Opt Math 6 - 2078 (Final)

PERFECT OPTIONAL

MATHEMATICS

Class 6

N B Khatakho

Vidyarthi Pustak Bhandar

Publisher and Distributor

Kamalpokhari/Bhotahity, Kathmandu, Nepal

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PERFECT OPTIONAL

MATHEMATICS

CLASS 6

Publisher Vidyarthi Pustak Bhandar

Copyright Kamalpokhari/Bhotahity, Kathmandu, Nepal
Edition Tel : 01-4245834, 4227246, 4423333
ISBN P.O. Box 12990, Email: [email protected]
Computer
Printed Publisher

First, 2005
Fifth Revised, 2021

978–99946–1–880–4

Dynamic Computer Services
Bhaktapur, 01-6614740

PREFACE

Perfect Optional Mathematics is recently revised series of school-level textbooks
based on the current pedagogical principles on mathematics prepared for
maintaining continuity in learning mathematics considering the maturity, needs and
interest of young learners of school age. This series provides ample opportunities
for the students to enhance their learning by encouraging/reinforcing creativity of
the students. For the purpose of making learning mathematics suitable as per the
capacity of the students of different age groups, appropriate exercises and activities
have been included presenting the concepts inductively through examples. Examples
have been particularly given at appropriate places for presenting mathematics as a
subject matter which can be learnt and applied in day-to-day life. The presentation
and exercises essential for developing students’creative thinking and problem solving
skills have been given an important place.

The materials have been presented with special attention to meaningful learning
of mathematics based on different researches on teaching-training experiences,
following the spirit of curriculum developed by CDC. Therefore, meaningfulness
and practicality of learning have been taken into consideration rather than searching
additional and complex subject matters. The current necessities of meaningful
learning and pragmatic use of mathematics for school-level children are the main
features of this series which can be integrated to their higher studies as well.

Perfect Optional Mathematics Class 6 prepared for the childere of class 6, taking
their capacity and interest into consideration, is a revised link book of this series,
which embodies all the features discussed above. This book is also believed to
make additional contribution to enhance the learning of school-level mathematics
meaningfully in the present text.

Mr. Shanta Gautam, the Managing Director of Vidyarthi Pustak Bhandar deserves
my thanks for taking the responsibility of publishing this book. Likewise, I extend
my thanks to Mr. Binod Kasula and Mr. Ramesh Konda for their contrubution in
course of currently revising and updating this book.

Any constructive suggestions for further improvement of this book will always be
highly acknowledged. Thanks!

Author



CONTENTS Pages

Topics 1 – 21
1. Algebra 1 – 11
12 – 17
Relation 18 – 21
Polynomials 22 – 28
Sequence and Series 22 – 28
2. Matrices 29 – 36
Concept of Limit 29 – 33
3. Matrices 34 – 36
Matrices 37 – 52
Operation of Matrices 37 – 41
4. Co-ordinate Geometry 42 – 46
Co-ordinate Axes 47 – 52
Application of Distance Formula 53 – 73
Section Formula 53 – 57
5. Trigonometry 58 – 60
Measurement of Angles 61 – 65
Pythagoras Theorem 66 – 70
Trigonometric Ratios 71 – 73
Values of Trigonometric Ratios 74 – 86
Heights and Distances 74 – 81
6. Vector 82 – 86
Scalar and Vector 87 – 100
Operation of Vectors 87 – 90
7. Transformation 91 – 95
Translation 96 – 100
Reflection 101 – 108
Rotation 102 – 104
8. Statistics 105 – 108
Arithmetic Mean 109 – 114
Median and Mode
Answers



1 ALGEBRA Algebra

Learning Objectives

state the ordered pairs and use its equality.
construct the cartesian product of two sets.
find the relation and its domain and co-domain.
identify the types of relation.
define function, its domain, co-domain and range.
define polynomials and find its degree.
perform the operations on polynomials : addition, subtraction and multiplication.
find the general term of a sequence.

1.1 Relation

Ordered Pairs

Following are the pairs of some information.

(i) Husbands Wives (ii) Students Obtained Marks

Anil Sangita Gopi 45

Suman Rabina Resma 85

Krishna Sushmita Sangita 75

Suresh Rekha Surendra 35

Rupesh Melina Ritesh 94

In the above table (i), 'husbands' and 'wives' are related to each other. (Anil, Sangita) is an ordered
pair in which 'Anil' is the first component and 'Sangita' is the second component. There are five
ordered pairs in table (i): (Anil, Sangita), (Suman, Rabina), (Krishna, Sushmita), (Suresh, Rekha),
(Rupesn, Melina).

First component (Husband, Wife) Second component
(Anil, Sangita)

Similarly, in table (ii), 'students' and 'obtained marks' are related to each other. The ordered pairs
in table (ii) are (Gopi, 45), (Resma, 85), (Sangita, 75), (Surendra, 35), (Ritesh, 94).

Let P = {3, 2}, Q = {2, 3} are examples of two sets. Both of these sets contain a pair of elements.

Relation 1

These two sets are equal as they contain the same elements. Here, the order of the occurrence of
the elements is not important. In some cases, if the order of occurcence of the pair of elements is
important then the pair of two elements is known as an ordered pair.

For example, in co-ordinate geometry two pairs (2, 7) and (7, 2) represent different points.
The ordered pair of two elements a and b is denoted by (a, b). The element a is called the first
component or x-component and element b is called the second component or y-component.

Thus, an ordered pair is a pair of objects whose components occur in a special order. The
components of an ordered pair are separated by a comma and enclosed by a pair of small brackets.

Equality of Ordered Pairs

Column A Column B

i) (2, 3) and (2, 3) i) (2, 3) and (3, 2)

ii) (5, 1) and (6 – 1, 3 – 2) ii) (5, 2) and (6 – 2, 3 – 1)

iii) (5 + 3, 2 + 3) and (8, 5) iii) (3 + 5, 1 + 7) and (8, 7)
iv) (7 – 1, 2 + 4) and (6, 8 – 2) iv) (6 – 1, 1 + 3) and (5 + 2, 6 – 1)

The ordered pairs in column 'A' are the equal ordered pairs.

For example, (5, 1) = (6 – 1, 3 – 2) or (5, 1) = (5, 1) Y
The ordered pairs in column 'B' are not equal ordered pairs.
8

For example, (5, 2) ≠ (6 – 2, 3 – 1) or (5, 2) ≠ (4, 2) 7 (3, 6)
6 (6, 3)
Two ordered pairs (a, b) and (p, q) are said to be equal if a = p 5
and b = q. Conversely, if c = m and d = n, then two ordered 4
pairs (c, d) and (m, n) are equal. 3

Two ordered pairs (3, 6) and (6, 3) are different as they are 2
1
located on different positions on the graph paper. Hence, the X'
ordered pairs (a, b) and (b, a) are not equal. O 1 2 3 4 5 6 7 8 X
Y'

Example 1:

Find the value of x and y if (x, y + 2) = (5, 3) .
Solution:

Here, (x, y + 2) = (5, 3)
First component of the first ordered pair = First component of the second ordered pair
\ x = 5
Second component of the first ordered pair = Second component of the second ordered pair
i.e. y + 2 = 3
or, y = 3 – 2 = 1

2 Perfect Optional Mathematics Grade 6

Cartesian Product of Two Sets Algebra

Consider two sets A = {1, 3} and B = {a, b, c}. Make a list of all possible ordered pairs using the
elements of set A and set B.

Ordered pairs from set A to set B Ordered pairs from set B to set A

a a 1 1 1
1 b2 ba 3
c b c
c 3 3

The possible ordered pairs are : The possible ordered pairs are :
(1, a), (1, b), (1, c), (2, a), (2, b), (2, c) (a, 1), (a, 3), (b, 1), (b, 3), (c, 1), (c, 3)

The set of ordered pairs is called cartesian The set of ordered pairs is called cartesian

product of set A and set B. product of set B and set A.

It is deoned by A×B. It is deoned by B×A.

A×B={(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} B×A= {(a, 1), (a, 3), (b, 1), (b, 3), (c, 1), (c, 3)}

Mapping diagram of A×B Mapping diagram of B×A
AB BA

1a a1

b b
3c c3

Let A and B be two non-empty sets. The cartesian product of A and B is denoted by A×B and is
read as 'A cross B' which is the set of all ordered pairs whose first element belongs to set A and
the second component belongs to set B. Using symbols we write, A×B = {(x, y) : x∈A and y∈B}

Example 2:

If P = {2, 3} and Q = {1, 4}, find P×Q.
Solution:

Here, P = {2, 3} and Q = {1, 4} are given two sets. For the cartesian product P×Q, we construct

the ordered pairs where the first component belongs to P and second component belongs to Q.

Such ordered pairs can be obtained in the following way.

P Q P×Q

1 (2, 1) PQ

2 21
4 (2, 4)

1 (3, 1) 3 4
3

4 (3, 4)

Hence, P×Q = {(2‚ 1)‚ (2‚ 4)‚ (3‚ 1)‚ (3‚ 4)}.

Relation 3

Example 3:

If A = {1, 2, 3}, B = {4, 5}, find A×B and B×A. Also represent them in the cartesian plane.
Solution:

Here, A = {1, 2, 3} and B = {4, 5}
A×B = {1, 2, 3}×{4, 5}
= {(1‚ 4)‚ (1‚ 5)‚ (2‚ 4)‚ (2‚ 5)‚ (3‚ 4)‚ (3‚ 5)}.
Again, B×A = {4, 5}×{1, 2, 3}
= {(4‚ 1)‚ (4‚ 2)‚ (4‚ 3)‚ (5‚ 1)‚ (5‚ 2)‚ (5‚ 3)}.
Representing them in the cartesian plane.

Y (A × B) X Y (B × A) X

8 8
7
7 6
5
6 4
3
5 2
1
4
X' O 1 2 3 4 5 6 7 8
3 Y'

2

1

X' O 1 2 3 4 5 6 7 8
Y'

Example 4:

If A = {x, y} and B = {a, b, c}, find A×B and B×A and represent them by an arrow diagram.
Solution:

Here, A = {x, y} and B = {a, b, c}, then

A×B = {x, y}×{a, b, c}

= {(x‚ a)‚ (x‚ b)‚ (x‚ c)‚ (y‚ a)‚ (y‚ b)‚ (y‚ c)}.

B×A = {a, b, c}×{x, y}

= {(a‚ x)‚ (a‚ y)‚ (b‚ x)‚ (b‚ y)‚ (c‚ x)‚ (c‚ y)}. A
AB B
x
xa a

bb

yc cy

Relation

The relation is any subset of the cartesian product of two sets. Let us consider the following
examples first. Let a couple of husband and wife, Rabin and Sita have two daughters Rina and
Nima. Let the set notation of the members of family be

A = {Rabin, Sita} and B = {Rina, Nima} then

A×B = {(Rabin‚ Rina)‚ (Rabin‚ Nima)‚ (Sita‚ Rina)‚ (Sita‚ Nima)}

4 Perfect Optional Mathematics Grade 6

Let the relation is “the father of.” Algebra

Rabin is the father of Rina and Rabin is the father of Nima.

So, R1 = {(Rabin‚ Rina)‚ (Rabin‚ Nima)}, which is the subset of A×B.
\ R1 is a relation from A to B.
A relation from set A to set A itself is said to be a relation on A.

Definition

A relation from set A to B is defined as any subset of cartesian product A×B. It is generally
denoted by R. Let us consider the following example first. Let a couple of husband and wife,
Manoj and Sita have two daughters named Rina and Nima and a son named Kishan. Let the set
notation of the members of family be

A = {Manoj, Sita} and B = {Rina, Nima, Kishan} then

The cartesian product A×B is given by

A×B B

Rina Nima Kishan

A Manoj (Manoj‚ Rina) (Manoj, Nima) (Manoj, Kishan)

Sita (Sita‚ Rina) (Sita, Nima) (Sita, Kishan)

A×B = {(Manoj‚ Rina)‚ (Manoj‚ Nima)‚ (Manoj, Kishan), (Sita‚ Rina)‚ (Sita‚ Nima), (Sita, Kishan)}

Let us consider the following two subsets of A×B.

i) R1 = {(Manoj‚ Rina)‚ (Manoj‚ Nima)‚ (Manoj, Kishan)}
ii) R2 = {(Sita‚ Rina)‚ (Sita‚ Nima), (Sita, Kishan)}
From the above two subsets, we conclude the following facts:

Subsets Property Relation
x-component is the father of y-component "is father of"
R1 x-component is the mother of y-component "is mother of"
R2

Example 5:

If A = {9, 16} and B = {3, 4}, find the relation from A to B such that the first element is the
square of the second element.
Solution:

Here, A = {9, 16} and B = {3, 4} then

A×B = {9, 16}×{3, 4}

= {(9‚ 3)‚ (9‚ 4)‚ (16‚ 3)‚ (16‚ 4)}.

As 9 is the square of 3 and 16 is the square of 4, the required relation from A to B is

R = {(9‚ 3)‚ (16‚ 4)}.

Relation 5

Example 6:

If P = {1, 4, 5} and Q = {2, 3} and a relation R from set P to Q is defined as the sum of
x-component and y-component is 7, represent the relation in mapping diagram and graph.
Solution:

Here, P = {1, 4, 5} and Q = {2, 3} then

P × Q = {(1‚ 2)‚ (1‚ 3)‚ (4‚ 2)‚ (4‚ 3)‚ (5‚ 2)‚ (5‚ 3)}. P Q
In the given relation, the sum of components is 7.
So, R = {(4‚ 3)‚ (5‚ 2)}. 1 2
4
Arrow diagram of the relation is 5 3

The graph form of the relation is Y X

Domain and the Range of a Relation 8
7
6
5
4
3
2
1

X' O 1 2 3 4 5 6 7 8
Y'

The set of all the first components of the ordered pairs of a relation R is known as the Domain of
a relation.

The set of all the second components of the ordered pairs of a relation R is known as the Range
of a relation.
Example 7:

Find the domain and range of the following relation:

R = {(1‚ 2)‚ (2‚ 3)‚ (3‚ 4)‚ (4‚ 5)}.
Solution:

Given relation R = {(1‚ 2)‚ (2‚ 3)‚ (3‚ 4)‚ (4‚ 5)}.
Domain of R = {1, 2, 3, 4}
Range of R = {2, 3, 4, 5}

6 Perfect Optional Mathematics Grade 6

Types of Relation Algebra

The relation from set A to B can be classified into following types:
a. One – one relation
b. Many – one relation
c. One – many relation

a. One – one relation

The relation from set A to B i.e. R: A→B is said to be one – one relation if different elements of
set A are related with different elements of B. Let A = {2, 5} and B = {4, 7} then

A×B = {(2‚ 4)‚ (2‚ 7)‚ (5‚ 4)‚ (5‚ 7)}.

and R1 = {(2‚ 4)‚ (5‚ 7)}. AB
In (2, 4), 2 is related to 4
24
In (5, 7), 5 is related to 7.
57
\M anRy1 is one – one relation

b. – one relation

A relation from set A to B i.e. R: A→B is said to be many-one relation if different elements of A

are related to single element of B.

Let P = {1, 2, 3} and Q = {2, 4}
then P×Q = {(1‚ 2)‚ (1‚ 4)‚ (2‚ 2)‚ (2‚ 4)‚ (3‚ 2)‚ (3‚ 4)} P Q

and R2 = {(1‚ 2)‚ (2‚ 2)‚ (3‚ 4)} 1 2
2 4
In relation R2, two elements 1 and 2 of set P are related with single 3
element of set Q.

In (1, 2), 1 is related with 2.

In (2, 2), 2 is related with 2.

In (3, 4), 3 is related with 4.

\OnRe2–ismmaannyy – one relation

c. relation

A relation from set A to B is said to be one – many relation if one element of A is related to more
than one element of B.

Let A = {a, b, c} and B = {p, q, r} then

A×B = {(a‚ p)‚ (a‚ q)‚ (a, r), (b‚ p)‚ (b‚ q)‚ (b, r), (c‚ p)‚ (c‚ q), (c, r)}.

and R3 = {(a‚ p)‚ (a‚ q), (b, r)}. AB
In (a, p), a is related to p.
In (a, q), a is related to q. ap
In (b, r), b is related to r. bq
\ R3 is one – many relation. cr

Relation 7

Inverse Relation

If we interchange the first and the second components of each ordered pair of the given relation
R, the new relation is the inverse relation of R and denoted by R– 1.

Example 8:

If R = {(a‚ b)‚ (p‚ q)‚ (x‚ y)}, find R– 1.
Solution:

Here, R = {(a‚ b)‚ (p‚ q)‚ (x‚ y)}.

then R– 1 = {(b‚ a)‚ (q‚ p)‚ (y‚ x)}.

Function

Consider two sets A = {2, 3} and B = {6, 7, 9}. Let a relation 'x is less than y by 4" from set A to
set B. Showing this relation in a mapping diagram:

AR B

2 6
9

37

The above ralation R = "x is less than y by 4" from set A to set B. So 2 is less than 6 by 4 and 3 is
less than 7 by 4. A relation R from set A to set B is said to be a function if

i. every element of set A is associated with only one element

ii. no element of A is associated with two or more elements of set B.

For any two non-empty sets A and B, A relation R from set A to set B is said to be a function if
each element of set A assigns with only one element of set B.

Study the following examples carefully.

a. Relation R1 from set A to set B is illustrated in the A R1 B
mapping diagram. Each element of set A is paired with
exactly one element of set B. a 1
b 2
\ R1 = {(a, 1), (b, 2), (c, 3)} c 3

b. Relation R2 from set P to set Q is illustrated in the mapping P R2 Q
diagram. Every element of set P is paired with one of the
elements of set Q. a p
b q
\ R2 = {(a, p), (b, p), (c, q), (d, r)} c r
d s

8 Perfect Optional Mathematics Grade 6

c. Relation R3 from set C to set D is illustrated in the C R3 D
mapping diagram. Two elements of set C have paired
with unique element of set D. 1 a Algebra
2
\ R3 = {(1, a), (2, b)} 3 b

d. Relation R4 from set X to set Y is illustrated in the X R4 Y
mapping diagram. Every element of set X is paired with
some elements of set B. p a
b
\ R4 = {(p, a), (p, b), (q, c), (r, d)} q c
d
r

Each element of the first set is matched with only one element of the second set in relations R1 and
R2. From the above definition of a function, the relations R1 and R2 are functions.

But, in relation R3, element '3' of the first set C is not associated with any element of the second
set D. In relation R4, element 'p' of the first set X is associated with two elements 'a' and 'b' in the
second set Y. So, by definition of a fuction, R3 and R4 are not functions.

Domain, Co-domain and Range of a Function

Look at the function given in the mapping diagram. All the A f B
elements of the set A are paired with the unique element of the set
B, and the element 8 is left in set B. But, it is a function. So, the 2 4
set of all the elements of set A is called a domain and the set of 3 5
all elements of set B is called a co-domain of the function f. Only 5 7
three elments 4, 5 and 7 of the set B are engaged in the mapping 8
diagram. The set of these elements of set is the range of function
f. Therefore,

The domain of f = {2, 3, 5}

The co-domain of f = {4, 5, 7, 8}

The range of f = {4, 5, 7}

Example 9:

Find the domain, co-domain and range of the function f : {(2, 3), (– 1, 0), (3, 4), (1, 2)}
Solution:

Here, given function is f = {(2, 3), (– 1, 0), (1, 2), (3, 4)}
The domain of f = {2, – 1, 1, 3}
The co-domain of f = {3, 0, 2, 4}

The range of f = {3, 0, 2, 4}

Relation 9

Exercise 1.1

1. Plot the following points on the cartesian plane:

a. (2, 4) b. (– 4, 3) c. (– 2, – 1) d. (3, – 5)

2. Which of the following ordered pairs are equal?

P(2, – 2), Q(– 2, 2), R(3, 5), S(2, – 2), A(3, 5), B(4, 3), C(4, 3), D(– 2, 2)

3. Find the values of x and y, if:

a. (x + 2, y) = (5, 4) b. (2x, 3y) = (6, – 9)

c. (x + 1, y – 3) = (2, 5) d. (2x + 1, 3y – 1) = (5, 11)

e. 2x , y + 1 = 3, 3 f. 3x , y2 + 1 = (6, 4)
3 4 2

4. a. If A = {1, 2} and B = {3, 4}, find A×B, B×A and show that A×B ≠ B×A.

b. If P = {a, b} and Q = {x, y, z}, find P×Q, Q×P and show that P×Q ≠ Q×R.

5. a. If A = {2, 5} and B = {1, 4}, find A×B and B×A. Represent them in the cartesian plane.

b. If M = {– 2, 3} and N = {4, – 1}, find M×N and N×M. Represent them in the cartesian plane.

6. a. If A = {a, b} and B = {m, n}, find A×B, B×A and A×A. Represent them in a mapping
diagram.

b. If P = {1, 2, 3} and Q = {4, 5}, find P×Q, Q×P and Q×Q. Represent them in a mapping
diagram.

7. From the following arrow diagram, find A×B.
a. A B
b. A B
a
1 4 1
b 2
3
c2 5

8. If A = {1, 9} and B = {1, 2, 3}, find the following relations from set A to set B.
a. The first element is the square of the second element.
b. The first element is less than the second element.
c. The first element is equal to the second element.
9. If A = {2, 4} and B = {1, 3, 5}, find the following relations from set A to B and represent them

in a mapping diagram and graph also.
a. The first element is greater than the second element.
b. The sum of the first element and the second element is 5.
10. If P = {– 1, 2} and Q = {1, 2, 3, 4}, find the following relations
a. The first component is less than the second component.
b. The sum of the first component and the second component is 3.

10 Perfect Optional Mathematics Grade 6

11. Find the domain and range of the following relations:

a. R1 = {(1‚ 2)‚ (2‚ 3)‚ (3‚ 4)}. Algebra
b. R2 = {(1‚ 4)‚ (2‚ 3)‚ (3‚ 2)‚ (4‚ 1)}.
c. R3 = {(a‚ b)‚ (c‚ d)‚ (e‚ f)‚ (x‚ y)}.
d. R4 = {(2‚ 4)‚ (2‚ 5)‚ (3‚ 4)‚ (3‚ 6)}.
12. Find the domain and range of the following relations:

a. A B b. P Q

1 32 1
2 43 3
3 56 4
4
B
13. Identify the types of the following relations:
2
a. P R1 Q b. A R2 0
3 3 4
4 2
4
54 3

c. P R3 Q d. A R4 B

a xx p

bq

cy yr

14. Find the inverse relations of the following relations:

a. R1 = {(2‚ 3)‚ (3‚ 4)‚ (5‚ 6)}
b. R2 = {(a‚ x)‚ (b‚ y)‚ (c‚ z)}
c. R3 = {(4‚ 6)‚ (3‚ 5)‚ (1‚ 3)‚ (2‚ 3)}
d. R4 = {(2‚ 3)‚ (3‚ 4)‚ (5‚ 6)‚ (3‚ 6)}
15. Find the domain, co-domain and range of the following functions.

a. A f B b. A f B
– 2
1 2 4
2 9
23 16
4 3 17
3 6
4 5 4

c. f : {(3, 4), (0, 1), (4, 5), (2, 3)} d. f : {(a, 1), (b, 1), (c, 2), (d, 3)}

Relation 11

1.2 Polynomials

Introduction

To define polynomial, we define some of the basic terms.

Variable

A symbol which may assume any value is known as a variable. The variables are represented by
u, v, w, x, y, z etc.

Constant

A symbol which assumes the same value throughout the mathematical problem is known as a
constant. For example, 2, 3, a, b, c, d etc.

Term

A quantity containing a constant and a variable or a product of constants and variables is known
as a term. For example, 5, 2x, abx, axy etc.

Coefficient

A real number or constant multiplying the variable of the term is known as the coefficient. In the
terms ax2 and – 4bx the coefficients are a and – 4b respectively.

Algebric Expression

A group of terms connected by + or – signs is said to be an algebric expression. For example, 2x2
+ 3x – 5, 2x + 4 etc. are the algebric expressions.

Like Terms

The terms which have the same variable factors are called like terms. For example, 2x2y, – 3x2y,
12x2y etc. are the like terms.

Unlike Terms

The terms which do not have same variable factors are called unlike terms. For example, 4xy,
– 3x2y, 5x3y2 etc. are the unlike terms.

Polynomials

A polynomial is a special type of algebraic expression. An algebraic expression in which the
powers of variables are all whole numbers is called a polynomial.
For example, 2x + 3, 3x2 + 2y, 12x3 + xy + 4, 4x2y + 5  2y2 etc. are polynomials.
However 3x2 –  4x– 3y, 4  x – 5 are not polynomials as at least one variable in the expression has a
power which is not a whole number.

12 Perfect Optional Mathematics Grade 6

Example 1:

Which of the following are polynomials? Algebra

a. 2x3 + 3  x – 7 b. 4x4 + 21 xy – 3 c. 4x3 + 2  3 x– 2
Solution:

a. The given expression is 2x3 + 3  x – 7. The powers of the variable of three terms of the given

expression are 3, 1 and 0. Here, 1 is not a whole number. Hence, the expression 2x3 + 3  x – 7
2 2
is not a polynomial.

b. The given expression is 4x4 + 12xy – 3. The powers of the variable of the terms of the given
expression are 4, 2 and 0. All the powers are whole
numbers. Hence, the expression 4x4 + 1
2
xy – 3 is a polynomial.

c. Given expression is 4x3 + 3  3x– 2. The powers of the variable of the terms of the given
expression are 3 and – 2. Here, – 2 is not a whole number. Hence, the expression 4x3 + 2  3
x– 2 is not a polynomial.

Degree of Polynomial

Observe the powers of each term in the following two polynomials:
(i) 3x2 + 4yx + 5 and (ii) 5x4 + 4y2x3 + 3xy
For the polynomial (i), the sum of the powers of variable in 3x2 = 2
the sum of powers of variables in 4yx = 1 + 1 = 2
the sum of powers of variables in 5 = 0
For the polynomial (ii), the sum of the powers of variable in 5x4 = 4
the sum of powers of variables in 4y2x3 = 2 + 3 = 5
the sum of powers of variables in 3xy = 1 + 1 = 2
Which are the highest powers in polynomials (i) and (ii) respectively?
The polynomial (i) has highest power 2, so 2 is the degree of polynomial (i).
The polynomial (ii) has highest power 5, so 5 is the degree of polynomial (ii).

Types of Polynomials

There are different types of polynomials on the basis of the number of terms.

Polynomials Definitions Names
A polynomial having only one term. Monomial
4x2 A polynomial having only two terms. Binomial
A polynomial having only three terms. Trinomial
2x3 + 3x
A polynomial having more than three terms. Polynomial
5x2 + 3x + 7
x5 + 3x3 + 7x – 3,
2x6 – 5x4 + 7x2 + x – 3

Polynomials 13

Example 2:

If a polynomial is f(x, y) = 3x3y + 5x2 – 7xy2, find the degree of the polynomial and identify
its type.
Solution:

Given polynomial is f(x, y) = 3x3y + 5x2 – 7xy2
Since degree of the first term is 3 + 1 = 4,
Degree of the second term is 2,
Degree of the third term is 1 + 2 = 3.
Hence, the degree of the polynomial is 4.
Since the polynomial has three terms, it is a trinomial.

Example 3:

In each of the following, which polynomials are monomials, binomials, trinomials or
polynomials?

a. 2x – 3 b. 3x2 + 12 xy – 3y2 c. 4x3y2
e. 3x4 – 4x2y f. 4x4 + 10x3y + 9x – 2
d. x3 + 3x2 + 52
Solution:

The monomial is a polynomial consisting only one term.

Hence, 4x3y2 (c) is a monomial from the above polynomials.

The binomial is a polynomial consisting only two terms.

Hence, 2x – 3 (a) and 3x4 – 4x2y (e) are binomials from the above polynomials.

The trinomial is a polynomial consisting only three terms.

Hence, 3x2 + 12 xy – 3y2 (b) and x3 + 3x2 + 5 (d) are trinomials from the above polynomials.
2

The polynomial given in (f) i.e. 4x4 + 10x3y + 9x – 2 is a polynomial.

Polynomial in the Standard Form

If the terms of a polynomial are arranged according to either ascending or descending order
of degrees, then the polynomial is said to be in the standard form.

2x4 + 3x2 – 7x + 1 and 2x + 8xy – 7x2y are the examples of standard polynomials.

Equal Polynomials

Two polynomials are said to be equal if they contain the same number of terms and corresponding
like terms are equal. For example,

f(x) = 3x2 + 7x – 8
g(x) = 7x + 26 x2 – 4 × 2 are equal polynomials.

14 Perfect Optional Mathematics Grade 6

Example 4: Algebra

Express the polynomial h(x) = 4x + 7x2 – 8 in the standard form.
Solution:

The given polynomial is h(x) = 4x + 7x2 – 8. Arranging the terms of polynomial in descending
order of degrees, the standard form of the polynomial is 7x2 + 4x – 8.
If we arrange the terms in ascending order of degree, the standard form of the polynomial is
– 8 + 4x + 7x2.
Example 5:

Identify whether the following two polynomials are equal or not.
f(x) = 2x3 + 6x2 – 8x + 4 and h(x) = 2(x3 + 3x2 – 4x + 2)
Solution:

Given polynomials are:
f(x) = 2x3 + 6x2 – 8x + 4
and h(x) = 2(x3 + 3x2 – 4x + 2)
= 2x3 + 6x2 – 8x + 4
Since the number of terms and the corresponding like terms of these two polynomials are equal
f(x) = h(x).

Operations on Polynomials

We can count and add together two or more like objects.

+ =     

3 apples + 2 strawberries 2 apples + 1 strawberry 5 apples + 3 strawberries

(3x + 2y) + (2x + y) = (3 + 2)x + (2 + 1)y = 5x + 3y

Addition of Two Polynomials

Addition of two polynomials is a new polynomial obtained by adding the coefficients of like
terms, the variable remaining the same. For example,

f(x) = 2x3 + x2 + 3 and h(x) = x3 + 3x – 8

Then, f(x) + h(x) = 2x3 + x2 + 3 + x3 + 3x – 8

= 3x3 + x2 + 3x – 5

Difference of Two Polynomials

The difference of two polynomials is a new polynomial obtained by subtracting the coefficients of
like terms, the variable remaining the same. For example,

f(x) = 5x3 – 2x2 + 7x – 5 and g(x) = 3x3 + 8x – 2

Polynomials 15

then, f(x) – g(x) = (5x3 – 2x2 + 7x – 5) – (3x3 + 8x – 2)
= 5x3 – 2x2 + 7x – 5 – 3x3 – 8x + 2
= (5 – 3)x3 – 2x2 + (7 – 8)x – 5 + 2
= 2x3 – 2x2 – x – 3

Multiplication of Polynomials (Product of Polynomials)

The product of two polynomials is a new polynomial obtained by multiplying each term of one
polynomial by each term of the other polynomial and getting the sum or the difference of the like
terms. For example, f(x) = 2x + 1 and g(x) = x – 3
Then f(x).g(x) = (2x + 1) (x – 3)
= 2x(x – 3) + 1(x – 3)
= 2x2 – 6x + x – 3
= 2x2 – 5x – 3.
Example 6:

If f(x) = 3x3 + 7x2 – 5x + 8 and g(x) = 2x3 – 4x – 8, then find f(x) + g(x) and f(x) – g(x).
Solution:

Here, f(x) = 3x3 + 7x2 – 5x + 8 and
g(x) = 2x3 – 4x – 8
then, f(x) + g(x) = 3x3 + 7x2 – 5x + 8 + 2x3 – 4x – 8
= 3x3 + 2x3 + 7x2 – 5x – 4x + 8 – 8
= 5x3 + 7x2 – 9x
and f(x) – g(x) = (3x3 + 7x2 – 5x + 8) – (2x3 – 4x – 8)
= 3x3 + 7x2 – 5x + 8 – 2x3 + 4x + 8
= 3x3 – 2x3 + 7x2 – 5x + 4x + 8 + 8
= x3 + 7x2 – x + 16
Example 7:

If f(x) = 2x – 1 and g(x) = x2 – x + 3, then find f(x) . g(x).
Solution:

Here, f(x) = 2x – 1 and g(x) = x2 – x + 3
f(x) . g(x) = (2x – 1) (x2 – x + 3)
= 2x(x2 – x + 3) – 1(x2 – x + 3)
= 2x3 – 2x2 + 6x – x2 + x – 3
= 2x3 – 2x2 – x2 + 6x + x – 3
= 2x3 – 3x2 + 7x – 3

16 Perfect Optional Mathematics Grade 6

Exercise 1.2

1. Identify, which of the following algebric expressions are polynomials? Algebra

a. 2x – 3 b. x2 – 7x + 3

c. x4 + 3x5/2 + 3x2 – 7 d. 4x5 – 7x3 + 3

e. x3 + 3x√x + 4x – 7 f. 4x6 – 7x4 + 5x3 – 4x + 2

g. x4 + 4x – 3 + 2x + 1 h. x2y + xy – 1 + 3y2

2. Find the degree of each of the following polynomials.

a. 4x3 + 7x2 – 3x + 2 b. x2y2 + 4xy2 + 7y + 3

c. 2x + 4x2y2 + 7x5 d. 2x2y – 3x2y3 + 5y4

3. Identify the following polynomials, whether they are monomial, binomial, trinomial or
polynomial.

a. 4x + 7y b. 2x2y + 3xy + 4

c. 5xy d. x2 + xy2 + 3y2 + 7

4. Express the following polynomials in the standard form.

a. 2x2 + 3x + 8 b. 2x2y2 – 4xy4 + x2 + 8

c. 2x – 3x2 + 4x3 – 9 + x d. 2xy + 3xy2 + 4y6 – x2y2

5. Identify which of the following pair of polynomials are equal.

a. f(x) = – 3x + 4x2 – 7, g(x) = 4x2 – 7 – 3x
51
b. f(x) = 2  x3 – 8x + 0.2, g(x) = 2.5x3 – 8x + 5 
2
c. f(x) = 5x2 – 3x + 4, g(x) = 5x2 – 3  x + 2 × 2
6
d. f(x) = 3x3 + 4x2 – 8x, g(x) = 2  x3 + 5x2 – 8x

6. For the following polynomials, find f(x) + h(x) and f(x) – h(x):

a. f(x) = 5x4 + 3x3 + 7x, h(x) = x4 + 3x3 + x2 + 3x + 2

b. f(x) = 3x3 + 2.5x2 – 7x + 2, h(x) = 2x3 – 3x2 + 2x – 3

c. f(x) = 5x + 3x2 + 4x3, h(x) = 4x3 – 2x2 + 3x + 4

d. f(x) = 4x3 + 7x2 – 8x – 5, h(x) = x3 – 4x2 – 3x

7. Find the product of following polynomials:

a. f(x) = 3x – 7 and g(x) = x – 4

b. f(x) = 2x2 and g(x) = 2x2 + x – 7

c. f(x) = 2x + 3 and g(x) = x2 + 3

d. f(x) = 2x2 + 3 and g(x) = 3x – 4

Polynomials 17

1.3 Sequence and Pattern of Numbers

Sequence of Numbers

Consider a few numbers : 1, 3, 5, 7, 9, ... ...
Let's find the difference between each consecutive pair of numbers.
i.e. 3 – 1 = 2, 5 – 3 = 2, 7 – 5 = 2, 9 – 7 = 2, ... ... etc.
Here, the difference between each consecutive pair of numbers is 2. The same difference 2 is also
known a common difference. So, these numbers are in a fixed pattern. The numbers 1, 3, 5, 7, 9,
... are said to be in a sequence.
Let's take another sequence of numbers : 2, 4, 8, 16, 32, ... ... In this sequence, the next
number of each consecutive pair is 2 times the earlier one.
For example, 2 × 2 = 4, 2 × 4 = 8, 2 × 8 = 16, 2 × 16 = 32, ... ... etc.
Here, the ratio between each consecutive pair of numbers is 2. The same ratio 2 is also known as
common ratio. So, these numbers are in a fixed pattern. The numbers 2, 4, 8, 16, 32, ... ... are said
to be in a sequence.

General term of a sequence

Consider a sequence 1, 3, 5, 7, ... ...
Here, the common difference of each consecutive pair of numbers is 2. Now, let's try to find
the rule to calculate the value of nth term of the sequence. Since, the common difference is 2, the
nth term of the sequence is either 2n + (a constant) or 2n – (a constant).
Now, the first term (t1) = 1 = 2 × 1 – 1
the second term (t2) = 3 = 2 × 2 – 1
the third term (t3) = 5 = 2 × 3 – 1
the fourth term (t4) = 7 = 2 × 4 – 1
... ... ... .... ... ... ... ... ...
the general term or nth term (tn) = 2 × n – 1 = 2n – 1

Again, consider another sequence: 2, 6, 10, 14, 18, ... ...
Here, the common difference of each consecutive pair of number is 4. Now, let's try to find
the rule to calculate value of nth term of the sequence. Since, the common difference is 4, the nth
term of the sequence is either 4n + (a constant) or 4n – (a constant).
Now, the first term (t1) = 2 = 4 × 1 – 2
the second term (t2) = 6 = 4 × 2 – 2
the third term (t3) = 10 = 4 × 3 – 2
the fourth term (t4) = 14 = 4 × 4 – 2
... ... ... .... ... ... ... ... ...
the nth term (tn) = 4 × n – 2 = 4n – 2

18 Perfect Optional Mathematics Grade 6

Again, consider another sequence: 2, 6, 18, 54, 162, ... ...

Here, the common ratio of each consecutive pair of numbers is 6 = 18 = 54 = 162 = 3. Let's Algebra
2 6 18 54

try to find the rule to calculate value of nth term of the sequence. Since, the common ratio is 3, the

nth term of the sequence is 3n × (a constant).

Now, the first term (t1) = 2 = 31 × 2
3
2
the second term (t2) = 6 = 32 × 3

the third term (t3) = 18 = 33 × 2
3
2
the fourth term (t4) = 54 = 34 × 3

... ... ... .... ... ... ... ... ...

the nth term (tn) = 3n × 2 = 2 × 3n
3 3

Example 1:

Find the nth term and 10th term of the following sequences.

a. 2, 5, 8, 11, 14, ... ... b. 3, 10, 17, 24, 31, ... ...

c. 1, 4, 16, 64, 256, ... ... d. 3, 6, 12, 24, 48, ... ...
Solution:

a. The given sequence is 2, 5, 8, 11, 14, ... ...

Here, the common difference of each consecutive pair of numbers is 5 – 2 = 8 – 5 = 3.

Now, the first term (t1) = 2 = 3 × 1 – 1
the second term (t2) = 5 = 3 × 2 – 1
the third term (t3) = 8 = 3 × 3 – 1
the fourth term (t4) = 11 = 3 × 4 – 1
... ... ... .... ... ... ... ... ...

the general term or nth term (tn) = 3 × n – 1 = 3n – 1
Hence, the 10th term (t10) = 3 × 10 – 1 = 30 – 1 = 29
b. The given sequence is 3, 10, 17, 24, 31, ... ...

Here, the common difference of each consecutive pair of numbers is 10 – 3 = 17 – 10 = 7.

Now, the first term (t1) = 3 = 7 × 1 – 4
the second term (t2) = 10 = 7 × 2 – 4
the third term (t3) = 17 = 7 × 3 – 4
the fourth term (t4) = 24 = 7 × 4 – 4
... ... ... .... ... ... ... ... ...

the general term or nth term (tn) = 7 × n – 4 = 7n – 4
Hence, the 10th term (t10) = 7 × 10 – 4 = 70 – 4 = 66

Sequence and Pattern of Numbers 19

c. The given sequence is 1, 4, 16, 64, 256, ... ...

Here, the common ratio of each consecutive pair of numbers is 4 = 16 = 64 = 256 = 4. Since,
1 4 16 64

the common ratio is 4, the nth term of the sequence is 4n × (a constant).

Now, the first term (t1) = 1 = 41 × 1
4

the second term (t2) = 4 = 42 × 1
4

the third term (t3) = 16 = 43 × 1
4

the fourth term (t4) = 64 = 44 × 1
4
... ... ... .... ... ... ... ... ...

the nth term (tn) = 4n × 1 = 4n
4 4

Hence, the 10th term (t10) = 410 × 1 = 410
4 4

d. The given sequence is 3, 6, 12, 24, 48, ... ...

Here, the common ratio of each consecutive pair of number is 6 = 12 = 24 = 48 = 2. Since,
3 6 12 24

the common ratio is 2, the nth term of the sequence is 2n × (a constant).

Now, the first term (t1) = 3 = 21 × 3
2

the second term (t2) = 6 = 22 × 3
2

the third term (t3) = 12 = 23 × 3
2

the fourth term (t4) = 24 = 24 × 3
2

... ... ... .... ... ... ... ... ...

the nth term (tn) = 2n × 3 = 3 × 2n
2 2

Hence, the tenth term (t10) = 210 × 3 = 3 × 210
2 2
Example 2:

If the nth term of a sequence is 2n – 1, find its 3rd and 6th term.
Solutin:

Here, the nth term of the sequence, tn = 2n – 1
So, 3rd term, t3 = 2 × 3 – 1 = 6 – 1 = 5
and 6th term, t6 = 2 × 6 – 1 = 17

20 Perfect Optional Mathematics Grade 6

Exercise 1.3

1. Find the common difference of the following sequences. Algebra

a. 7, 9, 11, 13, ... ... b. 7, 10, 13, 16, ... ...

c. 2, 7, 12, 17, ... ... d. 7, 11, 15, 19, ... ...

2. Find the common difference of the following sequences and also find the next two numbers.

a. 3, 5, 7, 9, ... ... b. 6, 9, 12, 15, ... ...

c. 5, 9, 13, 17, ... ... d. 7, 12, 17, 22, ... ....

3. Find the nth term of the following sequences and also find the indicated term.

a. 2, 5, 8, 11, ... ... (10th term) b. 5, 7, 9, 11, ... ... (9th term)

c. 1, 5, 9, 13, ... ... (12th term) d. 3, 8, 13, 18, ... ... (8th term)

4. Find the common ratio of the following sequences.

a. 1, 2, 4, 8, ... ... b. 3, 6, 12, 24, ... ...

c. 1, 3, 9, 27, ... ... d. 2, 8, 32, 128, ... ...

5. Find the common ratio of the following sequence and also find the next two numbers.

a. 2, 4, 8, 16, ... ... b. 3, 12, 48, 192, ... ...

c. 1, 5, 25, 125, ... ... d. 2, 10, 50, 250, ... ....

6. Find the nth term of the following sequences and also find the indicated term.

a. 1, 4, 16, 64, ... ... (5th term) b. 5, 10, 20, 40, ... ... (6th term)

7. Match the sequence with its nth term.

a. 3, 7, 11, 15, ... ... n+3

b. 4, 7, 10, 13, ... ... 2n

c. 2, 4, 8, 16, ... ... 7n

d. 3, 5, 7, 9, ... ... 4n – 1

e. 4, 5, 6, 7, ... ... 3n + 1

f. 7, 14, 21, 28, ... ... 2n + 1

8. Find the first four terms of each of the following sequences, whose nth term is given.

a. n + 1 b. 3n

c. 2n – 1 d. 2n + 2

9. Find the mentioned terms of each of the following sequences, whose nth term is given.

a. 2n + 3 (1st, 2nd, 5th) b. 7n – 5 (3rd, 5th, 8th)

c. 5n + 2 (1st, 3rd, 6th) d. 8n – 3 (4th, 6th, 8th)

Sequence and Pattern of Numbers 21

2 LIMITS

Learning Objectives

find the limit of activities of daily life.
find the pattern of numbers.
find the pattern of diagrams.
find the value of function.
find the limit of simple function.

2.1 Concept of Limit

Introduction

Let's consider water flowing in a natural river. What is the last destination of the water flowing in
the river? Obviously, its last destination is sea. A bottle is filled with milk. If we make a very small
hole at its bottom, what happens? Obviously, at last all the milk will be flown on the floor and the
bottle will be empty. Study one more practical example below for the concept of limit. Dinesh has
a chocolate of 20cm length. After one minute, he eats half of the given chocolate. Again he eats
half of the remaining chocolate after next one minute.

20cm

10cm

5cm

2.5cm 1.25cm 0.625cm

1st minute 2nd minute 3rd minute 4th minute 5th minute

Length of Chocolate Remaining

He continues the process of eating the chocolate five times, how long chocolate will he eat in third
minute?

22 Perfect Optional Mathematics Grade 6

The length of chocolate eaten in the first minute = 1 × 20cm = 10cm
2
1
The length of chocolate eaten in the second minute = 2 × 10cm = 5cm

The length of chocolate eaten in the third minute = 1 × 5cm = 2.5cm
2

If he eats the chocolate a large number of times, what happens?

Concept of Limits by Pattern of Numbers Limits

Let us study the following number patterns:

0, 10, 20, 30, 40, 50, ..., ..., ...
The numbers can be located in the following number line.

 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Can you add two more terms in the above number pattern? The values of the terms goes on
increasing by 10 from the previous term. In this pattern, the next two terms are 110 and 120.
If we continue filling the numbers in this pattern, what happens?
When we fill the numbers in this patten a large number of times, we reach very large number as
its term. This is known as the limit of this pattern of numbers. Hence, the limit of the pattern of
numbers is infinity.

Example 1:

Find the 5th, 6th and 7th term of a sequence 12, 14, 18, 116, ..., ...
Solution:

The first term of the sequence is 1 = 1
2 21

The second term of the sequence is 1 = 1
4 22

The third term of the sequence is 1 = 1
8 23

The fourth term of the sequence is 1 = 1
16 24

The fifth term of the sequence is 1 = 1
32 25

The sixth term of the sequence is 1 = 1
64 26

And the seventh term of the sequence is 1 = 1
128 27

Example 2:

Find the 5th, 6th and 7th term of a sequence 5.9, 5.99, 5.999, 5.9999, ..., ... . Also find the
limit of the sequence.

Concept of Limit 23

Solution:

The first term of the sequence is 5.9.
The second term of the sequence is 5.99.
The third term of the sequence is 5.999.
The fourth term of the sequence is 5.9999.
The fifth term of the sequence is 5.99999.
The sixth term of the sequence is 5.999999.
Hence, the seventh term of the sequence is 5.9999999.

If we fill the numbers in the given sequence a large number of times, we get the number very close
to 6. Hence, the limit of this sequence is 6.
Some more examples of pattern of numbers (sequence)

S.N. Pattern of numbers Limit of pattern (Approaches to)

1. 2, 4, 6, 8, 10, ..., ... ∞

2. 1, 5, 9, 13, 15, ..., ... ∞

3. 0.1, 0.01, 0.001, 0.0001, 0.00001, ..., ... 0

4. 2.1, 2.01, 2.001, 2.0001, 2.00001, ..., ... 2

5. 8.9, 8.99, 8.999, 8.9999, 8.99999, ..., ... 9

Concept of Limit by Sequence of Diagarms

Here, we have an equilateral triangle of side 20cm. The second triangle is inscribed by joining
the midpoints of the sides successively. In the second triangle, we repeat the process, inscribing
the third triangle.

10cm

5cm

20cm

Here, the length of side of the first triangle = 20cm
The length of side of the second triangle = 10cm
The length of side of the third triangle = 5cm
The length of side of the fourth triangle = 2.5cm
From the above information, we can prepare the sequence of the sides of triangles as follows:
20cm, 10cm, 5cm, 2.5cm, 1.25cm, 0.75cm, 0.375cm, ..., ....

24 Perfect Optional Mathematics Grade 6

As, we fill the triangles a large number of times, the length of the sides becomes closer to zero. Limits
Hence, the limit of the side of the triangles is 0.
Again, consider the sequence of the perimeters of the triangles.
The perimeter of the first triangle = 3 × 20cm = 60cm
The perimeter of the second triangle = 3 × 10cm = 30cm
The perimeter of the third triangle = 3 × 5cm = 15cm
The perimeter of the fouth triangle = 3 × 2.5cm = 7.5cm
From above information, we can prepare the sequence of the perimeters of triangles as follows.

60cm, 30cm, 15cm, 7.5cm, 3.75cm, 2.25cm, 1.125cm, ..., ....

As, we fill the triangles a large number of times, the length of perimeter becomes closer to zero.
Hence, the limit of the perimeter of the triangles is 0.

Value of Function

Consider the function f given by y = f(x) = x + 2.

If we put different values of variable x, we get different values of f(x). The values of x are known
as inputs and the values of y are known as outputs.

Filling the inputs and corresponding outputs for the given function f(x) = x + 2:

x1 2 3 4 5 6

f(1) = 1 + 2 f(2) = 2 + 2 f(3) = 3 + 2 f(4) = 4 + 2 f(5) = 5 + 2 f(6) = 6 + 2
f(x)

=3 =4 =5 =6 =7 =8

Limit of Function

For any function, different inputs change its outputs. Suppose the input gets closer and closer to
some number. Its corresponding output gets closer and closer to some number, then that number
is called the limit of the function at the input.

Consider the function f given by y = f(x) = 3x + 1.

Take x = 2 as an input of this function. Suppose we select input numbers for x closer and closer

to the number 2 from its left side and right side. Look at the corresponding output numbers of the

function f(x) = 3x + 1. x approaches 2 from its right side

x approaches 2 from its left side

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3

When x approaches 2 from its left side, we select some input numbers of x closer to 2 as follows.
Also find the corresponding outputs.

x 1.6 1.7 1.8 1.9 1.99 1.999

f(x) f(1) = 4.8 + 1 f(1) = 5.1 + 1 f(2) = 5.4 + 1 f(3) = 5.7 + 1 f(4) = 5.97 + 1 f(5) = 5.997 + 1

= 5.8 = 6.1 = 6.4 = 6.7 = 6.97 = 6.997

Concept of Limit 25

When x approaches 2 from its right side, we select some input numbers of x closer to 2 as follows.
Also find the corresponding outputs.

x 2.5 2.4 2.3 2.1 2.01 2.001

f(x) f(1) = 7.5 + 1 f(1) = 7.2 + 1 f(2) = 6.9 + 1 f(3) = 6.3 + 1 f(4) = 6.03 + 1 f(5) = 6.003 + 1

= 8.5 = 8.2 = 7.9 = 7.3 = 7.03 = 7.003

From the above tables, when the input is x = 1.999 (closer to 2 from its left side), the value of
function f(x) is f(x) = 6.997 (closer to 7)

Similarly, when the input x = 2.001 (closer to 2 from its right side), the value of the function f(x)

is f(x) = 7.003 (closer to 7)

Hence, the limit of the function f(x) = 3x + 1 at x = 2 is 7.

Exercise 2

1. a. An 800 ml glass is full of milk. A girl drinks half of the milk in the glass first time. Again

she drinks the half of the remaining milk in the second time.

1st time 2nd time 3rd time 4th time

400ml 200ml 100ml 50ml

She continues the process of drinking milk five times, how much milk will she drink in
b. the 4th and 5th time? If she drinks the milk a large number of times, what will happen?

Ramesh has a chocolate of 10cm length. After one minute, he eats half of the given
chocolate. Again he eats half of the remaining chocolate after next one minute.

1st minute 5cm remained

2nd minute 2.5cm remained

3rd minute 1.25cm remained

4th minute 0.675cm remained

If he continues the process of eating the chocolate five times, how long chocolate will he
eat in the 4th and 5th minute? If he eats the chocolate a large number of times, what will
happen?

26 Perfect Optional Mathematics Grade 6

2. Locate the following numbers in a number line and fill two more numbers.

a. 5, 10, 15, 20, ..., ... b. 3, 8, 13, 18, ..., ...

c. 10, 8, 6, 4, 2, ..., ..., ... d. 50, 40, 30, 20, ..., ...

3. Find the 5th, 6th­and 7th term of the following sequences. Also find the limit of each sequence.

a. 0, 5, 10, 15, 20, ..., ... b. 50, 70, 90, 110, 130, ..., ... Limits

c. 7.9, 7.99, 7.999, 7.9999, ..., ... d. 7.1, 7.01, 7.001, 7.0001, ..., ...

e. 2.31, 2.301, 2.3001, 2.30001, ..., ... f. 6.01, 6.001, 6.0001, 6.00001, ..., ...

4. Here, we have an equilateral triangle of side 40cm. The second triangle is inscribed by
joining the midpoints of the sides successively. In the second triangle, we repeat the process
inscribing the third triangle.

20cm

10cm

40cm

a. Find the length of the side of the fourth triangle.
b. Find the length of the side of the fifth triangle.
c. Find the sequence of the lengths of the sides of the triangles.
d. Find the limit of the side of triangles.

5. Here, we have an equilateral triangle of side 30cm. The second triangle is inscribed by
joining the midpoints of the sides successively. In the second triangle, we repeat the process
inscribing the third triangle.

15cm

7.5cm
30cm

a. Find the length of the perimeter of the first three triangles.
b. Find the length of the perimeter of the fourth triangle.
c. Find the sequence of the lengths of the perimeters of the triangles.
d. Find the limit of the perimeter of triangles.

Concept of Limit 27

6. A ball is dropped from a height of 256cm. In each rebound, it rises to a half of its previous fall.

256 cm

a. At what height does the ball approach in 4th rebound?

b. Write a sequence of the heights of successive rebounds.

c. What is the limit of the height of the foot ball?

7. Find the value of f(0), f(2), f(3), f(6) and f(2) + f(7) in each of the folloiwng functions.

a. f(x) = x + 5 b. f(x) = 2x – 3

c. f(x) = x2 + 4 d. f(x) = 2x + x2

8. a. Fill the outputs of the function f(x) = 2x + 3 for the input numbers for x closer and closer
to the number 3 from its left side and right side in the following tables.

x 2.6 2.7 2.8 2.9 2.99 2.999

f(x) f(2.6) = ... f(2.7) = ... f(2.8) = ... f(2.9) = ... f(2.99) = ... f(2.999) = ...

x 3.5 3.4 3.2 3.1 3.01 3.001

f(x) f(3.5) = ... f(3.4) = ... f(3.2) = ... f(3.1) = ... f(3.01) = ... f(3.001) = ...

From the above tables, find the limit of the function at x = 3.
b. Fill the outputs of the function f(x) = 4x – 3 for the input numbers for x closer and closer

to the number 2 from its left side and right side in the following tables.
x 2.6 2.7 1.8 1.9 1.99 1.999

f(x) f(1.6) = ... f(1.7) = ... f(1.8) = ... f(1.9) = ... f(1.99) = ... f(1.999) = ...

x 2.5 2.4 2.2 2.1 2.01 2.001
f(x) f(2.5) = ... f(2.4) = ... f(2.2) = ... f(2.1) = ... f(2.01) = ... f(2.001) = ...
From the above tables, find the limit of the function at x = 2.

28 Perfect Optional Mathematics Grade 6

3 MATRICES Matrices

Learning Objectives

arrange the values in the matrix form.
identify the order of matrix.
find the components of matrix.
idenity the types of matrices.
perform addition and subtraction of matrices.
multiply matrix by a scalar.

3.1 Matrices

Introduction to Matrix

The collection of numerical data arranged in rows and columns enclosed by round or square
brackets i.e. or is known as a matrix. The plural form of matrix is ‘matrices’.

Matrices are denoted by capital letters A, B, C etc. For example, the number of students in classes
6, 7 and 8 with the number of boys and girls can be shown in the following table.

Classes

Boys VI VII VIII
60 55 40 Row 1

Girls 45 50 55 Row 2

Column Column Column
1 2 3

In the above example, there are two rows and three columns. The first row indicates the number
of boys and the second row indicates the number of girls. The three columns indicate the number
of students in three different classes. The information is represented in the matrix form as given
below.

60 55 40
P = 45 50 55

Matrices 29

Order of a Matrix

The number of rows and the number colunms of a matrix indicates the order or size of a matrix.

3 5 7
For example, in A = 4 6 8 , there are two rows and three columns in matrix A, so its order

3 5 7 1 2
is 2×3. It is also written as A2×3 or 4 6 8 . Similarly, Q = is a matrix of order 2×2
3 4
2×3

1 2 .
and is written as Q2×2 or 3
4 2×2

Components of a Matrix

3 5 7
A = 4 6 8 is a matrix of order 2×3. The numbers included in this matrix 3, 5, 7, 4, 6,

8 are known as elements or components of matrix A. These elements of the matrix A are also
represented by a11, a12, a13, a21, a22 and a23 respectively. So, we can denote the above matrix as

A= a11 a12 a13 .
a21 a22 a23

Indicating the elements of a matrix in this form, the address of the element is exactly mentioned.
For example, a12 is an element that belongs to the first row and the second column. a23 is an
element belongs to the second row and the third column.

Example 1:

2 6 8
Find b11, b12, b13, b21, b22 and b23 of the matrix B = –  6 0 – 7 .

Solution:

2 6 8
Here, the given matrix is B = –  6 0 – 7 .

We can denote the above matrix as B = b11 b12 b13 .
b21 b22 b23

Hence, comparing these two matrices, we have

b11 = 2, b12 = 6, b13 = 8, b21 = – 6, b22 = 0 and b23 = – 7

Various types of Matrices

There are various types of matrices according to thier size and nature.

1. Row matrix

If there is only one row in a matrix, the matrix is known as a row matrix.
For example, A = [3 5] is a 1×2 row matrix and 0 21 1 is a 1×3 row matrix.

30 Perfect Optional Mathematics Grade 6

2. Column matrix

A matrix having only one column is known as a column matrix. For example, B = 6 is a
7
10

2×1 column matrix and C = 15 is a 3×1 column matrix.

20

3. Zero / null matrix 0

The matrix having each element zero is known as a zero matrix. For example, A = 0 is a
2×1 zero matrix.

0 0 0 Matrices
B = 0 0 0 is a 3×3 zero matrix.

0 0 0

C= 0 0 0 is a 2×3 zero matrix.
0 0 0

4. Rectangular matrix

The matrix having different numbers of rows and columns is known as a rectangular matrix.

For example,

a b c p s is a rectangular matrix
A = d e f is a rectangular matrix of order 2×3 and B = q t
of order 3×2.
r u

5. Square matrix

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

a b p q r

Here, A = c d is a 2×2 square matrix and B = s t u is a square matrix of order
3×3. v w x

6. Equal matrices

Two matrices, A and B of same order are said to be equal if their corresponding elements are equal.

1 4 5 1 4 5 a d 3 0
If A = 3 6 8 and B = 3 6 8 then A = B. In C = b e and D = –  2 5 , if

c f 4 6

a = 3, b = – 2, c = 4, d = 0, e = 5 and f = 6, then C = D.

Example 2:

Construct a matrix of order 2×3 whose elements are given by aij = 3i + 2j.
Solution:

Let A = a11 a12 a13 be a matrix of order 2×3 whose elements are given by
a21 a22 a23

Matrices 31

aij = 3i + 2j

\ a11 = 3 × 1 + 2 × 1 = 3 + 2 = 5

a12 = 3 × 1 + 2 × 2 = 3 + 4 = 7

a13 = 3 × 1 + 2 × 3 = 3 + 6 = 9

a21 = 3 × 2 + 2 × 1 = 6 + 2 = 8

a22 = 3 × 2 + 2 × 2 = 6 + 4 = 10

a23 = 3 × 2 + 2 × 3 = 6 + 6 = 12

Hence, A = a11 a12 a13 5 7 9
a21 a22 a23 = 8 10 12 .

Example 3:

2 x 2 3
If x + y 0 = 4 0 , find the values of x and y.
Solution:

2 x 2 3
As x + y 0 = 4 0 , equating corresponding elements

x = 3 and

x + y = 4

or, 3 + y = 4

or, y = 4 – 3 = 1

Hence, x = 3, y = 1 Exercise 3.1

2 4
1. a. If A = 5 – 3 , then write the order of matrix A and find the values of a21 and a22.

2 4
b. If A = 0 6 , then write the order of matrix A and find the values of a11, a12, a21 and a22.

0 3 – 2
c. If B = 7 8 – 3 , then write the order of matrix B and find the values of b12, b21, b23.

2. Write the order of the following matrices.

2 3 4 5 1 6 10
a. 6 7 8 9 b. 2 7 11

3 8 12

1 2 0 3
c. 23 d. 1 2 – 3

4

32 Perfect Optional Mathematics Grade 6

3. Identify the square and rectangular matrix from the following.

1 3 1 3 4 a b c
a. 2 4 b. 2 5 6 c. e f g d. 1 2 3

4. a. Construct a matrix of order 2×2 whose elements are given by aij = 3i – 2j.

b. Construct a matrix of order 2×2 whose elements are given by aij = 2i + 3j.

c . Construct a matrix of order 2×3 whose elements are given by aij = 5i + 2j.

d . Construct a matrix of order 2×3 whose elements are given by aij = 4i – j.

2 3 2 3
5. a. Are  – 4 0 and –   4 0 equal? Give the reason.
Matrices
0 0 0 0 0

b. Are 0 0 and 0 0 0 equal? Give the reason.
0 0 0 0 0

2 x 2  – 4
6. a. If  y 6 =  0 6 , find the values of x and y.

3 2 3 2
b. If  a + 2 b =  5 7 , find the values of a and b.

2 4 z – 4 2 4 2
c. If x y 8 = 5 3 8 , find the values of x, y and z.

5 6 x – 4 5 6 – 4
d. If y 3 z + 2 = 5 3 8 , find the values of x, y and z.

7. Find the values of x and y under the following conditions.

x 5 4 5 x + y 4
a.  2 x – y = 2 3 b. x – y = 2

y x – y 2 3 3x + y 7
c.  1 7 = 1 7 d. 2x – y = 3

Matrices 33

3.2 Operation of Matrices

Addtion of Matrices

If A and B are two matrices of same order then we can find the sum of A and B. The sum of
matrices A and B is denoted by A + B which is a matrix obtained by adding the corresponding
elements of A and B.

a b c p q r
If A = d e f and B = s t u then

a b c p q r a + p b + q c + r
A + B = d e f + s t u = d + s e + t f + u

Example 1:

2 0 –   2 5
If A = 1 3 and B = 4 – 1 , find A + B.

Solution:

2 0 – 2 5
A + B = 1 3 + 4 – 1

2 + (– 2) 0 + 5 2 – 2 5 0 5
= 1 + 4 3 + (– 1) = 5 3 – 1 = 5 2

Multiplication of Matrix by a Scalar

If A is a matrix of any order and k is a scalar quantity, then we can multiply the matrix A by k.
The product of k and A is denoted by kA and is obtained by multiplying each element of A by k.

p q r kp kq kr
If A = a b c then kA = ka kb kc .

As multiplication is the repeated addition, we have the following addition.

If A = 5 1 , then 2A = A + A = 5 1 + 5 1
3 4 3 4 3 4

= 5 + 5 1+1 = 10 2 = 2 × 5 2×1
3 + 3 4+4 6 8 2 × 3 2×4

Example 2:

If A = 3 – 2 , find 2A and – 6A.
5 6

34 Perfect Optional Mathematics Grade 6

Solution:

Here, given matrix is A = 3 – 2 then
5 6

3 – 2 2 × 3 2 × (– 2) 6 – 4
2A = 2  5 6 = 2 × 5 2 × 6 = 10 12 and

– 6A = – 6  3 – 2 = – 6 × 3 – 6 × (– 2) = – 18 12 .
5 6 – 6 × 5 – 6 × 6 – 30 – 36

Subtraction of Matrices

If A and B are two matrices of the same order then we can find the difference of A and B. Matrices
Difference of matrices A and B is denoted by A – B which is a matrix obtained by subtracting the
corresponding elements of A from that of B.

a b c p q r
If A = d e f and B = s t u then

a b c p q r a – p b – q c – r
A – B = d e f – s t u = d – s e – t f – u

Example 3:

5 0 1 3
If A = 7 – 3 and B = –   2 4 , find A – B.

Solution:

5 0 1 3
Here, A = 7 – 3 and B = –   2 4 then

5 0 1 3
A – B = 7 – 3 – –   2 4

5 – 1 0 – 3 4 – 3 4 – 3
= 7 – (– 2) – 3 – 4 = 7 + 2 – 7 = 9 – 7

Example 4:

If A = 2 – 3 and B = 5 3   , find 3A + 2B and 2A – B.
5 1 –  1 6

Solution:

Here, A= 2 – 3 and B = 5 3 then
5 1 –   1 6

3A + 2B

= 3  25 –1 3 + 2  5 3 = 3 × 2 3 × (– 3) + 2 × 5 2×3
–   1 6 3 × 5 3×1 2 × (– 1) 2×6

= 6 – 9 + 10 6 = 6 + 10 – 9 + 6 = 16 – 3
15 3 –  2 12 15 – 2 3 + 12 13 15

Operation of Matrices 35

Now, 2A – B

= 2  52 – 3 – 5 3
1 –   1 6

= 2 × 2 2 × (– 3) – 5 3
2 × 5 2×1 –   1 6

= 4 – 6 – 5 3
10 2 –  1 6

= 4 – 5 – 6 – 3
10 + 1 2–6

= – 1 – 9
11 – 4

Exercise 3.2

1. a. If A = 6 2 and B = 1 – 2 , find A + B.
3  – 4 3 5

b. If M = 2 – 4 and N = 3 6 , find M + N.
6  3 –  1 7

c. If P = –   4 2 and Q = 5 9 , find 4P + Q.
2  7 6 – 1

2. a. If P = –  2 – 3 , find 3P and – 5P.
4 5

b. If M = –  1 9 , find 2M and – 4M.
5 6

c. If D = – 8 – 6 , find 21 D and – 23 D.
10 4

3. a. If A = 1 3 and B = –   2 3 , find A – B.
2  – 5 6 8

b. If P = 3 – 1 and Q = 9 2 , find P – Q.
8 9 –   2 8

c. If C = 6 – 4 and D = 5 2 , find 2C – 3D.
3 8 7 – 6

4. Find A + B and A – B in the following cases.

3 0 2 4 1 – 2 3 3 4 2
a. A = 2  – 1 , B = 5 3 b. A = 6 0 4 , B = 3 5  – 5

  – 2 3 2 0 2 1 3 3 0 1

c. A = 5 6 , B =  – 3 7 d. A = 7 8 – 4 , B = –  2 6  – 5
0 4 6 – 4

36 Perfect Optional Mathematics Grade 6

4 CO-ORDINATE GEOMETRY

Learning Objectives

find the distance between two points on the co-ordinate plane.
use the application of distance formula.
use the section formula.

4.1 Co-ordinate Axes

Number Lines C Co-ordinate

Let AB be the horizontal number line, the positive numbers are located towards 4
the right side of number 0 and the negative numbers towards the left side of 3
number 0. 2
1
AB 0
– 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 – 1
– 2
Similarly, CD be the vertical number line; the positive numbers are located – 3
above the point representing number 0 and the negative numbers below the – 4
number 0.
D
Rectangular Co-ordinate Axes

Let XOX' and YOY', the horizontal and vertical number lines which intersect each other at point

O; O is called the origin. On the number line XOX', the numbers on the right of the point O are positive

numbers and on the left are negative numbers. 2nd Quadrant Y Y-axis 1st Quadrant
x → negative x → positive
Similarly, on YOY' positive numbers are taken y → positive 5 y → positive
above O and the negative numbers below O. 4
x, y 3
Two lines XOX' and YOY' divide the plane into
four parts, each part is known as a quadrant. The –, + 2 Origin x, y
four quadrants are 1
+, +
X-axis

XOY → 1st quadrant X ' – 5 – 4 – 3 – 2 – –1  1 O 1 2 3 4 5X
YOX' → 2nd quadrant 3rd Quadrant 4rd Quadrant
X'OY' → 3rd quadrant x → negative – 2
Y'OX → 4th quadrant x → positive

y → negative – 3 y → negative

x, y – 4 x, y

–, – – 5 +, –

Y'

Co-ordinate Axes 37

Every point on the plane can be expressed as (x, y) where x represents the x-co-ordinate
and y represents the y-co-ordinate. The signs of x-co-ordinate and y-co-ordinate on different
quadrants are shown in the above figure.

Example 1:

Locate the points A(2, 5) and B(– 3, 4) on the cartesian plane.
Solution:

The given points are A(2, 5) and B(– 3, 4). For point A(2, 5), since x-co-ordinate is 2 and

y-co-ordinate is 5, move 2 units from the origin O on the right and then 5 units up will give the

point A(2, 5). Similarly, move 3 units left from origin and 4 units up will give the point B(– 3, 4).

B(– 3, 4) 5 Y A(2, 5)

4

3

2
1

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

– 2
– 3
– 4
– 5

Y'

Distance between Two Points on a Horizontal Line

If A(2, 4) and B(5, 4) be two points on a horizontal X' Y B(5, 4) X
line, then A(2, 4) N
OM = x co-ordinate of A = 2.
ON = x co-ordinate of B = 5 OM
Then distance between two points A and B
= AB = MN
= ON – OM
= 5 – 2
= 3 units

Y'
If A(x1, y1) and B(x2, y2) are two points on the horizontal line then the distance between two points
A and B is denoted by AB. If x2 > x1, AB = x2 – x1 and if x1 > x2, AB = x1 – x2.

38 Perfect Optional Mathematics Grade 6

Distance between Two Points on a Vertical Line

If P(3, 5) and Q(3, 1) be two points on a plane lying on Y P(3, 5) X
the same vertical line then
OD = QN = y co-ordinate of Q = 1 C Q(3, 1)
OC = PN = y co-ordinate of P = 5 N
Now, distance between P and Q is D
PQ = PN – QN = 5 – 1 = 4 units X' O
If P(x1, y1) and Q(x2, y2) be two points on the vertical
line then the distance between two points is denoted Y'
by PQ.
If y2 > y1, PQ = y2 – y1 and if y1 > y2, PQ = y1 – y2.

Distance between Two Points on the Plane

Let A(2, 2) and B(6, 5) be two points on the plane. Y

Draw AM^OX, BN^OX and AC^BN as shown in Co-ordinate

figure. B(6,5)

Then OM = 2, AM = 2 C
NX
ON = 6, BN = 5

Now, AC = MN = ON – OM = 6 – 2 = 4 A(2,2)
BC = BN – CN = BN – AM = 5 – 2 = 3

As ABC is a right angled triangle,

AB2 = AC2 + BC2 (Pythagoras theorem) X' O M
= 42 + 32

= 16 + 9 = 25 Y'

\ AB = 5 units
Distance between Two Points

Let A(x1, y1) and B(x2, y2) be two points on a plane. Y
Draw AM^OX, BN^OX and AC^BN as shown.
A B(x2,y2)
Then AC = MN = ON – OM C
(x1,y1) NX
= x2 – x1
BC = BN – CN = BN – AM X' O M

= y2 – y1 Y'
From right angled triangle ABC,

AB2 = AC2 + BC2

AB2 = (x2 – x1)2 + (y2 – y1)2
AB = (x2 – x1)2 + (y2 – y1)2

Distance of point P(x, y) from origin O(0, 0) is given by
OP = x2 + y2

Co-ordinate Axes 39

Example 2:

Find the distance between the following pairs of points:

a. (2, 3) and (5, 7) b. (– 3, 6) and (4, – 2)
Solution:

a. Given points are (2, 3) and (5, 7) b. Given two points are (– 3, 6) and (4, – 2)

Let, (x1, y1) = (2, 3) and (x2, y2) = (5, 7) then, Let (x1, y1) = (– 3, 6) and (x2, y2) = (4, – 2) then,
distance between two points distance between two points
d2 = (x2 – x1)2 + (y2 – y1)2 d2 = (x2 – x1)2 + (y2 – y1)2
or, d2 = (5 – 2)2 + (7 – 3)2 or, d2 = {4 – (– 3)}2 + (– 2 – 6)2
or, d2 = 32 + 42 or, d2 = (4 + 3)2 + (– 8)2
or, d2 = 9 + 16 or, d2 = 72 + 64
or, d2 = 25 or, d2 = 49 + 64
or, d2 = 113
\ d = 5 units.

\ d = 113 units

Example 3:

If P(– 2, 5), Q(– 1, 3) and R(1, – 1) are three points then prove that PR = 3PQ.
Solution:

For PQ, P(– 2, 5) and Q(– 1, 3).

Let (x1, y1) = (– 2, 5) and (x2, y2) = (– 1, 3), then
PQ2 = (x2 – x1)2 + (y2 – y1)2
or, PQ2 = {– 1 – (– 2)}2 + (3 – 5)2
or, PQ2 = (– 1 + 2)2 + (– 2)2
or, PQ2 = 12 + 4
or, PQ2 = 5

\ PQ = 5 units
For PR, P(– 2, 5) and R(1, – 1)

Let (x1, y1) = (– 2, 5) and (x2, y2) = (1, – 1)
PR2 = (x2 – x1)2 + (y2 – y1)2
or, PR2 = {1 – (– 2)}2 + (– 1 – 5)2
or, PR2 = (1 + 2)2 + (– 6)2
or, PR2 = 32 + 36
or, PR2 = 9 + 36
or, PR2 = 45

\ PR = 45 = 3 × 3 × 5 = 3 5 units
Hence, PR = 3PQ.

40 Perfect Optional Mathematics Grade 6

Exercise 4.1

1. Find the co-ordinates of points A, B, C, D and Y A
E from the following graph:
5
4

3 E

B2

2. Plot the following points on the cartesian plane : 1

a. (4, 2) b. (3, 4) X' – 5 – 4 – 3 – 2 – –1  1 O 1 2 3 4 5 X

c. (– 1, 3) d. (3, – 2) – 2 D
– 3
e. (– 1, – 2) f. (1, – 7) – 4
– 5
g. (0, 3) h. (– 4, 0) C
Y'
3. Find the distance between the following pairs of points:

a. P(0, 2) and Q(3, – 2) b. A(2, 7) and B(5, 6) Co-ordinate

c. M(6, 6) and N(2, 3) d. R(– 4, – 3) and S(– 10, 5)

e. K(3, – 5) and L(7, 3) f. E(8, 4) and F(3, – 8)

g. G(3a, 6a) and H(– a, 2a) h. T(a, b) and R(b, a)

4. From the figure given below, find the co- Y A
ordinates of the points A, B, C, D, E and F.
Also find the distance between: 5

C4

3

a. A and B b. C and D 2
c. E and F
1 B

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

– 2 F

D – 3

– 4

E – 5

Y'

5. a. If A(2, 6), B(10, 0), C(4, – 6) and D(1, – 2) are four points, show that AB = 2CD.
b. If A(4, 2), B(1, 8) and C(2, 6) are three points, prove that AB = 3BC.
6. a. Show that the points (3, 4), (– 4, 3), (– 4, – 3) are equidistance from the origin.
b. Show that the points (4, 6), (0, 6) and (4, 0) are equidistance from the point (2, 3).

Co-ordinate Axes 41

4.2 Application of Distance Formula

Using the distance we can verify, the properties of different triangles and quadrilateral. For the
verifications, study the following properties.

Names Properties to Show
Scalene Triangle All three sides are of different length.
Isosceles Triangle Any two sides are equal.
Equilateral Triangle All three sides are equal
Right Angled Triangle Verify the pythagoras theorem.
Parallelogram Opposite sides are equal.
Opposite sides are equal and diagonals are equal.
Rectangle or Opposiste sides are equal and an angle is 90°.

Example 1:

If A(– 1, 2), B(1, 6) and C(4, 2) are three vertices of a triangle, show that DABC is an

isosceles triangle.
Solution:

Here the vertices of DABC are A(– 1, 2), B(1, 6) and C(4, 2).

For the length of side AB, And for the length of side BC,

let A(– 1, 2) = (x1, y1) and B(1, 6) = (x2, y2) let B(1, 6) = (x1, y1) and C(4, 2) = (x2, y2)

The length of side AB = (x2 – x1)2 + (y2 – y1)2 The length of side BC = (x2 – x1)2 + (y2 – y1)2
= (1 + 1)2 + (6 – 2)2 = (4 – 1)2 + (2 – 6)2
= 22 + 42 = (3)2 + (– 4)2
= 4 + 16 = 9 + 16
= 20
= 25
= 2×2×5

\ AB = 2 5 units \ BC = 5 units

For the length of side AB, let A(– 1, 2) = (x1, y1) and C(4, 2) = (x2, y2)

The length of side AC = (x2 – x1)2 + (y2 – y1)2

= {4 – (– 1)}2 + (2 – 2)2

= (4 + 1)2 + 02

= 52 + 0

= 25
\ AC = 5 units

As, AC = BC = 5 units, DABC is an isosceles triangle.

42 Perfect Optional Mathematics Grade 6

Example 2:

Show that the points (3, 2), (5, 7), (– 3, 3) and (– 5, – 2) are vertices of a prallelogram.
Solution:

Let A(3, 2), B(5, 7), C(– 3, 3) and D(– 5, – 2) are given points.

Then using distance formua, d = (x2 – x1)2 + (y2 – y1)2, we have

For the length of side AB, For the length of side CD,

let A(3, 2) = (x1, y1) and B(5, 7) = (x2, y2) let C(– 3, 3) = (x1, y1) and D(– 5, – 2) = (x2, y2) Co-ordinate
AB = (x2 – x1)2 + (y2 – y1)2 CD = (x2 – x1)2 + (y2 – y1)2
= (5 – 3)2 + (7 – 2)2 = {– 5 – (– 3)}2 + (– 2 – 3)2
= 22 + 52 = (– 5 + 3)2 + (– 5)2
= 4 + 25 = (– 2)2 + 25
\ AB = 29 units = 4 + 25
\ CD = 29 units
For the length of side BC, And for the length of side AD,

let B(5, 7) = (x1, y1) and C(– 3, 3) = (x2, y2) let A(3, 2) = (x1, y1) and D(– 5, – 2) = (x2, y2)
BC = (x2 – x1)2 + (y2 – y1)2 AD = (x2 – x1)2 + (y2 – y1)2
= (– 3 – 5)2 + (3 – 7)2 = (– 5 – 3)2 + (– 2 – 2)2
= (– 8)2 + (– 4)2
= (– 8)2 + (– 4)2

= 64 + 16 = 64 + 16

\ BC = 80 units \ AD = 80 units

From the above calculation of lengths of sides of quadrilateral ABCD, we have

AB = CD = 29 units BC = AD = 80 units
\ Opposite sides are equal. i.e. ABCD is a parallelogram.
Example 3:

If A(5, 2), B(8, 3) and C(3, 8) are the vertices of DABC, then prove that DABC in a right
angled triangle.
Solution:

Here, vertices of DABC are A(5, 2), B(8, 3) and C(3, 8).
For the length of side AB,

let A(5, 2) = (x1, y1) and B(8, 3) = (x2, y2)

Application of Distance Formula 43

The length of side AB = (x2 – x1)2 + (y2 – y1)2
= (8 – 5)2 + (3 – 2)2
= 32 + 12
= 9 + 1
= 10
\ AB2 = 10
For the length of side BC,
let B(8, 3) = (x1, y1) and C(3, 8) = (x2, y2)
The length of side BC = (x2 – x1)2 + (y2 – y1)2
= (3 – 8)2 + (8 – 3)2
= (– 5)2 + 52
= 25 + 25
= 50
\ BC2 = 50
For the length of side AC,
let A(5, 2) = (x1, y1) and C(3, 8) = (x2, y2)
The length of side AC = (x2 – x1)2 + (y2 – y1)2
= (3 – 5)2 + (8 – 2)2
= (– 2)2 + 62
= 4 + 36
= 40
\ AC2 = 40
Now, BC2 = AB2 + AC2
i.e. 50 = 10 + 40
i.e. 50 = 50
which is true. Hence, ®BAC = 90° and DABC is a right angled triangle.

44 Perfect Optional Mathematics Grade 6