Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

Example 5. If o = oo and o o o oo
Solution: OA 2 i +6 j OB = 2 i+ j find the angle between OA and OB.

Example 6. o o oo o o
Solution: Here, OA = 2 i + 6 j , OB = 2 i + j .

Example 7. oo
Solution: Let T be the angle between OA and OB . Then,

oo oo
a. b OA . OB
cosT = =
| oa || o OoA || o
b | | OB |

oo o o oo
But, OA . OB = (2 i + 6 j ) . (2 i + j )

= 2 × 2 + 6 × 1 = 10

o
| OA | = 22 + 62 = 4 + 36 = 40 = 2 10

o
| OB | = 22 +12 = 4+1 = 5

? cosT = 2 10 5 = 10 2 = 1 = cos45°
10. 10 2

? T = 45°

Hence, the angle between them is 45°.

If oo = 12, T = 45°, | oa | = 4, find | ob |.
a. b

Here, o o = 12, o = 4
a. b | a|

By definition of dot product of two vectors,

o . o = | oa | | ob | cosT
a b

o
or, 12 = 4 | b | cos45°

or, o 1 o
3=| b| 2 ? b =3 2

oo o o oo
If | a + b | = | a - b | , then, prove that a and b are perpendicular to

each other.

o o oo
Here, | a + b | = | a - b |

Squaring on both side, we get

oo oo
| a + b |2 = | a - b |2

or, (oa + ob )2 = (oa - ob )2 (?|oa |2 = a2)

oo oo
or, a2 + 2 a . b + b2 = a2 - 2 a . b + b2

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oo oo
or, 4 a . b = 0 ? a.b =0

oo
Hence, a and b are perpendicular to each other.

Example 8. If P(0,2), Q(2,5) and R(8,1) are the vertices of a triangle ∆PQR, prove that
Solution: ∆PQR is a right angled triangle

oo o
Here, we have OP = (0,2), QR = (2, 5), OR = (8, 1)

o oo
Now, PQ = OQ - OP = (2, 5) - (0,2) = (2, 3)

o oo
QR = OR - OQ = (8, 1) - (2, 5) = (6, -4)

o oo
RP = OP - OR = (0, 2) - (8,1) = (-8, 1)

oo
PQ . QR = (2, 3) . (6,- 4) = 2.6 + 3.(-4) = 12 - 12 = 0

oo
? PQ . QR = 0, angle between PQ and QR is 90°.

Hence, ∆ PQR is a right angled triangle.

o oo
Example 9. Find the angle made by a = 6 i - 8 j with X - axis.

Solution: oo
We know that unit vectors along X-axis and Y-axis are i and j respectively.

oo
Let b = unit vector along X-axis = (1, 0) = i

o oo
Given a = 6 i - 8 j

oo
Let T be the angle between a and b .

oo o o .oi
a. b (6 i - 8 j )
cosT oo = = 6 = 3
62 + (-8)2 12 + 02 100 50
| a || b |
( )?
T = cos-1 3 = 86.57°
50
o o o
Example 10. If p+ q+ r = (0, 0), | o o o 127 , find the angle
oo p| = 6, | q | =7,| r|=
between p and q .
oo
Solution: Let T be the angle between p and q .

oo o
| p | = 6, | q | = 7, | r | = 127
o oo
Here, p + q + r = 0

oo o
or, p + q = - r

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Squaring on both sides,

oo
( p + q )2 = r2

oo
or, p2 + 2 p . q + q2 = r2

oo
or, 62 + 2| p | | q | cosT + 72 = 127

oo
or, 2 × 6 × 7 × cosT = 127 - 85, where T is angle between a and b .

or, cosT = 42
84
1
or, cosT = 2 = cos60° ? T = 60°

Hence, the angle between two vectors is 60°.

Example 11. Given that o + o and o - o are orthogonal vectors and o o are
(a 2 b) 5a 4b oo a, b

unit vectors. Find the angle between a and b .

oo oo
Solution: Here, a and b are unit vectors. | a | = 1, | b | = 1 .

Given vectors are orthogonal to each other. Hence, their dot product is zero.

o ooo
? ( a + 2 b ).(5 a - 4 b ) = 0

o o o oo o
or, a . (5 a - 4 b ) + 2 b .(5 a - 4 b ) = 0

oo oo
or, 5a2 - 4 a . b + 10 a . b - 8b2 = 0

o o oo
or, 5.1 - 4 a . b + 10 a . b – 8.1 = 0

oo
or, 6 a . b = 3

or, | o | o . cosT = 1 or, 1.1. cosT = 1
a| b| 2 2

or, cosT = 1 or, cosT = cos60°
2
? T = 60°

Hence, angle between them is 60°.

Exercise 10.1

Very Short Questions

1. (a) Define scalar product of two vectors.

(b) If T is the angle between oa and ob , what is the formula of dot product of o and
o a

b?

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(c) If o o = 0, what is the relation between o and ob ?
a. b a

(d) If o = o y1 o and o o y2 o find o o
a x1 i + j b= x2 i + j, a. b.

( ) ( )(e) If oa =a1 and o = b1 , find the value of oa . ob .
a2 b b2

oo
(f) Under which condition two vectors a and b are perpendicular to each other ?

oo
2. Find the dot product of a and b in following cases:

oo
(a) | a | = 2, | b | = 3, angle between them (T) = 60°

oo
(b) | a | = 2, | b | = 5, angle between them = 90°

oo
(c) | a | = 15, | b | = 25, angle between them = 45°

oo
3. Find the dot product of the vectors a and b in the following cases:

( ) ( )(a) a = 4 , b = 1
o 3o 2 oo oo o o
(b) a = i + 2 j , b = 3 i - j

o o oo o o o oo o
(c) a = 4 i + 2 j , b = - i + 2 j (d) a = 10 i , b = 5 j

oo o o oo oo oo
4. If a = (2, 1), b = (2, –3), find (i) a . b (ii) b . a (iii) show that a . b = b . a

o 2o 4o 2

( ) ( ) ( )5. If a = 3 , b = 2 , c = -3 , find:

(a) a2 (b) b2 (c) c2

oo oo oo
(d) a . b (e) b . c (f) a . c

Short Questions

6. Find the angles between given pair of vectors:

o o oo o o oo
(a) | a | = 4, | b | = 5, a . b = 10 (b) | a | = 3, | b | = 5, a . b = 7.5

( ) ( )(c) a = -3 , b = -2
o 2o 1 oo
(d) p = (3, 4), q = (-4, 3)

o o oo o o
(e) p = 4 i + 2 j , q = - i + 2 j

7. Show that following pair of vectors are parallel:

o 2o 4 ( ) ( )o 1 o -3
(b) p -3 , q = 9
( ) ( )(a) p = 3 , q = 6
o o oo o o
o o oo o o (d) a = i + 5 j , b = -2 i - 10 j
(c) a = 2 i - 3 j , b = 4 i - 6 j

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8. Show that pair of vectors are perpendicular to each other in the following cases :

oo oo
(a) a = (3, 6), b = (6, -3) (b) a = (1, 2), b = (-2, 1)

( ) ( )(c) p = 2 i - 3 j , q = 2 j + 3 i (d) p = 4 , q = 3 -4
o o oo o o o 3o

9. Find the value of k if each pair of the following vectors are orthogonal to each other:

oo
(a) a = (2, -3), b = (k, 4)

o o oo oo
(b) a = (4 i + k j ) , b = (3 i - 6 j )

o o oo o o
(c) a = 3 i - 2k j , b = 10 i - 6 j

o o oo o o
(d) p = 2k i +4 j , b = 3 i +2 j

Long Questions

oo oo o
10. Express PQ and RS in the form of x i + y j and show that PQ is perpendicular to
o
RS in the following cases.

(a) P(3, 5), Q(4, 6), R(7, 6), S(8,5) (b) P(3, -2), Q(5, 1), R(-1, 4), S(2, 2)

11. (a) In ∆PQR if o = o - o , o = o o prove that ∆PQR is a right
PQ 5i 9j QR 4i + 14 j

angled triangle.

o o o o oo
(b) If PQ = – 3 i + 4 j and PR = – 7 i + j prove that ∆PQR is an isosceles right

angled triangle.

oo
12. (a) Find the angle between 4 i – 3 j with x-axis.

oo
(b) Find the angles between 2 i + j and y - axis.

oo ooo
13. (a) Show that the angle between two vectors a and c is 90° if a + b + c = (0, 0),
ooo
| a | = 3, | b | = 5, | c | = 4.

o o ooo o o
(b) Find the angle between a and b if a + b + c = 0, | a | = 6, | b | = 7,
o
| c|= 127 o o o
ooo
(c) If a + b + c = 0, | a | = 3, | b | = 5, | c | = 7, show that the angle between
oo
| a | and | b | is 60°.

oo oo oo
14. If ( a + b ) and (2 a - b ) are orthogonal vectors, a and b are unit vectors, find the
oo
angle between a and b .

oo o o oo
15. If | a - 3 b | = | a + 3 b | , prove that a and b are orthogonal vectors.

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2. (a) 3 (b) 0 (c) 375 2
3. (a) 10 (b) 1 2
4. (i) 1 (ii) 1
(d) 14 (c) 0 (d) 0
(c) 13 (b) 60°
6. (a) 60° 5.(a) 13 (b) 20
(b) 2
9. (a) 6 (b) 63.43° (e) 2 (f) -5
12. (a) 71.56°
(c) 7.13° (d) 90° (e) 90°

(c) - 5 (d) - 4
2 3

13.(b) 60° 14. 180°

10.2 Vector Geometry

Review : A

From the given figures, answer the following questions :

oo
(a) Find the sum of AB and BC in the triangle ABC.

oo o B (i) C
(b) In ∆ABC, can we write BC + CA + AB = 0 ? S
R
oo
(c) In the given rectangle PQRS, write a single vector representing PQ + QR .

oo P
(d) Is PQ . QR = 0?

oo Q 10cm
(e) What is difference between PQ . QR and PQ.QR ? (ii)
(f) State the triangle law and parallelogram laws in vector addition.

Triangle Law of Vector Addition

"If the magnitude and direction of two vectors are represented by two sides of a triangle
taken in order, then the magnitude and direction of their sum is given by the third side taken
in reverse order."

Parallelogram Law of Vector Addition

"If two adjacent sides of a parallelogram through a point represent two vectors in magnitude
and direction, their sum is given by the diagonal of the parallelogram through the same
point in magnitude and direction."

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10.3 Theorems on Vector Geometry
THEOREM 1

Mid Point Theorem

oo
If a and b are the position vectors of two points A and B and M is the mid-point of the
o o o
line segment AB, the position vector of M is OM = 1 (a + b)
2
Proof : Let O be the origin and AB be a line segment.

Position Vector of A = o = o oB
OA a b

Position Vector of B = o = o
OB b

M is the mid-point of AB. O M
A
Then by using triangle law of vector addition, we have, o
a
o oo o 1 o
OM = OA + AM = OA + 2 AB

= o 1 ( o - o
OA + 2 OB OA )

= o 1 o - o
a+ 2 (b a)

= oo o oo
2 a+ b- a a +b

2 = 2

? o 1 o o Proved.
OM = 2 (a + b)

Alternative Method

Since, M is the mid point of AB

oo
AM = MB

oo oo
or, OM – OA = OB – OM

ooo
or, 2 OM = OA + OB

? o 1 o + o Proved.
OM = 2 (a b)

Note :

If A(x1, y1) and B(x2, y2) are any two points and M is the mid point of line segment AB, then
the position vector of M is given

( )o

OM =
x1 + x2 , y1 + y2
2 2

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THEOREM 2

Section formula for internal division.

oo
If a and b are the position vectors of two points A and B respectively and P divides the
line segment AB internally in the ratio m:n, the position vector of vector of P is

oo oo
OP = p = mb + na
m+n
Proof : Let O be the origin and AB be a line segment. P divides AB internally in the ratio

m:n. B

oo o n
Here, position vector of A = OA = a bo P

position vector of B = o = o p
OB b

position vector of P = o = o m
OP p
O oA
Here, AP = m a
PB n
o oo
Now, AP = m ( AP and PB are in the same direction)?
o n
PB Proved.

oo
or, n AP = m PB

oo oo
or, n( OP - OA ) = m ( OB - OP )

oo oo
or, n( p - a ) = m b - m p

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

oo
o mb + na
? p = m + n

Note :

o 1 ( o o where o = o o = o o = o
If P is the mid point of AB, then p = 2 a + b ), OP p, OA a, OB b

THEOREM 3

Section formula for external division

oo
If a and b are position vectors of two points A and B respectively and the point
P divides the line segment externally in the ratio m:n, the position vector of P is

oo
o mb – na
p = m–n

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Proof : Let O be the origin and AB be a line segment. The point P divides line segment AB

in ratio m:n externally. P

Here, position vector of A = o o o
OA = a po

position vector of B = o o bB
OB = b

position vector of P = o o O
OP = p
o
Now, AP = m a A
BP n
o
or, AP = m oo?
o n ( AP and BP have same direction)
BP
oo
or, m BP = n AP

oo oo
or, m( OP - OB ) = n ( OP - OA )

oo oo
or, m p - m b = n p - n a

o oo
or, (m - n) p = m b - n a

oo
o mb – ma
? p = m–n Proved.

THEOREM 4

Centroid Formula

oo o
If a , b , and c are position vectors of the vertices A, B and C of ∆ABC, the position
o o o o o
vector of centroid of triangle is OG = g = 1 ( a+ b+ c)
3
Proof : Let O be the origin and A,B and C the vertices of ∆ABC. o
a A

Then, oo O o
position vector of A = OA = a g

position vector of B = o = o o Go
OB b b c

position vector of C = o = o DC
OC c

Consider a median AD and G the centroid of the triangle, B

G divides AD into 2:1 ratio. i.e. AG : GD = 2:1 = m:n

Now, o 1 ( o + o = 1 ( oo
OD = 2 OB OC ) 2 b +c)

o oo 2× 1 ( oo o 1 o o o
OG m OD + n OA = 2 b +c) + 2. a 3 a b + c)
2+1
? = m+n = ( + Proved.

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Worked out Examples

oo oo
Example 1. The position vectors of A and B are 6 i + 2 j and 2 i + 4 j respectively.
Solution:
If M is the mid-point of line segment AB, find the position vector of M.
Example 2.
Solution: Let O be the origin.

oo o o
Then, position vector of A = OA = a = 6 i + 2 j

position vector of O ob B
oo o o M
oa
B = OB = b = 2 i + 4 j

M is the mid-point if line segment AB.

? position vector of M o 1 ( o + o ) A
= OM = 2 a b

= 1 ( 6 o + 2 o + 2 o o
2 i j i +4 j )

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

oo oo
The position vectors of A and B are respectively 4 i +7 j and 6 i - 2 j .

Find the position vector of C and D if

(a) C divides AB in 2:1 ratio internally. B

(b) D divides AB in 3:2 ratio externally. 1
C
(a) Let O be the origin . 2

Then, ooo A
positive vector of A = OA = 4 i + 7 j

positive vector of B = o = o o . O
OB 6i –2 j

C divides AB in ratio 4:2 internally. i.e. m:n = 4.2

oo
m OB + n OA
Now, position vector of C = m+n

2(6 o – 2 o )+ 1(4 o + 7 o )
i j i j
= 2+1
)o o o o
= 12 i - 4 j + 4 i +7 j
3
oo
16 i +3 j
= 3

= 16 o o
3 i +j

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(b) Here, D divides AB in 3:2 ratio externally. i.e. m:n = 3:2

oo
m OB – n OA
Now, position vector of D = m+n

3(6 o – 2 o )- 2(4 o + 7 o )
i j i j
= 3-2
)o o o o
= 18 i – 6 j - 8 i - 14 j
1
oo
= 10 i – 20 j

Example 3. If A(4, 4), B(3, 2), and C(3, 4) are the vertices of ∆ABC, find the position
Solution: vector of centroid the triangle.

Let O be the origin. Then we have,
o

position vector of A = OA = (4, 4)
o

position vector of B = OB = (3, 2)
o

position vector of C = OC = (3, 4)

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

o = 1 o oo
OG 3 { OA + OB + OC }

= 1 {(4, 4) + (3, 2) + (3, 4)}
3
1
= 3 (10, 10)

= 10 , 10
3 3

Exercise 10.2

Short Questions
oo

1. Let the position vectors of A and B be respectively a and b , AB is the line segment
joining them. Then, write the formula in the following conditions:
(a) position vectors of the mid point M of AB.
(b) position vector of the point P on AB which which divides AB in ratio of m:n
internally.
(c) position vector of the point P of AB which divides AB in ratio of m:n externally.

2. (a) The position vectors of A and B are respectively 4 and 6 , find the

( ) ( )5 3

position vector of the mid point of the line segment AB.

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oo oo
(b) The position vectors of P and Q are respectively. (4 i + 2 j ) and (3 i + 3 j ) ,

then find the position vector of the mid point of line segment PQ.

oo oo
(c) The position vector of M and N are respectively (3 i + 5 j ) and (- i + 3 j ), if
oo
MP = PN , find the position vector of P.

oo oo
(d) The position vectors of A and B are (2 a - 3 b ) and (5 a - 4 b ) respectively. Find

the position vector of mid point of AB.

oo
3. (a) If the position vector of the mid point of the line segment AB is (3 i - 2 j ), where

oo
the position vector of B is (5 i + 2 j ), find the position vector of A.

oo
(b) The position vector of the mid point of the line segment PQ is (4 i + 2 j ), where

oo
the position vector of P is (3 i - 7 j ). Find the position vector of Q.

o o o oo o
4. (a) The position vectors of A and B are respectively a =3 i +4 j and b = i - 2 j . If

C divides AB in the ratio of 3:2 internally, find the position vector of C.

oo oo
(b) The position vectors of A and B are respectively (4 i + 6 j ) and (2 i + 3 j ) .

Find the position vector of C which divides AB in the ratio 2:3 internally.

(c) The position vectors of A and B are 2 o - 3 o oo
a b and 5 a – 4 b respectively. Find

the position vector of the point which divides AB in ratio 3:2 internally.

oo oo
5. (a) The position vectors of A and B are respectively (3 i - 2 j ) and (3 i + 6 j ).

Find the position vector of P which divides AB in ratio 2:3 externally.

oo oo
(b) The position vector of A and B are respectively ( i + j ) and (3 i + 5 j ). Find

the position vector of P which divides AB in ratio of 2:1 externally.

6. (a) If the position vectors of the vertices A, B, and C of ∆ABC are respectively
o o oo oo
(3 i + 5 j ), (5 i - j ), and ( i + 8 j ), find the position vector of centroid

of the triangle.
oo oo
(b) The position vectors of P, Q and R of ∆PQR are respectively (3 i +4 j ), (4 i +5 j ),
oo
and (5 i + 6 j ), find the position vector of centroid of the triangle.

(c) In ∆LMN, o o o o = o + o and the position vector of
OL = 4i -5 j , OM 6i 4j
o oo o
centroid G, OG = 2 i + j . Find ON.

o o oo oo
(d) In ∆ABC, OA = 3 i - 5 j , OB = - 7 i + 4 j and centroid G,

o oo o
OG = 2 i + 6 j , find OC .

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(e) Find the position vector of centroid G of ∆ABC, with vertices A(-1, -1), B(-1, 5),
and C(5, 2).

Long Questions

oo
7. (a) If a and b are the position vectors of points A and B respectively. Find the

oo
position vector of C in AB produced such that AC = 3 BC

(b) OABC is a parallelogram. P and Q divide OC and BC in the ratio of
o oo o o
CP:PO = CQ:QB=1:3. If OA = a , OC = c , find the vector PQ and show that

PQ//OB.

8. (a) In the given figure if A

o 3 o oo
CB 2 AB AC = 3 AD
o = , then show that : 2 o +
o PQ ,
CD 1 o PB D C
(b) In the given figure PA = 6 p A

then show that o
Oa

( )o o
qQ
a=
1 oo
6 5 p+ q

9. (a) In ∆ABC, the medians AD, BE and CF are drawn from the vertices A, B and C
ooo
respectively. Then prove that , AD + BE + CF = (0, 0). A

(b) If G is centroid of ∆ABC, FE
then prove that G
o oo
GA + GB + GC = O B DC

Project Work

10. State the triangle law of vector addition. Is this theorem necessary to prove mid-point
theorem and section formulas in vector geometry? Illustrate with examples.

11. State differences between theorems proved in vector geometry and plane geometry.
Illustrate with examples.

2. (a) 5 (b) 7 i + 5 j (c) i + 4 j (d) 7 a – 7 b
4 2 2 2 2
9 2
3. (a) i – 6 j (b) 5 i + 11 j 4.(a) 5 i + 5 j

(b) 16 i + 24 j (c) 19 a – 18 b 5.(a) 3 i – 18 j (b) 5 i + 9 j
5 5 5 5 (d) 10 i + 19 j

6. (a) 3 i + 4 j (b) 4 i + 5 j (c) -4 i + 4 j

(e) (1, 2) 7.(a) 3 b – a
2

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10.4 Theorems Related to Triangles

THEOREM 5

The line segment joining the mid-point of two sides of a triangle is parallel to the third side
and is half of it. (prove vectorically).

Solution:

Let ABC be a triangle in which D and E are the mid-points of AB and AC respectively, EF is

the line segment joining E and F. A

To prove : o = 1 o o // o
DE 2 BC , DE BC

Proof : DE

o oo B C
Here, DE = DA + AE (by triangle law of vector additional)

or, o 1 o 1 o (E and F are mid-points of AB and AC)
DE = 2 BA + 2 AC

or, o 1 oo
DE = 2 ( BA + AC )

o 1 o
DE = 2 BC

oo o o 1
? DE // BC ( DE =k BC , k = 2 )
o o
? DE = 1 BC Proved.
2

THEOREM 6

The median to the base of an isosceles triangle is perpendicular to the base.

Solution:

Let ∆ABC be an isosceles triangle with AB = AC, D be the mid point of base BC. AD is
median drawn from vertex A to BC.

To prove : AD A BC A
Proof :

oo o
1. BC = BA + AC (by triangle law of vector addition)

o oo
AD = AB + AC (by mid-point theorem)
( )2. 1 BD C
2

oo
AC + AB
( )=1
2
oo o o oo
BC . AD = AC AB ) . AB + AC
( ( )3. - 1
2

oo oo oo
AC - AB AC + AB AC 2 - AB 2
( ) ( ) [( ) ( ) ]=1 . = 1
2 2
1 1
= 2 [AB2 - AB2] = 2 .0 =0 ( ? AB = AC)

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oo
Since, BC . AD = 0.

oo
? AB is perpendicular to BC . Proved.

THEOREM 7

The mid-point of the hypotenuse of a right angled triangle is equidistant from its vertices

Solution: A

Let ∆ABC be a right angled triangle with ‘ABC = 90°,

D the mid point of hypotenuse AC. i.e. AD = DC D
To prove : AD = BD = CD

Proof :

oo o B C

1. In ∆ABD, AB = AD + DB (by triangle law of vector addition)

o o o oo o o o o
2. In ∆BDC, BC = BD + DC = BD + AD = AD + BD ( AD = DC )

3. Since ‘B = 90°, so we can write,
oo
AB . BC = 0

oo o o
or, ( AD + DB ) . ( AD + BD ) = 0

or, ( o - o ) . ( o + o ) = 0
AD BD AD BD

or, ( o )2 = ( o )2
AD BD

or, AD2 = BD2

or, AD = BD

4. Again, D is mid-point of AC i.e. AD = DC

5. From (3) and (4), we get

AD = BD = CD Proved.

Worked out Examples

Example 1. Prove by vector method that the perpendicular drawn from the vertex to the
Solution: base of an isosceles triangle bisects the base.

Let ∆ABC be an isosceles triangle with AB = AC. AD is drawn perpendicular

to BC. A

To prove : BD = DC

Proof :

1. Since AD A BC B DC

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oo oo
then, AD . BD = 0 or, AD . DC = 0

ooo
2. AB = AD + DB (by using triangle law of vector addition)

ooo
AC = AD + DC

oo ( ABC is an isoceles triangle)
3. | AB | = | AC |

oo oo
or, | AD + DB | = | AD + DC |

squaring on both sides, we get

( o o )2 = ( o + o )2
AD + DB AD DC

oo oo
or, AD2 + 2 AD . DB + DB2 = AD2 + 2 AD . DC + DC2

or, 2.0 + DB2 = 2.0 + DC2 (by using 1)

or, DB2 = DC2 Ÿ BD = DC Proved.

Example 2. In the given ∆PQR, RM and QN are medians of ∆PQR such that RM = QN,
Solution: then prove vectorically that ∆PQR is an isosceles triangle.

In ∆PQR, medians RM = QN

To prove : ∆PQR is an isosceles P
Proof :

We have, QN = RM M
i.e
| o | = | o | N
QN RM R

squaring on both sides, we get, Q
oo

| QN|2 = | RM|2

oo oo
or, ( QP + PN )2 = ( RP + PM )2

oo oo
or, QP2 + 2 QP . PN + PN2 = RP2 + 2 RP . PM +PM2

QP2 +( ) ( )or,142=QPo43QP 2.P-R122PoPoQR.+ 1 PR 2 = RP2 +2 o 1 o + 1 PQ 2
o 2 PR2 - 1 PR2 o RP . 2 PQ 2
or, PR = 4
PQ2 -or, - PR o
. PQ

3 PQ2
4
or, PQ2 = PR2

oo
or, | PQ |2 = | PQ |2

? PQ = PR.

Hence, ∆PQR is an isosceles. Proved.

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Exercise 10.3 S T
Q
Short Questions R
1. (a) In the given figure S and T are the mid points of sides P

PQ and PR. Write the relation between ST and QR. Q MR
(b) In the given ∆PQR, PQ = PR, PM is a median. Write the P

relation between PM and QR. M
(c) In the given ∆PQR, M is the mid-point of PR and
QR
‘PQR = 90°. Write the relations among PM, MR and
QM.

(d) In the given ∆PQR if PM = MR = QM. Find the relation
between PQ and QR.

oo
(e) If O is origin and OA = a ,

o oo o oo A
OB = b , OC = c . Write the relation of a , b ,
oa
oo og
c and g , where G is the centroid of the triangle. O
G
ob C
oc
B
D

Long Questions P
2. (a) In ∆PQR, ‘PQR = 90°.

Prove that PR2 = PQ2 + QR2

QR

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(b) ∆ABC is a right angled triangle, ‘B = 90°. A
Let ‘C = T, AC = h, BC = b, AB = p, then prove B
that : h2 = p2 + b2

CT
A

(c) In the figure ∆ABC is an isosceles triangle. AD is a
oo
median, show that AD . BC = 0
B DC

A

3. In triangle ABC, M and N are the mid points of AB

and AC respectively. Then, prove vertically that MN
(a) MN//BC

(b) o = 1 o PB C
MN 2 BC

4. In the given triangle, PM is a median of an isosceles,

triangle PQR with PQ = PR. Then show that PM is

perpendicular to QR, vertically. Q MR

5. Prove vectorically that the mid point of hypotenuse of a right angled triangle is

equidistant from its vertices.

6. Prove vectorically that the line joining the vertex and the mid point of P

the base of an isoscles triangle is perpendicular to the base.

7. ∆PQR is an isosceles triangle with PQ = PR and PM A QR. Then

vectorically show that QM = RM. Q MR

oo o P
8. If a , b , and c are the position vectors of the vertices
G
A, B, C of ∆ABC respectively. Prove vectorically that the position MR

ooo
a+b+ c
o 1 Q
3
( )vector of centroid G of ∆ABC is g =

1. (a) ST // QR, o = 1 o (b) PM A QR (c) PQ A QR, ‘PQR = 90°
ST 2 QR

(d) PM = RM = QM (e) o = 1 o + o + o
OG 3 (a b c)

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10.5 Theorem on Quadrilateral and Semi - circle

THEOREM 8

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is
a parallelogram.

Solution: AS D

In the figure, P,Q,R and S are the mid-points of sides AB, BC, CD,

and AD respectively of a quadrilateral ABCD. Joining P,Q,R, and S P R
successively a quadrilateral PQRS is formed.

To prove : PQRS is a parallelogram. B QC
Construction : Join BD.

Proof :

1. In ∆ABD, o = 1 o
PS 2 BD (In ∆ABD, P and S are the mid-points of AB and AD)

2. In ∆BCD, o = 1 o (In ∆BCD, Q and R are the mid-points of BC and CD)
QR 2 BD

oo
3. PS = QR = [From statement (1) and (2)]

4. | PoS| = | QoR|, PQ // SR (Magnitudes and directions of equal vectors o and o equal)
PS QR

o o oo
5. Similarly, | SR | = | PQ | , SR // PQ . (as in statement (4))

6. ? PQRS is a parallelogram. (from (4) and (5) opposite sides are parallel).

THEOREM 9

The diagonals of a parallelogram bisect each other.

Solution: O
M
In the figure ABCD is a parallelogram. Diagonals D C
AC and BD are drawn. B

Let the mid point of diagonal BD be M.

To prove : M is the mid-point of AC. A

Construction : Let the position vectors of A,B,C,D
oooo o
and M be OA , OB , OC , OD and OM
respectively.

Proof : o o o
OM = ( OD + OB )
1. In ∆ OBD, 1 (by mid-point theorem)
2

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or, o = 1 o + o + o (In ∆ OAD, o = o o
OM 2 ( OA AD OB ) OD OA + AD )

or, o 1 o o o
OM = 2 ( OA + OB + BC )

? o = 1 o o ooo
OM 2 ( OA + OC ) (In ∆ OBC, OB + BC = OC )

2. ? M is the mid-point of AC. (By mid-point theorem o = 1 o o
OM 2 ( OA + OC ))

3. Hence, the diagonals the parallelogram bisect each other C from statement (1) and
(2). Proved.

Alternative Method

Let O be the origin. O

o ooo D C
Then OA , OB , OC , OD are the position vectors of A B
o oo o
A, B C and D respectively. Let OA = a , OB = b ,
oo oo
OC = c and OD = d

Proof

1. By using mid-point theorem, position vector of mid point of diagonal
o o o o
BD = 1 ( b + d ). Position vector of mid-point of diagonal AC = 1 ( a + c ).
2 2

oo
2. Now, AB = DC (opposite sides of a parallelogram)

o o oo
or, OB - OA = OC - OD

oo oo
or, b - a = c - d

oo oo
b+ d a+ c
or, 2 = 2 (dividing both sides by 2)

3. Position vector of mid-point of BD = position vector of mid point of AC, [from (2)].

4. Hence, the diagonals of a parallelogram bisect each other. Proved.

THEOREM 10

The diagonals of a rhombus bisect each other at right angle. Prove it vectorically.

Solution : D C
B
In the figure, let ABCD be a rhombus with diagonal AC and BD.

Then AB = BC = AD = DC.

To Prove : AC A BD, mid-point of AC = mid-point of BD. A

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Proof :

ooo (by using triangle law of vector addition)
1. In ∆ ABC, AC = AB + BC

2. In ∆ABD, (by triangle law of vector addition)
o oo oo
BD = BA + AD
oo ( AD = BC )
= BA + BC
oo
= BC – AB
oo

3. Taking dot product of BD and AC ,

oo oo o o
BD . AC = ( BC - AB ) . ( BC + AB )
oo
= BC2 – AB2 ( | BC | = | AB | )

= BC2 – BC2 = 0

4. BD A AC.
o oo oo oo o

5. Let OA = a , OB = b , OC = C , OD = d .

oo
a+ c
Position vector of the mid-point of AC = 2

oo
b+ d
Position vector of the mid-point of BD = 2

oo
6. AB = DC (Opposite sides of parallelogram)

oo oo
or, OB - OA = OC - OD

oo o o
or, b - a = c - d

oooo
? a+ c=b+d

oo oo
a+ c b+ d
? 2 = 2

i.e., Mid-point of diagonal AC = Mid-point of diagonal BD.

7. Hence, the diagonals of rhombus bisect each other at right angle. (From statement (4)
and (6).) Proved.

THEOREM 11

Prove vectorically that the diagonals of a rectangle are equal. D C
Solution: B

Let ABCD be a rectangle.

AC and BD are diagonals of the rectangle. A

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To prove : | o | = | o |
AC BD

Proof :

1. In ∆ABC, o = o + o (By triangle law of vector addition)
AC AB BC

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

oo oo o
or, ( AC )2 = ( AB )2 + 2. AB . BC + ( BC )2

oo
Since AB A BC, AB . BC = 0,

? AC2 = AB2 + BC2

2. In ∆ABD, o = o + o
BD BA AD

or, o = (- o ) + o . ( o = o )
BD AB BC AD BC

oo oo o
( BD )2 = (- AB )2 + 2( AB ). BC + ( BC ) 2

ooo
? ( DB )2 = ( AB )2 + ( BC )2

? DB2 = AB2 + BC2

3. From (1) and (2), we get.

AC2 = BD2

or, AC = BD

oo
? | AC | = | BD | Proved.

THEOREM 12

Prove vectorically that the angle at semi - circle is a right angle.
Solution : Let O be the centre of the circle and AB be a diameter.

‘ACB angle at semi-circle. OC is joined. C
To prove : ‘ACB = 90°

Proof : AO B
1. By triangle law of vector addition,

oo o
AC = AO + OC

o oo
CB = CO + OB

oo oo
= CO + AO ( AO = OB radii of the same circle)

oo
= AO - OC

oo
2. Taking dot product of AC and CB , we get,

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oo o o o o
AC . CB = ( AO + OC ). ( AO - OC )

oo
= ( AO )2 - ( OC )2

= AO2 - AO2 (OA = OC = OB, radii of same circle)

=0

3. Since o . o = 0, o A o
AC CB AC CB

Thus, shows that ‘ACB = 90° Proved.

Exercise 10.4

Short Questions

1. (a) In the adjoining figure P, Q, R, S are the mid-points of D SC
sides AD, AB, BC, and CD respectively. The points P, Q, R R

and S are joined. What type of quadrilateral is PQRS?

(b) What is the measure of circumference angle at semi- P
circle?

(c) Write relations of diagonals in each of the following A Q B
quadrilateral.

(i) Square (ii) Rhombus (iii) Parallelogram

Long Questions

2. In a quadrilateral ABCD, the mid-points of sides AB, BC, CD, and DA are respectively
P, Q, R and S. Prove vectorically that PQRS is a parallelogram.

3. Prove vectorically that the diagonals of parallelogrm PQRS bisect each other.

4. Prove vectorically that the diagonals of rhombus EFGH bisect each other at right angle.

5. Prove vectorically that the diagonals of rectangle PQRS are equal.

6. Prove vetcotically that the angle at semi-circle is right angle. R
7. In the figure O is the centre of the semi-circle. Prove that

‘PRQ is right angle.

8. If the diagonals of a quadrilateral bisect each other, prove by P OQ
vector method that is a parallelogram.
BLC

9. In the given figure, ABCD is a parallelogram. L and M are the M
mid-points of sides BC and CD respectively. D

Prove that o + o 3 o A R
AL AM = 2 AC S N
Q
10. In the given figure PQRS is a parallelogram M and N are the mid M
points of PS and QR, show that MQNS is a parallelogram. 297

P

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11. If a line is drawn from the centre of a circle to the mid-point of a chord, prove by vector

method the line is perpendicular to the chord.

12. In the given figure PQRS is a trapezium where PS//QR. M and N P S

are the mid-points of PQ and SR respectively. Prove vectorically N
that M

(i) o = 1 o + o Q R
MN 2 ( PS QR ) D C

oo
(ii) MN // QR

13. In the adjoining figure, PQRS is a parallelogram. G is the point O G
of intersection of the diagonals. If O is any point prove that

oooo A B
OA + OB + OC + OD SM R
o = 1
4 NQ
( )OG C

14. In the adjoining figure PQRS is a parallelogram. M and N are

two points on the diagonal SQ. If SM = NQ, then, prove by

vector method by PMRN is a parallelogram. P

oo D
15. ABCD is a parallelogram and O is the origin. If OA = a ,
o oo o o oo o
OB = b , OC = c find OD in terms of a , b and c .

Project Work A B
O
15. Write three statements of vector geometry. Show differences
between them in vector geometry and plane geometry.

1. (a) parallelogram (b) 90° 15. oa – ob + oc

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11Transformations

11.0 Review

Discuss the answers of the following questions:
(a) Name the four fundamental transformations that you studied in class 9 ?

(b) Can transformations be considered as functions?

(c) What is an invariant point in transformation ?

(d) Can you say any common characteristics between an object and image in reflection,
rotation and translation ?

(e) What is the image of point P(2, 4) when it is rotated through 90° about centre (1, 2) ?

(f) Give any two examples of enlargements used in our daily life.

(g) What type of transformation can be explained by rotating a wheel of bicycle ?

(h) What is the distance between the centre and a point on the circumference of a circle
called ?

(i) When an object is shifted through a certain distance in certain direction, what type of
transformation does it represent?

Review of formulae of transformation :
Let us review the transformation formulae that we studied in class 9.

Reflection : Let p(x,y) be any point and p'(x',y') be an image, of P under a reflection.

(a) Reflection on X - axis

P(x, y) X - axis P'(x, -y)

(b) Reflection on on Y-axis

P(x,y) Y - axis P'(-x, y)

(c) Reflection on line y = x

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

(d) Reflection on line y = -x or x = - y

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

(e) Reflection on line x = h

P(x, y) x=h P'(2h - x, y)

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(f) Reflection on line y = k

P (x, y) y=k P' (x, 2k - y)

Translation :

( )a

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

( )P(x, y) T= a P' (x + a, y + b)
b

Rotation :
Let P(x, y) be any point and P'(x1 , y1) be its image under rotation R, about a centre.

(a) Positive Quarter turn or rotation of + 90° about origin.

P(x, y) p'(-y, x)

Positive quarter turn is equivalent to rotation - 270° about the same centre.

(b) Negative quarter turn or rotation of - 90°

P(x, y) P' (y, -x)

Nagative quarter turn is equivalent to rotation + 270° about the same center.

(c) Half turn or rotation of 180° about origin.

P(x, y) P'(-x, -y) , P(x, y) R[(a, b), 180°] P'(2a – x, 2b – y)

Enlargement
Let E[(0, 0) k] denote the enlargement with centre origin and scale factor k. Let P(x ,y) be a
point. Then, its image is given by

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

If the centre of enlargement is (a, b) with scale factor k.
P(x, y) E[(a,b), k] P'(k(x - a) + a, k(y - b) +b)

11.1 Composition of Transformations/Combined Transformation

Introduction Y
P(4, 5)

Let us take a point P(4,5), reflect it on X-axis. Then,

P(4, 5) X-axis P'(4, -5)

Again, the image p' (4, -5) of(4, 5) is reflected on Y X' OX
- axis.
P"(–4, –5) P(4, –5)
P'(4, -5) Y-axis P''(-4, -5) Y'

Here, the image P''(-4, -5) is the image of P(4, 5) due
to combined reflections. Such type of transformation

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is called combined/composition of transformations. Now, let us answer the question.

-What is the combined transformation of above transformation ?
Rotation through 180° about centre origin O is the single transformation of above combined
transformations (reflections) .

Definition
When an object has been transformed, its image can again be transformed to form a new
image. Such transformation is called combination of transformation (or composition of
transformations). After a combination of transformations, the change from the single object
to the final image can be described by a single transformation.

Let R1 be a transformation which transforms a point P to the point P'. Again let R2 be
another transformation which transforms P' to the point P''. Then, the transformation which
transforms the point P to P'' is said to be the composition of R1 and R2. It is denoted by R2oR1
(read as R1 is followed by R2).
R2oR1 means that the first transformation is R1 and it is followed by R2. It means the first
transformation is due to R1 and , it is followed by R2.
R1oR2 means that the first transformation is R2, and it is followed by R1.
Hence, R1oR2 and R2oR1 have different meanings. There may be a single transformation
which represents the combination of given transformations.

11.2 Combination of Translation and Translation

( ) ( )2 4

Let P(1, 3) be a point and T1 = 5 and T2 = –4 be two translation vectors. Firstly
P is translated by T1. We get,
Y

( )P (1, 3)T1 =2 P' (3, 8)
5
P' (1 + 2, 3 + 5) = P' (3, 8)

Again, P' (3, 8) is translated by T2 we get, P"(7, 4)
X
( )P'(3, 8)T2 =4P'' (3 + 4, 8 – 4) = P'' (7, 4) P(1, 3)
-4 O
Y'
Also a single translation that represents the above two X'
translations is given by

T1 T2

2 46 P"

( ) ( ) ( )T = 5 + –4 = 1
( )A (1, 3)
T= 6 P'' (1 + 6, 3 + 1) = P''(7, 4) T2 n
1 P' m C

A translation followed by another translation: T2oT1
T1
( )a b

Let, T1 = b be a translation vector. P aA B

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( )m

T2 = n be another translation vector.
Let P(x, y) be any point.

T1 translates P to P' with 'a' units along the x-axis and 'b' units along the y - axis.

P(x, y) ( )T2 =a P'(x + a, y + b)
b

Again, T2 translates P' to P'' with 'm' units along the x-axis and 'n' units along the y-axis.

P'(x + a, y + b) ( )m P'' (x + a + m, y + b + n)

T2 = n

Now, let us find a single translation vector T which translates P to P''.

Now, x-component of T = PB = PA + AB = PA + P'C = a + m

y - component of T = BP'' = BC + CP'' = AP' + CP'' = b + n

( )a + m

T = T2 o T1 = b+ n
a m
n , the combined
( ) ( )Hence, if translation T1 = b is followed by translation T2 =

a+m

( )translation is given by T = T1 + T2 = b+ n .
Note :

( ) ( )If T1 =
am
b and T2 = n are two translation vectors, the combined translation
a+m
( )T2oT1 have the same effect i.e. T1oT2 = T2oT1 = b + n

Worked out Examples

( ) ( )1

Let T1 = 2
Example 1. –3
Solution: and T2 = 2 be two translation vectors. Then find the
image of point P(2, 3) under the combined translation T2 o T1
302
Given translation vectors are :

( ) ( )1 -3 Y

T1 = 2 and T2 = 2 P"(6, 10)
The combined translation vector of

T1 and T2 is given are 2 T2 P'(3, 5)
-3
( ) ( ) ( )T = T1 + T2 = T1 1
1 -3 -2
2 + 2 = 4 P(2, 3)

( )-2 X' O X

Now, P(2, 3) T = 4 P' (2 - 2, 3 + 4)

= O' (0,7) Y'

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Alternative Method

T2 o T1 means translation T1 is followed by T2

( )P(2, 3)T= 1 P' (2 + 1, 3 +2) = P'(3, 5)
2

( )-3 P' (3 - 3, 5 + 2) = P'(0, 7)

Again, P'(3, 5) T = 2

( ) ( )Let T1 =2 4
Example 2. 4 and T2 = 5 be two translations vectors, what point
Solution:
would have the image (9, 12) under the combined translation T2 o T1 ?

( ) ( )Here, T1 =2 4 , combined translation
4 T2 = 5

( ) ( ) ( )vector T of T2 o T1 = 4 + 5 i.e. T = 9
24 6

Let P(x, y) be the required point.

( )Then, P(x, y)T= 6 P' (x + 6, y + 9) But P'(9, 12)
9

? (9, 12) = ( x + 6, y + 9)

Then equating the corresponding elements, we get,

9 =x + 6 or, x = 3

and 12 = y + 9 or, y = 3

? P(x, y) = P(3, 3)

Example 3. ( ) ( )-1 2
Solution:
Let T1 = 3 and T2 = 3 be two translation vectors. Find the image
of the vertices of ∆ABC with A(2, 3), B(4, 5) and C(6, 3), under composite
translation T2 o T1. Show ∆ABC and ∆A'B'C' on the same graph.

-1( ) ( )Here, T1 = 2 Y
3 and T2 = 3 B'

Combined translation vector A' C'

( ) ( )-1 2 B

T = T1 + T2 = 3 + 3

= ( 1 ) A C
6 X
O
( )1 X' Y'

Under translation T2 o T1 = T = 6
( )A(2, 3)
T= 1 A' (2 + 1, 3 + 6) = A1 (3, 9)
6

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B(4, 5) T = (61 ) B' (4 = 1, 5 + 6) = B'(5, 11)
C(6, 3) T = (16 ) C'(6 + 1, 3 + 6) = C'(7, 9)

The graphs of ∆ ABC and ∆A'B'C' are plotted alongside.

Exercise 11.1

Very Short Question

1. (a) Define combined transformation.

(b) If T1 and T2 are two transformations, write a difference between T1oT2 and T2 o T1.

( ) ( )(c) Write the combination of two translations T1 = a m
b and T2 = n .

Short Questions

( ) ( )2 -5

2. If T1 = 1 and T2 = 6 are two translations, find the images of given points
under the given combined translations.

(a) T2 o T1 (2, 5) (b) T1 o T2 (6, 5)
(c) T12 (2, 4) (d) T22 (3, 5)

( ) ( )3. 3 1
Let T1 = 2 and T2 = 2 be two translations, find the image of the following

points under the combined translations T1 o T2 and T2 o T1

(a) A (4, 5), (b) B(-6, 7) (c) C(2, -5) (d) D(2, 3)

( )4. a
(a) The image of a point (3, 4) is the point (10, 10) under a translation b followed

( )by another translation 4 . Find the values of a and b.
5

( ) ( )2

(b) Let T1 = 3
4
and T2 = 4 be two translation vectors. What point would
have the image (-8, 12) under the combined translation T2 o T1.

( ) ( )(c) If T1 = 1 -2 and T1 o T2 (x, y) = (8, 8) find the values of
x and y. 2 and T2 = 4

( ) ( )(d) If P =4 -4 and ToP (x, y) = (6, -6), find the values of x and y.
5 , T= 2

(e) If T1 and T2 are two translations defined by T1(x, y) = (x + 6, y - 3), and
T2(x, y) = (x - 3, y + 4). Find the image of P(1, 3) under T2 o T1 .

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Long Questions

( ) ( )-1 -3

5. Let T1 = 2 and T2 = 4 be two translations. Translate ∆ABC with vertices
A(2, 2), B(-2, 2), and C(6, 6) under translation T1 and then T2.
(a) Find the image ∆A'B'C' of ∆ABC.
(b) Find the image ∆ A" B" C" of ∆ A'B'C' under translation T2.
(c) Draw graphs of ∆ABC, ∆A'B'C' and ∆A"B"C" on the same graph paper.

( ) ( )3 1

6. Let T1 = -2 and T2 = 1 be two translation vectors. If P(2, 3), Q(4, 5), R(6, 4),
S(7, 8) are the vertices of quadrilateral PQRS. Find the image of quadrilateral PQRS
due to transformation T2 o T1 . Show the quadrilateral PQRS and its image on the same
graph.

7. P(1, 2), Q(7, 2), R(7, 9) and S(1, 9) are the vertices of rectangle PQRS. Draw it on a graph
paper . Find the coordinates of the vertices of the image rectangle P'Q'R'S' under the

( ) ( )2 2

transformation T1 o T2 if T1 = 3 and T2 = -2 . Also draw the image on the same
graph paper.
8. If T1 and T2 be two translations defined by T1(x, y) = (x + 3, y + 2) and T2(x, y) =(x - 4,
y - 7). Find T1 o T2 (x, y) and T2 o T1, (x, y). Also find the images of A(1, 2), B(7, 2) and
C(4, 7) of ∆ABC under the composite translation T1 o T2 and T2 o T1. Are images due to
T1oT2 and T2oT1 same ? Show ∆ABC and it images on the ame graph paper.

( )1.
(c) a+m 2.(a) (-1, 12) (b) (3, 12) (c) (6, 6)
b+n 3.(a) (8, 9) (b) (-2, 11) (c) (6, -1)

(d) (-7, 17)

(d) (6, 7) 4.(a) a = 3, b = 1 (b) (-14, 5) (c) x = 9, y = 2

(d) x = 6, y = -13 (e) (4, 4)

5. A'(1, 4), B'(-3, 4), C'(5, 8), A"(-2, 8), B"(-6, 8), C"(2, 12)

6. P'(6, 2), Q'(8, 4), R'(10, 3), S'(11, 7)

7. P'(5, 3), Q'(11, 3), R'(11, 10), S'(5, 10) 8. A'(0, -3), B'(6, -3), C'(3, 2)

11.3 Combinations of Rotation and Rotation

A rotation followed by another rotation can be expressed as a single rotation. We have the
following two cases:

A. When two rotations have same centre.
A point or an object once rotated about a centre through a given angle can further be
rotated about the same centre through another given angle.

Let P(x, y) be any given point in the plane.

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Let it be rotated through positive 90° about origin.

P(x, y) P'(-y, x)

Again, let P' (-y, x) be further rotated through positive 90° about the same centre O.

P (- y, x) P' (-x, -y)

Here the final image of P is P".

But P" (- x, - y) is the image of the point P(x, y) under the rotation of 180° about the
origin.

Y

P'(–y, x)

P(x, y)

Dq Eq Dq

X' Eq X

O

P"(–x, –y)

Y'

Hence, a rotation of D° followed by another rotation of E° about the same centre is
equivalent to the rotation of (D° + E°) about the same centre.

Worked Out Examples

Example 1. Let P(4, 2), Q(6, 2) and R(5, -1) be the vertices of ∆PQR. Find the coordinates
Solution: of image ∆P'Q'R' under the rotation through 90° followed by a rotation
through 180° about the origin.

The first angle of rotation (D°) = 90° about origin O.

The second angle rotation (E°) = 180° about origin O.

Since the centre of rotation in both cases is same O.

Hence, combined rotation can be expressed as a single rotation (90° + 180°)
= 270° positive about the same centre O.

The vertices of image ∆ P'Q'R' due to combined rotations are as follows :
P(x, y) R[(0, 0), 270°] P'(y, –x)

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P(4, 2) P'(2, -4)
Q(6, 2) Q'(2, -6)
R(5, -1) R' (-1, -5)

Example 2. A(2, 3), B(4, 5) and C(5, 1) are the vertices of a triangle ABC. The triangle ABC
Solution: is first rotated about the origin through positive 90°. Find the coordinates of
vertices image ∆A'B'C'. Again ∆A'B'C' is rotated through 90° in anticlock. Find
the coordinates of image ∆A"B"C". Plot ∆ABC, ∆A"B"C" on the same graph,
∆A"B"C". what is the single transformation of the combined rotation.

The first angle of rotation = 90° about centre O.

The images due to rotation of +90° about centre O of A, B, C, are given below:

P(x, y) R[(0, 0), +90°] (–y, x) Y

A(2, 3) R[(0, 0), +90°] A'(-3, 2) C' B

B(4, 5) R[(0, 0), +90°] B'(-5, 4) B'
A
A'
C(5, 1) R[(0, 0), +90°] C'(-1, 5) C
X' X
C" O
Again, ∆A'B'C' is rotated through

+90° about the same centre O. A"

Its image vertices are given below: B"
A'(-3,2) R[(0, 0), +90°] A" (-2, -3) Y'
B'(-5, 4) R[(0, 0), +90°] B" (-4, -5)
C'(-1, 5) R[(0, 0), +90°] C" (-5, -1)

Graphs of ∆ABC, ∆A'B'C' and ∆A"B"C"are plotted on the graph.

Since, the centre of rotation is same in given two rotations, the combined
rotation is (90° + 90°) = 180° about the same centre O in positive 180°.

B. When two rotations are about the different centres.

A point or an object once rotated about a centre through a given angle can further be rotated
Y
about the different centre through another given angle.

Let R1 be the rotation through an angle x° about a centre A C
and R2 be another rotation through an angle y° about the C' B'
different centre. Then, the equivalent transformation B
(R2oR1) is again a new rotation through the angle (x°+y°) A' RP X
about the third centre. This centre is the meeting point of O
the perpendicular bisectors of the line segments joining X' B"
corresponding vertices of the object and its final image. In Q
other words, perpendicular bisectors of the line segments
joining corresponding vertices of the object and its final A"
image are concurrent and the point of concurrence is the C"
centre of new rotation (i.e., combined rotation).
Y'

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Let ABC be the given triangle (object). It is rotated to ∆ A'B'C' through 90° about the centre
P. The triangle A'B'C' is again rotated to ∆ A"B"C" through 90° about the centre Q. Then
∆A"B"C" is the image of ∆ ABC under the rotation through 180° about the third centre R.
Perpendicular bisectors of AA", BB", and CC" intersect each other at the point R.

Example 3. Let R1 = Rotation of 90° about origin in anti clockwise and R2 = rotation
Solution: of 180° in anti clockwise about centre (2, 3). Then, find the image of
P(5, 3) under combined transformation R2 o R1 (P).

Here, R1 = rotation of +90° about O.

R2 ritation of 180° about (2, 3)
Since (x, y) R[(0, 0), +90°] (-y, x)
and (x, y) R[(a, b), 180°] (2a - x, 2b - y)

Now, P (5, 3) P'(-3, 5)

Again P'(-3, 5) R[(a, b), 180°] P" (2.2 + 3,2.3 - 5) = P" (7, 1)

Combined transformation of Rotation and Translation.

Let P(x, y) be any point on the plane.

( )Let R = rotation of + 90° about origin and T = translation vector = a
b

We find the image due to ToR of P, under rotation of + 90° about O.

p(x, y) R[(0, 0), 90°] p' (-y, x)

( )Again, P'(-y, x) T= a P' (-y + a, x + b)
b

Similar relations can be established for RoT and other rotation followed by
translation of translation followed by rotation.

Example 4. ( )A(2, 1), B(7, 2) and C(3, 6) are the vertices of ∆ABC. Let T = 3 and
Solution: 2
R = rotation of + 90° about origin. Find the image of ∆ABC under RoT. Draw

graphs of ∆ABC and its image in the same graph paper.

( )Here, T = 3 , R = rotation of +90° about O. ∆ABC is translated by T, we
2
get the image ∆A'B'C'.

T=( )P(x, y) 3 (x + a, y + b)
2

T=( )A(2,1) 3 A'(2 + 3, 1 + 2) = A' (5, 3)
2

( )B(7, 2)T= 3 B'(7 + 3, 1 + 2) = B'(10, 4)
2

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( )C(3, 6) T= 3 C'(3 + 3, 6 + 2) = C' (6, 8)
2

Again, ∆A'B'C' is rotated about O B" Y
through +90°. C
C'

A'(5, 3) A"(-3, 5) C" A"
B'(10, 4) B"(-4, 10)
A' B B'
X
C'(6, 8) C"(-8, 6) X' A
O

Graphs of ∆ABC, ∆A'B'C' and Y'
∆A"B"C" are plotted.

Exercise 11.2

Short Questions

1. Let P(2, 3) be a point in plane and R1 = rotation of +90° about origin.
R2 = rotation of 180° about origin.
(a) Find the image of the point P under the following transformations:

(i) R1oR2 (ii) R2oR1

(b) Write down a single transformation of above combined.

2. Let, R1 = rotation of +90° about origin.
R2 = rotation of +90° about origin.

(a) Find the image of the following points under the following transformations:

(i) R1 o R2 (4, 5) (ii) R1 o R2 (-3, -2) (iii) R2 o R1(4, 6)

(b) Write down a single transformation of combined transformation.

(i) R2oR1 and (ii) R1 o R2

3. (a) Find the image of P(2, 3) due to combined transformation.

(i) R1oR2 where R1 = rotation of +90° about origin.

(ii) R2R1 R2 = rotation of + 90° about (2, 5)

(b) Find the image of M(2, 5) due to combined transformation

(i) R1oR2 where R1 = rotation of +90° about (1, 2)

R2 = rotation of 180° about O

(ii) R2 o R1

(c) Find the image of S(7, 5) under combined transformation R2 o R1, where,

R1 = rotation of 180° about (2, 3)

R2 = rotation of +90° about origin.

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( )4. (a) Let T =2 and R = rotation of +90° about O.
3

Then find the images of given points under.

(i) ToR (2, 7) (ii) RoT (4, 5) (ii) ToR (-2, -3)

( )(b) Let R = rotation of 180° about origin. T = -2
3

Find the images of A(2, 3) and B(6, 7) under (i) ToR (ii) RoT

Long Questions

5. A(2, 5), B(6, 5) and C(2, -3) are vertices of ∆ABC. Rotate ∆ABC through +90° about
origin and obtain image ∆A'B'C'. Again rotate ∆A'B'C' about the same centre origin
through 180° obtain images ∆A"B"C", plot the triangle ABC and its images on the same
graph paper. What is a single transformation for above transformation?

6. Points A(5, 2), B(4, 5) and C(8, 4) are the vertices ∆ABC. Find the image of the vertices
of ∆ABC under the combined rotation of - 180° followed by +90° about the centre
origin. Draw ∆ABC and its image in the same graph paper. Write a single transformation
of above combined transformation.

7. Let M(1, 1), N(-3, -6) and P(6, -1) be the vertices of ∆MNP. The vertices are transformed
by a single transformation obtained by the combination of the rotation [(0,0), + 90°]
and on the same direction of a rotation [(0,0), + 180°]. Find the coordinates of the
images of these points.

8. Draw a triangle with vertices A(1, 2), B(-2, 3) and C(2, 4) on a graph paper.

(a) Find the image ∆A'B'C' under rotation of - 90° about origin.

(b) Find the image ∆A"B"C" of ∆A'B'C' under rotation through+ 180° about the point (2.2).

(c) Find the single transformation to represent these two rotations.

9. If R1 represents a rotation of +90° about the origin and R2 represents a rotation of
- 270° about the origin. What transformation does R2oR1 represents ? Transform the
triangle with vertices A(-4, 0), B (-6, 2) and C(-4, 2) under R2oR1. Draw graphs of ∆ABC
and its images on the same graph paper.

( )10. ∆ABC having vertices A(3, 4), B(-2, 1) and C(4, 1) is translated by T = -1 . The
2
image ∆A'B'C' so formed again rotated through +90° about origin. Find the vertices of

final image. Draw graph of ∆ABC, ∆A'B'C' and ∆A"B"C" on the same graph paper.

11. ∆PQR having vertices P(-2, 2), Q(2, 2), and R(6, 6) is rotated through +90° about the

( )origin. The image ∆P'Q'R' so formed is again translated by T = 3 . Find the vertices
2
of ∆P'Q'R' and ∆P"Q"R". Draw graphs of ∆PQR and its images ∆P'Q'R' and ∆P"Q"R" on the

same graph paper.

Project Work

12. Search some examples of combination of a rotation followed by another rotation.
Prepare a report about them and present in your class room.

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1. (a) (i) (3, -2) (ii) (3, -2) (b) R[(0, 0), 270°]

2. (a) (i) (-4, -5) (ii) 3, 2) (iii) (-4, -6) (b) (i) R[(0, 0), 180°]

(ii) R[(0, 0), 180°] 3.(a) (i) (-5, 4) (ii) (5, 0) (b) (8, -1)

(ii) (-1, -6) (c) (-1, -3)

4. (a) (i) (-5, 5) (ii) A'(-8, 6), B'(5,1) (b) A'(-4, 0), B'(-8, -4)

5. A'(-5, 2), B'(-5, 6), C'(3, 2) A"(5, -2), B"(5, -6), C"(-3, -2)

6. A'(2, -5), B'(5, -4), C'(4, -8), R[(0, 0), -90°] 7. M'(1,-1), N'(-6,3), P'(-1,-6), R[(0, 0), 270°]

8. (a) A'(2, -1), B'(3, 2), C'(4, -2) (b) A"(2, 5), B"(1, 2), C"(0, 6)

(c) R[(0, 4), +90°] 9. R[(0, 0), -180], A'(4, 0), B'(6, -2), C'(4, 2)

10. A'(2, 6), B'(-3, 3), C'(3, 3), A"(-6, 2), B"(-3, -3), C"(-3, 3)

11. P'(-2, -2), Q'(-2, 2) and (-6, 6), P"(1, 0), Q"(1, 4), R"(-3, 8)

11.4 Combinations of Reflection and Reflection

A point or an object can be reflected and the image so formed can further be reflected. There

are two cases in combination of reflections.

A. When the axes of reflections are parallel Y

Let x = - 2 and x = 2 be two parallel lines represented l1 l2
by l1 and l2 respectively.

Let us take a point P(-5, 2). Firstly it is reflected on l1 i.e. (–5, 2) M P' N P"
x = -2 line we get, P O

P (–5, 2) x = –2 P'(1, 2) X' X

Again P'(1, 2) is reflected on l2 i.e. x = 2 line x = –2 x=2

P' (–5, 2) x = 2 P" (3, 2)

Hence, P (–5, 2) P" (3, 2) Y'

o
Let us take two points M on l1 and N on l2 and find MN .

o 4
0
( ) ( )Here, M(-2, 2) and N(2, 2) MN =
2+2 = .
2-2

o 4 = 8 .
0 0
( ) ( )But total displacement is 2 MN = 2

Also, P (–5, 2) T= 8 P' (–5 + 8, 2 + 0) = PI (3, 2)
0

But P"= P'

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This shows that combination of two reflections when the axes of reflections are parallel
is equivalent to a translation and the translation is twice the distance between the axes of
reflection and the direction is perpendicular to the axes of reflections.

Let l1 and l2 be two parallel lines which are represented by x = x1 and x = x2 respectively.
P is reflected on l1 line, we get,

P' x = x1 P"

Again p'is reflected on x = x2 and we get,

P' x = x2 P"

By definition of reflection,

PM = MP' , P'N = NP"

PP' = 2MP' and P'P" = 2P'N

o oo
Now, PP" = PP' + P'P"

oo
= 2 MP' + 2 P'N

oo
= 2( MP' + P'N )

( )= 2 x2-x1 , ( y2 - y1 = 0)
0

Hence if the axes of reflections are parallel, a reflection followed by another reflection is

equivalent to the translation. The translation is twice the distance between the axes of

reflection and the direction is perpendicular to the axes of reflection.

B. When the axes of reflections intersect at a poin. Y y=x
P (2, 6)
Let us consider two axes of reflection x-axis and A
line y = x. The angle between them is 45°.

Let us take P(2, 6) in the plane. Firstly P(2, 6) is
reflected on y = x line.

P (2, 6) y=x P' (6, 2)

Again P'(6, 2) is reflected on x - axis. P' (6, 2)
BX
P' (6, 2) x-axis P" (6, –2) X' T P"(6, –2)
O
Now, measure ‘POP" we get ‘POP" = 90° (-ve).
Y'
The point of intersection of y = x and x - axis is the
origin.

Hence, the combined transformation for P(2, 6) is

P' (2, 6) P' (6, –2)

? Combined transformation is equivalent to rotation and the angle of rotation is twice the
angle between the axes. The centre of rotation is the point of intersection of the axes.

Let l1 and l2 be two lines intersecting at O. Let the lines be the axes of reflections. The angle
between them is T i.e. ‘AOB = T

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Let P be reflected on line l1, we get,
P P'

Again P' is reflected on l2 we get,
P' P"

Now, join OP, OP' OP"

By simple plane geometry ‘POA = ‘AOP' , ‘P'OB = ‘BOP"

‘POP" = ‘POP' + ‘P'OP"

= 2‘AOP' + 2‘P'OB

= 2(‘AOP' + ‘P'OB) = 2T

Hence, if the axes of reflections intersect at a point, then combined reflection is equivalent to
a rotation about the centre, the point of intersection of axes, the angle of rotation is the angle
between the axes. The direction of rotation is direction from first axis (OA) to the second
axes (OB).

Worked out Examples

Example 1. Let r1 denote the reflection on line x = 2 and r2 denote the reflection on
Solution: line x = 4. Then find the image of the point P(1, 5) under the combined
transformation r2or1 and r1or2.

Here, r1 : x = 2 and r2 : x = 4 are parallel axes of reflection, So, the combined
reflection is equivalent to translation. For r2 o r1 .

( ) ( )translation vector T = 2 4-2 = 4
0 0
? P(1, 5) T = 40
P' (1 + 4, 5 + 0) = P' (5, 5)

( ) ( )For r2 o r1, translation vector T is given by T = 2 2-4 = -4
0 0

Image of P is given by,

P(1, 5) T = -04 P' (1 – 4, 5 + 0) = P' (–3, 5)

Alternatively

For r2 o r1 , first reflect P on r1 i.e. x = 2 Y
P P' P"
P(1, 5) x=2 P' (2 × 2 - 1, 5) = P'(3, 5)

and then P'(3, 5) is reflected on r2 i.e. x = 4

P' (3, 5) x=4 P"(2 × 4 - 3, 5) = P"(5, 5) X' O r1 r2 X
x=4
For r1o r2, first reflect P on r2 i.e. x = 4
x=2
P(1, 5) x = 4 P'(2 × 4 - 1, 5) = P'(7, 5)
Y'

Again P'(7, 5) is reflected on r1 i.e. x = 2

P'(7, 5) x=2 P'(2 × 2 - 7, 5) = P'(-3, 5)

Note : r1 o r2 z r2 o r1

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Example 2. If r1 represents the reflection on x-axis and r2 represents the reflection on y
Solution: axis, find the image of the point P(2, 5) under the combined transformation
r2 o r1
Example 3.
Solution: Here, r1 = reflection on x-axis

r2 = reflection on y-axis

X - axis and Y - axis intersect at the origin and the angle between them is 90°.

Hence, the combined reflection is rotation of (2 × 90) = 180° about O.

P(2, 5) R[(0, 0),180°] P'(-2, -5).

Alternatively

r2o r1 means first reflection on r1 (x - axis), then reflection on r2 (y - axis)

P (2, 5) x-axis P'(2, -5)

P' (2, - 5) y-axis P" (-2, -5)

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

Here, A(1, 2), B(4, -1) and C are the vertices of ∆ABC.

Reflecting the ∆ABC on line x = 5, we get,

P(x, y) x=h P'(2h – x, y) x=5
Y
A(1, 2) x=5 A'(2 × 5 - 1, 2)
C P" C'
= A'(9, 2)

B(4, -1) B'(2 × 5 - 4, -1) A A'
X
= B'(6, -1) X'
O B B'
C(2, 5) C'(2 × 5 - 2, 5) P y = –2
B"
= C'(8, 5)
Y' A"
Again reflecting ∆A'B'C' on y = -2, we get.
C"
A'(9, 2) y = –2 A"(9. 2 × (-2) - 2)

= A" (9. -6)

B'(6, -1) B"(6, 2 ×(-2) + 1)

= B"(6, - 3)

C'(8, 5) C"(8, 2 × (-2)- 5)

= C"(8, - 9)

∆ABC, ∆A'B'C' and ∆A"B"C" are plotted in the graph. The lines x = 5 and
y = -2 intersect at point P(5, -2). The angle between these two lines is 180°

Hence, the combined reflection is rotation of 180° about centre P(5, -2)

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Example 4. A triangle with vertices P(3, 5), Q(4, 1) and R(1, 0) is reflected successively
on the lines x =2 and x =6. Find by stating the coordinates and graphically
represent the images under transformations. Also write a single
transformation representing the above combined transformation.

Solution: Here, P(3, 5), Q(4, 1) and R(1, 0) Y x=6
are the vertices of ∆PQR. x=2
P"
Reflecting the ∆PQR on line x = 2, P' P
we get, P(x, y)x = h P'(2h - x, y)
P(3, 5) x = 2 P"(2 × 2 - 3, 5) X' Q' R Q"
O Q R' R" X
= P'(1, 5)
Q(4, 1) x = 2 Q'(2 × 2 - 4, 1)

= Q'(0, 1) Y'
x = 2 R'(2 × 2 - 1, 0)
R(1, 0)
=R'(3, 0)

Again, reflecting ∆P'Q'R' on the line x = 6, we get

P'(1, 5) x = 6 P"(2 × 6 - 1, 5) = P"(11, 5)

Q(0, 1) x = 6 Q"(2 × 6 - 0, 1) = Q" (12, 1)

R'(3, 0) x = 6 R"(2 × 6 - 3, 0) = R"(9, 0)

Since the axes of reflections are parallel, the combined reflection is a
translation, the translation is two times the displacement between the

( ) ( )axes. Here, translation T is given by 6-2 8
T1 = 2 0 = 0 .

Combined Transformation of Reflection and Translation

( )Let P(x, y) be any point on the plane, T = a be a translation vector. Then,
b
a
P(x, y) T= b P'(x + a, y +b)

Again P' is reflected on x-axis, we get,

P'(x + a, y + b) x-axis P" (x + a, - y - b)

This transformation is translation followed by reflection. Similar, transformation result can
be derived for reflection followed by translation.

i.e. P(x, y) x-axis P'(x, - y) T = ba P"(x + a, -y + b).

Composition of Reflection and Rotation

A point or an object once reflected can further be rotated to get a new image. This is also a
type of combined transformation.

Let P(x, y) be any point on the plane. Then P is reflected on x - axis, we get,

P(x, y) P'(x, - y)

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Again P' is rotated through -90° about origin.

P'(x, -y) P"(-y, -x)

Similar, combination of transformation results can be derived for rotation followed by
reflection.

Let, Rx = reflection on x-axis
Ry = reflection on y - axis
Ry = x = reflection on line y = x
Ry = -x = reflection on line y = - x
R1 = rotation through 90° about origin
R2 = rotation through - 90° about origin
R3 = Rotation through 180° about origin.
Then, discuss about the following combination of transformation.

(i) Rx o R1 (ii) Ry o R2 (iii) Ry ox R1 etc.

Example 5. Let Rx denote reflection on x - axis and R1 denote, the rotation of +90° about
Solution: the origin. Find the image of ∆PQR with P(2, 3), Q(6, 7) and R(0, 3) under
R1oRx .

P(2, 3). Q(6, 7) and R(0, 3) are the vertices of ∆PQR. R1 o Rx means Rx followed
by R1, where

R1 = rotation of +90° about O.

Rx = reflection on x-axis.

Under reflection on x-axis, we get,

P(x, y) x-axis P'(x, –y)

P(2, 3) P'(2, -3)

Q(6, 7) Q'(6, -7)

R'(0, 3) R"(0, -3)

Again, ' P'Q'R' is rotated through +90° about origin.
P(x, y) R[(0, 0), +90°] P'(–y, x)

P'(2, –3) P"(3, 2)

Q'(6, -7) P"(7, 6)

R'(0, -3) P"(3, 0)

Exercise 11.3

Short Question
1. (a) Write the combined reflection of reflection in x = 2 followed by reflection in x = 6

(b) Write the combined reflection of reflection in x = 0 followed by reflection in y = 0

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(c) Write the combined reflection of reflection in x = 4 followed by reflection in y = -2

(d) Write the combined reflection for reflection an y = x followed by reflection x - axis.

(e) What is the combined reflection when the axes of reflection are (i) parallel (ii)
intersecting at a point?

2. Let Rx = Reflection in x - axis
Ry = Reflection in y - axis
R1 = Reflection in line y = x
R2 = Reflection on y = -x

( )T =2 , translation vector.
3

r1 = rotation of +90° about origin.

r2 = rotation of -90° about origin

r3 = rotation of 180° about origin.

Find the image of points under the following combined transformations:

(a) RxoRy (2, 3) (b) RyoRx(2, 3) (c) R1oR2(2, 4)
(d) R2oR1(6, 2) (e) R1oT(2, 4) (f) ToRx(4, -4)
(g) r1or2(4, 5) (h) r2or1(-4, 5) (i) r1oRx(6, 8)
(j) T1oR2(2, 4) (k) T1o r2(-4, -5) (l) ToRx(-5, -5)
Long Questions

3. A(- 4, 0), B(-6, 2) and C(-4, 3) are the vertices of ∆ABC. The triangle ABC is reflected
successively on the line x = - 3 and x = 1. Find the final image and describe the single
transformation equivalent to the combination of these transformation.

4. If R is the reflection on the line y = x and Ry is the reflection the line x = 0. Then what
does RoRy represent ? By using the combined transformation of this result, find the
image of ∆ ABC whose vertices are A(3, 4), B(7, 0), and C(4, 0). Draw graph of ∆ABC
and its image on the same graph paper.

5. On a graph paper draw ∆ABC having the vertices A(5, 4), B(2, 2) and C(5, 2). Find the
image of ∆ABC by stating coordinates and graphing them after successive reflections on
x-axis following by reflection on the line y = x.

6. A triangle with vertices P(1, 4), Q(4, 1) and R(7, 5) is reflected successively in
y- axis followed by reflection on the line y = x. Find the final image of ∆PQR and
present the object and images in the same graph paper.

7. A triangle with the vertices A(1, 2), B(4, -1) and C(2, 5) is reflected successively in the
line x = 4 and y = -2. Find by stating coordinates and graphically represent the images
under there transformations given by the combination of these transformation.

8. A(4, 5), B(-2, -2) and C(4, 8) are the vertices of ∆ ABC. ∆ABC is rotated through +90°
about the origin followed by the reflection in x-axis. State the single transformation.
Draw ∆ABC and its images in the same graph paper.

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9. If R1 denotes rotation of +90° about origin and R2 denotes rotation of -270° about the
origin. What does R2oR1 represent? By using R2oR1, transform quadrilateral ABCD with
A(-4, 0), B(-6, 2), C(-2, 4) and D(-1, 1). Present the quadrilateral ABCD and its image on

the same graph paper.

10. O(0. 0) A(2, 0), B(3, 2) and C(1, 2) are the vertices of quadrilateral OABC. Translate the
0
quadrilateral by translation vector 2 .

Reflect the image so formed on the line x = 3. Represent the images and object in the

same graph.

11. A triangle having vertices A(2, 5), B(-1, 5) and C(4, 1) is rotated through +90°

about the origin. The image obtained is reflected on the line x = 0. Find the vertices

of image triangles. Show all the triangles in the same paper and also write the single

transformation to represent these two transformations.

12. State the single transformation equivalent to the combination of reflections on the

x-axis and y-axis respectively. Using this single transformation find the coordinates

of the vertices of the image of ∆ABC having vertices A(2, 3), B(3, -4) and C(-1, 2). Also

draw the object and its image on the same graph.

13. Draw ∆ABC having the vertices A(2, 0), B(3, 1) and C(1, 1) on a graph paper. It is rotated

about the origin through -90° and present ∆A'B'C' on the same graph paper. Then '

A'B'C' is reflected on the line x – y = 0 and plot 'A"B"C" on the same graph paper. Write

down the coordinates of vertices of ∆A"B"C".

Project Work

14. Search some examples of combination of reflection and their use in daily life. Prepare
a report and present in your classroom.

1. (a) T = 8 (b) R[(0, 0), 180°] (c) R[(4, -2), 180°]
0 (e) (i) Translation (ii) Rotation

(d) R[(0, 0), -90°]

2. (a) (-2, -3) (b) (-2, -3) (c) (-2, -4) (d) (-6, -2)

(e) (7, 4) (f) (6, 7) (g) (4, 5) (h) (-4, 5)

(i) (-8, 6) (j) (-2, 1) (k) (-3, 7) (l) (-3, 8)

3. 8 , A'(4, 0), B'(2, 2), C'(4, 3) 4. R[(0, 0), -90°], A'(4, -3), B'(0, -7), C'(0, -4)
0

5. R[(0, 0), +90°, A'(5, -4), B'(2, -2), C'(5, -2), A"(-4, 5), B"(-2, 2), C"(-2, 5)

6. P'(-1, 4), Q'(-4, 1), R'(-7, 5), P"(4, -1), Q"(1, -4), R"(5, -7)

7. A'(7, 2), B'(4, -1), C'(4, 8), A"(7, -6), B"(4, -3), C"(4, -12)

8. A'(-5, 4), B'(2, -2), C'(-8, 4), A"(-5, -4), B"(2, 2), C"(-8, -4), y = x

9. A'(0, -4), B'(-2, -6), C'(-4, -2), A"(4, 0), B"(6, -2), C"(2, -4), R[(0, 0), -180°]

10. O'(6, 2), A'(2, 2), B'(3, 4), C'(1, 4), O"(6, 2), A"(4, 2), B"(3, 4), C"(5, 4)

11. A'(-5, 2), B'(5, - 1), C'(-1, 4), A"(5, 2), B"(5, -1), C"(1, 4)

12. R[(0, 0), 180°], A'(-2, -3), B'(-3, 4), C'(1, -2)

13. A'(0, -2), B'(1, -3), C'(1, -1), A"(-2, 0), B"(-3, 1), C"(-1, 1)

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11.5 Combination of Enlargement and Enlargement

An object once enlarged (or reduced) can again be enlarged. The single enlargement (or
reduction) is equivalent to these two transformations is the combination of enlargements
and enlargement. There are two cases in the combination of two enlargements (or reduction)

A. When the centres of enlargements are same Y A"
Let AB be a given line segment and E1[O, k1] be an A'
enlargement with centre O and scale factor k1. Let A'B' be
the image of a given line segment AB under enlargement

E1. Then we write A'B' = k1AB A B"

Again let E2 [O, k2] be another enlargement with the B B'
same centre O and scale factor K2. Let A"B" be the image
of A'B' under enlargement E2. Then we write X' O X

A"B" = k2A'B' Y'

Finally, we get, A"B" = k1 A'B' = k1. k2 AB

Here, A"B" is the image of AB under the enlargement E[O, k1 k2].

Hence, we conclude that an enlargement with centre O and scale factor k1, followed
by another enlargement with the same centre O and scale factor k2 is equivalent to the
enlargement with the same centre with scale factor k1.k2 .

Note :
(i) If E1[(0, 0), k1] and E2[(0, 0), k2] are two enlargements, the combined enlargement is

E[(0, 0), k1k2].

(ii) If E2[(a, b), k1] and E2[(a, b), k2] are two enlargements, the combined enlargement is
E[(a, b), k1k2].

Example 1. If P(3, 2), Q(9, 2) R(9, 6) and S(3, 6) are the vertices of square PQRS. Find the
Solution: image of square PQRS under an enlargement with the centre at origin and
scale factor 2 followed by an enlargement with centre at origin and scale
factor 3.

The enlargement E1[0, 2] followed by enlargement E2 [0, 3] is equivalent to
the single enlargement E[0, 2 × 3] = E[0, 6] with the same centre O and scale
factor 6.

Then, the image of square PQRS due to E[0, 6] is P'Q'R'S'

Object Image

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

P(3, 2) P'(6 × 3, 6 × 2) = P'(18, 12)

Q(9, 2) Q'(6 × 9, 6 × 2) = R'(54, 12)

R(9, 6) R' (6 × 9. 6 × 6) = R'(54, 36)

S(3, 6) S'(6 × 3, 6 × 6) = S'(18, 36)

squares PQRS squares P'Q'R'S'.

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B. When the centres of enlargements are different. A" B"
Let us take a line segment AB and E1[O, k1] be the enlargement
with centre P and scale factor k1. A' B'
Then, A'B' = kAB
AB
Again, let E2[O', k2], then A"B" = k2A'B' = k2 k1AB. O O" O'
Join AA" and BB"
Fig. Combined enlargement with
Let the point of intersection of AA" and BB" be O". different centres.

Hence, A"B" is the image of AB under the enlargement
E[O", k1k2] with the third centre O" and scale factor k1k2 .

Example 2. Find the image of ∆ABC with vertices A(1, 1), B(2, 4) and C(2, -1) under
Solution: enlargement E1[(2, 3), 2] followed by another enlargement E2 [(3, 3), 3]

Under enlargement E1 [(2, 3), 2]
We have, P(x, y) E[(a, b), k] P'(kx - ka + a, ky - kb + b)

Now, A(1, 1) E1[(2, 3), 2] A'(2 × 1 - 2 × 2 + 2, 2 × 1 - 2 × 3 + 3) = A'(0, -1)

B(2, 4) B'(2 × 2 - 2 × 2 + 2, 2 × 4 - 2. 3 + 3) = B'(2, 5)

C(2, -1) C'(2 × 2 - 2 × 2 + 2, 2 × (-1) - 2 × 3 + 3) = C'(2, -5)

Again under E2 [(3, 3), 3], ∆A'B'C' is enlargement.

A'(0, -1) A"(3 × 0 - 3 × 3 + 3, 2 × -1 - 3 × 3 + 3) = A' (-6, -8)

B'(2, 5) B"(3 × 2 - 3.3 + 3, 3 × 5 - 3 × 3 + 3) = B"(0, 9)

C'(2, -5) C"(3 × 2 - 3 × 3 + 3, 3 × (-5) - 3 × 3 + 3) = C" (0, -21)

Combination of a transformation with enlargement :
Here, Combination of transformations may be

- Reflection and Enlargement
- Rotation and Enlargement
- Translation and Enlargement and other.

Combination of Reflection and Enlargement
Let P(x, y) be a point an the plane.

Let P be reflected on y-axis. P'(-x, y)
P(x, y) y-axis

Again, let P'(-x, y) be enlarged under E[(0, 0), k]. Then we write

P'(-x, y) P"(-kx, ky)

So, P"(-kx, ky) is the image of P(x, y) under reflection on y-axis followed by enlargement
E[(0,0), k]

Combination other reflections and enlargement can be derived in the similar ways.

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Combination of Rotation and Enlargement
Take a point P(x, y) on the plane.

Let P(x, y) be rotated through +90° about origin.

P(x, y) P'(-y, x)

Again let P' (-y, x) be enlarged under E [(0,0), k], we get,

P'(-y, x) P" (-ky, kx)

P" (-ky, kx) is the image of P(x, y) under rotation +90° about O combination of other rotations
followed by E[(0, 0), 3] with enlargement can be derived in similar ways.

Combination of Translation and Enlargement
Let P(x, y) be a point on the plane.

( )T =a be a translation vector then image of P under T is given by
b

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

Again P' is enlargement under E[(0, 0), k], we get,

P'(x + a, y +b) P" [k(x + a), k(y + b)]

P"[k (x + a), k(y + b)] is the image of P(x, y) under translation followed by enlargement E[(0, 0), k]

Example 3. A point P(7, 8) is reflected on the line y = -x. The image so obtained is
Solution: enlarged under E[(0,0), 2]. Find the coordinates of the images.

Given point is P(7, 8)

Under the reflection about the line y = - x, we get
P(x, y) y = -x P'(-y, -x)

P(7, 8) P'(-8, - 7)

Again P(-8, -7) is enlarged under E[(0,0), 2],

P'(-8, -7) P"[2 (-8), 2(-7)] = P"(-16, -14)

Example 4. A point P(-2, - 3) is rotated through +90° about origin. The image so obtained
Solution: is enlarged under E[(o,o), 2].

Example 5. Given point is P(-2, - 3),
Solution:
Under rotation through +90° about origin, we get,

P(x, y) P'(-y, x)

P(-2, -3) P'(3, -2)

Again, P'(3, 2) is enlarged under enlargement E[(o, o), 2]

P'(3, -2) P"(6, - 4)

A(1, 0), B(0, 2), and C(2, 1) are the vertices of ∆ABC. Enlarge ∆ABC under
E[(o,o), 3] and the image so formed is again reflected on y-axis. Draw the
graph of ∆ABC and its images on the same graph paper.

Here, A(1, 0), B(0, 2), and C(2, 1) are the vertices of ∆ABC. ∆ABC is enlarged

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under E[(0,0), 3] we get,
P(x, y) E[(0, 0), k] P'(kx, ky)
A(1, 0), o A'(3 × 1, 3 × 0) = A'(3, 0)
B(0, 2) o B'(0, 3 × 2) = B'(0, 6)
C(2, 1) o C'(3 × 2, 3 × 1) = C'(6, 3)
Again, ∆A'B'C' is reflected on Y - axis.
A'(3, 0) o A"(-3, 0)
B'(0, 6) o B"(0, 6)
C'(6, 3) o C"(-6, 3)

Y
B" B'

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

Y'

Exercise 11.4

Very Short Questions

1. (a) Transform point A(4, 5) under enlargement E[(0, 0), 2].

(b) Transform point G(2, 2) under enlargement E[(0, 0), -2].

(c) Find the image of P(5, 3) under enlargement E[(1, 1), 2].

(d) Find the image of M(2, -2) under enlargement E[(0, 4), - 2].

2. (a) Let E1[(0,0)2] and E2 [(0,0), 3] what is the single enlargement due to E1oE2?

(b) Let E1 [(0,0), 4] and E2 (0, 0), 1 , what is the single enlargement due to E2E2?
2

(c) Let E1 be [(2, 1), 3], and E2 be [(2, 1), 2]. Write the single transformation equivalent
to E2E1.

Short Questions

3. (a) Let E1[(0,0), 2] and E2 [(0, 0), 3] be two enlargements. Find the image of P(2, 5)
under E2oE1 .

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(b) Let E be [(1, 1), 3] and E2 be [(1, 1), 2]. Find the image of P(2, 6) under E2oE1.

(c) Find the image of N(2, 3) under enlargement E1[O, 2] followed by enlargement
1
E2 O, 2 2 .

(d) A point P(3, k) is the first transformed by E1[O, 2] and by E2 O, 3 so that the final
image is [9, 12], find the value of k. 2

4. Let E1 = enlargement about (0, 0) with scale factor 2.

E2 = enlargement about (0, 0) with scale factor 3 .
2
R1 = Rotation of + 90° about 0.

R2 = Rotation of 180° about 0.

[ ]T = 2 , translation vector.
3

Find the image of the following points the combined transformation given below:

(a) E1 o E2 (2, 4) (b) E1 oT (4, 5) (c) E2 o T(2, 5)
(d) ToE1 (-2, -3) (e) R1 o E1 (4, 5) (f) R2 oE2 (1, 2)
(g) E2 o R1 (4, 4)
Long Questions

5. P(3, 0) , Q(4, 2), R(2, 4), and S(1, 2) are the vertices of parallelogram PQRS. Draw PQRSon

graph paper. Enlarge the parallelogram under E1 [(0,0), 2] followed by E2 (0, 0), 1 1 .
2

Find the images P'Q'R'S' and P"Q"R"S". Draw P'Q"R'S' and P"Q"R"S" on the same graph paper.

6. The vertices of ∆ABC are A(2, 3), B(4, 5), and C(6, 2). If E1 [(0, 0),2] and E2 [(0, 0)2] are
two enlargements, find the coordinates of the image of ∆ABC under the enlargement
E2oE1. Draw both triangles on the same graph.

7. Find the image of ∆ABC with the vertices A(2, 1), B(3, 5) and C(5, 4) under enlargement
E1 [(0, 0), 3/2] followed by another enlargement E2 [(0,0), -2]. Draw ∆ABC and its images
on the same graph paper.

8. P(3, 0), Q(4, 2), R(2, 4) and S(1, 2) are the vertices of parallelogram PQRS. Enlarge the
parallelogram PQRS E1 [(1, 2), 2] and by enlargement E2 [(1, 2), 3]. Draw both the object
triangle and image triangle on the same graph paper.

9. A(2, 5), B(-1, 3), and C(4, 1) are the vertices of ∆ABC. Find the coordinates of the
image of the ∆ABC under the rotation of positive 90° about the origin followed by the
enlargement E[(0,0), 2]

10. Let E denote enlargement about centre (-3, - 4) with scale factor 2 and R denote a
reflection of the line y = x. ∆ABC with vertices A(2, 0), B(3, 1), and C(1, 1) is mapped
under the combined transformation RoE. Find the image of ∆ABC and draw both figures
on the same graph.

11. A(2, 0), B(3, 1), and C(1, 1) are the vertices of a triangle ABC. Find the coordinates of
the vertices of the triangle ABC under the reflection on the line x = y following the
enlargement E[(-3, -4), 2]

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-2

( )12. ∆PQR having the vertices P(3, 4), Q(2, 1), and R(4, 2) is translated by T = 3 . The
image so formed is enlarged by E[(0, 0), 2]. Writing the coordinates of the vertices of
images thus obtained of the ∆PQR and its images in the same graph paper.

13. The vertices of ∆ABC are A(1, 1), B(3, 1), and C(2, 3) ∆A'B'C' is obtained by rotating
∆ABC about the origin half turn. The image formed is enlarged about the same centre
with scale factor 2 to get ∆A"B"C". Find the coordinates of the final image and represent
all three triangles on the same graph paper.

14. M(3, 4), N(1, 1), and P(4, 1) are the vertices of ∆MNP. Find the image of ∆MNP. Under
the enlargement with centre (1, -1) and scale factor -2 followed by the rotation about the
origin through negative quarter turn. Also show the images on the same graph paper.

1. (a) A'(8, 10) (b) G'(-4, -4) (c) P'(9, 5) (d) M'(-4, 16)

2. (a) E[(0, 0), 6] (b) E[(0, 0), 2] (c) E[(0, 0), 6]

3. (a) P'(12, 30) (b) P'(7, 31) (e) N'(10, 15) (d) k = 4

4. (a) (6, 12) (b) (12, 16) (c) (8, 16) (d) (-2, -3)

(e) (-10, 8) (f) - 3 , -3 (g) (-6, 6)
2

5. P'(6, 0), Q'(8, 4), R'(4, 8), S'(2, 4), P"(9, 0), R"(6, 12), Q"(12, 6), S"(3, 6)

6. A'(8, 12), B'(16, 20), C'(24, 8) 7. A"(-6, -3), B"(-9, -15), C'(-15, -12)

8. P"(13, -10), Q"(19, 2), R"(7, 14), S"(1, 2)

9. A'(-5, 2), B'(-3, -1), C'(-1, 4), A"(-10, 4), B"(-6, -2), C"(-2, 8)

10. A'(7, 4), B'(9, 6), C'(5, 6), A"(4, 7), B"(6, 9), C"(6, 5)

11. A'(0. 2) . B'(1, 3), C'(1, 1), A"(3, 8), B"(5, 10), C"(5, 6)

12. P'(1, 7), Q'(0, 4), R'(2, 5), P"(2, 14), Q"(0, 8), R"(4, 10)

13. A'(-1, -1), B'(-3, -1), C'(-2, -3), A"(-2, -2), B"(-6, -2), C"(-4, -6)

14. M'(-3, -11), N'(1, -5), P"(-5, -5), M"(-11, 3), N"(-5, -1), P"(-5, 5)

11.6 Inversion Transformation and Inversion Circle

Introduction

Let us discuss answer of the following questions in group.

(a) What is the equation of a circle with centre at the origin and radius 5 units?

(b) Find the centre and radius of the circle whose equation is C
OB
(x - 4)2 + (y + 3)2 = 25 ? A
(c) In the given circle O is the centre of a circle, find the value of ‘ACB.

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