Shu OPT matheamtics Book 8 2077 final com (1)

Approved by Government of Nepal, Ministry of Education,Science and Technology,
Curriculum Development Center(CDC), Sanothimi, Bhaktapur, Nepal

Infinity

Optional Mathematics

8Grade

Authors
Nil Prasad Ghimire
Shakti Prasad Acharya

Sujit Shrestha

Editors
Ramesh Subedi
Jibnath Sharma

Shubharambha Publication Pvt.Ltd.
Kathmandu, Nepal

Published by:

Shubharambha Publication Pvt. Ltd.

Kathmandu, Nepal
URL: www.shubharambhapublication.com.np
E-mail: [email protected]
www.facebook.com/shubharambhapublication

Book : Infinity Optional Mathematics-Book 8

Authors : Nil Prasad Ghimire

Shakti Prasad Acharya
Sujit Shrestha

Layout Design : Zeeta Computer Service Pvt. Ltd.

Ghantaghar, Kathmandu

Mobile No. 9841418545

Copyright © : Publisher

Edition : First: 2077 B.S.



© : Publisher

No part of this book may be reproduced or transmitted by means (electronic,
photocopying, recording or otherwise) without prior written permission from the
publisher. Any breach of this condition will entail legal action and prosecution.

Printed in Nepal

Preface

Infinity Optional Mathematics is a set of 5 books tailored to teach optional mathematics
to the students in the grades between 6 and 10. It is developed in strict compliance with
the recent school syllabus prescribed by the Curriculum Development Centre (CDC),
Ministry of Education, Science and Technology, Government of Nepal.

The text-books in the series are a result of a painstaking effort from a team of dedicated
and hard-working authors who have devised the lesson units with fitting ingenuity
and the acumen derived from years of experience as teachers.

The chapters in the books feature a fundamental theory and sufficiently-worked-out
examples right at the beginning as a ready reference for students. The classification
and the order of the chapters in each book are systematic and are put in a proper
sequence so that students can learn better and comprehend the progression of topics
and subject-matter with clarity. The authors are also confident that the books in their
present form are comprehensive, and helpful to the students in their preparation,
which is consistent with the exam requirements of the new question model (grid).

We thank the management team of Shubharambha Publication for their help in the
publication of this series. We also remain indebted to Mr Ananda Prasad Poudel,
Managing Director of Zeeta Computer Service, and his team for their expertly typing,
layout and cover design.

While every effort has been made to keep the series error-free, it is still possible for
some errors to have unwittingly crept into the work. If the beneficiaries of these books,
the reviewers and the others concerned bring to our notice any discrepancies, or areas
for potential improvement, we shall be extremely grateful to you. We would love to
receive your valuable suggestion, feedback, or queries via email at shubharambha.
[email protected].

Once again, we hope that this new edition of Infinity Optional Mathematics does
more than meet the expectations of students and teachers alike.

Authors

Contents

Unit 1 RELATION 5

1.1 Order Pair 5

1.2 Cartesian Product 7

1.3 Relation 12

Unit 2 POLYNOMIAL 16

2.1 Polynomial 16

2.2 operations on Polynomials 20

Unit 3 SEQUENCE AND SERIES 24

3.1 Sequence and Series 24
3.2 Sigma Notation 28

Unit 4 MATRIX 31

4.1 Introduction to Matrix 31

4.2 Operation on Matrix 36

Unit 5 CO–ORDINATE GEOMETRY 40

5.1 Coordinate Geometry 40

5.2 Distance between two points (Distance formulae) 48

5.3 Section Formulae 58

5.4 Locus and Equation 63

Unit 6 MEASUREMENT OF ANGLES 69

6.1 System of Measurement of Angles 69

6.2 Polygon and Clock 77

Unit 7 TRIGONOMETRY 82

7.1 Trigonometry 82
7.2 Trigonometric Ratios 87
7.3 Operation Trigonometric Ratios 91
7.4 Trigonometrical Ratios of Some Standard Angles 95
7.5 Relation of Trigonometric Ratios from Pythagorean ... 102
7.6 Complementary Angles 109
7.7 Solution of a Right Angled Triangle 113
7.8 Height and Distance 115

Unit 8 VECTORS 120

8.1 Vector 120
8.2 Types of Vectors 132
8.3 Operations of Vectors 137

Unit 9 TRANSFORMATION 144

9.1 Transformation 144

9.2 Rotation 153

9.3 Translation 159

9.4 Enlargement 163

Unit 10 STATISTICS 167

10.1 Measure of Central Tendency 167

10.2 Median and Quartiles 173

10.3 Mode 181

Answer Sheet 183

UNIT

1 RELATION

1.1 Order Pair

Review

Discuss the following questions.
(a) Are {2, 3} and {3, 2} equal?
(b) Are (1, 5) and (5, 1) equal? Plot the points (1, 5) and (5, 1) on the graph paper

and discuss.
(c) Plot the points (2, 3), (1, 5), (–3, 2) and (–5, – 7) on the graph paper.

Ordered pair

The table given below shows the marks obtained by 5 students in mathematics.

Name of Students Ram Shyam Sita Rita Gita

Marks Obtained 95 85 73 92 80

To represent the above information, it can be written in the following form as:

(Ram, 95), (Shyam, 85), (Sita, 73), (Rita, 92) and (Gita, 80) which are called ordered
pairs.

In the ordered pair (Ram, 95), Ram is called first component or abscissa and 95 is
called second component or ordinate.

A set containing of any two elements in a definite order is called ordered pair. In
the ordered pair, the elements are separated by comma ( , ) and are enclosed by
( ). For example (2, 3) is an ordered pair.

Y

Since the order of the elements in the ordered pair 6 (2,3)
are fixed (2,3) and (3, 2) are not equal which can be 5 (3, 2)
illustrated in the graph alongside. 4
3
Equality of ordered pair 2

1

Observe the following ordered pairs. O 1 2 3 4 5 6X

1) A(2, 5) and B(2, 5). Are the points A and B same?

Infinity Optional Mathematics Book - 8 5

Plot them on the graph and discuss.

2) P(4, 6) and Q 12 , 12 . Are the points P and Q same?
32

Plot them on the graph and discuss.

3) E(2, 5) and F(8, –2). Are the points E and F same?

Plot them on the graph and discuss.

4) X 3, 16 and Y(5 + 1, 8 – 2). Are the points X and Y same?
2

Plot them on the graph and discuss.

The ordered pairs given in the examples (1) and (2) are equal. That is A(2, 5) =

B (2, 5) and P (4, 6) = Q 12 , 12 . Both the points have same position on the graph.
32

The ordered pairs given in the examples (3) and (4) are not equal. That is E(2, 5) ≠

F(8, –2) and X 3, 16 ≠ Y(5 + 1, 8 – 2). Both the points have different position on
2

the graph.

Two ordered pairs are said to be equal if the corresponding components of both
the ordered pairs are equal.

WORKED OUT EXAMPLES

1. If (2x – 5, 4) = (9, y + 3) then find the values of x and y
Solution:
Here, (2x – 5, 4) = (9, y + 3)
By equating the corresponding elements, we get,

2x – 5 = 9 4=y+3
or, 2x = 14 or, 4 – 3 = y
or, x = 7 or, y = 1

∴ The values of x and y are 7 and 1 respectively,

2. Make all the possible ordered pairs from the set P = {1, 2}
Solution:
All the possible ordered pairs from the set P = {1, 2} can be obtained by the

6 Infinity Optional Mathematics Book - 8

following way.

1 (1, 1) 1 (2, 1)
(2, 2)
1 2

2 (1, 2) 2

Hence, the required ordered pairs are (1, 1), (1, 2), (2, 1) and (2, 2)

Exercise 1.1

Section 'A'

1. (a) Define ordered pair with an example.

(b) Under what condition the two ordered paris (w, x) and (y, z) are equal?

2. In each of the following ordered pairs, write down its abscissa and ordinate.

(a) (2, 6) (b) (–8, 5) (c) (9, –1)

(d) (x, y) (e) (p, q) (f) (5 + a, 36)

3. Which of the following ordered pairs are equal?

(a) (2, 5) and (5, 2) (b) (–3, 8) and (–3, 8)

(c) (7, 3) and (7, 3) (d) (6, 8) and (5 + 1, 10 – 2)

(e) (4, 1) and 12 , 2 (f) (4 – 2, 5 + 1) = 3, 16
32 2

Section 'B'

4. Find the values of x and y if (b) (–3, y) = (x, 4)
(a) (x, 5) = (2, y) (d) (4, y–5) = (x + 3, 1)
(c) (x–2, y + 4) = (2, 3)

(e) (5x–3, x+y) = (2, –6) (f) or2xde, ryed+ 6 = (–3, 8)
5. From each of the following sets possible pairs.
make all the

(a) P = {5, 6} (b) Q = {a, b} (c) R = {1, 2, 3}

1.2 Cartesian Product

Consider two sets P = {a, b} and Q = {1, 2}. Let's form all possible ordered pairs
such that first component of each ordered pair is from the set P and the second
component of each order pair is from then set Q. The all possible ordered pairs
formed are (a, 1), (a, 2), (b, 1) and (b, 2).

The set of all the possible ordered pairs i.e. {(a, 1), (a, 2), (b, 1), (b, 2)} is called

Infinity Optional Mathematics Book - 8 7

Cartesian Product and it is written as P × Q.

Let A and B be two given sets. Then the set of all possible ordered pairs (a, b) such
that first element i.e. a belongs to the set A and second element i.e. b belongs to
the set B is called the Cartesian product of the set A and B. It is denoted by A × B.

So, P × Q = {a, b} × {1, 2} = {(a, 1), (a, 2), (b, 1), (b, 2)}

Q × P = {1, 2} × {a, b} = {(1, a), (1, b), (2, a), (2, b)}

Ways of representing cartesian product.

A Cartesian product from A to B can be represented by the following methods.

(a) Tabulation method

(b) Tree diagram method

(c) Arrow diagram method

(d) Graphical method.

Consider the sets A = {1, 2} and B = {3, 4}. Then

A × B = {1, 2} × {3, 4}

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

The cartesian product A × B can be represented by tabulation method as follows.

Sets B

×3 4

A 1 (1, 3) (1, 4)

2 (2, 3) (2, 4)

∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

The cartesian product A × B can be represented by tree diagram method as follows.

A B A×B
3 (1, 3)

1

4 (1, 4)

3 (2, 3)

2

4 (2, 4)

∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
8 Infinity Optional Mathematics Book - 8

The Cartesian product A × B can be represented by arrow diagram method as
follows.

AB

13

24

A×B
∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
The cartesian product A × B can be represented by the graphical method as follows:

Y

6
5
4
3
2
1

0 1 2 3 4 5X

WORKED OUT EXAMPLES

1. If A = {a, b} and B = {m, n}, then find A × B, B × A, A × A and B × B.
Solution: Here,

A = {a, b}
B = {m, n}
A × B = {a, b} × {m, n}
= {(a, m), (a, n), (b, m), (b, n)}
B × A = {m, n} × {a, b}
= {(m, a), (m, b), (n, a), (n, b)}
A × A = {a, b} × {a, b}
= {(a, a), (a, b), (b, a), (b, b)}
B × B = {m, n} × {m, n}
= {(m, m), (m, n), (n, m), (n, n)}

Infinity Optional Mathematics Book - 8 9

2. If P = {Bindu, Subi} and Q = {Nepal, India}, find P × Q and Q × P and
present them by arrow diagram method

Solution: Here,
A = {Bindu, Subi} and B = {Nepal, India}

A × B = {Bindu, Subi} × {Nepal, India}

= {(Bindu, Nepal}, (Bindu, India), (Subi, Nepal), (Subi, India)}

B × A = {Nepal, India} × {Bindu, Subi}

= {Nepal, Bindu), (Nepal, Subi), (India, Bindu), (India, Subi)}

Representing A × B in arrow diagram. Representing B × A in arrow diagram.

A B B A



Bindu Nepal Nepal Bindu

Subi India India Subi

A×B B×A

Exercise 1.2

Section 'A'

1. (a) Define cartesian product.
(b) If A × B = {(1, 2), (1, 5), (2, 2), (2, 5), (3, 2), (3, 5)}
then find the sets A and B.

2. From the following arrow diagram, find A × B.

(a) A B (b) A B
a
1
Ram Sita

2 Shyam Rita

b

3

10 Infinity Optional Mathematics Book - 8

3. From the following graphs, find P × Q.

(a) Y (b) Y

3 3

2 2
1
1
0 1 2 3 4 5X
0 1 2 3 4 5X P×Q
P×Q

Section 'B'

4. Find A × B, B × A, A × A and B × B from the following sets.

(a) A = {1, 2}, B = {3} (b) A = {2, 4}, B = {5, 10}

(c) A = {1, 2, 3}, B = {2, 3} (d) A = {a, b}, B = {p, q, r}

5. (a) If P = {p, q} and Q = {a, b}, find P × Q and Q × P. Also show that

P × Q ≠ Q × P

(b) If M = {m, n} and N = {x, y}, find M × N and N × M. Also show that

M × N ≠ N × M.

6. (a) If A = {1, 2} and B = {–2, 3} then find A × B and also present it (i) by
tabulation method (ii) by graphical method.

(b) If Y = {Krishna, Radha} and Z = {Tea, Coffee} then find Y × Z and present
it (i) by tree diagram method (ii) by arrow diagram method.

Infinity Optional Mathematics Book - 8 11

1.3 Relation

Let A = {Kathmandu, Delhi} and B = {Nepal, India}.

Then A × B = {(Kathmandu, Nepal), (Kathmandu, India), (Delhi, Nepal),

(Delhi, India)}

Consider a new set say R satisfying the condition "is capital of". Therefore
R = {(Kathmandu, Nepal), (Delhi, India)} Which is a subset of A × B i.e R ⊂ A × B.
So R is relation from A to B. It is denoted as R: A→B

For example, P = {1, 2, 3} and Q = {2, 3}. Then,

P × Q = {1, 2, 3} × {2, 3}

= {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}

A relation R1 satisfying condition "is equal to" is {(2, 2), (3, 3)}
A relation R2 satisfying the condition "is half of" is {(1, 2)}
A relation R3 satisfying the condition "is less than" is {(1, 2), (1, 3), (2, 3)} etc.

Ways of representing relation
A relation from sets A to B can be represented by the following method.

• Set of ordered pairs method

• Tabulation method

• Arrow diagram method

• Graphical method.

Consider A = {1, 4, 9} and B = {1, 2, 3}. Then

A × B = {1, 4, 9} × {1, 2, 3}

= {(1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3), (9, 1), (9, 2), (9, 3)}

A relation R from A to B satisfying the condition "is square of" can be written in
ordered pairs form as:

R = {(1, 1), (4, 2), (9, 3)}

A relation R from A to B satisfying the condition "is square of" can be represented
by tabulation method as:

x 14 9
y 12 3

12 Infinity Optional Mathematics Book - 8

A relation R from A to B satisfying the condition "is square of" can be represented
by arrow diagram method as:

R

A B

1 1
4 2
9 3

A relation R from A to B satisfying the condition "is square of" can be represented
by graphical method as:

Y

4
3
2
1

O 1 2 3 4 5 6 7 8 9 10 X

Domain and Range of a Relation
Consider a relation R = {(1, 1), (4, 2), (9, 3)}. Then the set of all x-components in R =
{1, 4, 9} = domain of R and the set of all y-components in R = {1, 2, 3} = range of R.

If R is a relation from A to B then the set of x-components of all ordered pairs in R is
called domain of R and the set of y-components of all ordered pairs in R is called range
of R.

Inverse relation

Consider a relation R = {(1, 2), (2, 4), (3, 6)}. Then a new relation R-1 obtained by

interchanging the components of each ordered pairs of the relation R i.e. R–1 =

{(2, 1), (4, 2), (6, 3)} is called inverse relation of R. R–1
R

1 2 2 1
2 4 4 2
3 6 6 3

A new relation obtained by interchanging the components of each ordered pair of the
given relation R is called inverse relation of R. It is denoted by R–1.

Infinity Optional Mathematics Book - 8 13

WORKED OUT EXAMPLES

1. If A = {1, 2, 3} and B = {1, 2, 3}, find the relation from set A to set B
satisfying the following conditions.

(a) is equal to (b) is double of

Solution: Here,

A = {1, 2, 3} and B = {1, 2, 3}

So, A × B = {1, 2, 3} × {1, 2, 3}

= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

(a) A relation from A to B satisfying the condition 'is equal to' is
{(1, 1), (2, 2), (3, 3)}

(b) A relation from A to B satisfying the condition 'is double of' is {(2, 1)}

2. From the relation given below.

R1 Y R2 X

1 2 4
2 3 3
3 4 2
5 1

O 12 34

(i) Write each relation is set of ordered pairs. R1
(ii) Find its domain and range.
(iii) Find its inverse relation. 1 2
Solution: 2 3
(a) The given relation is at the along side. 3 4
5
(i) R1 = {(1, 2), (2, 3), (3, 4)}
(ii) Domain of R1 = {1, 2, 3} Y R2 X
Range of R1 = {2, 3, 4}
(iii) R1–1 = {(2, 1), (3, 2), (4, 3)} 4
(b) The given relation is at the along side. 3
(i) R2 = {(1, 1), (2, 1), (3, 2), (4, 2)} 2
(ii) Domain of R2 = {1, 2, 3, 4} 1
Range of R2 = {1, 2}
(iii) R2–1 = {(1, 1), (1, 2), (2, 3), (2, 4)} 0 12 34

14 Infinity Optional Mathematics Book - 8

Exercise 1.3

Section 'B'

1. (a) Define relation.
(b) Define inverse relation.

2. Given that A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}. Find the
relation defined by following

(a) is equal to (b) is half of

(c) is less than (d) is double of

3. Find the domain, range and inverse relation of the following relations.

(a) R1 = {(1, 2), (1, 3), (1, 4)} (b) R2 = {(1, 1), (2, 4), (3, 9), (4, 16)}

4. From the relations given below.

(a) R1 (b) R1

a p 3 1
b q 4 9
c r 5 16
25
Y R4
(c) Y R3 (d)

OX OX

(i) Write each relation in the set of ordered pairs.
(ii) Find its domain and range.
(iii) Find its inverse relation.

5. If A = {2, 4} and B = {1, 2}. Find the relation from A to B satisfying the
following conditions.

(a) is double of (b) is greater than

Represent each of the relation in the following methods.

(i) Arrow diagram (ii) Tabulation (iii) Graphical.

6. If P = {1, 2, 3} and Q = {2, 3} then find the relation from P to Q defined
by the following conditions.

(a) x = y (b) x + y ≤ 4 (c) x < y

Also find the domain and range of each of the relation

Infinity Optional Mathematics Book - 8 15

UNIT

2 POLYNOMIAL

2.1 Polynomial
Review:

• Discuss the examples of constant and variable.

• Define algebraic expression with examples.

• What is the power of x in 4x3, 1 and 3 x2­?
x3

• What is the degree of 2x3y?

Basic terms of polynomial.

Let's discuss some basic terms which help to define polynomial.
Variable
A variable is a quantity which takes different values. For examples x, y, z etc.
Constant
A constant is a quantity which has a fixed numerical value.
For examples 1, 2, 3, a, b etc.
Term
A term is a quantity containing a constant or a variable or a product of constant and
variable. For example 5, x, 5x etc.
Algebraic expression
An algebraic expression is a combination of terms which are connected by the signs
(+) and (–). For example:
2x + 5, 5x2 – 3x + 1, 5x2y + 3xy + y2 etc.
Like term
Like terms are the terms having same variables and same powers. For example:

16 Infinity Optional Mathematics Book - 8

(5x and – 3x), (xy2 and 8xy2), (6x3 and 2x3) etc.

Unlike terms

The terms which are not like are called unlike terms.
For example –3x, 8xy2, 2x3 etc.

Polynomial
Consider some examples of algebraic examples.
(i) 6x2 + 4x – 1

In this expression, the numerical coefficients are 6, 4 and – 1 and powers of
variable in each term are 2, 1 and 0. So it is a polynomial.

(ii) 4x3 3 x2 + 3 x + 1
5 3
1
In this expression, the numerical coefficients aarpeo4ly, n–5o3m, ia3l. and 3 and powers
of variable in each term are 3, 2, 1, 0. So it is

(iii) 8x3 – 3x2/3 + 4 – 5
x

In this expression, the numerical coefficients are 8, –3, 4 and – 5 and the

powers of variable in each term are 3, 32, –1, 0. So it is not a polynomial,

From the above examples, it is concluded that

An algebraic expression in which the power of variables in each term is non
negative integers (whole numbers) is called as polynomial

Degree of a Polynomial

If the polynomial is of one variable, the highest power of the variable used in the
polynomial is called the degree of the polynomial. For example: f(x) = 4x3 – 5x2 +
3x – 1 is a polynomial. In this polynomial, the highest power of the variable is 3. So
the degree of the polynomial f(x) is 3.

If the polynomial has two or more variables, than the highest degree of a term
among the terms of the polynomial is called the degree of the polynomial. For
example: f(x, y) = 3xy + 4x2y3 – 6x2y2 is a polynomial. In this polynomial, the
degree of the first term i.e. 3xy = 1 + 1 = 2, the degree of the second term i.e. 4x2y2
= 2 + 3 = 5 and degree of the third term i.e. 6x2y3 = 2 + 2 = 4. The highest degree of
the second term is 5. So, the degree of the polynomial f(x, y) is 5.

Infinity Optional Mathematics Book - 8 17

Types of Polynomial

The types of Polynomial on the basis of the number of terms are

1) Monomial
A polynomial having only one term is called a monomial.
For example 5, x, 3x, 8x2y etc.
2) Binomial
A polynomial having two terms is called a binomial.
For example x – 3, 5x2 + 2x, x3 + y3 etc.
3) Trinomial
A polynomial having three terms is called a trinomial. For example
3x2 – 2x + 1, 5x2y2 – 4xy + 2, x4 + x2 + 1 etc.
4) Zero polynomial
A polynomial in which the coefficients of all the terms of the polynomial are

zero, is called zero polynomial. For example
0.x, 0.x3 – 0.x2 + 0, 0.x2 + 0.x + 0 etc.

Standard forms of polynomial

If the degree of the terms of a polynomial are arranged either in ascending or in

descending order then the polynomial is said to be in standard form.

For example: f(x) = 3x3 – 4x2 + 2x + 8 (in descending order)

g(x) = 6 – 8x + 2x2 – x3 (in ascending order)

WORKED OUT EXAMPLES

1. Identify which of the following algebraic expressions are polynomial.

(a) x – 2x + 5 (b) 5x2 + 8x – 9

Solution: Here,

(a) x – 2x + 5 is not polynomial because the power of x in first term is 1
which is not a whole number. 2

(b) 5x2 + 8x – 9 is a polynomial because the power of x in each term are
whole number.

2. Find the degree of each of the following polynomial

(a) 3x3 – 8x2 + 4 (b) x3 – 4x2y2 + y3
Solution: Here,

(a) 3x3 – 8x2 + 4. Here, the highest power of x is 3. So the degree of the given
polynomial is 3.

18 Infinity Optional Mathematics Book - 8

(b) x3 – 4x2y2 + y3. Here, the degree of x3 is 3, the degree of 4x2y2 is 2 + 2 =
4 and the degree of y3 is 3. So the degree of the given polynomial is 4.

3. Identify the following polynomials on the basis of the number of
terms.

(a) 5x2 – x + 3 (b) 2x (c) x4 – 1
Solution: Here,

(a) 5x2 – x + 3

Since the given polynomial has three terms, it is a trinomial.

(b) 2x

Since the given polynomial has only one term, it is a monomial.

(c) x4 – 1

Since the given polynomial has two terms, it is a binomial.

4. Rewrite the following polynomials in standard form.

(a) 5x – 3x2 + 8x3 – 1 (b) 2x3 – x4 + 6x.
Solution: Here,

(a) 5x – 3x2 + 8x3 – 1

Writing given polynomial in the standard form as 8x3 – 3x2 + 5x – 1

(b) 2x3 – x4 + 6x

Writing given polynomial in the standard form as – x4 + 2x3 + 6x.

Exercise 2.1

Section 'A'

1. (a) Define polynomial with an example.

(b) Define binomial with an example.

2. Identify which of the following algebraic expressions are polynomial?

(a) x2 – y2 (b) 5x 3 – 6x + 4

1 (d) 8x2yz + 3 xy – 1
8
(c) 8x3 – 2x2 + 1 5
(f) 6x2 – x + 2
(e) x4 – 4x3 + 3x2 – 5

3. Write the degree of each of the following polynomials.

(a) 6x3 (b) – 3

(c) 3 x5 + 6x4 – 8x + 2 (d) 4x2 – 8x3 + 6

(e) – 9x2yz + x3 (f) x2y2 + 3x3y3 + xy.

Infinity Optional Mathematics Book - 8 19

4. Rewrite the following polynomials in the standard form.

(a) 5x3 – 8 + 4x – x2 (b) 5x2 + x3 – 6x + 1
8
y4
(c) –2y + y3 + 3 + 7 (d) y6 – 2y + 3y4 + 6y2

(e) 3x – x2 + x3 (f) 18 + 7x2 – 4x4 + 3x
5 7

5. Classify the following polynomials as monomial, binomial and

trinomial.

(a) x2 – y2 (b) – 6x3 + 4x – 6

(c) – 8x2y (d) 5 x2 + 32 x – 21
(f) x3 – 1
(e) 3

2.2 Operations on Polynomials

Two or more polynomials can be added, subtracted and multiplied.

Addition and Subtraction of polynomials
Let f(x) = 5x2 – 8x + 2 and g(x) = 3x2 + 5x – 9 be the polynomials. The addition of f(x)

and g(x) can be done as follows.
f(x) + g(x) = (5x2 – 8x + 2) + (3x2 + 5x – 9)
= (5 + 3)x2 + (–8 + 5)x + (2 – 9)
= 8x2 – 3x – 7

Similarly, the subtraction of f(x) and g(x) can be done as follows.
f(x) – g(x) = (5x2 – 8x + 2) – (3x2 + 5x – 9)
= 5x2 – 8x + 2 – 3x2 – 5x + 9
= (5 – 3)x2 + (– 8 – 5)x + (2 + 9)
= 3x2 – 13x + 11

While adding and subtracting the polynomials, following steps to be followed.

• Write polynomials first in standard form.
• Combine the coefficients of the like terms.
• Add or Subtract the coefficients of the like terms to get the coefficient of each

term in new polynomial.

Multiplication of polynomials
Let f(x) = (2x – 3) and g(x) = (4x2 – 5x – 1) be two polynomials. The product of f(x)

and g(x) can be done as follows.
f(x). g(x) = (2x – 3) . (4x2 – 5x – 1

20 Infinity Optional Mathematics Book - 8

= 2x (4x2 – 5x – 1) – 3(4x2 – 5x – 1)
= 8x3 – 10x2 – 2x – 12x2 + 15x + 3
= 8x3 + (–10 – 12)x2 + (– 2 + 15)x + 3
= 8x3 – 22x2 + 13x + 3
While multiplying the polynomials, following steps to be followed.
– Multiply each term of one polynomial by each term of the other polynomial by

using the relation am × an = am+n
– Combine the coefficients of the like terms.
– Add or subtract the coefficients of the like terms to get the coefficient of each

term in the new polynomial.

WORKED OUT EXAMPLES

1. Find the sum of the following polynomials f(x) = 6x4 – 2x3 + 5x – 2 and
g(x) = –4x4 + 6x + 8

Solution: Here,
f(x) = 6x4 – 2x3 + 5x – 2
g(x) = – 4x4 + 6x + 8

f(x) + g(x) = ?
Now, f(x) + g(x) = (6x4 – 2x3 + 5x – 2) + (– 4x4 + 6x + 8)
= 6x4 – 2x3 + 5x – 2 – 4x4 + 6x + 8
= (6 – 4)x4 – 2x3 + (5 + 6)x + (–2 + 8)
= 2x4 – 2x3 + 11x + 6

2. Find p(x) – q(x) if p(x) = 2x5 – 4x4 + 3x3 – 8x + 1 and q(x) = 2x4 – 3x + 2

Solution: Here,

p(x) = 2x5 – 4x4 + 3x3 – 8x + 1

q(x) = 2x4 – 3x + 2

p(x) – q(x) = ?

Now, p(x) – q(x) = (2x5 – 4x4 + 3x3 – 8x + 1) – (2x4 – 3x + 2)

= 2x5 – 4x4 + 3x3 – 8x + 1 – 2x4 + 3x – 2

= 2x5 + (– 4 – 2)x4 + 3x3 + (– 8 + 3)x + (1 – 2)

Infinity Optional Mathematics Book - 8 21

= 2x5 – 6x4 + 3x3 – 5x – 1

3. Find the product of the following polynomials.
f(x) = (x2 + 5) and g(x) = (3x2 – 5x + 7)

Solution: Here,

f(x) = x2 + 5

g(x) = 3x2 – 5x + 7

f(x). g(x)= ?

Now, f(x) . g(x) = (x2 + 5). (3x2 – 5x + 7)

= x2(3x2 – 5x + 7) + 5(3x2 – 5x + 7)

= 3x4 – 5x3 + 7x2 + 15x2 – 10x + 35

= 3x4 – 5x3 + (7 + 15)x2 – 25x + 35

= 3x4 – 5x3 + 22x2 – 25x + 35

4. What must be subtracted from 3x2 – 4x + 2 to get x2 + 8x – 7?

Solution: Here,

Let required polynomial to be subtracted be k.

According to question,

(3x2 – 4x + 2) – k = (x2­ + 8x – 7)

or, (3x2 – 4x + 2) – (x2 + 8x – 7) = k

or, 3x2 – 4x + 2 – x2 – 8x + 7 = k

or, (3 – 1)x2 + (–4 – 8)x + (2 + 7) = k

or, 2x2 – 12x + 9 = k

∴ k = 2x2 – 12x + 9

The required polynomial is 2x2 – 12x + 9.

22 Infinity Optional Mathematics Book - 8

Exercise 2.2

1. Find the sum of the following polynomials.
(a) f(x) = 5x – 3 and g(x) = – 8x + 1
(b) f(x) = x2 – 8x + 5 and g(x) = 3x2 – 2x + 1
(c) f(x) = – 6x2 – 2x – 9 and g(x) = x2 + 7x – 6
(d) f(x) = 8x4 – 5x3 + 3x2 – 8 and g(x) = 8x3 + 5x – 12
(e) f(x) = 8x7­– 2x3 + 1 and g(x) = 4x6 + 3x3 – 6

2. Find p(x) – q(x) from the following polynomials

(a) p(x) = 4x + 5 and q(x) = 8x – 1

(b) p(x) = 6x2 – 3x + 2 and q(x) = –x2 + 2x – 9

(c) p(x) = –9x2 + 2x + 8 and q(x) = 2x2 – 7x + 1

(d) p(x) = 4x3 – 2 x2 + 21x + 1 and q(x) = –51x3 + 4x2 + 1 x – 5
5 3 6 3 2 6
(e) p(x) = 3x5 – 8x4 + 6x3 – 4x + 2 and q(x) = 2x5 – 3x3 + 9

3. Find the product of the following polynomials.

(a) f(x) = 3x2 and g(x) = (5x – 8)

(b) f(x) = – 5x and g(x) = (4x + 3)

(c) f(x) = (2x – 3) and g(x) = (4x2 + 6x + 9)

(d) f(x) = (7x + 2) and g(x) = (2x2 – 5x – 1)

(e) f(x) = (x2 – 3x + 2) and g(x) = (4x2 + 6x – 7)

4. (a) What should be added to 4x – 3 to get – 5x + 1 ?

(b) What should be added to 3x2 – 8x – 3 to get x2 + 2x – 8 ?

(c) What should be subtracted from – 8x + 2 to get 6x – 3?

(d) What should be subtracted from 4x2 – x – 3 to get 6x2 + 2x – 5?

(e) What should be subtracted from the sum of x3 + 2x2­ – 1 and 3x2 – x + 2
to get 1 ?

Infinity Optional Mathematics Book - 8 23

UNIT SEQUENCE AND SERIES

3

3.1 Sequence and Series

Discuss the following questions.
(a) If f(x) = x + 2 then find the values of f(1), f(2), f(3) f(4) and f(5).
(b) If f(x) = 2x then find the values of f(1), f(2), f(3), f(4), f(5).
(c) Observe the pattern of figures given below.

,, , ..., ..., ..., ...,

(i) Add one more figure of above pattern
(ii) Express the above pattern in terms of numbers.

Sequence (progression) 7, 9, ..............
Let's consider a set of numbers ↓ ↓
3rd term 4th term
3, 5,
↓↓
1st term 2nd term

In this set of numbers, each number (except first) is obtained by adding 2 to the
preceding term. So the number next to 9 is 9 + 2 = 11. So, 3, 5, 7, 9 ..... is a sequence.

Let's consider another set of numbers

1, 3, 6, 10, ..............
↓↓ ↓ ↓
1st term 2nd term 3rd term 4th term

Hence the given numbers 1, 3, 6, 10 is also sequence.

Here,

t2 – t1 = 3 – 1 = 2
t3 – t2 = 6 – 3 = 3
t4 – t3 = 10 – 6 = 4
Similarly,

t5 – t4 = 5

24 Infinity Optional Mathematics Book - 8

or, t5 – 10 = 5
∴ t5 = 15
Similarly,
t6 – t5 = 6
or, t6 – 15 = 6
∴ t6 = 21

From the above examples, it is concluded that

A set of numbers each of which is formed according to a certain rule is called a
sequence. Each element of a sequence is called its term.

Series

Look at the following examples.

Sequence Series

2, 4, 6, 8, 10, 12 2 + 4 + 6 + 8 + 10 + 12

1, 3, 6, 12, 24, 48, ....., ..... 1 + 3 + 6 + 12 + 24 + 48 + ......

Thus, when the terms of a sequence are separated by the sign of + or – signs, the

corresponding series is formed.

If a1, a2, a3, ..., an, ... is a sequence then its corresponding series is a1 + a2 + a3 + ... an + ...

nth term or General Term of a sequence.
Consider on sequence 1, 3, 5, 7, 9, ... . Can you find 5th term of the given sequence?
Can you find 15th term of the sequence? Can you find 60th term of that sequence?
Can you find 100th term of the sequence?

The terms 5th and 15th of the sequence can be obtained just by counting numbers. But
60th term and 100th term of that sequence are difficult to get by counting numbers.
For this, at first we find the rule of the sequence as the specific term i.e. nth term

Let t1, t2, t3, ..., tn, ... be a sequence then the nth term is denoted by tn. The nth
term of the given sequence is also known as the general term of that sequence.

Infinity Optional Mathematics Book - 8 25

For example, 7, 9, 11, 13, 15, ... is an infinite sequence. The general term of the
above sequence can be obtained as follows.

Here the given sequence is 11, 13, 15
7, 9,

2 2 2 2

Here difference between two successive terms is 2. So,

first term t1 = 7 = 2 × 1 + 5
t2 = 9 = 2 × 2 + 5
t3 = 11 = 2 × 3 + 5
t4 = 13 = 2 × 4 + 5
........... ........... .............

∴ tn = 3n + 5

WORKED OUT EXAMPLES

1. Draw one more shape in the following sequence of figures and present
in a numerical sequence.

,, , ..., ..., ..., ...,

Solution:
One more figure from the given sequence of figures is

The numerical sequence from the given sequence of figures is 1, 3, 5, 7, ..., ...

2. Find the next three terms of each of the following sequnce.

(a) 5, 6, 7, 8, ..., ... (b) 8, 5, 2, – 1, ..., ...

Solution:

(a) The given sequence is 5, 6, 7, 8, ..., ...

The next three terms of the given sequence are 9, 10 and 11.

26 Infinity Optional Mathematics Book - 8

(b) The given sequence is 8, 5, 2, – 1, ..., ...

Since the given sequence is decreasing by – 3, the next three terms of the
given sequence are – 4, – 7 and – 10.

Exercise 3.1

Section 'A'

1. (a) Define a sequence with an example.

(b) Define a series with an example.

2. Identify the following set of numbers form a sequence or not. Give
reasons.

(a) 1, 2, 3, 4, ..., .... (b) 2, 4, 6, 8, ..., ....

(c) 1, 3, 6, 10, ..., ... (d) 6, 2, – 2, – 6, ..., ...

(e) 1, 3, 9, 27, ..., ... (f) 2, 3, 5, 8, 11, ..., ...

3. Identify which of the following are sequence or series.

(a) 1, 3, 5, 7, ..., ... (b) 2 + 5 + 8 + 11 + ...

(c) –2 + 6 – 10 + 14 – 18 (d) – 1, 2, – 3, 4, –5, 6

(e) 1, 12, 14, 18, ... (f) –1 + 2 – 3 + 4 –, ...
3 5 7 9

Section 'B'

4. Draw one more shape in the following sequence of figures and present
in a numerical sequence.

(a)

,, , ......, ......
, ......, ......
(b) ,
,,

(c)

,, , ......, ...... , ......, ......
(d) ,

,

Infinity Optional Mathematics Book - 8 27

5. Find the next three terms of each of the following sequence.

(a) 1, 3, 5, 7, ..., ... (b) 10, 5, 0, – 5, ..., ...
(c) 3, 9, 27, ..., ...
(d) –3 , 47, –85, 69, ..., ...
Section 'C' 2

6. Write the first five terms of the sequence whose general term (tn) is
given by

(a) tn = 3n + 2 (b) tn = 4n – 1

(c) tn = n (d) tn = 1
n+1 (f) tn = (2–n1+1)n+1

(e) tn = (–1)n (n + 5) 2n+5

7. Find the nth term of the given sequences and use it to find 10th term.

(a) 2, 3, 4, 5, ..., ... (b) 5, 7, 9, 11, ..., ...

(c) 20, 15, 10, 5, ..., ... (d) 11, 18, 25, 32, ..., ...
(e) 12, 14, 16, 18, ..., .... (f) 1, 4, 16, 25, ..., ...

3.2 Sigma Notation

Consider a series 3 + 5 + 7 + 9 + ... to 10 terms.

Here, t1 = 3 = 2 × 1 + 1
t2 = 5 = 2 × 2 + 1
t3 = 7 = 2 × 3 + 1
t4 = 9 = 2 × 4 + 1
........ ........ ..........

tn = 2 × n + 1 = 2n + 1

Thus, the sum of first 10 terms is represented as 3 + 5 + 7 + 9 + ... to 10 terms =

Σ10
(2n + 1).
n=1

If a1, a2, a3, ..., an be a sequence of n terms then its corresponding series is a1 + a2 + a3 +

... + an. This sum can be written in Sigma notation as:

Σn

a1 + a2 + a3 + ... + an = k = 1ak

28 Infinity Optional Mathematics Book - 8

WORKED OUT EXAMPLES

Σ5

1. Evaluate: (3n – 1)
n=1

Solution: Here,

Σ5

(3n – 1) = (3 × 1 – 1) + (3 × 2 – 1) + (3 × 3 – 1) + (3 × 4 – 1) + (3 × 5 – 1)
n=1

= 2 + 5 + 8 + 11 + 14

= 40.

2. Rewrite the following series using Sigma notation

2 + 4 + 6 + 8 + 10 + 12 + 14

Solution: Here,

2 + 4 + 6 + 8 + 10 + 12 + 14

Now, t1 = 2 = 2 × 1
t2 = 4 = 2 × 2
t3 = 6 = 2 × 3
t4 = 8 = 2 × 4
..... ..... ......

tn = 2 × n = 2n 7
Σ∴ 2 + 4 + 6 + 8 + 10 + 12 + 14 = 2n
n=1

Infinity Optional Mathematics Book - 8 29

Exercise 3.2

1. Evaluate: Section 'B' Σ(c) 8
(2n + 5)
Σ(a) 5 Σ(b) 4 n=4
(n + 5) (3n – 1)
n=1 n=1

Σ(d) 6 (e) Σ6 Σ(f) 5
(20 – m) (–1)k(2k + 3)
k=1 3k m= 1 k=2

2. Rewrite the following series using Sigma notation

(a) 1 + 2 + 3 + 4 + 5 (b) 2 + 4 + 6 + 8 + 10 + 12

(c) 3 + 5 + 7 + 9 + 11 (d) 0 – 5 – 10 – 15

(e) 12 + 2 + 3 + 45 (f) 1 + 1 + 1 + 116+ 1 + 1 + 1
3 4 4 9 25 36 49

30 Infinity Optional Mathematics Book - 8

UNIT

4 MATRIX

4.1 Introduction to Matrix

The table given below shows the marks obtained by Bindu and Subi in Mathematics,
Science and Social.

Name of Students Mathematics Science Social
Bindu 85 90 72
Subi 95 87 84

Expressing the above information in the rectangular form by omitting the name of

students and subjects and enclosed by round or squared brackets as shown below.
85 90 72
85 90 72 or 95 87 84
95 87 84

In the above rectangular form, the numbers are arranged in horizontal lines (called
rows) and in vertical lines (called columns). Such rectangular arrangement of
numbers is called a matrix. For example

85 90 72 → 1st row (R1)

95 87 84 → 2nd row (R2)
↓↓ ↓

1st column 2nd column 3rd column

A rectangular arry of numbers arranged in rows and columns and enclosed by
round (or square) brackets is called a matrix.

Components and Notation of a matrix

Matrices are denoted by capital letters of English alphabet as A, B, C ... X, Y, Z and
their components (also called elements) by the small letters of english alphabet as
a, b, c, ... x, y, z. For example

A= a b c
x y z

The matrix and its elements can also be denoted by following way.

A= a11 a12 a13
a21 a22 a23

Infinity Optional Mathematics Book - 8 31

Where,

a11 is the element in the 1st row and 1st column.
a12 is the element in the 1st row and 2nd column.
a13 is the element in the 1st row and 3rd column.
a21 is the element in the 2nd row and 1st column.
a22 is the element in the 2nd row and 2nd column.
a23 is the element in the 2nd row and 3rd column.
In general, aij is an element lies in i row and j column.

Order or Size or Dimension of a Matrix

12
Consider a matrix A = 3 4

56

In the matrix A, there are 3 rows and 2 columns. So the order of the matrix A is
3 × 2. The order of matrix is obtained as number of rows of the matrix followed by
the number of columns of the matrix.

For example, M = 3 7 and N = (0 –3 – 6).
–3 4

In the matrix M, there are 2 rows and 2 columns. So the order of M is 2 × 2. In the

matrix N, there are 1 row and 3 columns. So, the order of N is 1 × 3. They are also

written as M = 3 7 2×2 and N = (0 – 3 – 6)1×3
–3 4

Types of Matrices

Row matrix

A matrix having only one row is called a row matrix. For example,

A = (1 – 3)1×2, B = (a b c)1×3 are row matrices.
Column matrix

A matrix having only one column is called a column matrix. For example,
p
3
P = –2 2×1, Q = q 3×1 are column matrices.
r

32 Infinity Optional Mathematics Book - 8

Rectangular matrix

A matrix whose number of rows and number of columns are not equal is called a
12
3 5 1 2 × 3, B = 34
rectangular matrix. For example : A = 2 4 6 5 6 3×2

Square matrix

A matrix having same number of rows and columns is called a square matrix.

For example, E = (5)1×1, F = a b 2×2 are the square matrices
c d

Zero or Null matrix

A matrix having each of its elements equal to zero is called a zero or null matrix.

For example, A = 0 0 2×2, B = (0 0 0)1×3 are the Zero or Null matrices.
0 0

Equal Matrices

Two matrices are said to be equal if they are of same order and their corresponding
–4 –4
elements are equal. For example, if A = 1 8 2×2 and B = 1 8 2×2 then A = B.
0 0

WORKED OUT EXAMPLES

1. Find the order of the following matrices.
1
(a) A = 2 (b) 3 0 –9
B= 1 24
–3

Solution: 1
(a) Here, A = 2
–3

The matrix A has 3 rows and 1 column. So, the order of A is 3 × 1.

3 0 –9
(b) Here, B = 1 24

The matrix B has 2 rows and 3 columns. So, the order of B is 2 × 3.

2. Construct 2 × 2 matrix A whose element aij is given by aij = 2i + j.

Solution:

Here, aij = 2i + j.

Infinity Optional Mathematics Book - 8 33

Let the 2 × 2 matrix be a11 a12
a21 a22

Put i = 1 and j = 1. Then a11 = 2 × 1 + 1 = 3

Put i = 1 and j = 2. Then a12 = 2 × 1 + 2 = 4

Put i = 2 and j = 1. Then a21 = 2 × 2 + 1 = 5

Put i = 2 and j = 2. Then a22 = 2 × 2 + 2 = 6

∴ The required 2 × 2 matrix is 34
56

3. Find the values of a, b, c and d if a–3 4b – 8 = 1 4
2c – 3 d+5 –7 2

Solution:

Here, a–3 4b – 8 = 1 4
2c – 3 d+5 –7 2

Equating the corresponding elements of the equal matrices, we get

a–3=1 4b – 8 = 4 2c – 3 = – 7 d+5=2

or, a = 1 + 3 or, 4b = 8 + 4 or, 2c = – 7 + 3 or, d = 2 – 5

∴ a=4 or, b = 12 or, 2c = – 4 ∴ d=–3
4 ∴ c=–2
∴ b=3

∴ The required values of a, b, c and d are 4, 3, –2 and – 3 respectively.

Exercise 4.1

Section 'A'

1. (a) Define matrix with an example

(b) Define the following matrices with an example

(i) Row matrix (ii) Column matrix

(iii) Rectangular matrix (iv) Square matrix

(v) Null matrix

2. (a) If A = 3 –5 then find the values of a11, a12, a21 and a22
1 9

(b) If B = 1 23 then find the values of b11, b13, b22 and b23.
4 0 –5

3. Write down the order of the following matrices.

(a) A= 1 (b) B = (10 –5 4) (c) C = (– 5)
2
34 Infinity Optional Mathematics Book - 8

a b 11 (f) –26 –02 –14
c d
(d) D= (e) 2 2
3 3
4. State the type of the following matrices.

1 2 3 0 0 –1
4 5 6 0 0
(a) (b) (c) –2
–3

(d) –13 5 (e) (– 5 3) (f) (2)
2

5. Construct the following matrices.

(a) 1 × 3 row matrix (b) 3 × 2 rectangular matrix

(c) 2 × 2 square matrix (d) 2 × 3 zero matrix

(e) 2 × 1 column matrix (f) 1 × 1 square matrix

6. State which of the following matrices are equal.

P= 1 2 , Q= 2 , R= 1 0 –3 , S = (4 – 3 6)
–3 6 5 4 –5 3
14
A = (2 5), B = (4 –3 6), C= 1 2 , D = 0 –5
–3 6
–3 3

Section 'B'

7. Construct 2 × 2 matrix A whose elements aij are given by

(a) aij = i + j (b) aij = 2i – j

(c) aij = ij (d) aij = 3i – 2j

8. Find the values of p and q in each of the following case

(a) p = 4 (b) (3 + p 2q) = (5 –6)
q –2 (d) (p + q q) = (– 8 – 3)

(c) 33pq = –12
6

Section "C"

9. Find the values of a, b, c and d.

(a) a –3 = –2 b (b) a3–c3 d+b1 = –2 –1
4 d –c 1 4 0

(c) 42b++a5 6 = a+3 d (d) –2a 36 = 2 3 d–1
4c+7 4 –5 2a + b b+c 0 3 –4 0

Infinity Optional Mathematics Book - 8 35

4.2 Operations on matrices

Addition, Subtraction and multiplication are the operation of matrices. Here we
discuss only addition and subtraction of the matrices.

Addition of matrices

Consider two matrices A = 1 2 2×2 and B = 3 0 2×2.
3 4 2 6

So, A + B = 1 2 + 3 0
3 4 2 6

= 1+3 2+0 = 42
3+2 4+6 5 10 2×2

Here, the matrix (A + B) is also a matrix of the same order as that of A or B.

If A and B are two matrices of the same order, then the sum (A + B) is obtained
by adding corresponding elements of A and B. The order of (A + B) is same as that
of A or B.

Subtraction of matrices

Consider two matrices A = 3 4 2×2 and B = 1 2 2×2 .
5 6 3 4

Here, the order of both the matrices A and B are same.

So, A – B = 3 4 – 1 2
5 6 3 4

= 3–1 4–2 = 2 2
5–3 6–4 2 2 2×2

i.e the matrix (A – B) is also a matrix of the same order as that of A or B.

If A and B are two matrices of the same order, then the difference (A – B) is
obtained by subtracting the corresponding elements of A and B. The order of
(A – B) is same as that of A or B.

Transpose of a matrix.

12

Consider a matrix A = 34
5 6 3×2
The matrix formed by interchanging the rows and columns of the matrix A i.e.

1 3 5 2 × 3 is denoted by AT or A'.
2 4 6

Therefore, AT = 1 3 5 .
2 4 6

36 Infinity Optional Mathematics Book - 8

The new matrix obtained by interchanging rows and columns of a given matrix A
is called the transpose of the matrix A. It is denoted by A' or AT.

Note

The transpose of the transpose of a matrix is the matrix itself (A')' = A.

Multiplication of a matrix by a scalar

Consider a matrix A = 1 3 . Then 3A = 3 1 3
–2 5 –2 5

= 3×1 3×3 = 39
3×(–2) 3×5 –6 15

If a matrix of any order is multiplied by a scalar then every element of the matrix
is multiplied by that scalar

WORKED OUT EXAMPLES

1. Add the following matrices. 5 –3 + 6 2
0 2 3 –5

Solution: we have,

5 –3 62 5 + 6 –3 + 2 11 –1
0 2 + 3 –5 = 0 + 3 2+(–5) = 3 –3

2. 2 3 –2 3
If P = 1 5 and Q = 0 –5 then find 3P – 2Q.

Solution: Here,

23 –2 3
P = 1 5 and Q = 0 – 5 , 3P – 2Q = ?

2 3 = 3×2 3×3 = 6 9
Now, 3P = 3 1 5 3×1 3×5 3 15

–2 3 2×(–2) 2×3 = –4 6
2Q = 2 0 – 5 = 2×0 2 × (–5) 0 –10

69 –4 6
∴ 3P – 2Q = 3 15 – 0 –10

Infinity Optional Mathematics Book - 8 37

= 6 –(–4) 9–6 = 6+4 9–6 = 10 3
3–0 15 –(–10) 3–0 15 + 10 3 25 .



3. If 9 3x – 3 2 6 1
10 –7 7 6y = 3 2 then find the values of x and y.

Solution: Here,

9 3x – 3 2 = 6 1
10 –7 7 6y 3 2

9–3 3x – 2 = 6 1
or, 10 – 7 – 7 – 6y 3 2

6 3x – 2 = 6 1
or, 3 –7 – 6y 3 2

By Comparing the corresponding elements,

3x – 2 = 1 – 7 – 6y = 2

or, 3x = 3 or, – 6y = 2 + 7

∴ x=1 or, –6y = 9

or, y = –3
2

Hence, the values of x and y are 1 and –3 respectively.
2

Exercise 2.2

Section 'A'
1. (a) Under what condition that two matrices can be added or subtracted?

a
(b) If A = b then find kA, where k = 2.

(c) If P = 1 2 , then find 2P.
3 4

2. Find the transpose of the following matrices.

(a) A = (2 – 3) 3 (c) C= 1 3
(b) B = 6 5 –1

(d) D= 2 5 –3 –3 6 123
3 3 6 (e) E = 2 5 (f) F = 8 7 6

01 543

38 Infinity Optional Mathematics Book - 8

Section 'B'

3, Add the following matrices.

(a) (– 2 3) + (4 1) (b) 8 –2
0+3
6 –5

5 2 + 1 3 (d) 34 –03 + 1 2
(c) 1 1 8 2 3 4

1 2 4 2 3 4 2 –4 10
(e) –5 3 0 1 2 3
+ (f) –3 8 + 2 –3
1 –3 –4 –2
4. Subtract the following matrices.

(a) (6 4) – (2 1) 68
(b) 4 – 6

52

1 9 – 1 5 (d) 52 –13 – 0 –5
(c) –8 2 2 –6 21

2 3 4 –1 3 0 4 8 69
1 2 4 –5 1 –6 1 52.
(e) – (f) 2 5 – 13
3

5. (a) If A = (1 3 5) and B = (–2 4 3) then find 3A + 2B.

(b) If X = 2 and Y = 6 than find A – 3B
–3 1

(c) If P = 2 0 and Q = 2 0 , find P + 2Q
2 1 –1 2

(d) If M = –1 5 and N = –4 –1 . find 5M – 3N.
3 2 –2 3

Section 'C'

x3 –3 z 61
6. (a) If 5 y + –2 –1 = 3 0 then find the values of x, y and z.

–3 4 1y –7 –6
(b) 2 5 x – 0 1 = 5 2 , find teh values of x and y.

4x 13 z –4
(c) If 3 0 1 + 2 y 2 = 8 7 than find the values of x, y and z.

x3 3 –2 74
(d) If 2 6 y–3 + 1 2 = 7 14 then find the values of x and y.

Infinity Optional Mathematics Book - 8 39

UNIT

5 COORDINATE GEOMETRY

5.1 Coordinate Geometry
Review

1. In an ordered pair (–2, 3)
(i) What does –2 represent?
(ii) What does 3 represent?
2. In the given figure

X' X
E' D' C' B' A' O A B C D E

O is the initial point and XX the horizontal line through O. Let, A, B, C,D, E,

..... and A', B', C', D', E', ............ be the points on XX' such that OA = AB = BC

= CD = DE = ............ 1 unit, then OB = 2 units, OC = 3 units, OD = 4 units and

OE = 5 units.

Now, AD = OD – OA = 4 – 1 = 3 units. This distance is measure in horizontal

line so this distance is known as the run. Similarly, the points on the left of O

have the negative real numbers.

Now, OA' = – 1 OB' = – 2, OC' = – 3 OD' = – 4, OE' = – 5

A'D' = OD' – OA'

= – 4 – (–1)

=–4+1

=–3 E
3. Similarly if A, B, C, D, E, ........ and A', B', C', D', E', ..... be the
D
points on the vertical line OA = AB = BC = CD = DE = ...... = 1 units C
Let's do an activity. Follow the instruction given Below. B

i. Stand at O. A
ii. Move 2 units above where are you now?
iii. Move 3 units above from O. Where are you now? O
iv. Move 4 units above from O. Where are you now?
A'
B'
C'
D'
E'

40 Infinity Optional Mathematics Book - 8

v. Move 5 units above from O. Where are you now?

vi) Move 5 units down from O. Where are you now?

vii) Move 3 units down from O. Where are you now?

viii) Can you measure the length of AC and A'E'? Vertical
Hence, from above discussion, we hope that E

the students must have understood how to D
C

move along a horizontal line and vertical E' D' C' B' A' B
line. If we put both these line together in AA B C D E
such a way that both O of vertical line and
O A' Horizontal
B'

horizontal line meet, it would look like Intercession of both O. C'
figure given below which we say a plane or D'
E'

Cartesian plane.

The distance like AC, AD etc are called rise. The points below the initial points O
have the negative real numbers.
Introduction
The Coordinate system was invented by Rene Des cartes, (1516 to 1650) that's why it
is also known as plane Cartesian Geometry. It has been said that once in his dream,
he dreamt about the marriage between Algebra and Geometry. He woke up from
his dream and searched the relation between them. Finally he succeeded and came
to the overlapping portion of Algebra and Geometry which we now say Coordinate
Geometry.
Many other similar stories also exist, but whatever they are, Coordinate Geometry is
the branch of mathematics which deals with the combination of algebra and geometry
using a pair of numbers called coordinates to fix the position of points.
Rectangular Coordinate axes and the Signs of the Coordinates
In Coordinate geometry, we use two perpendicular number lines to create a method
of locating points. So, XOX' and YOY' are two fixed lines (number lines) interesting
each other at point O in the same plane. The line XOX' is horizontal and called X-axis
or axis of X. The line YOY' is obviously vertical and called y-axis or axis of Y. The
point of intersection O is called the origin of the Coordinates or Simply the origin.
The Coordinates of the origin are (0, 0). The coordinate axes divides the plane into
four parts which are called first, second, third and fourth quadrant.

Infinity Optional Mathematics Book - 8 41

(a) For distance along x-axis, the values of x are measured from origin O along the

axis of X. Y
i) The positive values are measured to the

right of the origin. Second quadrant First quadrant

ii) The negative values are measured to the (–, +) (+, +)

left of the origin. X' X
(b) For the distance along the y-axis, the values Third quadrant 0 Fourth quadrant

(–, –) (+, –)

of y are measure from origin along the axis of Y'
y.

i) The positive values are measured upwards from origin.

ii) The negative values are measured downward from origin.

The following two sentences can also be remembered for determining the sign of

coordinates.

"Plus to the right minus to the left, positive height, negative depth"

Coordinates (ordered pair) Y
Let us consider a point P on a

plane. To locate the point P, draw
PM perpendicular to X'OX and PN N x P (x, y)

perpendicular to YOY'.
Then NP = OM = x and y y

PM = ON = y
X' MX
The distance NP or OM along OX is O x

called x-coordinate or abscissa of P. The

distance MP or ON along OY is called Y' Y
– Coordinate or ordinate of P.

The abscissa and ordinate x and y together are known as Cartesian coordinate of P.

We will use notation P (x, y) to represent the point P with coordinates (x, y).

42 Infinity Optional Mathematics Book - 8

Plotting Points in Coordinate Plane.

Let, A (2, 3), B (–3, 4), C (–4, –3) D (3, – 1), E (4, 0), F (0, –3) are the points which are

going to be plotted in the graph.

Plot in a graph.

The point A (2, 3) lies in the Scale: 1 box = 1 unit Y
first quadrant. To plot the B (–3, 4)
point count 2 units along OX A (2, 3)

to the right side of O and then

count 3 units upwards and

mark the points thus obtained E (4, 0)
as shown in the figure and X' 0X

write A (2, 3) near to it D (3, –1)

The point B (–3, 4) lies in

second quadrant. To plot the C (–4, –3) F (0, –3)
point, count 3 unit along OX'
to the left of O and then count Y'

4 units upwards. Mark the points thus obtained as shown in the figure and then

write B (–3, 4) near to it.

The point C (–4, –3) lies in the third quadrant. Count 4 units from O along OX' to the

left of O and then 3 units downward. Mark the point so obtained as shown in figure.

The point D (3, – 1) lies in the fourth quadrant. Count 3 units from O along

OX to the right side of O and then 1 unit downward. Mark the point so obtained as

shown in the figure.

Where does the point E (4, 0) lies? Where does the point F (0, – 3) lies? Since

in E (4, 0) Y – coordinate is zero (0) and X – coordinate is plus 4, so it lies 4 units

right from origin on X - axis. Similarly in F (0, –3), X - coordinate is zero (0) and Y -

coordinate is minus 3 so it lies 3 units below from X-axis on Y-axis.

Note:-

1) If a point having X-coordinate zero (0) and any value of Y- coordinate, that

point lies on Y-axis.

2) If a point having Y-coordinate zero (0) and any value of X coordinate, that point

lies on X-axis.

Infinity Optional Mathematics Book - 8 43

WORKED OUT EXAMPLES

1. Plot the ordered pairs and name the quadrant or axis in which the
point lies. A (5, 2), B (–3, 3), C (–4, –1), D (–4, 0) and E (0, 3)

Solution: Here, Y Scale : 1 box = 1 unit
Plotting the given points A (5,2),
B (–3, 3) C (–4, –1), D (–4, 0) and B (–3, 3) E (0, 3)
E (0,3) on a graph paper.
A (5, 2)

The point A (5, 2) lies in first X' D (–4, 0)

quadrant 0 X

The point B (–3, 3) lies in second C (–4, –1)

quadrant Y'
The point C (–4, –1) lies in third

quadrant

The point D (–4, 0) lies in x-axis (Negative).

The point E (0, 3) lies in y-axis (Positive).

2. Find the coordinates of the point E, F, G, H and I from the given graph.

Y Scale: 1 box = 1 unit

E
F

X' 0 IX

GH

Y'

Solution: Here,
From the given graph the coordinates of the points E(–3, 3), F(4, 2), G(–2, – 3),
H(6, – 3) and I(7, 0).

44 Infinity Optional Mathematics Book - 8

3. The coordinates of three vertices of a rectangle are A (1, –3), B (1, 3)
and C (4, 3). Find the coordinates of its fourth Vertex D using graph
paper.

Solution: Here, Let, fourth vertex be D(x, y)



Y Scale: 1 box = 1 unit

B (1, 3) C (4, 3)

X' 0 X

A (1 – 3) D (x, y)

From the above graph, the required coordinates of the fourth vertex is D (x, y)
= D (4, –3)

4. Plot the points given below join them to identity the shape give reason
for your answer.

(a) O (0, 0) A (–4, 0) and B (0, –4) (b) K (1, 5), L (–2, 3) and M (–5, 1)

(c) A (0, 2), B (5, –2), C (7, 7) and D (3, 5)

Solution: Here, Y
Scale: 1 box = 1 unit

K (1, 5)

D (5, –2)

L (–2, 3)

A (0, 2)

M (–2, 3)

'X A (–4, 0) 0 O (0, 0) X

B (5, –2)

B (0, –4)

Y'

Infinity Optional Mathematics Book - 8 45

Exercise 5.1

Section A

1. (a) Write the names of the quadrants in which the following points lie
(i) (4, 3) (ii) (–2, 3) (iii) (–4, –7) (iv) (–6, –7)
(v) (–7, 3) (vi) (–8, –10)

(b) Write the name of axis in which the following points lie.

(i) (4, 0) (ii) (–3, 0) (iii) (0, 6) (iv) (0, –8)

2. Plot the following set of points and name the quadrant or axis in which the

point lie.

(i) (–3, 4) (ii) (6, 1) (iii) (–7, –1) (iv) (5, 0)

(v) (2, –8) (vi) (0, –8)

3. Write the coordinates of the following points A, B, C, D, E, F, G, H, I &
J from the given graph.

Y Scale: 1 box = 1 unit

B C
A D

X' 0 EX
F
J

I G
H

Y'

4. Draw lines which join the following pair of points using graph.

(i) (1, 0) and (5, 1) (ii) (–3, –4) and (5, 4)

(iii) (–3, 4) and (4, –3) (iv) (0, 0), (2, 2) & (–2, –2).



46 Infinity Optional Mathematics Book - 8

Section : B
5. Plot the given points on a square paper then join them in order. What

type of polygon have you drawn? Write it
(i) A (3, 3), B (–4, 0) and C (4, 6)
(ii) A (1, 1), B (4, 4), C (4, 8) and D (1, 5).
(iii) P (0, 4), Q (4, 0), R (–4, 0) and S (0, –4)
(iv) M (2, 3), N (2, 0), O (–1, 0) and P(–1, 3)
6. Plot the given three vertices of a rectangle on a square paper and join

them in order. Find the coordinates of the fourth vertex D or S.
(i) A (–2, 0), B (–2, 2), C (3, 2).
(ii) P (1, 1), Q (1, –5), R (–2, –5).
(iii) A (1, 1), B (6, 1), C (6, – 5)
7. Plot the given three vertices of a square on a graph paper and join

them in order. Find the coordinates of the fourth vertex D or H.
(i) A (0, 0), B (3, 0), C (3, 3).
(ii) E (–1, 2), F (2, 2), G (2, –1)
(iii) E (–3, 2), F (–3, 5), G (0, 5)

Section C
8. Find a point that lies on the x-axis and y-axis by joining the points

(–3, –3) (1, 1), (2, 2) and (–1, –1).
9. (a) EFGH is a rectangle having the coordinates of two opposite corners F and

H are (1, 2) and (7, 6) respectively. By drawing them in a graph find the
coordinates of E and G.
(b) ABCD is a square having the coordinates of two opposite vertices A and
C are (2,2) and (5, 5) respectively. Find the coordinates of B and D by
plotting them on the graph.

Infinity Optional Mathematics Book - 8 47

5.2 Distance Between Two Points (Distance Formula)

Let, P (x1, y1) and Q (x2, y2) be two given points. Let 'd' Y Q (x2, y2)
be the distance between P and Q. C y2
d BX
To find the distance between P and Q, draw PA P (x 1, y 1)
y1
perpendicular to OX, QB perpendicular to OX and PC 0 x1 A

⊥ QB. Then, x2
Y'
OA = x1, OB = x2, PA = y1, QB = y2 X'
Now,

OA + AB = OB ( ∴ Whole part axiom)

or, x1 + AB = x2
∴ AB = x2 – x1
Also, AB = PC = x2 – x1 ...... (i)
Also,

BC + CQ = BQ [ ∴ Whole part axiom]
CQ = BQ – BC

= y2 – AP [ ∴BC = AP]

∴ CQ = y2 – y1 .......... (ii)

Now, in right angled PCQ, by pythagoras theorem

PQ2 = PC2 + CQ2 [ h2 = p2 +b2]

or, d2 = (x2 – x1)2 + (y2 – y1)2 [ ∴ From given, (i) & (ii) relation]
∴ d = (x2 – x1)2 + (y2 – y1)2
Hence, the distance between two points

P (x1, y1) and Q (x2, y2) is
PQ = d = (x2 – x1)2 + (y2 – y1)2
Which is also written as

d = (x1 – x2)2 + (y1 – y2)2 [ (a – b)2 = (b – a)2]

Note :- Distance of a point from the origin.

The distance of the point P (x, y) from origin (0, 0) is given by
OP = d = (x – 0)2 + (y – o)2
OP = d = x2 + y2

48 Infinity Optional Mathematics Book - 8

WORKED OUT EXAMPLES

1. Find the distance between the following points

(i) (–4, 5) and (6, 5) (ii) (a, b) and (b, a)

Solution: (i) Here,

Let, (–4, 5) = (x1, y1) , (6, 5) = (x2, y2)
By using distance formula

d = (x2– x1)2 + (y2 – y1)2 A (–4, 5) B (6, 5)

= (6 + 4)2 + (5 – 5)2

= 100+ 0

d = 10 units

 Distance between two given points is 10 units

(ii) Solution: Here,

Let (a, b) = (x1, y1) d B
(b, a) = (x2, y2) A (a, b) (b, a)
By using distance formula

d = (x2– x1)2 + (y2 – y1)2
= (b – a)2 + (a – b)2

= b2 – 2ab + a2 + a2 – 2ab + b2
= 2a2 – 4ab + 2b2

= 2(a2 – 2ab + b2)

= 2(a – b)2 = 2 (a – b) units.

Hence the distance between two given points is 2 (a – b) units.

2. P (2, – 5), Q (4, 1) and R (8, – 3) are three points show that PQ = PR.

Solution:- Here,

P (2, – 5), Q (4, 1) and R (8 – 3) are three given points

To show, PQ = PR P (2, –5)
For PQ :- Let,

P (2, – 5) = (x1, y1) Q (4, 1) R (8, –3)
Q (4, 1) = (x2, y2)
By using distance formula

Infinity Optional Mathematics Book - 8 49

d= PQ= (x2– x1)2 + (y2 – y1)2 = (4 – 2)2 + (1 + 5)2 = 4 + 36 = 40 = 2 10 units
Similarly, for PR:-

Let, P (2, –5) = (x1, y1) R (8, – 3) = (x2, y2)

By using distance formula

PR = (x2– x1)2 + (y2 – y1)2
= (8 – 2)2 + (–3 + 5)2 = 36 + 4 = 40 = 2 10 units

∴ PQ = PR = 2 10 units. Proved.

3. If the distance between A (5, 5) and B (x, 0) is 5 units, find the possible
values of x.

Solution :- Here,

Let, A (5, 5) = x1, y2) 5 units B (x, 0)
B (x, 0) = (x2, y2) A (5, 5)
AB = d = 5 units

To find: The value of x.

By using distance formula

AB = d= (x2– x1)2 + (y2 – y1)2

or, 5 = (x – 5)2 + (0 – 5)2

or, 5 = x2–2.x.5 + 52 + (–5)2
or, 5 = x2– 10x + 25 + 25

Squaring on both sides we get

(5)2 = ( x2 – 10x + 50 )2

or, 25 = x2 – 10x + 50

or, 0 = x2 – 10x + 50 – 25

or, 0 = x2 – 10x + 25

or, 0 = x2 – (5 + 5)x + 25

or, 0 = x2 – 5x – 5x + 25

or, 0 = x (x – 5) – 5 (x – 5)

or, 0 = (x – 5) (x – 5)

Either, OR

x – 5 = 0 x–5=0

x = 5 x=5

∴ The value of x = 5

50 Infinity Optional Mathematics Book - 8