Prime Optional Mathematics 10

10. The angle of elevations of the top of a cloud from a point

10m above from the water level of a pond from another

side is 30° and angle of depression to its reflection is 45°. A
E
Find the height of the cloud from the water level of the B
F
pond.
A
Solution : 30° C
45° B
Let, BC be the water level of the pond. D

A be the position of the cloud. x 4x 10m
F be the reflection of the cloud.

D be the position of the observer. C

Given : CD = EB = 10m
AB = BF (object and its reflection)
\ ADE = 30°
\ EDF = 45°

Now, In right angled DAED

Tan 30° = AE G
DE

or, 1 = AE
3 DE

\ DE = 3AE .... (i)

Again, In right angled DDEF

Tan45° = EF
DE

or, EF = DE

or, EB + BF = 3AE[from equation ..... i]

or, 10 + AB = 3(AB – EB)

or, AB( 3 – 1) = 10( 3 + 1)

or, AB = 10 × 2.732
0.732

\ AB = 37.32 m

11. A pole is divided by a point in the ration 1:4 where both D q
parts makes equal angles at a point on the ground q
respectively at a distance of 10m. Find the height of the 10m
pole.
Solution :

Let, AB be the height of a pole which is divided by a point
‘C' in the ratio 1:4.
D be the point on the ground.

Given :

\ ADC = \ CDB = q (say)

BC = x(say)

AC = 4x

BD = 10m

246 PRIME Opt. Maths Book - X

Now, In right angled DCBD,

Tan q = BC
BD

or, Tanq = x ..... (i)
10

Again,

In right angled DABD,

Tan (q + q) = AB
BD

or, 2Tani = 5x
1 – Tan2 i 10

x

or, 5x = 2 10
10 5

1 a x 2
10
k

x

or, x = 2
2 x2
1 – 100

x x 100 20
or, 2 = 5 × 100 – x2

or, 100 – x2 = 40
or, x2 = 160
\ x = 12.65m
\ Height of the pole AB = 5 × 12.65 = 63.25 m

Steps of solving heights and distances
• Draw the suitable ray diagram according to question.
• Use meanings for the rays the rays used in diagram.
• Use right angled triangle so formed in diagrams and use

suitable trigonometrical rations for two right angled triangles

separately.
• Simplify and solve for unknowns.

PRIME Opt. Maths Book - X 247

Exercise 5.6

1. Find the value of the following as indicated in the following diagrams.
A P
a) b) 30°

60°
20m
30° 45° B 12cm Q R 30m S
D ?C ?d) C

c) A ?
60m
?E 30° A 45° E
C 30° D
B 20cm
e) ?
60° D
B
A ?

D

60° 12m
CB

2. i) The angle of elevation of the top of a pole 20m high is found to be 45° from a point
on the ground and elevation changes to 60° on walking towards the pole. Find the
distance traveled by him from first observation point to second observation point.

ii) The angle of elevation of the top of a house 24m high from the bank of a river is 30°
and from the another bank towards house is 60°. Find the width of the river.

iii) The angle of depressions of the two edges of a road in front of a house from the top
of the house are found to be 60° and 45° respectively. Find the height of the house
where width of the road is 20m.

iv) The angle of depression of the two stones situated on the ground at opposite side of
a pole from the top of a pole 16m high are found to be 30° and 60° respectively. Find
the distance between the stones.

248 PRIME Opt. Maths Book - X

v) The length of the shadow of a tree becomes 30m longer when sun's altitude changes
from 60° to 30°. Find the height of the tree.

3. i) The angle of depression of a boat coming towards a light house of 100m high is 30°.
After 20 seconds depression changes to 45°. Find the speed of the boat.

ii) The angle of elevation of the top of a cliff situated at the sea-shore from a ship
coming towards the sea-shore is 30°. The elevation changes to 60° after traveling
20 seconds. At what time, the ship reach to the sea–shore?

iii) An aeroplane is descending to the ground from a height of 100 3 m by 60° and
comes in rest after traveling 5 seconds on the ground where the angle of elevation
of the aeroplane from the stop before descending was 30°. Find the average speed
of the aeroplane on the ground.

iv) The angle of elevation of the top of a column from any two points on the ground
lying in a straight line which are at a distance of 18m and 50m far from the foot of
the column are found to be complementary. Find the height of the column.

v) The angle of depression of the two places on the ground which are on opposite side
of a pole at a distance of 72m and 128m from the top of the pole are found to be
complementary. Find the height of the pole.

4. i) A flagstaff of height 6ft is at the top of a pole where angle of elevation of the foot
of the flagstaff from a point on the ground is 30° and then angle formed during
observing to the top is 15° respectively. Find the height of the pole.

ii) The angle of elevation of the foot of the chimney and angle formed during observing
the top of the chimney situated at the top of a house 20ft high are found to be 45°
and 15° respectively from a point on the ground. Find the height of the chimney.

iii) One pole is double than the other which are at a distance of 200m. The angle
of elevations of their tops from mid-way between their bases are found to be
complementary. Find the height of the taller pole.

iv) An aeroplane is flying horizontally from a height of 4000m the angle of elevation of
the aeroplane from a point on the ground is 60° and change to 45° after 5 seconds.
Find the speed of the aeroplane.

v) A 8 m long ladder is taken against an electric pole where the top of the ladder
touches the pole 8m below from the top of the pole. If angle of elevation of the top
of the pole from the foot of the ladder is 60°, find the height of the pole.

5. i) The angle of elevation of the top of a house from the top of a column 12m high is
found to be 30° and from its foot is 60°, find the height of the house.

ii) The angle of elevation of the top of a tower from the foot of a house 40m high is 60°
and angle of elevation from the foot of the tower to the top of the house is 45°, find
the height of the tower.

iii) The angle of depression of the top and bottom of a pole from the top of a house 36
m high are found to be 30° and 60° respectively, find the height of the pole.

iv) The angle of elevation and depression of the top and bottom of a tower from the top
of a house 18m are found to be 30° and 60° respectively, find the height of the tower.

v) The angle of elevations of the window and roof of a house from the eye of an
observer 2m tall standing on the ground are found to be 45° and 15° respectively
where height of roof from the window is 5m, find the height of the house.

PRIME Opt. Maths Book - X 249

6. PRIME more creative questions

a. i) A rope is tied at the top of the equal posts of height 10m where a rope dancer is

standing on the rope and parts of the rope makes the angles 30° and 45° with the

horizontal line at 6m above from the ground. Find the length of the rope.

ii) A bird is at the top of a tower standing at the bank of a pond where angle of elevation

of the bird from 10m above in opposite side is 30° and angle of depression to its
reflection is 60°, find the height of the tower from the water level.
iii) A pole is divided by a point in the ratio 1:9 where both parts of the pole makes equal
angles at a point on the ground 20m far from the foot, find the height of the pole.
iv) The angle of elevation of the top of a mountain from its foot is found to be 45°. On

walking up 1.5 km towards the top making an angle of 30°, the elevation changes to

60°. Find the height of the mountain.
v) A 2m tall boy finds the angle of elevation of the top of a tree from one bank of a river

to the opposite bank is 30° and angle of depression to the image of the top of the

tree into the river is found to be 60°. Find the height of the tree.

b. i) A screen is kept at the one corner of a rectangular cricket ground of 60m × 20 3m
ii) at a certain height where angle of elevation of it from the opposite corner is 30°, find
iii) the height of the screen from the ground.
iv)
v) The angle of depression of the foot of a column from the top of a house is double than

that of the angle of elevation of the top of the column. If the height of the column and

height of the house are in the ratio of 4:3 and distance between them is 8 3m. What
will be the height of them? Also find the angles so formed.
The angle of depressions of the top and bottom of a lamp post from the top of a

house 60m high are found to be complementary where distance between them is
20 3m. Find the height of the lamp post. Also find the angles so formed.
The angle of elevation of the top of a pole from a point on the ground is double the

angle formed by another point. Find the height of the pole where the points are on a

same straight line at a distance of 10 3m and 30 3m respectively from the foot of
the pole. Also find the angles so formed.
The angle of elevation of the top of a house 15m high from the foot of the tree is half
of the angle of depression from the top of the tree of height 60m, find the distance
between the tree and the house. Also find the angles so formed.

7. Project Work
Prepare a chart of trigonometric formulae in chart paper and paste in your classroom.

250 PRIME Opt. Maths Book - X

Answer

1. i) 14.64 m ii) 25.98 m iii) 30 m iv) 38.03 m, 21.96 m v) 18 m

2. i) 8.45 m ii) 27.71m iii) 53.76 m iv) 18.47 m v) 25.98 m

3. i) 3.66 m/s ii) 10 sec iii) 40 m/s iv) 30 m v) 96 m

4. i) 8.2 ft ii) 14.64 ft iii) 200 m iv) 338.1 m/s v) 12 m

5. i) 18 m ii) 69.28 m iii) 12 m iv) 24 m v) 8.83 m

6.a. i) 13.656 m ii) 20 m iii) 10 5 m iv) 2049 m v) 4 m

b. i) 40 m ii) 32m, 24m, 30° & 60° iii) 40 m, 30°, 60°

iv) 30m, 30° & 60° v) 15 3 m, 30°, 60°

PRIME Opt. Maths Book - X 251

Trigonometry : 1

Unit Test

Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. If CosA = 1 , find the value of Cos2A.
2

3 3 +1
2. a) If Cos330° = 2 , prove that Cos165° = – 2 2 .

b) If Cos A = 1 (m + 1 ), find CosA.
3 2 m

c) Find the value of 2Cos r .
16

3. a) Prove that : Sinq. Sin(60° – q).Sin(60° + q) = 1 Sin3q
4

b) Prove that : Sin2q – Cos2q.Cos2b = Sin2b – Cos2b.Cos2q

4. If 2Tan q = 3 Tanb, prove that : Tan(q + b) = 5Sin2b
5Cos2b – 1

Trigonometry : 2

Unit Test

Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:

1. Prove that : Cos80° + Cos40° = Cos20°

2. a) Prove that : Sin10° + Cos40° – Sin70° = 0

b) Solve for (0 ≤ q ≤ 180°) : Sinq – Cosq = 0
SinA + SinB A+B
c) Prove that : CosA + CosB = Tan 2

3. a) Prove that : = Tan(a + b)

b) Solve : 3Cos2q – Sin2q = 3 (0 ≤ q ≤ 360°)
4. The angle of elevation of the top of the poles from mid-way between their bases are found

to be complementary where one pole is double than the other and distance between the
poles is 200m, find the height of the poles.

252 PRIME Opt. Maths Book - X

Unit 6 Vector Geometry

Specification Grid Table

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

No. of Questions 1 2 – 1 4 10 18
Weight 1 4 – 5

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

Questions, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to define scalar product and able to find angle between

any two vectors.
• Students are able to use condition of perpendicular vectors and parallel

vectors.
• Students are able to solve the geometrical theorems using vector method.
• Students are able to know the position vector in different situations in daily

life.

Materials
• Directed line segment in graph.
• Ways of informations given by RADAR at the airport.
• Geometrical shapes with the position vector.
• Chart of triangle law, mid-point formula & section formula.
• Graph paper.
• Graph board.

PRIME Opt. Maths Book - X 253

6.1 Vector

Enjoy the recall:
i. Vector quantity:

The quantity having magnitude as well as direction is called vector quantity.
Examples : Velocity, acceleration, force, power etc.

ii. Magnitude of a vector:
The modulus of a vector which is the length of the vector too is called its magnitude.
|a| =

iii. Direction of a vector.
The angle made by a vector to the horizontal axis which is calculated using tangent of
angle is called its direction.
For direction 'q'
y – component
Tanq = x– component

iv. Equal vectors:
Any two vectors having same direction and equal magnitude are called equal vectors.
AB = AB

v. Opposite vectors (Negative vectors):
Any two vectors having opposite direction but equal magnitudes are called opposite
vectors.

AB = – BA

vi. Parallel vectors:
Any two vector a & b which are in the condition of a = mb or b = na are called parallel
vectors.
i.e. having same directions but different magnitudes.

vii. Unit vectors:
The vector having magnitude always unit (one) is called unit vector.
Unit vector of a (a) = 1 (a)
a

viii.Triangle law of vector addition:
Sum of any two vectors AB & BC results the resultant vector AC of the shortest distance
between such vectors is called triangle law of vector addition.

i.e. AB + BC = AC C
or, AB + BC – AC = 0 B
\ AB + BC + CA = 0

A

254 PRIME Opt. Maths Book - X

ix. Quadrilateral law: D C
According to triangle law, A B
AC = AB + BC [in DABC]
AD = AC + CD [in DACD]
Then,
AD = AC + CD
or, AD = AB + BC + CD
or, AB + BC + CD – AD = 0

\ AB + BC + CD + DA = 0

x. Parallelogram law:
Sum of any two vectors having same initial point is equal to the diagonal vector of a
parallelogram completed in such vectors.
BC

i.e. OA + OB = 0 bc
\ a+b=c

OA

xi. Position vector: a

The vector of a point in which origin is taken as the initial point is called position vector

of the point. Y
A(x, y)
i.e. OA = d x2 – x1 n
y2 – y1

x–0
= d n
y –0

x X' O X
= dn
y

Y' Y

xii. Unit vector along x- axis is i where,

1
i = dn
0
X'
Oi X
Y'

xiii. Unit vector along y - axis is j where,

0
j = dn
1 Y

j X
X' O
255
Y'

PRIME Opt. Maths Book - X

Xiv. i and j vector for a vector x is a vector.
It can be written as, =d n
a y

1 n + y 0 n
a = x d d
0 1

= xi + yj

xv. For any vector joining the points PPQ(xo1n, yx1-)axainsd Q(x2, y2)
x-component of PQ = projection of
Y
= M1N1
= ON1 – OM1
= x2 – x1
N2 Q
N1
y component of PQ = projection of PQ on y-axis M2 P
X' O M1
= M2N2
= Oy2N–2 y–1OM2 Y'
= X

\ PQ = dM1 N1n
M2N2

= d x2 – x1 n
y2 – y1

= (x2 – x1)i + (y2 – y1)j

Dot product of any two vectors (scalar product of two vectors)

The dot product of any two vectors a and b can be written as a . b which is also called
the scalar product of two vectors.

B

b

q a A
O

The dot product of any two vectors having same initial
point is the product of their magnitude and cosine of angle
between them.

i.e., a . b = a b Cosq PRIME Opt. Maths Book - X
OA . OB = OA OB Cosq

256

Conditions : b
i) To find angle between any two vectors q

Cosq = a . b a
ab
b
ii) For the orthogonal vectors (perpendicular).
a . b = a b Cos 90° = 0
a.b =0

iii) For the parallel vectors, a
a . b = a b Cos 0° = a b
a.b = a b b
a
iv) i . i = |i ||i |Cos0°
=1×1×1 (1, 0)
=1

j . j = |j ||j | Cos0° (0, 1)
=1×1×1
=1

i . j = i j Cos90° (0, 1)
=1×1×0 (1, 0)
=0

v) If a = d x1 n , b = d x2 n , then Y
y1 y2

a .b = dx1 n . d x2 n (x1, y1)
y1 y2
q1 (x2, y2)
= _x1 i + y1 j i. _x2 i + y2 j i X' q2 X
= x1 x2 i . i + x1 y2 i . j + x2 y1 j . i + y1 y2 j . j
\ ab = x1 x2 ×1 + x1 y2 ×0 + x2 y1 ×0 + y1 y2 ×1
= x1 x2 + y1 y2
= x1 x2 + y1 y2

Y'

PRIME Opt. Maths Book - X 257

vi) a2 = a . a a

= a . a Cos0°
= a 2#1

= a2

` a2 = a 2 = a2

Worked out Examples

1. If a = d32n and b = d15n , find the angle between a and b .

Solution:

2 = 5
= d n , b dn
a 3 1

Here,

|a| = x2 + y2 = 22 + 32 = 13 units

|b| = x2 + y2 = 52 + 12 = 26 units

25
a. b = d n . dn
3 1

=2×5+3×1

= 13.

Then, for the angle between a & b is ‘q'.

Cosq = a . b
ab

or, Cosq = 13

13 × 26

131
or, Cosq =

13 2

or, Cosq = Cos45°
\ q = 45°

2. Prove that the vectors a = d43n and b = d–34n are orthogonal.
Solution:

3 –4
a = dn , b= d3 n
4

Now,

3 –4
a. b = d n . d n
4 3

= (3 × (–4) + (4) × 3

= –12 + 12

=0

\ a. b = 0
i.e They are orthogonal vectors.

258 PRIME Opt. Maths Book - X

3. If position vectors of A and B are 3 i – 2j and 6 i – mj which are parallel to each other,

find the value of ‘m'.

Solution:

3
= 3i – 2j = dn
OA –2

6
= 6i – mj = dn
OB –m

Now,

OA = x2 + y2 = 32 + (–2)2 = 13 units

OB = x2 + y2 = 36 + m2

OA . OB = OA OB

(2m + 18)2 = 13 , m2 + 36
Squaring on both sides,
(2m + 18)2 = _ 13 (m2 + 36 i2
or, 4m2 + 72m + 324 = 13m2 + 468
or, 9m2 – 72m + 144 = 0
or, m2 – 8m + 16 = 0
or, (m – 4)2 = 0

or, m – 4 = 0

\ m=4

4. If a + b + c = 0, and a , b & c are unit vectors find the angle between a and b .
Solution,
a+b+c =0
|a| = |b| = |c | = 1 units. [unit vectors]
Now,
Taking,
a+b+c =0
or, a + b = –c
Squaring on both sides,

or, ^a + bh2 = ^–c h2

or, a2 + 2a . a + b2 = c2

or, a 2 + 2|a|.|b| Cosq + b 2 = c 2

or, 2 × 1 × Cosq = 12 – 12 – 12

or, Cosq = – 1
2

or, Cosq = Cos120°
\ q = 120°

PRIME Opt. Maths Book - X 259

Exercise 6.1

1. Answer the following questions.
i) Write down the formula to find scalar product of a and b.

ii) If a = dx1 n and b = d x2 n , find the result of a .b.
y1 y2

iii) Find the value of i . i and i .j .
iv) Prove that a2 = a2.
v) In what condition any two vectors a and b. becomes perpendicular?

2. Find the dot product of the given vectors from the followings.

i. a = 3 and b = 4
dn dn
2 –3

ii. p = 1 and q = 3
dn dn
–2 2

iii. a = 2 i + 3j and b = 4 i – j
iv. c = 3 i – 2j and d = 5 i – 3j
v. OA = 2 3 i + j and OB = 3 i – j

3. Find the scalar product of the given pairs of vectors.
i. |a| = 13 units, |b| = 26 units, q = 45°
ii. |a| = 2 units, |b| = 2 units, q = 30°
iii. |p| = 50 units, |q| = 5 units, q = 135°.
iv. OP = 2 units, OR = 6 units, q = 60°.
v. |c | = 6 units, |a| = 2 units, q = Cos–1 ( 2).

4. Find the angle between the following pairs of vectors.
i. |a| = 2 units, |b| = 2 2 units, a. b = 4

ii. |c | = 5 units, |d| = 4 units and c. d = –10.
iii. |p| = 4 units |q| = 3 units and p. q = 6.
vi. |a| = 6 units, |b| = 4 units and a. b = 12 3.
v. op = 3 units OR = 3 2 unit and OP. OR = –9

5. Find the angle between the vectors a & b from the followings:

i. a = 1 and b = 3
dn dn
2 1

ii. a = e 3o and b = d 1
1 n

3

iii. a= 2 and b = 3
dn 33
0

260 PRIME Opt. Maths Book - X

iv. a = 5 i + j and b = – 2 i – 3j
v. a = –i – 7j and b = 3 i – 4j

6. Find the angle between the vectors from the followings.

i. Find \ AOB where 3 and 4 n
= dn = d .
OA 5 OB 1

ii. Find \ POQ where OP = 4 i + j and OQ = –3 i – 5j .

iii. Find \ AOB where position vectors of A and B are 3 i + 3 3 j and –2i

iv. Find \ POQ where position vectors of P and Q are 3 i – j and j – 3 j

v. If A(1, –1), B(3, 1) and C(3, –1) are the angles of a triangle, find \ A, \ B and \ C.

7. Prove that the following pair of vectors are orthogonal.
i. a = 3 i – 4j and b = 4 i + 3j
ii. b = 6 i + 8j and c = 4 i – 3j

iii. p = –2 and q = 9
dn
e3 2o
6
2

iv. OA = e–2 3 o and OB = e 3 o

66

v. AB and PQ where, A(3, –2), B(–1, 4), OP(–6, –5), and Q(12, 7) are the points.

8. Prove that the following pairs of vectors are parallel.
i) OA = 2 i – 4j and OB = 6 i – 12j

ii) a = ` 4 j and b 2
2 =d n

1

–3

iii) OP = ` –6 j and OQ =e 2 o
4 1

iv) c = 8 i – 6j and d = 4 i – 3j

v) AB and CD where A(2, 1), B(1, 3), C(7, –2) and D(1, 2) are the four points.

9. i) If a =ap p 1 k and b 4 are perpendicular vectors, find the value of ‘p'.
+ =d n

–3

ii) If OA = 6 i – 8j and OB = (m + 1) i + 3j are orthogonal vectors, find the value of
‘m'.

iii) If p= a and q= 4 are parallel vectors, find the value of ‘a'.
dn dn
3 6

iv) If OP = 3 i – 2j and OQ = 6 i – nj are parallel vectors, find the value of ‘n'.
3 p
v. If angle between the vectors a = a 2 k and b = ` 1 j is 45°, find the value of ‘p'.

10. PRIME more creative questions:
a. i) If a, b and c are the unit vectors and a – b + c = 0, find the angle between

a and b.

PRIME Opt. Maths Book - X 261

ii) If a+ b + c = 0, |a| = 3, |b| = 5 and |c | = 7, find the angle between a and b.
iii) If a + b = a – b , prove that a and b are orthogonal vectors.
iv) If a + 2b and 5a – 4b are perpendicular vectors where a and bare unit vectors,

find the angle between a and b.
v) If p and q are the unit vectors, a = 2p – 3q , b = 3p – 2q and a.b = 12, prove that p

and q are perpendicular vectors.

b. i) If angle between the vectors 2i – pj and i – 3j is 45°, find the value of p.
ii) Prove that a and b are orthogonal vectors from the relation |4a + b| = |4a – b|.
If a + b + c , |b| = 5, |c | = 6 and b.c = –18, find |a|.
iii) Find the angles of a triangle having vertices A(1, 3), B(1, –4) and C(–6, –4).
iv) If mi + (m + 1)j and 12i + 15j are parallel vectors, find ‘m'.
v)

Answer

1. Show to your teacher.

2. i) 6 ii) – 1 iii) 5 iv) 21 v) 5
iii) –25 iv) 6
3. i) 13 ii) 2 3 iii) 60° iv) 30° v) 12
iii) 60° iv) 135°
4. i) 45° ii) 120° iii) 120° iv) 150° v) 135°
iv) 4
5. i) 45° ii) 30° iii) 2 v) 45°
iv) 60° v) 4
6. i) 45° ii) 135° iv) 45°, 45°, 90° v) 45°

7. i) 3 ii) 3 v) –5 or 1
10. a. i) 60° ii) 60° 5
iii) 5 units
b. i) 1 or –4

262 PRIME Opt. Maths Book - X

6.2 Vector Geometry:

Position vector of a point p(x, y) can be taken as the Y
O
following geometrically as given in diagram :

Here, PM^OX. P(x, y)
Then, Position vector of P(x, y) is, MX

OP = = ` OM j = ` x j
PM y

We can prove geometrical properties by using
vector operations as the application of the vector
in geometry is called the vector geometry.

Here, we are discussing about the theorems and their theoretical proof by using vector
geometry.

Application of position vector in vector geometry: O

Theorem 1 : Position vector of mid - point of a line segment.

Solution: b
Let ‘M' be the mid - point of a line segment AB where the ?
position vectors of end points are
a M B

OA = a

OB = b

OM = m = ? A

Here,
Since, M be the mid point between A and B, we have

AM = MB

or, AM = MB
or, AO + O M = MO + OB [\ Using triangle law]

or, – a + OM = – OM + b

or, 2 OM = a + b

\ OM = a + b
2

i.e. OM = OA + OB
2

Theorem 2 : Position vector of a point which cuts the line segment in the ratio m:n internally.
Solution :
Let, a point ‘P' cuts a line segment AB in the ratio m:n internally, where position vectors
of A and B are
OA = a

OB = b

OP = ?

PRIME Opt. Maths Book - X 263

Here, As the concept of section point, O
b
AP = m
PB n

or, n AP = m PB a ? B
or, n AP = m PB mP

or, n ^AO + OPh = m ^PO + OBh [ a Using triangle law ] n

or, n ^–a + OPh = m ^–OP + bh A

or, m p + n OP = m b + n a

\ OP = mb +na
m+n

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

Theorem 3 : Position vector of a point which cuts a line segment in the ratio m:n externally.
Solution :
Let, a point P cuts the line segment AB in the ratio m:n externally, where the position

vectors are, O
OA = a

OB = b ?
OP = ?

Here, as the concept of section point, ab P

AP = m n
BP n
B

or, n AP = m BP Am

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

or, n (– a + OP) = m (–b + OP)

or, –n a + nOP = –mb + mOP

or, m b – na = OP (m – n)

\ OP = mb – na
m–n

i.e. OP = mOB – nOA
m–n

Corollary 1 : Position vector of a point ‘p' which cuts a line segment AB as 2AP = 3PB.

Solution : O

Here, A point ‘P' cuts a line segment AB as 2AP = 3PB, where b

the position vectors are,

OA = a a? B

OB = b

OP = ? P
A
Now, Taking,
2AP = 3PB

or, 2 AP = 3 PB

or, 2 (AO + OP) = 3 (PO + OB) [ a Using triangle law ]

or, 2(– ^–a + OPh = 3 (–OP + OB)

or, – 2 a + 2 OP = – 3OP + 3OP

264 PRIME Opt. Maths Book - X

or, 5 OP = 2a + 3b

\ OP = 2a + 3b
5

Corollary 2 : Find the position vector of ‘D' of a parallelogram ABCD where OA = a , OB = b

and OC = c AD
Solution:

Here, a
ABCD is a parallelogram, where the position vectors are,

OA = a BC
OB = b

OC = c bc
O
OD = ?
Now, By using triangle law,

OD = OC + CD [ a Opposite sides of a parallelogram ]
= OC + BA

= OC + (BO + OA)

OD = C + ( – b + a)

= a–b+c

\ OD = a–b+c

A

Corollary 3 : Position vector of centroid of a triangle. R GQ
Solution:
Let ‘G' be the centroid of a DABC, where AP, BQ and CR are
the medians.

The position vectors are, BP C
OA = a

OB = b O

OC = c

OG = g = ?

Now, Using position vector of mid - point P of side BC,
1
OP = 2 (OB + OC)

= 1 (b + c)
2
Again,
We have, centroid ‘G' cuts the median AP in the ratio 2 : 1,

Then,

OG = mOP + nOA
m+n

=

= b+c +a = a +b +c
3 3

\ g = a +b +c
3

i.e. OG = OA + OB + OC
3

PRIME Opt. Maths Book - X 265

Theorem 4 : Prove that AC2 = AB2 + BC2 in right angled triangle at \B = 90°

Solution: A
In a right angled 3ABC, \B = 90°
Now, Using triangle law,

AC = AB + BC

Squaring on both sides,
or, AC 2 = (AB + BC)2
or, AC 2 = AB 2 + 2AB BC + BC 2 B C

or, AC 2 = AB 2 + 2 AB.BC + BC 2

or, AC2 = AB2 + 2 x 0 + BC2 [ a AB = BC and AC 2 = AC 2 = AC2]
\ AC2 = AB2 + BC2 is Proved

Exercise 6.2

1. Answer the following questions.
i) Write down the triangle law of vector addition.
ii) Write down the position vector of mid-point of a line segment AB.
iii) Write down the position vector of centroid of DABC.
iv) Write down quadrilateral law of vector addition.
v) Write down the position vector of a point P which cuts a line segment AB in the ratio
m:n internally.

2. i) If 2 i + j and 4 i – 3j are the position vectors of A and B of a line segment AB. Find
the position vector of mid - point ‘P' of AB.
ii) If position vector of P and Q are 3 i – j and i + 5j , Find the position vector of middle
point ‘M' of PQ.
1
iii) Find the position vector of mid - point of a line segment AB where OA = a 2 k and

iv) FOiBnd=t`h34e j. vector of mid - pint of AB where A(1, – 2) and B(3, – 2) are the
position

points.
v) If position vector of mid - point of a line segment joining the points P(1, – 3) and Q(a,
b) is (–1, 2), find the co - ordinate of Q.

3. i) Find the position vector of a point which cuts the line segment joining the points
(5, – 2) and (– 1, 4) in the ratio 1 : 2.

ii) Find the position vector of a point which cuts the line segment joining the points
(– 3, –4) and (7, – 4) in the ratio 3 : 2.

iii) The position vector of A and B are 2 i + 3j and – 4 i + 3j respectively. Find the
position vector of a point which cuts AB in the ratio 2 : 1.

iv) Find the position vector of a point which divides a line segment PQ in the ratio 1 : 3
externally where OP = 3 i – j and OQ = 5 i + j .

v) The position vectors of M and N are 2 i + 3j and i – 2j respectively. Find the
position vector of a point which cuts MN in the ratio 2:1 externally.

266 PRIME Opt. Maths Book - X

4. i) Find the position vector of centroid of a triangle having vertices A(1, 3), B(2, –1) and
C(3, 1).

ii) Find the position vector of centroid of DABC where OA = 2 i – j , OB = 3 i – 4j and,

OC = i – 4j .
iii) The position vectors of the vertices of TPQR are i + 4j , 2 i – 3 j and 3 i – 7j , find

the position vector of its centroid.
iv. Find the position vector of centroid of TABC where the position vector of ends of

median AP are 4 i + 2j and i – 7j respectively.
v. The vertex A(2, –1) and mid-point of opposite side is M(2, –4) of DABC. Find the

position vector of centroid of DABC.

5. i) If OA = a , OB = b in DOAB where M is the mid - point of AB, prove that
1
OM = 2 (a + b) .

ii) If OA = a, OB = b and a point ‘p' cuts AB in the ratio 5 : 2 externally, prove that
1
OP = 3 (5b – 2a) .

iii) A point ‘C' cuts AB as 2AB = 3CB, prove that OC = 1 (2a + b)
O 3

b

a? B

C
A

iv) If PA = 1 PQ in the adjoining diagram, prove that a = 1 _3p + q i .
4 4

Q

q A

b
O

pP

v) Prove that AB + BC + CD + DE + EF + FA = 0 in a hexagon ABCDEF.
PB

A

Q
C

PRIME Opt. Maths Book - X 267

6. PRIME more creative questions :
i) If AB2 + BC2 = AC2 in a triangle prove vectorally that, it is a right angled triangle at B.
ii) If AP : PB = AQ : QC = 3 : 1 in the adjoining diagram, find PQ and prove that PQ ' BC .
1
iii) If CP : PO = CQ : QB = 2 : 3 in a parallelogram OABC, prove that PQ = 4 OB and

PQ // OB where OA = a and OC = c .
CQ
B

P

OA

iv) If D, E and F are the mid - points of sides BC, AC and AB respectively of DABC, prove
that AD + BE + CF = 0.

v) If P, Q and R are the mid - points of sides BC, AC and AB of DABC, prove that
OA + OB + OC = OP + OQ + OR for any point ‘O'.

Answer

1. Show to your teacher.

2
2. i) 3 i – j ii) 2 i – 2j iii) dn iv) (2, –2) v) (–3, 7)
3

3 3 –2
3. i) dn ii) dn iii) dn iv) 2 i – 2j v) – 7j
0 –4 3 iv) 2 i – 4j

2 2
4. i) dn ii) 2 i – 3j iii) 2 i – 2j v) dn
1 3

268 PRIME Opt. Maths Book - X

6.3 Theorems on vector geometry

Theorem 5 : Prove that line joining the mid - point of any two sides A
of a triangle is equal to half at the third side and parallel to it.

Solution: P Q
In DABC, P is the mid - point of AB. B C
Q is the mid - point of AC.
1
To prove : PQ = 2 BC and PQ ' BC

Proof :

By using triangle law,

PQ = PA + AQ

= 1 BA + 1 AC [In DAPQ]
2 2 [In DABC]

= 1 ^BA + ACh
2

= 1
2 BC

\ PQ = 1 BC
2

It is same as the a = mb for a // b .

So, PQ ' BC also

Theorem 6 : Prove that a line joining the vertex and mid - point of base of an isosceles triangle is

perpendicular to the base. A
Solution:

In an isosceles DABC,
AB = AC = a (say)
D is the mid - point of base BC.

To prove : AD = BC .
Proof : By using position vector of mid - point ‘D',
1
AD = 2 ^AB + ACh ....................................(i)

By using triangle law, B DC

BC = BA + AC ..................................(ii)
Taking the dot product of (i) and (ii),
1
AD . BC = 2 ^AB + ACh. ^BA + ACh

= 1 ^AB + ACh ^–AB + ACh
2

= 1 ^AC 2 – AB 2h
2

= 1 ^ AC 2 – AB 2h
2

= 1 ^AC 2 – AB 2h
2

= 1 ^a2 – a2h [ a sides of isosceles triangle.]
2

= 1 #0
2

\ AD . BC = 0 i.e. AD = BC

PRIME Opt. Maths Book - X 269

Theorem 7 : Prove that line joining the mid - point of adjacent A SD
sides of a quadrilateral forms a parallelogram. R

Solution : QC
In a quad. ABCD P, Q, R, S are the mid - points of adjacent
sides. P

Join a diagonal AC.
To prove : PQRS is a parallelogram.
B
Proof :
By using triangle law,

PQ = PB + BQ [In DPBQ]

= 1 AB + 1 BC
2 2

= 1 ^AB + BCh
2

= 1 AC ........................(i) [In DABC]
2

Again, = SD + DR [In DSDR]
SR

= 1 AD + 1 DC
2 2

= 1 ^AD + DCh
2

= 1 AC ........................(ii) [In DADC]
2

From the relation (i) and (ii), we get
PQ = SR

Also,
PQ ' SR [ a Equal vectors are parallel also ]
Similarly as above :
PS = QR and PS//QR
Here, Opposite sides are parallel to each other.
Hence, PQRS is a parallelogram.

Theorem 8 : Prove that diagonals of a parallelogram are bisected to A D
each other at a point.

Solution: P

In a parallelogram ABCD, B C
AC and BD are the diagonals.
To prove : Diagonals AC and BD are bisected at point P.
Proof : Let, ‘P' be the mid - point of a diagonal BD.
Then, Using position vector of mid - point ‘P' of diagonal BD,
1
AP = 2 ^AB + ADh

= 1 ^DC + BCh [ a Opposite sides of parallelogram ]
2

= 1 ^– CD – CBh
2

270 PRIME Opt. Maths Book - X

=– 1 ^CD + CB h
2

= – CP [ a Position vector of mid - point of BD ]

= PC

\ AP = PC
i.e. P is the mid - point of diagonal AC also.
Hence, Diagonals AC and BD of a parallelogram ABCD are bisected to each other at point
'P'.

Theorem 9 : Prove by vector method that diagonals of a rectangle are equal.

Solution: A D
ABCD is a rectangle C
AC and BD are its diagonals.

To prove : AC = BD B
Proof : Using triangle law,
AC = AB + BC

squaring on both, sides,
or, AC2 = AB2 + 2 AB . BC + BC2

or, AC 2 = AB 2 + 2 × 0 + BC 2 [ a AB ^ BC and a2 = a 2 = a2]

or, AC2 = AB2 + BC2 [ a Opposite sides of parallelogram]
or, AC2 = DC2 + BC2

or, AC2 = BD2 [ a h2 = p2 + b2 in DDCB]

\ AC = BD.

Theorem 10 : Prove that diagonals of a rhombus are bisected to each other at right angle.

Solution : A D
ABCD is a rhombus.
AC and BD are the diagonals intersected at point O. O
To prove : Diagonals AC and BD are bisected at right angle at O.

Proof : B C

Let us consider O is the mid point of a diagonal BD.

i.e. BO = OD
using position vector of mid-point,
1
AO = 2 ( AB + AD )

= 1 ( –BA – DA )
2

= 1 ( –CD – CB ) [ a Opposite sides of a parallelogram]
2

= – 1 ( CD + CB )
2

= –[ 1 ( CD + CB )]
2

= – CO

= OC

AO = OC
\ i.e. O is mid point of diagonal AC also.

PRIME Opt. Maths Book - X 271

Again, [ a Using triangle law]
Taking

AC . BD = ( AB + BC ). ( BC + CD )

= ( AB + BC ).( BC – DC ) [ a Opposite sides of a parallelogram]

= ( AB + BC ).( BC – AB ) [ a a2 = a 2 ]
= BC2 – AB2 [ a Sides of rhombus are equal]
= BC 2 – AB 2

= a2 – a2

=0

\ AC ^ BD
From above two relations, we get
Diagonals AC and BD are bisected at right angle at 'O'.

Theorem 11 : Prove that inscribed angle formed in semi-circle is a right angle.

Solution : P
O is the centre of semi-circle.
AB is a diameter
\ APB is an inscribed angle.

To prove : \ APB = 90° A O B
Proof :
P and O are joint
Then, using triangle law of vector addition,

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

PB = PO + OB

= PO + AO [ a AO = OB]

= PO – OA .................................. (ii)
Taking dot product (i) and (ii)

PA . PB = ( PO + OA ).( PO – OA )

= ( PO )2 – ( OA )2 [ a a2 = a 2 ]
= PO 2 – OA 2 [PO = OA = radius (r)]

= r2 – r2

=0

\ PA ^ PB

i.e. \ APB = 90°

Theorem 12 : Prove that mid-point of hypoteneous of a right angled

triangle is equidistant from the vertices. A

Solution: P
In right angled TABC , P is the mid-point of hypotenuse AC. C

\ B = 90°

To prove: AP = BP = CP.
Proof: Taking position vector of mid-point P of side AC,
B

BP = 1 (BA + BC)
2

272 PRIME Opt. Maths Book - X

Squaring on both sides

or, ^BPh2 = :12 ^BA + 2

BC hD

or, BP2 = 1 ( BA2 + 2BA . BC + BC2 )
4

or, BP 2 = 1 ( BA 2 +2×0+ BC 2 ) [ a BA = BC and a2 = a 2 = a2]
4

or, BP2 = 1 (BA2 + 2 × 0 + BC2)
4

or, BP2 = 1 × AC2 [ a AC2 = AB2 + BC2]
4

or, BP2 = 1 × (2AP)2 [ a P is mid-point of AC]
4

or, BP2 = AP2
\ BP = AP ................................................(i)

Also, AP = CP from given .......................(ii)
From relation (i) and (ii) we get
AP = BP = CP

Corollary 4 : Prove that perpendicular drawn from the centre of a circle to A O B
a chord bisects the chord. P

Solution :

O is the center of a circle
AB is a chord.
OP ^ AB.
AO and BO are joint.
To prove : P is mid-point of chord AB.
Proof : By using triangle law,
OA = OP + PA

Squaring on both sides,

OA2 = OP2 + 2 OP . PA + PA2

OA 2 = OP 2 + 2 × 0 + PA 2 [ a OP = PA and a2 = a 2 = a2]

or, OA2 = OP2 + 2 × 0 + PA2

or, OA2 = OP2 + PA2 .................................. (i)
Similarly as above,

OB2 = OP2 + PB2 ....................................... (ii)
we have OA = OB = radius of cirlce ................................... (iii)

From equation (i) and (ii), we get
or, OP2 + PA2 = OP2 + PB2
or, PA2 = PB2

\ PA = PB
i.e. P is mid-point of chord AB.

PRIME Opt. Maths Book - X 273

Corollary 5 : Prove that a2 = b2 + c2 – 2bcCosA in a DABC. A

Solution :

In DABC,

AB = c

BC = a

AC = b

To prove : a2 = b2 + c2 + 2bcCosA BC

Proof :
Using triangle law of vector addition,

BC = BA + AC

or, BC = AC – AB
Squaring on both sides,

or, BC2 = ( AC – AB )2
or, BC2 = AC2 – 2 AC . AB + AB2
or, BC 2 = AC 2 – 2 AC . AB CosA + AB 2
or, BC2 = AC2 – 2AC.AB.CosA + AB2
or, a2 = b2 – 2bcCosA + c2
\ a2 = b2 + c2 – 2bcCosA.

Exercise 6.3

1. i) Prove that line joining the mid-point of non-parallel sides of a trapezium is half of
the sum of its parallel sides and parallel to them.

ii) Prove that OA + OB + OC + OD = 4 OG in a parallelogram ABCD where 'G' is the
intersecting point of diagonals and O is any point.

iii) Prove that the quadrilateral having bisected diagonals is a parallelogram.
iv) In a parallelogram ABCD, BP = QD, prove that APCQ is also a parallelogram.

AD
Q

P

BC
v) Prove that inscribed angle formed in a semi-circle is a right angle.

2. i) Prove that the parallelogram having equal diagonals is a rectangle.
ii) Prove that line joining the centre of a circle to the mid-point of a chord is
perpendicular to the chord.
iii) In the given diagram PS = QS = RS. Prove that DPQR is a right angled triangle at Q.
P

S

QR

274 PRIME Opt. Maths Book - X

iv) Prove that the triangle is an isosceles triangle where perpendicular drawn from the

vertex to the base bisects the base.
v) In a parallelogram PQRS, A and B are the mid point of sides RS and QR respectively.
3
Prove that PA + PB = 2 PR

S A R

B
PQ

3. PRIME more creative theorems:
yax=+mbyx
i) Prove by vector method that = 1 as the equation of straight line.
ii) Prove by vector method that + c as the equation of straight line.

iii) Prove that in a trapezium ABCD where P and Q are the mid-point
of diagonals BD and AC respectively.
AB

PQ

D C
iv) Prove that b2 = a2 + c2 – 2acCosB in DABC.

v) In the adjoining diagram O is the centre of semi-circle and PQ^AB prove that
PQ2 = AQ.QB
P

AQ O B

4. Project work
Prepare the charts of basic theorems of vector geometry and paste in your classroom.

General concept of solving vector geometry.
• Suitable diagram for the statement.
• Explanation of statements in the diagram.
• Using of triangle law of vector addition.
• Using position vector of mid-point and section-point for

specific problems.
• Proper use of vectors according to the concept of direction

during simplification of geometry.

PRIME Opt. Maths Book - X 275

Vector

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Write down the condition of the any two vectors being orthogonal and parallel.

2. a) If position vector of the points A & B are 5i + j and 2i + 3j , find the angle \ AOB.
b) If 2a + 4b and 10a – 8b are perpendicular vectors where a & b are the unit vectors,
find the angle between a and b.
c) If a point C cuts a line segment AB as 2AC = 3CB, find OC in terms of a and b. Where
OA = a and OB = b

3. a) Prove vectorally that the line joining the centre of a circle to the mid point of a chord
is perpendicular to the chord.

b) Find the angles of a triangle having vertices A(–2, 4), B(–2, –3) and C(5, –3) using
vector method.

4. Prove that PQRS is a parallelogram where P, Q, R and S are the mid-point sides AD, BD, BC
and AC respectively in a quadrilateral ABCD.
A PD

QS C
BR

276 PRIME Opt. Maths Book - X

Unit 7 Transformation

Specification Grid Table

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

No. of Questions 1 – 1 1 3 10 15
Weight 1 – 4 5

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

Questions, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to combine the any two transformations from reflection,

rotation, translation and enlargement.
• Students are able to find the image of an objects using combined

transformation.
• Students are able to define inversion & inversion circle & to find image of

a point.
• Students are able to use matrix for the transformations & to find image

using matrix.

Materials
• Collection of patterns printed in clothes and other objects.
• Formula charts of different transformations.
• Chart of concepts of geometrical theorems.
• Chart of the 2 × 2 matrix which represents the transformations.
• Graph paper and graph board.
• Geo board.

PRIME Opt. Maths Book - X 277

7.1 Transformation

The process of changing the position or size of an object under any
geometrical conditions is called transformation.
There are reflection, rotation, translation and enlargement in the
process of transformation.

The transformations mentioned above are already introduced in grade IX in details. They are
useful for various field to decorate or to generate the patterns of objects to make attractive
cloths, art, Thanka art etc which has economic importance too.
Patterns of figures can be drawn by using the process of transformation as follows which can
be taken as the sample to understand transformation. Such patterns are printed in clothes,
carpet, bed-sheet, curtains etc.

Such type of transformations is called the composition or combination of transformations
which have to be discussed in grade ‘X' as the application of transformation.
In grade ‘X', the transformation chapter gives an idea about the composition of transformations,
inversion transformation (inversion circle) and transformation using matrix.

Enjoy the recall: Formulae P(x, y) → P'(x, –y)
About x- axis P(x, y) → P'(–x, y)
S.N. Transformations About y- axis P(x, y) → P'(y, x)
1. Reflection about an About y = x P(x, y) → P'(–y, –x)
About y = -x P(x, y) → P'(2h–x, y)
axis (mirror line) P(x, y) → P'(x, 2k–y)
About x = h
About y = k

278 PRIME Opt. Maths Book - X

2. Rotation about the About +90°(–270) P(x, y) → P'(–y, x)
P(x, y) → P'(y, –x)
centre as origin About –90°(+270°) P(x, y) → P'(–x, –y)
P(x, y) → P'(x, y)
About 180° P(x, y) → P'[–(y – b) + a, (x – a) + b]
P(x, y) → P'[(y – b) + a, –(x – a) + b]
Rotation about the About 360° P(x, y) → P'[–(x – a) + a, –(y – b) + b]
centre as (a, b) About +90° P(x, y) → P'[(x – a) + a, (y – b) + b]
About –90°

About ±180°

About ±360°

3. Translation about a a
T = <F
vector b

T = AB = <aF = <x2 – x1F P(x, y) → P'(x + a, y + b)
b y2 – y1

4. Enlargement about E[(0, 0), k] P(x, y) → P'(kx, ky)
centre and scale P(x, y) → P'[k(x – a) + a, k(y – b) + b]
E[(a, b), k]
factor

Combination of transformation

Let us see the examples to understand the combination of transformations having different
properties.

Y

X' A
A'' X

A'

Y'

Here, A is transferred to A' and then to A'' where there is reflection on x-axis for A and
reflection on y- axis for A'.
\ Two reflections R1 and R2 combined together and the single reflection is taken as R2oR1.

PRIME Opt. Maths Book - X 279

Y
A' B
B' A
X' X
A''
B''

Y'
Here, In second example, A line segment AB is rotated to A'B' under rotation about + 90°
followed by rotated to A''B'' under rotation about again +90° with centre origin.
iTmwaogerowtahteiorensthRe1iransidngRl2earroetactoimonbiisneRd2otRo1g. ive image of an object AB to A''B'' as the final

Above examples give an idea about the combined transformations
which can be done between the same type of transformations as well
as different types like reflection and reflection, reflection and rotation,
rotation and translation, reflection and enlargement etc.

1. Reflection and Reflection: Y
Here, R1 is the reflection about B
x axis.
i.e. R1 (x, y) → (x, – y) 7

Ri.e2.isRt2h(ex,rye)fl→ect(i–oxn, about y- axis, 6
y)
5

4C

Where, DABC transferred to 3 A
DcDoAAm''B'Bb'C'i'nC' e''udnirsdeeflurencRdti1eornfosR.ll2owaesd to 2
the
1

X' –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 7 X
–1

C'' A'' –2 A'
–3

–4 C'

–5

–6

B'' –7 B'
Y'

280 PRIME Opt. Maths Book - X

The single transformation R2oR1 is of reflections R1 & R2
which can be defined as,
R2oR1 →→→(RR–22x[[,R(–x1y, ()–xy, )y])]

\ P(x, y) R2oR1 (–x, –y)]
It is rotation about 180° with Centre (0, 0).

2. Rotation and Rotation:

Y

7 B'
A'' 6

5
4 A'

3

B'' C'' 2 C'
1

X' –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 7 X

–1 C B
–2

–3

–4

–5
–6 A

Y'

Here, F is the rotation about +90° with centre origin biiy..ee..DRRA21'(B(xx',C,yy'))i→s→t(r(–a–yny,s,xfxe))rred to
G is the rotation about +90° with centre origin
Where, DABC is transferred to DA'B'C' under F followed
DA''B''C'' under G as the combination of transformation on rotations.

The single transformation F and G is GoF which can be defined as,
GoF → G[F(x, y)]

→ G(–y, x)
→ (–x, –y)]
\ P(x, y) GoF P'(–x, –y)]
It is rotation about 180°.

PRIME Opt. Maths Book - X 281

3. Translation and translation :

Y

6

5 B'
4

B 3C' A' B''
2

X' C 1 A C'' 7 8 A''
–4 –3 –2 –1 O 1 2 3D'4 5 6 9X
–1

D –2 D''
–3

–4

Y'

Here, T1 = 3 is the first translation.
<F
2

i.e, T1 (x, y) → (x + 3, y + 2)

T2 = 4 is the second translation,
<F
–2

Wi.eh.,eTr2e(,xq, yu)adrilatera→l (x + 4, y– 2) to A'B'C'D' under T and quad A'B'C'D' is
ABCD is translated
translated to A''B''C''D'' under T2 as the combination of translations.

The single transf→→→or(mTTx22a+[(Ttxi31o+(+nx3,4o,y,fy)yT]+1+2a2)n–d T2 is T2oT1 which can be defined as,
T2oT1 2)

→(x + 7, y)

\ P(x, y) T2oT1 P'(x + 7, y)

7
It is translation about T2oT1 = <F
0

It also can be expressed as,

347
T2oT1 = T1 + T2 = <F + <F = <F
2 –2 0

\ p(x, y) T2oT1 p'(x + 7, y)

7
It is translation about T = <F
0

282 PRIME Opt. Maths Book - X

4. Enlargement and Enlargement:

Y
9
8

7 A''
6

5 A'
4

3A
2

1
X' –9 –8 –7 –6 –5 –4 –3 –2 –1 O B 1 B2' 3C 4 5 C'6 7 8 9 10 11 12 13 14 X

–1

–2 B'' C''
–3

–4

–5

Y'

Here, Enlargement DuAn'd'Be'r'CE''1[a0s, 0] enlarged DABC to DA'B'C' followed by E2[(–2, 2), 2]
enlarged DA'B'C' to the combined enlargement

i.e. E1(x, y) → (2x, 2y)
E2 (x, y) → {k(x - a) + a, k(y - b) + b}

→ {2(2x + 2) – 2, 2(2y – 2) + 2}

→ (2x + 2, 2y – 2)

The single transformation of the enlargements E1 and E2 is
which can be defined as,
E2oE1 →→→ EE(222[((E221xx(,)x2+, yy2)),] 2(2y)
E2oE1

→ (4x + 2, 4y – 2) – 2)

\ P(x, y) E2oE1 P'(4x + 2, 4y – 2)

PRIME Opt. Maths Book - X 283

5. Reflection and Rotation:

Y

9

8

7

6B
5

4A
3

2

X' 1C

–9 –8 –7 –6 –5 –4 –3 A–2'' –1 O 1 2 3 4 C'5 6 7 89 X
–1

–2

C'' ––34A'
–5
B''

–6 B'

–7

–8

–9
Y'

Here, F is the reflection on x-axis.
i.e. F(x, y) → (x, –y)
G is the rotation under –90° with centre origin.
i.e. G(x, y) → (y, –x)
DABC is transfered to DA'B'C' under reflection about x-axis ‘F' and DA'B'C' is transferred
to DA''B''C'' under rotation about –90° ‘G' as the combination of transformation.

The single transformation of the reflection F and rotation
G is GoF which can be defined as,
GoF → G[F(x, y)]
→ G(x, –y)
→ (–y, –x)
\ P(x, y) Gof P'(–y, –x)
It is reflection about y = – x.

284 PRIME Opt. Maths Book - X

6. Reflection and translation:

Y

9

8

C 7 C'
6

5 C''
4

A 32A'

B1 2 B' B'' 6 7 8 9 X
X' 3 45

–9 –8 –7 –6 –5 –4 –3 –2 –1 O 1
–1

–2 A''

–3

–4

–5

Y'

Here, DABC is reflected to DA'B'C' under reflection on y-axis ‘R'.
i.e. R(x, y) → (–x, y)

DA'B'C' is translated to DA''B''C'' under translation about T = 3
<F
–2

i.e. T(x, y) → (x + 3, y – 2)

The combined transformation is ToR after reflection followed by translation as the
combination of transformation.

The single transformation of the reflection R and
translation T is ToR which can be defined as,
ToR → T[R(x, y)]

→ T(–x, y)
→ (–x + 3, y – 2)
\ P(x, y) ToR P'(–x + 3, y – 2)

PRIME Opt. Maths Book - X 285

7. Rotation and enlargement:

Y

13

12

C'' 11
10

9

8

7
B'' 6

5 C'

A'' 4
3
C B' A' 2

X' –9 –8 –7 –6 –5 –4 –3 A 1 2 3 4 5 6 7 89 X
–2 –1 O 1
–1

B –2
–3

–4

Y'

Here, DABC is rotated with R to DA'B'C' under rotation about –90° with centre origin.
i.e. R(x, y) → (y, –x)
DABC is enlarged to DA'B'C' with E under enlargement about E[(3, 0), 2]
i.e. E(x, y) → [k(x – a) + a, k(y – b) + b]

→ (2x – 3, 2y)
The combined transformation is EOR after rotation ‘R' followed by enlargement ‘E' is
EoR as the combination of transformation.

The single transformation of rotation R and enlargement E
is EoR which can be defined as,
EoR → E[R(x, y)]

→ E(y, –x)
→ (2y – 3, – 2x)
\ P(x, y) EoR P'(2y – 3, –2x)

286 PRIME Opt. Maths Book - X

8. Translation and enlargement

Y

16

15

14

A'' 13
12

11

10

9

8
7 A'

6

5
A4

3 C''
B'' 2
X' B' 1 2 3 C' 6 7 8 9 X
45
–9 –8 –7 –6 –5 –4 –3 –2 –1 O 1
–1

B C –2
–3

Y'

Here, DABC is translated to DA'B'C' with T under translation about T = 4
<F
3

i.e. T(x, y) → (x + 4, y + 3)

DA'B'C' is enlarged to DA''B''C'' with E under enlargement about E under enlargement
about E [(1, – 1), 2]

i.e. E(x, y) → [k(x – a) + a, k(y – b) + b] = (2x – 1, 2y + 1)

The combined transformation is T followed by E is taken as DA'B'C' followed by DA''B''C''
as the images.

The single transformation of T and E of translation and
enlargement respectively is EoT which can be defined as,
EoT → E[T(x, y)]
→ E(x + 4, y + 3)
→ (2(x + 4) – 1, 2(y + 3) + 1)
→ (2x + 7, 2y + 7)
\ P(x, y) EoT P'(2x + 7, 2y + 7)

PRIME Opt. Maths Book - X 287

9. Rotation and translation :

Y

D7 C
6

5

4

A 3 B
2

1

X' 1 23 45 6789 X
–9 –8 –7 –6 –5 –4 –3 –2 –1 O
–1

D' –2 A'
––D34''
A''

–5

C' B–'6
C'' –7
B''
–8

Y'

Here, 4ABCD is rotated to 4A'B'C'D' under a rotation about 180°.
i.e. R(x, y) → (–x, –y)

4A'B'C'D' is translated to 4A''B''C''D'' under a translation T = 3
<F .
–2

i.e. T(x, y) → (x + 3, y – 2)

The 4A''B''C''D'' is the final image of 4ABCD under rotation followed by under

translation as the combination of transformation.

The single transformation represented by rotation R and
translation T is ToR where,
ToR → T[R(x, y)]
→ T(–x, –y)
→ (–x + 3, –y + 2)
\ P(x, y) ToR P'(–x + 3, –y + 2)
3
It is rotation about 180° with centre ( 2 , 1)

288 PRIME Opt. Maths Book - X

Worked out Examples

1. Find the image of a point A(–2, 3) under rotation about 180° following by reflection about
y – 2 = 0.
Solution :
Under rotation about 180°
P(x, y) → P'(–x, –y)
A(–2, 3) → A'(2, –3)

Again, under reflection about y – 2 = 0
P(x, y) → P'(x, 2k – y)
→ P'(x, 4 – y)
\ A(2, –3) → A''(2, 4 + 3)
→ A''(2, 7)


\ A(–2, 3) → A''(2, 7)

2. Find the image of a point P(3, –4) under a combined transformation FoG where F is

reflection under y = x and G is translation about T = <–32F .
Solution :
Under reflection F about y = x,
F(x, y) → P'(y, x)

Under translation G about T = –2
<F

3

G(x, y) → T'(x – 2, y + 3)

Then, the combined transformation is
FoG → F[G(x, y)]
→ F(x – 2, y + 3)
→ (y + 3, x – 2)
\ P(x, y) FoG P'(y + 3, x – 2)
\ P(3, –4) → P'(–4 + 3, 3 – 2)
→ P'(–1, 1)


\ P(3, –4) → P'(–1, 1)

3. If a point ‘P' is transformed to P'(–2, 6) under reflection about x + 2 = 0 followed by rotation
about + 90° with centre origin, find the co-ordinate of the point ‘P'.
Solution : Under reflection about x + 2 = 0
P (x, y) → P'(2h – x, y)
→ P'(–4 – x, y)

Again, under rotation about + 90°
P(x, y) → P'(–y, x)
\ P'(– 4 –x, y) → P''(–y, – 4 – x)

From given, → P''(–2, 6)
P(x, y)

PRIME Opt. Maths Book - X 289

By equating the image points p''.
– y = – 2 and – 4 – x = 6
\ y = 2 and x = – 10
\ Co-ordinate of point P is (–10, 2)

4. The image of a point A(–1, 3) is A'(4, –2) under a combined transformation of translation

about T and rotation about –90° with centre origin. Find the translation vector T.

Solution :

Under a translation T = a
<F
b

T(x, y) → T'(x + a, y + b)
Under rotation about –90° with centre origin.
R(x, y) → (y, –x)
Then, combined transformation is,
ROT → R[T(x, y)]
→ R(x + a, y + b)
→ (y + b, – x – a)]
\ P(x, y) → P'(y + b, – x – a)
A(–1, 3) → A'(3 + b, – 1 – a)

From given
A(–1, 3) → A'(4, – 2)

By equating the corresponding elements of image,
3+b=4 and –1 – a = – 2
\ b=1 and a=1
1
\ Translation vector is T = 1

5. Find the image of a triangle having vertices A(–3, 2), B(–1, 5) and C(0, –3) under reflection
about x – 2 = 0 followed by rotation about – 90° with centre origin. Also plot them in graph.
Solution :
Under reflection about x – 2 = 0
p (x, y) → p'(2h – x, y)
→ p'(4 – x, y)
A(–3, 2) → A'(7, 2)
B(–1, 5) → B'(5, 5)
C(0, –3) → C'(4, –3)

Again,
Under rotation about –90° with centre origin
p (x, y) → p'(y, – x)

A'(7, 2) → A''(2, –7)
B'(5, 5) → B''(5, –5)
C'(4, –3) → C''(–3, –4)

290 PRIME Opt. Maths Book - X

Y

8

7

B 6 B'
5

4

3

A2 A'

1

X' X
–6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 7
–1

–2

–3 C'
C'' C –4 B''

–5

–6

–7 A''
–8

Y'

6. Find the image of DPQR under reflection about y = x followed by reflection about y = 2

where the vertices are A(–2, 1), B(–3, 4) & C(3, 3). Also plot them in graph and find the

single transformation shown by it.

Solution :
Under reflection about y = x.
R1(x, y) → (y, x)
Y

A(–2, 1) → A'(1, –2) 8
B(–3, 4) → B'(4, –3) 7A'' B''
C(3, 3) → C'(3, 3) 6
5

Again, B 4C
Under reflection about y = 2, 3 C'
AR'2(1(x, ,–y2)) → (x, 2k – y) = (x, 4 – y) 2
→ A''(1, 4 + 2) = A''(1, 6)
B'(4, –3) → B''(4, 4 + 3) = A''(4, 7) A 1 C''
C'(3, 3) → C''(3, 4 – 3) = C''(3, 1) X' –4 –3 –2 –1 O 1 23 4 5 X

–1

Again, for single transformation of R1 & R2, –2 A'
R2oR1 (RRy22,[(4Ry,1–x(x)x,)y)] –3
→ B'
→ –4
→
Y'
\ P(x, y) R2oR1 P'(y, 4 – x)

PRIME Opt. Maths Book - X 291

7. Find the image of quadrilateral having vertices A(1, 3), B(2, 5), C(5, 7) & D(6, 1) under

translation about T1 = <32F followed by T2 = <32F . Also find the single translation & plot them
in graph.
Y
Solution : 12 B'' C''
11
Under translation about T1 = 3
<F
2

T1 (x, y) → (x + a, y + b) = (x + 3, y + 2) 10 B' C'
A(1, 3) → A'(1 + 3, 3 + 2) = A'(4, 5) 9 A''
B(2, 6) → B'(2 + 3, 6 + 2) = B'(5, 8) 8 D''
C(5, 7) → C'(5 + 3, 7 + 2) = C'(8, 9) 7C B
D(6, 1) → D'(6 + 3, 1 + 2) = D'(9, 3) 6

Again, under translation about T2 = 2 5 A'
<3F 4

AT'(2(4x,,5y)) → AT'''2((4x + 2, y + 3) 3A D'
→ + 2, 5 + 3) 2
= A'' (6, 8)
B'(5, 8) → B''(5 + 2, 8 + 3) = B''(7, 11) 1 D

C'(8, 9) → C''(8 + 2, 9 + 3) = C''(10, 12) X' –1 O 1 2 3 4 5 6 7 8 9 10 11 X
D'(9, 3) → D''(9 + 2, 3 + 3) = D''(11, 6) –1

For single tr→→→anT(Tsxl22a[(+tTxio13+n(+3xo,,2fyy,T)y+]1+2&)2T+2, –2
T2oT1 Y'

→ (x + 5, y + 5) 3)

\ P(x, y) T2oT1 (x + 5, y + 5)

It is translation about T = 5
5

8. Find the image of DABC having vertices A(–5, –1), B(–1, 3) & C(–6, 4) under rotation about

+ 270° following by translation about AB . Also plot them in graph.

Solution :
Under rotation about + 270° (–90°)
R(x, y) → (y, –x)
\ A(–5, –1) → A'(–1, 5)
B(–1, 3) → B'(3, 1)
C(–6, 4) → C'(4, 6)

Again, under translation T = AB = d x2 – x1 n
y2 – y1

= e–1 + 5o
3+1

\ T(x, y) → T'(x + a, y + b) = 4
→ T'(x + 4, y + 4) 4

292 PRIME Opt. Maths Book - X

\ A'(–1, 5) → A''(–1 + 4, 5 + 4) = A''(3, 9)
B'(3, 1) → B''(3 + 4, 1 + 4) = B''(7, 5)
C'(4, 6) → C''(4 + 4, 6 + 4) = C''(8, 10)

Y

11 C''

10
9 A''
8

7

A' 6 C'

5 A''
C B4
3
2 B'

1
X' –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 7 8 9 X

–1
A –2

–3

Y'

9. Find the image of DPQR having vertices P(3, –2), Q(1, 2) and R(–2, –1) under an enlargement

E1[0, 2] followed by enlargement about E2[(1, –1), 2]. Also plot them in graph.
Solution :
Under enlargement about EP1''[(02,x2, ]2y)
P(x, y) → P'(kx, ky) =

\ P'(3, –2) → P'(2 × 3, 2 × (–2)) = P'(6, –4)
Q'(1, 2) → Q'(2 × 1, 2 × 2) = Q'(2, 4)
R'(–2, –1) → R'(2 × (–2), 2(–1)) = R'(–4, –2)

Again
Under enlar→ge mP'e[kn(txa–boau)t+Ea2,[(k1(x, ––1b),)2+]
P(x, y) b]
→ P'[2(x – 1) + 1, 2(x + 1) – 1]
→ P'(2x – 1, 2x + 1)

\ P'(6, –4) → P''(2 × 6 – 1, 2 × (–4) + 1) = P''(11, –7)
Q'(2, 4) → Q''(2 × 2 – 1, 2 × 4 + 1) = Q''(3, 9)
R'(–4, –2) → R''(2(–4) – 1, 2(–2) + 1) = R''(–9, –3)

PRIME Opt. Maths Book - X 293

Y B''
9

8

7

6

5 B'
4

3B
2

1

X' 1 2 3 4 5 6 7 8 9 10 11 12 X
–9 –8 –7 –6 –5 –4 –3 –2 –1 O –1

C' C –2 A
–3
C''
–4 A'

–5

–6

–7 A''
–8

–9
Y'

10. Find the image of ABC having vertices A(–2, 3), B(–3, –2) & C(–4, 4) under a combined

reflections x = 2 and x = 1. Also plot them in graph.

Solution :
Under reflection about x = 2,
r1(x, y) → (2h – x, y) = (4 – x, y) C' C Y
Under reflection about x = 1
5
4

r2(x, y) → (2h – x, y) = (2 – x, y) B' B 3

Then, the combined reflection is, 2

r2or1 → r2 [[r41(–xx, ,yy)]] 1 X
→→ r[22 – 4 + x, y] X' –7 –6 –5 –4 –3 –2 –1 O 1
= (–2 + x, y)
–1

A' A –2
–3
Then,
P(x, y) r2or1 P'(–2 + x, y)
Y'

Then,
\ A(–2, 3) → A'(–2 – 2, 3) = A'(–4, 3)

B(–3, –2) → B'(–2 – 3, –2) = B'(–5, –2)
C(–4, 4) → C'(–2 – 4, 4) = C'(–6, 4)

294 PRIME Opt. Maths Book - X

11. Find the image of 4PQRS having vertices P(–1, 2), Q(2, 2), R(3, –1) & S(0, –1) under a
3
combined enlargement of E1[0, – 2 ] and E2[0, + 2]. Also plot them in graph.

Solution :

Under enlargement about E1[0, – 3 ]
E1(x, y) → (kx, ky) 2

E1(x, y) → (– 3 x, – 3 y)
2 2

Under enlargement about E2[0, 2]

EE22((xx,, y) → (kx, ky) Y
y) → (2x, 2y) 5

Then, the combined enlargement is, R' S' 4
→ E2[E1(x, y)] 3Q
E2OE1 P 2

→ E2(– 3 x, – 3 y) 1
2 2

→ [–2( 3 x), –2( 3 y)] X' –9 –8 –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 X
2 2
–1
\ P(x, y) → (–3x, –3y) S –2 R
→ P'(–3x, –3y)
–3

\ P(–1, 2) → P'(3, –6) –4
→ Q'(–6, –6) –5
Q(2, 2) → R'(–9, 3) Q' –6 P'
R(3, –1) → S'(0, 3)
S(0, –1) Y'

12. Find the image of DABC having vertices A(3, –2), B(5, –4), & C(7, 2) under the combined

transformation of translation about AB and rotation about 180°. Also plot them in graph.

Solution :

Under translation about AB = d x2 – x1 n = 2 n
y2 – y1 d
–2
P (x, y) → (x + a, y + b)
P (x, y) → (x + 2, y – 2)

Under rotation about 180°
R(x, y) → (–x, –y)

Then, the combined transformation is,
RoT → R[T(x, y)]

→ R(x + 2, y – 2)
→ (– x – 2, – y + 2)

\ P (x, y) RoT P'(– x – 2, – y + 2)
A(3, –2) → A'(–3 –2, 2 + 2) = A'(–5, 4)
→ B'(–5 –2, 4 + 2) = B'(–7, 6)
B(5, –4) → C'(–7 –2, – 2 + 2) = C'(–9, 0)

C(7, 2)

PRIME Opt. Maths Book - X 295