of ∆PQR under
enlargement E1oE2. Draw both triangles on the same graph.

7. (a) Let E1[(1, 2), – 1] and E2 [(1, 2), 2] be two enlargements. Find the
co-ordinates of image of ∆XYZ having vertices X(–2, 3), Y(4, 0) and

Z(1, 5) under enlargement E1oE2. Draw ∆XYZ and its image ∆X'Y'Z' on
the same graph.
(b) Let E1 [(–3, 0), 23] and E2[(–3, 0), – 6] be the enlargements. Find the
co-ordinates of image of ∆ABC with vertices A(–2, –3), B(–1, 2) and
C(–1, 1) under the enlargement E2oE1. Draw ∆ABC and its image
∆A'B'C' on the same graph.
8. (a) A triangle with vertices A(1, 0), B(2, 1) and C(3, –1) is transformed by
enlargement with centre (0, 0) and scale factor – 2 and draw its image
in the graph. Again its image is transformed by the enlargement with
centre (–1, 2) and scale factor 1 and draw its image on the same graph.
(b) Draw a quadrilateral OABC having vertices O(0, 0), A(2, 0), B(3, 1)
and C(1, 1) on a graph paper. Find the images of the vertices of the
quadrilateral OABC under E1[(0, –3), 2] followed by E2 [(0, 0), 3] and
draw the images on the same graph paper.

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