Vedanta Excel in Opt. Mathematics Book 7 Final (2078)

Trigonometry

3.8 Operation of Trigonometric Ratios

Addition, subtraction, multiplication, and division are basic operations on
trigonometric ratios as in algebra.

Addition

In algebra, the coefficients of like terms are added. In the similar manner,
trigonometric ratios are added.

For examples:

1. x + x = 2x sinT + sinT = 2sinT

2. 2x + 3x = 5x 2sinT + 3sinT = 5sinT

Subtraction

We can subtract the trigonometric ratios as we subtract in algebra.

For example:

1. 2x – x = x 2sinT – sinT = sinT

2. 7y – 5y = 2y 7tanT – 5tanT = 2tanT

Multiplication and division

For multiplication trigonometric ratios also follow the laws of indices. Let’s review
am 1
laws of indices once: (i) am.an = am + n (ii) an = am – n (iii) a–m = am (iv) (am)n = amn

(a) x2 . x = x3, sin2T. sinT = sin3T

(b) x3 = x sin3T = sinT
x2 sin2T

Note: sin2T = (sinT)2

Some Basic Algebraic formulas

S.No. Formula Expanded form Factorized form
1. (a + b)2 a2 + 2ab + b2 (a + b) (a + b)
2. (a – b)2 a2 – 2ab + b2 (a – b) (a – b)
3. a2 – b2 – (a + b) (a – b)
4. a2 + b2 (a + b)2 – 2ab –
or
5. (a + b)3 (a – b)2 + 2ab (a + b) (a + b) (a + b)
a3 + 3a2b + 3ab2 + b3

or

a3 + b3 + 3ab (a + b)

vedanta Excel in Additional Mathematics - Book 7 53

Trigonometry a3 – 3a2b + 3ab2 – b3 (a – b) (a – b) (a – b)

6. (a – b)3 or (a + b) (a2 – ab + b2)
(a – b) (a2 + ab + b2 )
7. a3 + b3 a3 – b3 – 3ab (a – b)
8. a3 – b3 (a + b)3 – 3ab (a + b)
(a – b)3 + 3ab (a – b)

Worked out Examples

Example 1. Add the following :
Solution:
(a) sinT + 4 sinT (b) 2 cos2T + 3 cos2T + 6 cos2T

(c) 4 cosec3T + secT + 5 cosec3T

(a) Here, sinT + 4 sinT

= sinT (1 + 4)

= sinT (5)

= 5 sinT

(b) Here, 2 cos2 T + 3 cos2 T + 6 cos2 T

= cos2 T (2 + 3 + 6)

= cos2 T (11)

= 11 cos2 T

(c) Here, 4 cosec3 T + sec T + 5 cosec3 T

= 4 cosec3 T + 5 cosec3 T + sec T

= cosec3 T (4 + 5) + sec T

= cosec3 T (9) + sec T

= 9 cosec3 T + sec T

Example 2. Subtract :
Solution: (a) 6 tanT – 4 tanT
(b) 5 sinT + 2 cosT from 7 sinT – 3 cosT
(a) Here, 6 tanT – 4 tanT

= tan T (6 – 4)
= tan T (2) = 2 tanT
(b) Here, 5 sin T +2 cos T from 7 sin T – 3 cos T
= (7 sin T – 3 cos T) – (5 sin T + 2 cos T)

54 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

= 7 sin T – 3 cos T – 5 sin T – 2 cos T
= 7 sin T – 5 sin T – 3 cos T – 2 cos T
= 2 sin T – 5 cos T

Example 3. Simplify :
Solution: (a) 4sin3T + 2cos2T – 2tanT + 4tanT – 5cos2T + 2sin3T
(b) cot2T – 3tanT – 4cot2T + 4tanT + sec2T – 7sec2T
(a) Here, 4sin3T + 2cos2T – 2tanT + 4tanT – 5cos2T + 2sin3T

= 4sin3T + 2sin3T + 2cos2T – 5cos2T – 2tanT + 4tanT
= 6sin3T – 3cos2 T + 2tanT
(b) Here, cot2T – 3tanT – 4cot2T + 4tanT + sec2T – 7sec2T
= cot2T – 4cot2T – 3tanT + 4tanT + sec2T – 7sec2T
= – 3cot2T + tanT – 6sec2T

Example 4. Find the product of :
Solution:
(a) sin2T × sin3T (b) (tanT – cotT) (tanT + cotT)

(a) Here, sin2T × sin3T

= sin5T

(b) Here, (tanT – cotT) (tanT + cotT)

= tan2T + tanT . cotT – cotT . tanT – cot2T

= tan2T – cot2T

Example 5. Divide :
Solution:
(a) 24 sin2T.cos3T by 8 sinT.cosT

(b) (sec2T – tan2T) by (secT + tanT)

(a) Here, 24 sin2T.cos3T by 8 sinT.cosT

= 24 sin2T . cos3T
8 sinT . cosT

= 3 sinT . cos2T

(b) Here, (sec2T – tan2T) by (secT + tanT)

= (sec2T – tan2T)
(secT + tanT)

= (secT – tanT) (secT + tanT)
(secT + tanT)

= secT – cotT

vedanta Excel in Additional Mathematics - Book 7 55

Trigonometry

3.9 Factorization

Study the following illustrated examples.

3 u 5 = 15 (3 and 5 are the factors of 15)

2 u 2 u 3 = 12 (2, 2 and 3 are the factors of 12)

Similarly,

2 u x = 2x (2 and x are the factors of 2x)

x u x2 = x3 (x and x2 are the factors of x3)

2x (x + 1) = 2x2 + 2x (2x and (x + 1) are the factors of 2x2 + 2x)

(x + y) (x – y) = x2 – y2 (x + y and x – y are the factors of x2 – y2)

Thus, when two or more algebraic expressions are multiplied, the result is called
the product and each expression is called the factor of the product.

The process of finding out factors of an algebraic expression is known as factorisation.

Example 6. Factorise :
Solution: (a) 3 sinT . cosT + 2 cos2T . sin3T (b) sec2T – cot2T
(a) Here, 3 sinT . cosT + 2 cos2T . sin3T

= sinT . cosT(3 + 2 cosT . sin2T)
(b) Here, sec2T – cot2T

= (secT + cotT) (secT – cotT)

Example 7. Expand: (sinT + cosT)2
Solution: Here, (sinT + cosT)2

= sin2T + 2 sinT . cosT + cos2T

Exercise 3.3

Short Questions :
1. Add the following :

(a) 2 tan T + 5 tanT
(b) cos2 T + 3 cos2 T
(c) sec3 T + 9 cosec2 T + 3 sec3 T – 9 cosec2 T
(d) sec2 T + 7 tan2 T + 5 sec2 T – 3 tan2 T
(e) sin T + cos T + 2 sin T + 4 cos T

56 vedanta Excel in Additional Mathematics - Book 7

2. Subtract : Trigonometry
(a) 5 sin T from 15 sin T
(b) 3 sin T – 5 cos T from 9 sin T – 2 cos T 57
(c) 5 tan2 T + 9 tan T + 2 from 9 tan2 T – 18 tan T + 7
(d) 7 sin2 T + 6 cos2 T from 12 sin2 T – 4 cos2 T
(e) 4 cosec2 T + 8 cot2 T from 8 cosec2 T – 5 cot2 T

3. Multiply :
(a) 80 tan T by 4 tan2 T
(b) 4 cos2 T . sin2 T by 2 cos T . sin T
(c) (tan T + cot T) (tan T – cot T)
(d) (1 + cos T) (1 – cos T)
(e) (sin T – cos T) (sin2 T + sin T . cos T + cos2 T)

4. Divide :
(a) 27 sec T by 9 sec T
(b) 24 cos2 T by 12 cos T
(c) cot8 T ÷ cot2 T
(d) (tan4 T – cot4 T) ÷ (tan2T + cot2 T)
(e) (cosec2 T – cot T) ÷ (cosec T – cot T)

5. Factorise the following :
(a) cos2 T + 2 cos T . sin2 T
(b) sin3 D . cos3 D + sin4 D . cos2 D
(c) sin2 T – sec2 T
(d) cot3 T – tan3 T
(e) sec4 T – tan4 T
(f) tan3 T + cot3 T

6. Expand the following :
(a) (tan T + cot T)2
(b) (cot T – cosec T)2
(c) (tan T – 2)2
(d) (cot T – cosec T)3
(e) (sin T + cos T)3

vedanta Excel in Additional Mathematics - Book 7

Trigonometry

1. (a) 7tanT (b) 4cos2T (c) 4sec3T (d) 6sec2T + 4tan2T

(e) 3sinT + 5cosT 2.(a) 10sinT (b) 6sinT + 3cosT

(c) 4tan2T – 27tanT + 5 (d) 5sin2T – 10cos2T (e) 4cosec2T – 13cot2T

3. (a) 320tan2T (b) 8cos3T . sin3T (c) tan2T – cot2T (d) 1 – cos2T

(e) sin3T – cos3T 4.(a) 3 (b) 2cosT (c) cot6T

(d) tan2T – cot2T (e) cosecT + cotT 5.(a) cosT(cosT + 2sin2T)

(b) sin3D . cos2D(cosD + sinD) (c) (sinT + secT) (sinT – secT)

(d) (cotT – tanT) (cot2T + tanT . cotT + tan2T)

(e) (secT + tanT) (secT – tanT) (sec2T + tan2T)

(f) (tanT + cotT) (tan2T – tanT . cotT + cot2T)

6. (a) tan2T + 2tanT . cotT + cot2T (b) cot2T – 2cotT . cosecT + cosec2T

(c) tan2T – 4tanT + 4 (d) cot3T – 3cot2T . cosecT + 3cotT . cosec2T – cosec3T

(e) sin3T + 3sin2T . cosT + 3sinT . cos2T + cos3T

3.10 Fundamental relation between trigonometric ratios

Let us consider a right angled triangle PQR with right angle at Q and reference

angle PRQ = T. P

In the figure Perpendicular (P)

hypotenuse (h) = PR Hypotenuse (h)
perpendicular (p) = PQ
base (b) = RQ

From the combination of above three sides and R T Q
reference angle we have the following six ratios. Base (b)

(a) sin T = p (b) cos T = b (c) tan T = p
h h b

(d) cosec T = h (e) sec T = h (f) cot T = b
p b p

(a) Reciprocal Relations

We know,

sinT = p = 1 × p = 1 × p = 1 = 1
h h 1 h h cosecT
p

58 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

? sinT = 1 ........................................................... (i)
cosecT

or, cosecT × sinT = 1 .................................................... (ii)

or, cosecT = 1 .......................................................... (iii)
sinT

(i), (ii), and (iii) are the formula for the reciprocal relation between sinT and

cosecT.

Again, we know

cosT = b = 1 × b = 1 = 1
h h h secT
b
So, cosT = 1 .............................................................. (iv)
secT

or, cosT × secT = 1 ........................................................ (v)

or, secT = 1 ............................................................... (vi)
cosT

(iv), (v), and (vi) are the formula for the reciprocal relation between cosT and

secT.

Similarly,

tanT = p = 1 × p = 1 × p = 1 = 1
b b 1 b b cotT
p
So, tanT = 1 .............................................................. (vii)
cotT

or, tanT × cotT = 1 ......................................................... (viii)

or, cotT = 1 ............................................................... (ix)
tanT

(vii), (viii) and (ix) are the formula for the reciprocal relation between tanT and

cotT

Table showing reciprocal relations:

sinT × cosT = 1 cosT × secT = 1 tanT × cotT = 1

sinT = 1 cosT = 1 tanT = 1
cosecT secT cotT

cosecT = 1 secT = 1 cotT = 1
sinT cosT tanT

vedanta Excel in Additional Mathematics - Book 7 59

Trigonometry

(b) Quotient Relations

We know,

sinT = p and cosT = b
h h

So, we if divide sinT by cosT , we get,

sinT p p h p
cosT h h b b
= b = × = = tanT

h

? tanT = sinT ......................................................... (x)
cosT
Similarly,

Dividing cosT by sinT, we get,

csionsTT = b b h b
h h p p
p = × = = cotT

h

? cotT = csionsTT ......................................................... (xi)
(x) and (xi) are called the quotient relations of tanT and cotT respectively.

Hence, we have,

tanT = sinT
cosT

cotT = cosT
sinT

(c) Trigonometric Ratios from Pythagoras Theorem
From Pythagoras theorem we have,

h2 = p2 + b2

dividing by 'h2' on the both the sides, we get

h2 = p2 + b2
h2 h2 h2

or, 1= p 2+ b2
h h

or, p 2+ b 2 =1
h h

or, (sinT)2 + (cosT)2 = 1

or, sin2T + cos2T = 1 ............. (xii)

60 vedanta Excel in Additional Mathematics - Book 7

we also have, Trigonometry

sin2T = 1 – cos2T [On transposition] 61

or, cos2T = 1 – sin2T

? sinT = 1 – cos2T

? cosT = 1 – sin2T

Similarly, in the same right angle triangle,

dividing by 'b2' on both sides, we get,

h2 = p2 + b2
b2 b2 b2

or, h 2= p 2+1
b b

or, h 2– p 2 =1
b b

or, (secT)2 – (tanT)2 = 1

or, sec2T – tan2T = 1 ................... (xiii)

we also have,

sec2T = 1 + tan2T [On transposition]

or, tan2T = sec2T – 1

or, tanT = sec2T – 1

and, sec2T =1 + tan2T

or, secT = 1 + tan2T

On the same way,

In the same right angled triangle,

dividing by 'p2' on both the sides, we get,

h2 = p2 + b2
p2 p2 p2

or, h 2=1+ b2
p p

or, h 2– b 2 =1
p p

or, (cosecT)2 – (cotT)2 = 1

or, cosec2T – cot2T = 1

we also have,

cosec2T = 1 + cot2T

vedanta Excel in Additional Mathematics - Book 7

Trigonometry

or, cot2T = cosec2T – 1
? cosecT = 1 + cot2T
? cotT = cosec2T – 1

3.11 Trigonometric Identities

The relations which involve trigonometric ratios are called trigonometric identities.

Example : (i) sin2T + cos2T = 1

(ii) sinT . cosecT = 1

(iii) tanT = sinT , etc.
cosT

These trigonometric identities are satisfied by every value of T.

3.12 Proving Trigonometric Identities

To prove the given trigonometric identities, we show both sides of the identities
equal to each other by using different techniques and known identities.

Techniques for proving Trigonometric Identities :
Any one method of the following techniques can be used to prove the trigonometric
identities:

(a) Start from Left Hand Side (LHS) and reduce it to the Right Hand Side (RHS) if
LHS is more complicated.

(b) Start from RHS and reduce it to the LHS if RHS is more complicated.

(c) If LHS and RHS are equivalent, reduce both of them to the simplest form.

(d) If complicated, transpose or apply the method of cross – multiplication to
change the given form of identities. Then prove the new LHS = new RHS

It is to be noted that more priority is given to the proving of original identities. We
should choose the side in which algebraic formula can be applied. The mathematical
operations such an LCM, reciprocal of double fraction, multiplication by conjugate
etc, are also to be applied to prove the given identities.

Worked out Examples

Example 1. Using the basic trigonometric ratios involving p, b, h, prove the
Solution: following :

(a) sin2T + cos2T = 1 (b) tanT . cotT = 1

(c) cotT . sinT = cosT

(a) sin2T + cos2T = 1

62 vedanta Excel in Additional Mathematics - Book 7

LHS = sin2T + cos2T Trigonometry

= p 2+ b2 63
h g

= p2 + b2
h2 h2

= p2 + b2
h2

= h2 [ h2 = p2 + b2]
h2

= 1 = RHS Proved.

(b) tanT . cotT = 1

LHS = tanT . cotT

= p × b
b p

= 1 = RHS Proved.

(c) cotT . sinT = cosT

LHS = cotT . sinT

= b × p
p h

= b
h

= cosT = RHS Proved.

Example 2. Prove the following identities:
Solution: (a) sin2T . cos2T + sin4T = sin2T
(b) sin4T – cos4T = 2sin2T – 1
(c) cot2T + cosec2T = 2cot2T + 1
(a) Here, sin2T . cos2T + sin4T = sin2T

LHS = sin2T . cos2T + sin4T
= sin2T(cos2T + sin2T)
= sin2T(sin2T + cos2T)
= sin2T . 1
= sin2T = RHS Proved.

(b) Here, sin4T – cos4T = 2sin2T – 1
LHS = sin4T – cos4T
= (sin2T)2 – (cos2T)2

vedanta Excel in Additional Mathematics - Book 7

Trigonometry

= (sin2T + cos2T) (sin2T – cos2T)
= 1{sin2T – (1 – sin2T)}
= sin2T – 1 + sin2T
= 2sin2T – 1 = RHS Proved.
(c) Here, cot2T + cosec2T = 2cot2T + 1
LHS = cot2T + cosec2T
= cot2T + (1 + cot2T)
= 2cot2 + 1 + cot2T
= 2cot2T + 1 = RHS Proved.

Example 3. Prove the following :
Solution:
(a) sinA + cosA = cosecA . secA
cosA sinA

(b) 1 sinA + cosA = cosecA
+ cosA sinA

(c) 1 + sinT – 1 – sinT = 4tanT . secT
1 – sinT 1 + sinT

(a) Here, sinA + cosA = cosecA . secA
cosA sinA

LHS = sinA + cosA
cosA sinA

= sin2A + cos2A
sinA . cosA

= sinA 1
. cosA

= 1 . 1
sinA cosA

= cosecA . secA = RHS Proved.

sinA cosA
(b) Here, 1 + cosA + sinA = cosecA

LHS = 1 sinA + cosA
+ cosA sinA

sinA × sinA + cosA (1 + cosA)
= (1 + cosA) sinA

sin2A + cosA + cos2A
= sinA (1 + cosA)

(sin2A + cos2A) + cosA
= sinA (1 + cosA)

64 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

(1 + cosA)
= sinA (1 + cosA)

= 1
sinA

= cosecA = RHS Proved.

(c) Here, 1 + sinT – 1 – sinT = 4tanT . secT
1 – sinT 1 + sinT
1 + sinT 1 – sinT
LHS = 1 – sinT – 1 + sinT

= (1 + sinT)2 – (1 – sinT)2
(1 – sinT) (1 + sinT)

= 1 + 2sinT + sin2T – 1 + 2sinT – sin2T
1 – sin2T

= 4sinT
cos2T

= 4 . sinT . 1
cosT cosT

= 4tanT . secT = RHS Proved.

Exercise 3.4

Short Questions :
1. Using the basic trigonometric ratios involving p, b, h, prove the following:

(a) tanT × cosT = sinT
(b) cosT × cosecT × tanT = 1
(c) secT × cotT = cosecT
(d) sec2T – tan2T = 1
(e) cosT = 1 – sin2T
(f) sec2T + tan2T = 2tan2T + 1
(g) 1 + tan2T = secT
(h) sinT ÷ tanT = cosT
(i) cosec2T – cot2T = 1
(j) tanT . cosecT = secT
(k) cotT . secT = cosecT
2. Prove the following :
(a) sinT × cosecT × cotT × secT = cosecT

vedanta Excel in Additional Mathematics - Book 7 65

Trigonometry

(b) sinT × cotT × secT = 1

(c) sin2T.cosT + cos3T = cosT

(d) sin2T – cos2T = 2sin2T – 1

(e) cosec2T + cot2T = 2cosec2T – 1

(f) (1 – cos2T) (1 + tan2T) = tan2T

(g) (sec2T – 1) (1 + cot2T) = sec2T

(h) 1 – sin2T . cosecT = cotT

(i) cosT . sec2T – 1 = sinT

3. Prove the following:

(a) sin4T + sin2T . cos2T = sin2T

(b) sec4T – sec2T . tan2T = sec2T

(c) cosec4T – cosec2T . cot2T = cosec2T

(d) sin2T . cos2T + cos4T = cos2T

4. Prove the following :

(a) secT – tanT =1
cosT cotT
cosecT cotT
(b) sinT – tanT =1

(c) sec2T – 1 = sin2T
sec2T
11
(d) 1 – sinD + 1 + sinD = 2sec2D
1 + sinT cosT
(e) cosT + + sinT = 2secT
1
11
(f) 1 + cosT + 1 – cosT = 2 cosec2T
1
(g) secT – tanT = secT + tanT

(h) 1 cotT = cosecT + cotT
cosecT –
sinT + cosT
(i) secT + cosecT = sinT . cosT

(j) tanT + tanD = tanD . tanT
cotT + cotD

66 vedanta Excel in Additional Mathematics - Book 7

Coordinate Geometry

4Coordinate Geometry

4.0 Introduction

A set of values that shows an exact position of a Y
point on a graph is called coordinates. It is usually
2 (2, 5)

a pair of numbers, the first number shows the

distance along and the second number shows the 5

distance up or down. For example, the point (2, 5) X' O X
shows that it is 2 units right from origin and 5 units

up from X-axis. Y'

Coordinate geometry is the branch of mathematics

which deals with the combination of algebra and geometry. In this level, we discuss

with cartesian coordinates in a plane.

First of all, let's see how we locate a point in a number line. After that we are
comfortable with a plane. In this level, we discuss with a plane. The line below is
a number line. We are standing at 0. If we move on the right side its positive and
negative if we move on the left side.

4.1 An activity on number line

Let us draw two lines horizontal and vertical as shown in the figure. Then, mark the
points to indicate numbers as shown in the lines. Then do the following activities:

X' X Y4
3
–5 –4 –3 –2 –1 0 1 2 3 4 5 2
1
Fig (i) 0
–1
(a) Discuss the numbers in the horizontal and vertical lines. –2
–3
(b) What does O (zero) represent? –4

(c) In horizontal line, what are the meanings of –4 and 4? Y'

(d) In vertical line, what are the meanings of –4 and 4? Fig (ii)

(e) Arrange the two lines together in such a way that both the zeros of
the lines intersect at zero perpendicularly.

Now, replace zero by O (capital O). We get the figure as shown below:

vedanta Excel in Additional Mathematics - Book 7 67

Coordinate Geometry

Now, the horizontal line is called X-axis, written as XOX' and vertical line is
called Y-axis, written as YOY'.

(f) Then the four parts XOY, YOX', X'OY', and Y'OX are formed and named as the
the first, the second, the third and the fourth quadrants.

Y
5

4

Second Quadrant 3 First Quadrant
2

1

X' – 5 – 4 – 3 – 2 – 1 O –11 2 3 4 5 X

Third Quadrant –2
Fourth Quadrant

–3

–4

Y' –5

Note :

1. The ordered pair (a, b) is called coordinates.

a is called x-coordinate or abscissa.

b is called y-coordinate or ordinate.

2. The coordinates of O are always (0, 0).

3. When XOX' and YOY' cut at O at right angles, the plane formed is called
Cartesian plane.

4. Signs of coordinates

Quadrants x-coordinate y-coordinate
1st +ve +ve
2nd –ve +ve
3rd –ve –ve
4th +ve –ve

68 vedanta Excel in Additional Mathematics - Book 7

Coordinate Geometry

Worked out Examples

Example 1. Plot the ordered pairs and name the quadrant or axis in which the
Solution: given points lie: P(2, 5), Q(– 2, 2), R(–4, – 5), S(4,0) and T(0, 7)

P(2, 5) lies in quadrant I.

Q(–2, 2) lies in quadrant II. R(–4, –5) lies in quadrant III.

S(4, 0) lies in X - axis. T(0, 7) lies in Y-axis.

8
7 T(0, 7)

6
5 P(2, 5)

4

Q(–2, 2) 3
2

-6 -5 -4 1 S(4, 0)
-3 -2 -1-1 1 234 5 67

-2

-3

-4

R(–4, –5) -5
-6

-7

Example 2. Find the co-ordinate of the points P, Q, R, S, T, and U.

8

7

6

5

4

R3

2U Q
234
-6 -5 -4 1 1 5 67
T P

-3 -2 -1-1

-2

-3

-4
S

-5

-6

-7

vedanta Excel in Additional Mathematics - Book 7 69

Coordinate Geometry

Solution: Since point P corresponds to 2 along the X-axis and – 3 along the
Y-axis, P' s ordered pair coordinates are (2, – 3)

Since point Q corresponds to 3 along the X-axis and 2 along the Y-axis,
Q's coordinates are (3, 2)

Since point R corresponds to – 2 along the X-axis and 3 along the
Y-axis, R's coordinates are (– 2, 3)

Since point S corresponds to – 3 along the X-axis and – 4 along the
Y-axis, S's coordinates are (– 3, – 4)

Since point T corresponds to – 3 along the X-axis and 0 along the
Y-axis, T's coordinates are (– 3, 0).

Since point U corresponds to 0 along the X-axis and 2 along the Y-axis,
U's coordinates are (0, 2).

Example 3. Plot the given points and join them. Then, identify the shapes formed.
Solution:
(a) P(2, 3), Q(4, 5), and R(5, 6) Y
(b) O(0, 0), A(4, 0), and B(0, 3)
(c) A(2, 3), B(–2, –3), and C(0, 7) R(5, 6)
(a) Plotting three points P(2, 3), Q(4, 5), Q(4, 5)

and R(5, 6) in graph, joining them, we X' P(2, 3)
get a straight line PR.
OX
Here, the points P, Q, and R are called Y'
Y

collinear points. B(0, 3)

(b) Plotting the points O(0, 0), A(4, 0), and X' O(0, 0) A(4, 0)
B(0, 3) in a graph, joining them, we get
a right angled triangle OAB. X

Y'
Y

C(0, 7)

(c) Plotting the points A(2, 3), B(–2, –3), A(2, 3)
and C(0, 7) in a graph, joining them
OX
we get a triangle ABC. X'

B(-2, -3)

Y'

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Exercise 4.1

1. (a) From the given graph, find the coordinates Y
of P, Q, R, and S:

(b) Name the quadrants in which the given Q P
S
points lie: (ii) (– 6, 4) X' O X
(i) (2, 2)
R
(iii) (2, – 4) (iv) (2, – 6)

(v) (–6, – 2) (vi) (–5, 7) Y'
(vii) (–7, –3)

2. In a graph paper draw X-axis and Y-axis meeting at origin O(0, 0) and taking a
scale of 1 units = 5 smallest division, plot the following point given below:

(a) A(2, 3) (b) B(– 4, 5) (c) C(5, 5) (d) D(– 5, – 5)

(e) E(6, 0) (f) F(– 8, 0) (g) G(0, 5) (h) H(0, – 4)

3. Find the vertices of given geometrical shapes from the graph:
Y
R

C

DP Q
M
B
N L

X' AG L X
O J
71
E

H

I

G

F
Y'

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4. Plot the points given below in graph. Join them to get a shape. Identify the
shapes.
(a) (i) O(0, 0), A(4, 4), B(– 5, – 5) (ii) A(6, 0), B(5, – 2), C(7, 2)
(iii) P(0, 8), Q(– 4, 2), R(– 5, 0)
(b) (i) O(0, 0), A(– 5, 0), B(0, – 5) (ii) A(– 7, 4), B(2, 6), C(8, 2)
(iii) P(– 3, 5), Q (3, 5), R(0, 6)
(c) (i) P(2, 3), Q(4, 0), R(8, 5), S(6, 6) (iii) O(0, 0), A(5, 0), B(5, 5), C(0, 5)
(iii) A(2, 2), B(4, 8), C(–5, 7), D(– 5, 3)

1, 2, 3. Show to your teacher

4. (a) (i) straight line (ii) straight line (iii) straight line

(b) (i) triangle (isosceles right angled triangle)

(ii) triangle (iii) triangle

(c) (i) quadrilateral (ii) square (iii) quadrilateral

4.2 Distance Between Two Points

The distance between two points can be determined by how much a point has
moved from one point to another point. If a point moves along coordinate axes, it
is easy to calculate the distance between two points. If a point moves arbitrary, we
need a distance formula.

Y

R

X' QO P X

72 S
Y'

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If a point is at the origin and it moves four units right to the point P. So, the distance
between O and P is 4 units.

i.e. OP = 4 units

Similarly, if the point P moves 8 units to left from P to Q, then, the distance between
P and Q is 8.

So, PQ = 8 units

Again, a point is at the origin, it moves 8 units above the origin to R. Then, the
distance between O and R is 8 units.

So, OR = 8 units

If a point moves arbitrarily, we proceed as follows. Let us consider two points A(1,

2) and B(7, 10). Y

If the point A(1, 2) moves to B(7, 10), it has moved B(7, 10)
6 units along - horizontally and 8 units vertically.

In the figure, let A(x1, y1) = (1, 2), B(x2, y2) = (7, 10). 8
x-component of AB = AC = 6

i.e. AC = x2 – x1 A(1, 2) 6 C
=7–1
=6 X' O X

y-component of AB = CB = 8 Y'
i.e. AB = y2 – y1

= 10 – 2

=8

We observe that 'ABC is a right angled triangle.

Now, by using Pythagoras theorem,

AB2 = AC2 + CB2

= (x2 – x1)2 + (y2 – y1)2
or, AB = (x2 – x1)2 + (y2 – y1)2
? AB = (x2 – x1)2 + (y2 – y1)2
In the above graph,

A(x1, y1) = A(1, 2), B(x2, y2) = B(7, 10)
Now, AB = (7 – 1)2 + (10 – 2)2

= 62 + 82

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Coordinate Geometry

= 36 + 64
= 100
= 10 units.

Distance Formula Y
B(x2, y2)
We can also find out the x-component and
y-component as follows. Let A(x1 , y1) and y2 - y1

B(x2, y2) be any two points.

For the line segment AB

x - component = x2 – x1 A(x1, y1) C
y - component = y2 – y1 x2 - x1
X' X
O

So, (AB)2 = (x component)2 + (y - component)2 Y'

(AB)2 = (x2 – x1)2 + (y2 – y1)2
? AB = (x2 – x1)2 + (y2 – y1)2
If AB is denoted by d. Then, d = (x2 – x1)2 + (y2 – y1)2

Worked out Examples

Example 1. Find the distance between the points:
Solution:
(a) O(0, 0) and A(6, 0) (b) A(– 3, 0) and B(4, 0)

(c) C(0, 6) and D(0, – 2) (d) E(1, 3) and F(4, 7)

(a) O(0, 0) and A(6, 0)

Here, let O(0, 0) be (x1, y1) and A(6, 0) be (x2, y2)

Now, x-component of OA = x2 – x1 Y

=6–0

=6

y-component of OA = y2 – y1 X' O A(6, 0) X
=0–0 Y'
=0

OA = 6 units (As it has travelled is along axis)

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(b) A(–3, 0) and B(4, 0)

Here, let A(–3, 0) be (x1, y1) B(4, 0) be (x2, y2) Y
Now, x-component of AB = x2 – x1

= 4 – (– 3)

=4+3

=7 A(–3,0) B(4,0)

y-component of AB = y2 – y1 X' O X

=0–0 Y'

=0

As, AB is a line segment on X-axis

AB = 12 units

(c) C(0, 6) and D(0, –2) Y
Here, Let C(0, 6) be (x1 , y1)

D(0, – 2) be (x2, y2)

Now, x-component of CD = x2 – x1 D(0,–2) C(0, 6) X

=0–0 X' O

= 0 Y'

y-component of CD = y2– y1

=–2–6

=–8

Since CD only moves along Y-axis,

CD = 10 units (distance is always positive)

Example 2. If O(0 , 0), A(4, 4), and B(–4, –4) are three points, show that OA = OB.

Solution: Here, for OA,

O(0, 0) = (x1 , y1)
A(4, 4) = (x2, y2)
Now, x-component = x2 – x1

=4–0

=4

y-component = y2 – y1

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Coordinate Geometry

=4–0
=4
Now, using formula, we get,
OA = (x2 – x1)2 + (y2 – y1)2
= 42 + 42
= 16 + 16
= 32
? OA = 4 2 units
Again, for OB,
O(0, 0) = (x1, y1)
B(–4, –4) = (x2 , y2)
Now, x-component = x2 – x1
=–4–0
=–4
y-component = y2 – y1
=–4–0
=–4
Now, using formula, we get,
OB = (x2 – x1)2 + (y2 – y1)2
= (–4)2 + (–4)2
= 16 + 16
= 32
? OB = 4 2 units.
? OA = OB Proved.
Example 3. A(2, 2), B(5, 4), C(3, 2), and D(6, 4) are four points. Show that AB = CD
Solution: Here, A(2, 2), B(5, 4), C(3, 2), and D(6, 4) are four points.
For AB,
Let A(2, 2) be (x1, y1)
B(5, 4) be (x2 , y2)

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Now, x-component = x2 – x1
=5–2 =3

y-component = y2 – y1
=4–2 =2

Now, using distance formula,

AB = (x2 – x1)2 + (y2 – y1)2
= 32 + 22

= 9+4

? AB = 13 units

Again for CD,

Now, x-component = x2 – x1
=6–3=3

y-component = y2– y1
=4–2=2

Now, using distance formula,

CD = (x2 – x1)2 + (y2 – y1)2
= 32 + 22

= 9+4

? CD = 13 units

? AB = CD Proved.

Example 4. A is a point on X-axis whose abscissa is 6 and B is a point on Y-axis
Solution: whose ordinate is –4. Find the length of AB.

Here, A is a point on X-axis whose abscissa is 6.

So, the coordinates of A are (6, 0)

Again, B is a point on Y-axis whose ordinate is –4

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So, the co-ordinate of B is (0, – 4) , Y

Now, to find the length of AB,

Let A(6, 0) be (x1 , y1) X' O A(6, 0) X
B(0, – 4) be (x2, y2)

? x-component = x2 – x1 B(0, –4)
=0–6=–6
Y'
y-component = y2 – y1

=–4–0=–4

Now, (AB)2 = (x-component)2 + (y-component)2

= (– 6)2 + (– 4)2

= 36 + 16

= 52

or, AB = 52

or, AB = 4 × 13

? AB = 2 13 units.

Hence the length of AB is 2 13 units.

Exercise 4.2

Short Questions :
1. (a) Write formula to find distance between points P(x1, y1) and Q(x2, y2).

(b) Find distance between points (0, 0) and (a, b).
(c) In a right angled triangle write relation between p, b, and h.
2. Find the distance between the two points:
(a) O(0, 0) and A(6, 8)
(b) A(4, 5) and B(3, 4)
(c) C(3, 0) and D (0, 4)
(d) E(– 2, – 2) and F(– 5, – 6)
(e) G(8, 4) and H(6, 4)
(f) P(5, 4) and Q (– 5, – 6)

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3. Find the distance between the origin and the following points.

(a) (3, 4) (b) (–8, 6)

(c) (5, –12) (d) (–12, –9)

Long Questions :

4. (a) If O(0, 0), B(6, 0), and C(0, 6) are three points, prove that OB = OC.

(b) If A(2, 3), B (2, 0), and C (–1, 0) are three points, show that AB = BC.

5. (a) A(4, 4), B(2, –1), C(–5, – 2), and D (– 3, 3) are any four points. Show that
AB = CD.

(b) Prove that PQ = QR = RS = PS if the coordinates of points are P(2, 3),
Q(2, 0), R(–1, 0), and S(– 1, 3).

6. (a) A is a point on X-axis whose abscissa is 4 and B is a point on Y-axis whose
ordinate is 3. Find the distance between AB.

(b) P is a point on Y-axis whose ordinate is – 8 and Q is a point on X-axis whose
abscissa is – 6. Determine the length between PQ.

Project Work

7. A post office box is at P(8, 9). Two letters from two places A(4, 6) and B(0, 8) are
to be posted on the post office box. Which place is nearer to the post office box?

(Take, 1 unit = 1 km)

1. (a) d = (x2 – x1)2 + (y2 – y1)2 (b) d = a2 + b2 (c) h2 = p2 + b2
(c) 5 (d) 5
2. (a) 10 (b) 2

(e) 2 (f) 10 2

3. (a) 5 (b) 10 (c) 13 (d) 15

6. (a) 5 (b) 10

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5Matrices

5.0 Introduction

Let us consider three students Hary, Krish, and Jasmin of class 6.

Hary got 58 marks in Science, 68 in Maths and 70 in English. Krish got 70 marks in
Science, 90 in Maths and 64 in English. Jasmin got 60 marks in Science, 75 in Maths
and 70 in English.

We can arrange the above information as shown in the table below :

Subjects Science Maths English

Students 58 68 70
Hary 70 90 64
Krish 60 75 70
Jasmin

We can represent it in the form given below :

Hary Science Maths English
Krish 58 68 70
Jasmin 70 90 64
60 75 70

Here, the first, the second and the third rows represent the marks obtained by Hary,
Krish and Jasmin respectively. Similarly, the first, the second and the third columns
show the marks obtained students in Science, Maths, and English respectively.

Thus rectangular arrangement of numbers in called matrix.

5.1 Definition

A rectangular arrangement of numbers arranged in rows and columns enclosed
within large brackets or round brackets is known as matrix. Usually, matrix is
represented by a capital letters. Its plural form is matrices.

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For example : A = 2 4 B= 0 3 4 –2 5
5 6 2 5 7 C= 3 7
8
7

5.2 Rows and Columns of a Matrix

A matrix is a rectangular arrangement of numbers which contains elements in the
form of rows and columns. A row is usually a horizontal line and column is a
vertical line.

For example : A = 2 6 R1
3 7 R2

C1 C2
This matrix A has two rows and two columns.

Order of a matrix

The order of a matrix is defined as the "number of rows × number of columns"
present in the matrix.

For example : A = 2 6 5 R1
3 7 7 R2

C1 C2 C3
Matrix A has 2 rows and 3 columns So, its order is 2 × 3. It is read as 2 by 3.

It is also denoted by A2 × 3.

Worked Out Examples

Example 1. Determine the order of the following matrices.

(a) A= a b 2 6
c d (b) B = 4 7
8
6

23 4 (d) D = [2]
(c) C = 5 6 7

8 9 10

Solution: (a) A= a b
c d
Here, number of rows present in matrix A = 2

number of column present in matrix A = 2

So, order of A is 2 × 2.

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26
(b) B = 4 7

68

Here, number of rows present in matrix B = 3

number of columns present in matrix B = 2

So, order of B is 3 × 2

23 4
(c) C = 5 6 7

8 9 10

Here, number of rows present in matrix C = 3
number of columns present in matrix C = 3

So, order of C is 3 × 3
(d) D = [2]

Here, number of row in matrix D = 1
number of column in matrix D = 1

So, order of D is 1 × 1

Example 2. Construct a matrix with the order specified. (You can use any
Solution: numbers in the matrices you farmed.)

(a) A, order 2 × 3 (b) B, order 2 × 2

(c) C, order 3 × 3 (d) D, order 3 × 2

(e) E, order 1 × 3

(a) A, order 2 × 3

It means A should have 2 rows and 3 columns.

A= 2 6 4
3 7 5

(b) B, order 2 × 2

It means matrix B should have 2 rows and 2 columns.

B= 1 2
2 4

(c) C, order 3 × 3

It means matrix C must have 3 rows and 3 columns.

adg
C= b e h

cf i

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(d) D, order 3 × 2.
It means matrix D must have 3 rows and 2 columns.

14
D= 2 5

37
(e) E, order 1 × 3

It means matrix E must have 1 row and 3 columns
E = [1 2 3]

5.3 Types of Matrices

(a) Row Matrix

A matrix having only one row is called a row matrix. It can have any number of
columns.

Example : A = [p q r]

It has only one row. Its order is 1 × 3.

B = [4 6]

It also has only one row. Its order is 1 × 2.

(b) Column Matrix

A matrix having only one column but any number of rows is called a column
matrix.

1 B= 5
Example : A = 2 6

3

Here, A and B both matrices have only one column.

(c) Rectangular Matrix

A matrix having unequal number of rows and columns is called a rectangular
matrix.

Example : A= 2 3 4
5 6 7
Here, matrix A has 2 rows and 3 columns which is unequal. So it is rectangular

matrix.

1 4
B= 2 5
6
3

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Matrices

(d) Zero Or Null Matrix

A matrix having each of the elements zero (0) is called a null or zero matrix.

Example: 0 = 0 0 0
0 0 0

This is a null matrix which is of order 2 × 3.

(e) Square Matrix

A matrix whose number of rows and columns are equal is known as a square matrix.

Example : A = p r 147
q s B= 2 5 8

369

Here, in matrices A and B, the number of rows and columns both are equal. So,
they are square matrices.

Example 3. State the types of matrices given below and also state their orders.
Solution:
(a) A = [2 3] (b) B= 4
5

(c) C= 4 7 (d) O= 0 0
6 8 0 0

26
(e) P = 3 8

4 10
(a) A = [2 3]

It is a row matrix.

Its order is 1 × 2.

(b) B= 4
5

It is a column matrix.

Its order is 2 × 1.

(c) C= 4 7
6 8

It is a square matrix.

Its order is 2 × 2.

(d) O= 0 0
0 0

It is a null matrix.

Its order is 2 × 2.

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26
(e) P = 3 8

4 10
It is a rectangular matrix.

Its order is 3 × 2.

5.4 Equal Matrices

Two matrices having same order and equal corresponding elements are said to be
equal matrices.

For example : A = 4 6 B= 4 6
5 7 5 7
Here, matrices A and B are said to be equal matrices as their order is same as well

as their corresponding elements are equal.

Example 4. Find the values of p, q, r, and s if the matrices are equal.
Solution:
(a) p 6 = 5 r 4 2p 4r 2
4 q s 8 (b) 3q 1 = 1 3s

p 6 5 r 83
4 q s 8
(a) Here, =

As the matrices are equal, their corresponding elements are also
equal.

So, p = 5, q = 8, r = 6 and s = 4

4 2p 4r 2
(b) Here, 3q 1 = 1 3s

83
Since the matrices are equal their corresponding elements are
also equal,

So, 4 = 4r

or, r = 4 = 1 ? r=1
4

2p = 2

or, p = 2 = 1 ? p=1
2
1
3q = 3

or, q = 3 1 3 = 1 ? q = 1
× 9 9
1
3s = 8

or, s = 8 1 3 ? s = 1
× 24
19, 1
? p = 1, q = r = 1 and s = 24 .

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Example 5. Find the values of x and y if

p–3 4 = 0 4
5 q+2 5 3
Solution: Since the matrices are equal, their corresponding elements are also equal.

So, p – 3 = 0

or, p = 3

Again, q + 2 = 3

or, q = 3 – 2

or, q = 1

? p = 3 and q = 1

Exercise 5.1

1. State the orders of the following matrices:

(a) [p q r] 14 (c) 44
(d) [10] (b) 2 5 44

36 25 8
(f) 3 6 9
(e) 4
6 4 7 10

2. Construct a matrix with the order specified below. (You can use any numbers

in the matrices you farmed.)

(a) A1 × 1 (b) B1 × 2 (c) C2 × 1

(d) D2 × 2 (e) E2 × 3 (f) F3 ×2

(g) G1 × 3 (h) H3 × 1 (i) I3 × 3

3. State the type of matrices given below and also state its order:

6

(a) [8] (b) [a b] (c) 8

9

(d) 4 –6 (e) 1 2 3 (f) 00
5 7 4 6 7 00

4. Find the values of p, q, r and s from the pair of equal matrices given below.

(a) p 0 = 5 r
q –2 3 s

(b) 2 2p = 6r –8
3q 4 12 4s

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Matrices

pq

(c) 2 3 = 8 6
r s 16 25

45

5. Find the values of x and y if-

(a) 5x + 5 5 = 15 5
7 3y – 2 7 13

(b) x 6 = 5 6
–7 3 –7 y–3

(c) x–1 = 5
y+2 8

(d) x–6 8 = 3x – 8 8
2 7 2 y+3

1. (a) 1 × 3 (b) 3 × 2 (c) 2 × 2 (d) 1 × 1

(e) 2 × 1 (f) 3 × 3 2. Show to your teacher

3. (a) square matrix, order 1 × 1 (b) row matrix, order 1 × 2

(c) column matrix, order 3 × 1 (d) square matrix, order 2 × 2

(e) rectangular matrix, order 2 × 3 (f) null/zero matrix, order 2 × 2

4. (a) p = 5, q = 3, r = 0, s = –2 (b) p = –4, q = 4, r = 1 , s = 1
(c) p = 16, q = 18, r = 64, s = 125 3

5. (a) x = 2, y = 5 (b) x = 5, y = 6 (c) x = 6, y = 6 (d) x = 1, y = 4

5.5 Operation on Matrices

In this level we do only addition and subtraction of matrices.

Two matrices of same order can be added or subtracted. We add or subtract the
corresponding elements when we add or subtract two matrices.

A= 2 4 and B= 2 3
3 5 1 0
A+B=?

Here, as the order of both the matrices is 2 × 2. We can add them,

So, A + B = 2+2 4+3
3+1 5+0

= 4 7
4 5

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Again, A= 4 7 , C = 2
4 5 5

A+C=?

4 7 + 2
4 5 5
As the order of these two matrices A and C are not the same we cannot add them.

For addition and subtraction of two matrices, their orders must be same and the
corresponding elements are added or subtracted.

Is B – A possible?
Yes, the order is same. So its possible.

Now,

B–A=?

So, B – A = 2 3 – 2 4
1 0 3 5

= 2–2 3–4
1–3 0–5

= 0 –1
–2 –5

Worked out Examples

Example 1. Can the following matrices be added or subtracted ? Give reasons for
Solution: your answer.

(a) A = [4 5], B = [2 6] (b) P= 2 4 , Q = 2 4
6 2 6 2
2 4 2
(c) X= 8 2 , Y = 4

(a) Here, A = [4 5], B = [2 6]

Yes, they can be added and subtracted as their orders are the
same.

(b) Here, P = 2 4 , Q = 2 4
6 2 6 2
Yes, they can be added and subtracted as their orders are the

same.

(c) Here, X = 2 4 , Y = 2
8 2 4
No, they cannot be added or subtracted as their orders are

different.

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Example 2. Perform the indicated operations:
Solution:
(a) A= 3 , B = 2 (b) P= 2 3 , Q = 2 6
4 1 4 5 3 2

Find A + B, A – B and B – A Find P + Q and P – Q

(a) A= 3 , B = 2
4 1

A+B = 3 + 2
4 1

= 3+2
4+1

= 5
5

A–B = 3 – 2
4 1

= 3–2
4–1

= 1
3

B–A = 2 – 3
1 4

= 2–3
1–4

= –1
–3

(b) Here, P = 2 3 , Q = 2 6
4 5 3 2

P and Q are of the same order matrices.

So, P + Q and P – Q are possible

P+Q = 2 3 + 2 6
P–Q 4 5 3 2

= 2+2 3+6
4+3 5+2

= 4 9
7 7

= 2 3 – 2 6
4 5 3 2

= 2–2 3–6
4–3 5–2

= 0 –3
1 3

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Exercise 5.2

1. Can the following matrices be added or subtracted ? Give reasons for your
answer.

(a) A = [2 3], B = [4 2] (b) C = [6 3], D = [1 2 3]
(c) E = [6
8], F = 5 (d) G= 2 2 , H = 2 3
8 2 2 2 5

(e) I= 4 5 6 , J = 4 3
3 2 1 6 2
3 2

2. Perform indicated operations of matrices.

(a) A = [2 6], B = [1 2], find A + B, A – B.

(b) P= 2 , Q = 2 , find Q – P, Q + P.
3 4

(c) A= 4 6 , B = 2 5 , find A + B , A – B, B – A.
3 2 6 3

1 6 42

(d) X = 2 7 , Y = 5 3 , find X + Y and X – Y.

3 8 87

(e) P= 1 2 3 , Q = 2 5 3 , find P + Q and P – Q.
4 5 6 4 8 2

Project Work

3. Write the marks of your marksheet of class 6 in matrix form.

1. Show to your teacher (b) 0 , 4
2. (a) [3 8], [1 4] 1 7

(c) 6 11 , 2 1 , –2 –1 5 8 –3 4
9 5 –3 –1 3 1 (d) 7 10 , –3 4
1
3 7 6 –1 –3 0 11 15 –5
8 13 8 0 –3 4
(e) ,

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6Transformation

6.0 Introduction

A process in which an object changes its shape, size, or position is called a
transformation. A transformation forms an image which is congruent or similar to
given object. In a transformation each point of an object has exactly one image point
and each image point has exactly one pre-image in the same plane. In daily life, we
come across phenomenon like our image formed on mirror, rotation of wheels in
vehicles, photographs taken by smart mobile cameras, etc. These are some examples
of transformation.

Transformation sometimes leaves certain points unchanged. These points are said
to be invariant. For example, in a rotation centre of rotation is invariant point.

6.1 Types of Transformation

Transformations may or may not change the size of the given objects: on the basis of
it, the transformation is divided into following two types :

(a) Isometric Transformation

(b) Non-isometric Transformation

(a) Isometric Transformation
A transformation in which the size of an object does not change but the position
of the object changes is known as isometric transformation. Reflection, rotation,
and translation are isometric transformations. Isometric transformations are
also called congruent transformations.

(b) Non-isometric Transformation
The transformations in which the size of an object changes is known as non-
isometric transformation. In this transformation, the distance between two
points of an object and the distance between their corresponding images after
transformation are unequal Enlargement and reduction are examples of non-
isometric transformation.

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Transformation

Four Fundamental Transformations

The following are the four fundamental transformations:

(a) Reflection (b) Rotation (c) Translation (d) Enlargement

In this class, we study only about reflection and rotation.

6.2 Reflection

Let us observe the object and its image below and try to write some notes from the
figures.

P (Object)

P P'

Q Q'

Mirror (Object) (Image)
(Axis of reflection) R R'

P' (Image) Mirror
Fig. i
(Axis of reflection)

Fig. ii

Note :

(i) The image lies on the opposite side of the object with respect to the mirror
(axis of reflection)

(ii) The object distance and the image distance from the axis of reflections are
equal.

(iii) The object size and image size are equal.

(iv) This an isometric transformation.

Definition: Reflection is a rule which shifts an object to the image of the same shape
and size each being at an equal distance from a fixed line. The fixed line is called
the mirror or axis of reflection.

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Worked Out Examples

Example 1. Find the image of the objects given below after reflection in the
mirror line M.

(a) M (b) P R

P

Q

Solution: M M

(a) Here, we follow the following steps.

(i) Draw perpendiculars from PO to the

axis of reflection or mirror M.

(ii) Produce PO to opposite side of the P O P'
(Object) (Image)

mirror such that PO = OP'.

(iii) P' is called the image of P.

(b) Here, we follow the following steps. P R
C
(i) Perpendicular PA, QB and RC are drawn R'

from object points P, Q and R to the mirror A
P'
line M.

(ii) PA, QB and RC are produced to the Q
opposite side of the mirror such that PA = B
Q'
M
AP', QB = BQ' and RC = CR'.

(iii) Join P'Q', Q'R' and R'P' successively.

(iv) 'P'Q'R' is the image of 'PQR.

(v) Shade the image triangle P'Q'R'.

6.3 Use of Coordinates in Reflection

We can use coordinates to locate an object in the graph paper and its image can be
found reflecting on X-axis, Y-axis or y = x line etc.

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Transformation Y

(a) Reflection on X-axis P(4, 5)

Take a point P(4, 5) on a graph paper. Then O MX
draw a perpendicular PM to X-axis. Let it be
reflected on X-axis. Produce PM to P' such that P'(4, –5)
PM = MP'. Find the coordinates of P'. We get X'
P'(–4, 5) which is the image of P. Y'

P(4, 5) X-axis P'(4, –5)

Similarly, we get
R(–4, 5) X-axis R'(–4, –5)
N(–4, –5) X-axis N'(–4, 5)

? P(x, y) X-axis P'(x, –y)

Example 2. If P(–2, 3), Q(2, 1), and R(4, 4) are the vertices of 'PQR. Plot 'PQR on
a graph paper and find its image reflecting it on X-axis.

Solution: Here, P(–2, 3), Q(2, 1) and R(4, 4) are the vertices of triangle PQR.
Reflecting it on X-axis. we get,
P(x, y) X-axis P'(x, –y)

Now, P(–2, 3) X-axis P'(–2, –3)
Q(2, 1) X-axis Q'(2, –1)
R(4, 4) X-axis R'(4, –4)

'PQR and its image 'P'Q'R' are plotted on a graph paper and
image is shaded as shown in the graph.

Y

R(4, 4)

P(–2, 3)

Q(2, 1)

X' O X

Q'(2, –1)

P'(–2, –3)

R'(4, –4)

Y'

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Transformation

(b) Reflection on Y-axis P'(–4, 5) Y P(4, 5)

Take a point P(4, 5) on a graph paper. Then M X

draw a perpendicular PM to Y-axis. Let it be O

reflected on Y-axis. Produce PM to P' such that

PM = MP'. Find the coordinates of P. We get X'
P'(–4, 5) which is the image of P.

P(4, 5) Y-axis P'(–4, 5)

Similarly, we get, Y'
R(–4, 5) Y-axis R'(4, 5)
N(–4, –5) Y-axis N'(4, –5)

? P(x, y) Y-axis P'(–x, y)

Example 3. If P(2, –2), Q(5, 3), and R(1, 5) are the vertices of 'PQR. Reflect 'PQR
Solution: on Y-axis and find the image 'P'Q'R'. Plot both 'PQR and 'P'Q'R' on
the same graph.

Here, P(2, –2), Q(5, 3) and R(1, 5) are the vertices of ∆PQR. Reflecting
it on Y-axis, we get
P(x, y) Y-axis P'(–x, y)

Now, P(2, –2) Y-axis P'(–2, –2)
Q(5, 3) Y-axis Q'(–5, 3)
R(1, 5) Y-axis R'(–1, 5)

∆PQR and its image ∆P'Q'R' are plotted on a graph paper and the
image is shaded as shown in the graph.

Y

R'(–1,5) R(1, 5)

Q'(–5,3) Q(5, 3)

X' O X

P'(–2,–2) P(2,–2)

Y'

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Transformation

Exercise 6.1

Short Questions :

1. Find the image of P(a, b) when :

(a) reflected on X-axis (b) reflected on Y-axis

2. Complete the following by drawing the images with l as the axis of reflection.

(a) P (Object) (b) l
P

l
Q

(c) Q (d) S

P R P
l Q
Rl

3. Find the images of the following points separately under the reflection

(a) X-axes (b) Y-axis

(i) P(x, y) (ii) Q(1, 2) (iii) R(–2, 4) (iv) S(– 4, – 4)

(v) T(–5, 6) (vi) M(4, 5) (vii) U(6, – 7) (viii) V(8, –4)

Long Questions :

4. (a) P(3, 8), Q(3, 4), and R(–6, 7) are the vertices of 'PQR. Reflect it on X-axis and
find the image vertices of 'PQR. Plot 'PQR and 'P'Q'R' on the same graph.

(b) A(1, 2), B(5, 5), and C(1, 8) are the vertices of 'ABC. Find the coordinates of
the image vertices of 'ABC when it is reflected on X-axis. Plot both 'ABC and
its image 'A'B'C' on the same graph.

(c) Let M(–2, 4), N(–1, 8), and P(–8, 4) are the vertices of 'MNP. Reflect it on X-axis
and find image 'M'N'P'. Plot 'MNP and its image 'M'N'P' on the same graph.

5. (a) If A(4, 8), B(1, 1), and C(8, 2) are the vertices of 'ABC. Reflect it on Y-axis
and find the image vertices of 'ABC. Plot both 'ABC and its image 'A'B'C'
on the same graph.

(b) If P(–2, 2), Q(–4, 8), and R(–6, 1) the vertices of 'PQR, find the coordinates
of the image vertices of the triangle PQR. Plot both the triangles on the
same graph.

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(c) A(2, –2), B(2, 2), and C(8, 6) are the vertices of 'ABC. Reflect the triangle on
Y-axis and find the image vertices of the triangle. Plot both of the triangle
on the same graph.

Project Work

6. List any five uses of reflection and discuss them in your class.

1. (a) P'(a, –b) (b) P'(–a, b) 2. Show to your teacher.

3. (a) (i) P'(x, –y) (ii) Q'(1, –2) (iii) R'(–2, –4) (iv) S'(–4, 4)

(v) T(–5, –6) (vi) M'(4, –5) (vii) U(6, 7) (viii) V'(8, 4)

(b) (i) P'(–x, y) (ii) Q'(–1, 2) (iii) R'(2, 4) (iv) S'(4, –4)

(v) T'(5, 6) (vi) M(–4, 5) (vii) U'(–6, –7) (viii) V'(–8, –4)

4. (a) P'(3, –8), Q'(3, –4), R'(–6, –7) (b) A'(1, –2), B'(5, –5), C'(1, –8)

(c) M'(–2, –4), N'(–1, –8), P'(–8, –4) 5.(a) A'(–4, 8), B'(–1, 1), C'(–8, 2)

(b) P'(2, 2), Q'(4, 8), R'(6, 1) (c) A'(–2, –2), B'(–2, 2), C'(–8, 6)

6.4 Rotation - Introduction

Let us take a point P on a plane paper and rotate it on clockwise and anti-clockwise

direction.

(a) Here, the point P is said to be rotated through 90° in anti- P'

clockwise direction or positive quarter turn about centre O. The

anti-clockwise direction is also called positive direction. P' is 90°

the image of P due to the rotation. OO P
P
(b) Here, the point P is said to be rotated through 90q in clockwise
direction or negative quarter turn about the centre O. The 90°

clockwise direction also called negative direction. P' is the image

of P due to the rotation. P'

(c) Here, the point P is said to be rotated through 180° in

anti-clockwise direction about centre O. It is also 180°
O
called positive half turn about O. P' is the image of P P' O P
–180° P
due to the rotation.

P'

(d) Here, the point P is said to be rotated through 180q in

clockwise about centre O. It is also called negative half

turn. P' is the image of P due to the rotation.

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Transformation

Definition: Rotation is a rule which shifts each point of the given object through an

equal angular displacement about a point in the given direction.

To rotate a geometrical figure, following three conditions are required:

(i) Centre of rotation (ii) Angle of rotation (iii) Direction of rotation

A figure can be rotated in two directions.

1. Anti-clockwise direction (Positive direction)

In anti-clockwise direction, an object is rotated in opposite direction of the
rotation of the hand of a clock. It is also called positive direction.

2. Clockwise direction (Negative direction)

In clockwise direction, an object is rotated in the same direction of the rotation
of the hand of a clock. It is also called negative direction.

Now, let’s learn to draw the image of a figure when it is rotated through the given
angle in the given direction about a given centre of rotation.

6.5 Rotation through 90° in anti-clockwise direction

(Positive Quarter turn about origin)

Let 'ABC be a given triangle. It is to be rotated B' Anti-clockwise
through 90° in anti-clockwise direction about O.

The following instructions are useful for this type C' C
of rotation.

(i) Join each vertex of the figure to the centre of A'

rotation O with dotted lines. B

(ii) On each dotted line, draw 90q at O with the O
help of a protractor in anti-clockwise direction. A

(iii) With the help of compasses, cut off OA' =OA,

OB' = OB and OC' = OC.

(iv) Join A'B', B'C', and C'A' successively, we get 'A'B'C'.

Here, 'A'B'C' is the image of 'ABC formed due to the rotation through 90° in anit-
clockwise direction about O.

6.6 Rotation through 90° in clockwise direction

(Negative Quarter Turn) O C

Let 'ABC be a given triangle. It is to be rotated through B
90° in clockwise direction about centre O. The following Clockwise
instructions are useful for this type of rotation. A

(i) Join each vertex of the figure to the centre of rotation A’ C’
O with dotted line.

(ii) With each dotted line draw 90° with the help of a B’
protractor in clockwise direction.

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Transformation

(iii) With the help of a compass, cut off OA' = OA, OB' = OB, OC' = OC.
(iv) Join A'B', B'C' and C'A' successively, we get triangle A'B'C.
Here, 'A'B'C' is the image of 'ABC due to the rotation through 90° in clockwise
direction.

6.7 Rotation through 180° about origin

(Half Turn) A' C
C' B
Let 'ABC be a given triangle. It is to be rotated
through 90° in anti-clockwise direction about OA
centre O. The following instructions are useful for B'
this type of rotation.

(i) Join each vertex of the figure to the centre of
rotation O with dotted line.

(ii) With each dotted line, draw 180° with the help of a protractor in clockwise
direction.

(iii) With the help to a compass, cut off OA' = OA, OB' = OB and OC' = OC.

(iv) Join A'B', B'C', and C'A' successively we get triangle A'B'C'.

Here, 'A'B'C' is the image of 'ABC due to the rotation of through 180° in clockwise
direction.

Worked out Examples

Example 1. Rotate the point P through an angle of positive 60° about the center

of rotation O. A'

Solution: Here, we join OA with centre O and make
angle 60° with OA in anti-clockwise
direction.

i.e. ‘AOA' = 60° 60°
A' is the image of A.
O A

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Transformation

Example 2. Rotate 'ABC through 90° in clockwise direction about O.
Solution:
(i) Join each vertex of the figure to the C
centre of rotation O with dotted line.

(ii) With each dotted line draw 90° with O B
Clockwise
the help of a protractor in clockwise A

direction. A’

(iii) With the help of a compass, cut off C’
OA' = OA, OB' = OB, OC' = OC.

(iv) Join A'B', B'C', and C'A' successively, B’
we get triangle A'B'C.

Here, 'A'B'C' is the image of 'ABC due to the rotation through 90° in
clockwise direction.

6.8 Use of Coordinates in Rotation

Let us take different points on a graph paper and rotate through 90° anti-clockwise,
clockwise, or 180° about centre O. Graphically, we can derive formula of rotations.

(a) Rotation through 90° in anti-clockwise direction about origin

P'(-2,4) Y Y YY

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

XO X' X O X' X O X' X O X'

Y' P'(-2,-3) P(-3,-2) Y' P(3,-4)
P'(2,-3)
P(4, 2) o P' (–2, 4) Y' P (3, –4) o P' (4, 3)
Y'
P (–3, 2) o P' (–2, –3)
P (–3, –2) o P' (2, –3)

? P (x, y) o P' (–y, x) ? P (–x, y) o P' (–y, –x) ? P (–x, –y) o P' (y, –x) ? P (x, –y) o P' (y, x)

Fig (i) Fig (ii) Fig (iii) Fig (iv)

Let us take P(4, 2) in a graph and rotate it in 90° in anti-clockwise direction about
origin O. Then, we get,

P(4, 2) o P'(– 2, 4) [in figure (i)]

Similarly, P(–3, 2) o P'(–2, –3) [in figure (ii)]

P(–3, –2) o P'(2, –3) [in figure (iii)]

P(3, –4) o P'(4, 3) [in figure (iv)]

? P(x, y) o P'(–y, x)

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(b) Rotation through 90° in clockwise direction about origin

Y YYY

P(2,3) P(-2,1) P'(1,2) P'(-2,2)

X' O X X' O X X' O X X' OP(4,-1) X

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

P'(-1,-4)

Y' Y' Y' Y'

P(2, 3) o P' (3, –2) P (–2, 1) o P' (1, 2) P (–2, –2)o P'(–2, 2) P (4, –1) o P' (–1, –4)

? P (x, y) o P' (y, –x) ? P (–x, y) o P'(y, x) ? P (–x, –y) o P' (–y, x) ? P (x, –y) o P' (–y, –x)

Fig (i) Fig (ii) Fig (iii) Fig (iv)

Let us take a point P(2, 3) in a graph paper. Rotate it through 90° in clockwise
direction about origin. Then we get,

P(2, 3) o P'(3, –2) [in figure (i)]

Similarly, P(–2, 1) o P'(1, 2) [in figure (ii)]

P(–2, –2) o P'(–2, 2) [in figure (iii)]

P(4, –1) o P'(–1, –4) [in figure (iv)]

? P(x, y) o P'(y, –x)

(c) Rotation through 180° about origin

When a point is rotated through 180q in anti-clockwise or clockwise direction about
origin the image point has the same coordinates in both cases.

Y Y Y Y
P(4,3) P(-4,2) P'(3,3) P'(-3,2)

X' O X X' O X X' O X X' O X

P'(-4,-3) P'(4,-2) P(-3,-3) P(3,-2)
Y' Y' Y' Y'
P (–4, 2) o P' (4, –2) P (3, –2) o P' (–3, 2)
P (4, 3) o P" (–4, –3) P (–3, –3) o P' (3, 3)

? P (x, y) o P' (–x, –y) ? P (–x, y) o P' (x, –y) ? P (–x, –y) o P' (x, y) ? P (x, –y) o P' (–x, y)

Fig (i) Fig (ii) Fig (iii) Fig (iv)

Let us take a point P(4, 3) and rotate it through 180° in anti-clockwise or clockwise
direction about the origin. Then we get,

P(4, 3) o P'(– 4, –3) [in figure (i)]

P(–4, 2) o P'(4, – 2) [in figure (ii)]

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Transformation

P(–3, –3) o P'(3, 3) [in figure (iii)]
P(3, –2) o P'(–3, 2) [in figure (iv)]

? P(x, y) o P'(–x, –y)

Worked out Examples

Example 1. Find the image of the point A(5, 6), B(7, 8), C(–5, –6), D(2, –5) under
the rotation through the following angles about the origin:

(a) 90° (b) –90°

(c) 180°

Solution: (a) Under the rotation through +90° about origin,

P(x, y) P'(–y, x)

A(5, 6) A'(–6, 5)

B(7, 8) B'(–8, 7)

C(–5, –6) C'(6, –5)

D(2, –5) D'(5, 2)

(b) Under the rotation through –90° about origin,

P(x, y) P'(y, –x)

A(5, 6) A'(6, –5)

B(7, 8) B'(8, –7)

C(–5, –6) C'(–6, 5)

D(2, –5) D'(–5, –2)

(c) Under the rotation through 180° about origin,

P(x, y) P'(–x, –y)

A(5, 6) A'(–5, –6)

B(7, 8) B'(–7, –8)

C(–5, –6) C'(5, 6)

D(2, –5) D'(–2, 5)

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