Prime Optional Mathematics 6

Grade
VI

PRIME Optional
Mathematics

Pragya Books &
Distributors Pvt. Ltd.

Author Editors

Dirgha Raj Mishra LN Upadhyaya
Rajkumar Mathema

DN Chaudhary
Narayan Shrestha

Khem Timsina
J.N. Aryal

Kadambaba Pradhan
Dinesh Silwal

Pragya Books & Distributors Pvt. Ltd.

Lalitpur, Nepal
Tel : 5200575
email : [email protected]

© Author

Author Dirgha Raj Mishra

Editors LN Upadhyaya
Rajkumar Mathema
DN Chaudhary
Narayan Shrestha
Khem Timsina
J.N. Aryal
Kadambaba Pradhan
Dinesh Silwal

First Edition 2076 B.S. (2019 A.D.)
Revised Edition 2077 B.S. (2020 A.D.)

ISBN 978-9937-9170-4-9

Typist Sachin Maharjan
Sujan Thapa

Layout and Design Desktop Team

Printed in Nepal

Preface

Prime Optional Mathematics series is a distinctly outstanding mathematics
series designed according to new curriculum in compliance with Curriculum
Development Centre (CDC) to meet international standard in the school level
additional mathematics. The innovative, lucid and logical arrangement of the
contents make each book in their series coherent. The representation of ideas in
each volume makes the series not only unique but also a pioneer in the evaluation
of activity based mathematics teaching.

The subject is set in an easy and child-friendly pattern so that students will
discover learning mathematics is a fun thing to do even for the harder problems.
A lot of research, experimentation and careful graduation have gone into the
making of the series to ensure that the selection and presentation is systematic,
innovative, and both horizontally and vertically integrated for the students of
different levels.

Prime Optional Mathematics series is based on child-centered teaching
and learning methodologies, so that the teachers can find teaching this series
equally enjoyable.

I am optimistic that, this series shall bridge the existing inconsistencies
between the cognitive capacity of children and the subject matter.

I owe an immense dept of gratitude to the publishers (Pragya Books team)
for their creative, thoughtful and inspirational support in bringing about the
series. Similarly, I would like to acknowledge the tremendous support of editors
team, teachers, educationists and well-wishers for their contribution, assistance
and encouragement in making this series a success. I would like to express my
special thanks to Sachin Maharjan (Wonjala Desktop) for his sincere support
of designing part of the book and also Mr. Gopal Krishna Bhattarai to their
memorable support to prepare this series.

I hope this series will be another milestone in the advancement of teaching
and learning Mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that I can refine and improvise the series
in the future editions.

– Author

Contents Page

S.N. Units 1
2
1. Algebra 5
1.1 Ordered pair 11
1.2 Cartesian product 18
1.3 Polynomials
1.4 Surds 23
24
2. Matrices 30
2.1 Introduction
2.2 Operation on matrices 39
40
3. Co-ordinate Geometry 50
3.1 Rectangular co-ordinate axis
3.2 Mid-point of a line segment 55
56
4. Trigonometry 61
4.1 Measurement of angles 70
4.2 Solution of right angled triangle 77
4.3 Trigonometric Ratios 83
4.4 Trigonometric identities 87
4.5 Conversion of Trigonometric ratios
4.6 Trigonometric Ratios of some standard angles 99
100
5. Vector 108
5.1 Introdcution
5.2 Vector operations. 115
117
6. Transformation 123
6.1 Reflection
6.2 Translation 129
130
7. Statistics 134
7.1 Arithmatic mean (Average) 139
7.2 Median 139
7.3 Mode
7.4 Range 144

Model Questions

1 Algebra

1. Algebra

1.1 Ordered pairs
1.2 Cartesian product
1.3 Polynomials
1.4 Surds

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
1– 3 7 16
No. of Questions 1 1 4–

Weight 12

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

Objectives : At the end of the lesson
• Students are able to know the ordered pairs and Cartesian products.
• Students are able to represent Cartesian product in arrow diagram

and graph.
• Students are able to know the number system including surds.
• Students are able to operate the surds.
• Students are able to know the polynomials, types and degree of polynomials.

Materials Required:

• Chart paper
• Chart of number system.
• Chart of types of surds.
• Chart of types polynomials & properties of addition and multiplication.

1.1 Ordered pair:

Let us discuss the pair values of any two elements taken in ordered
as (2, 4) where seconds component is double the first. In the example
(Chandragadi, Jhapa) where chandragadi is the head quarter of Jhapa
district.
The two examples discussed above of pair of two values, give an idea
of the two elements taken in pair. From where we find there is the
certain order between them is called ordered pair.

The pair of any two elements taken in the
form of (a, b) where there is a kind of order
between them is called ordered pair.

Here, (a, b) is called ordered pair.
‘a’ is called 1st component (Antecedent)
‘b’ is called 2nd component (Consequence)

Equal ordered pairs:

Let us compare any two ordered pairs (1, 2) and (3 – 2, 5 – 3).
The first component of second ordered pair is 3 – 2 = 1 which is same
an the first component of first ordered pair. Also, second component of
second ordered pair 5 – 3 = 2 which is same as the second component
of first ordered pair.

Such ordered pairs having same components antecedent as well as
consequence are called equal ordered pairs.
If any two ordered pairs (x, y) and (a, b) are equal, then there will be
x = a and y = b.

2 PRIME Opt. Maths Book - VI

Worked out Examples

1. Write down the ordered pair of first two vowel alphabets
according to their position in counting numbers:

Solution:
The first two vowel alphabets are a and e.
Their position in counting number are 1 and 5.
∴ Ordered pairs are (a, 1)and (e, 5)

2. Write down any four ordered pairs of district with respect to
headquarters.
Solution:
The name of any 4 districts are Lalitpur, Chitawan, Rautahat and
Dang, the headquarters are respectively Patan, Bharatpur, Gaur,
Ghorahi.

The Ordered pairs are:
(Patan, Lalitpur) (Bharatpur, chitwan),

(Gaur, Rautahat), (Dang, Ghorahi)

3. If (x + 2, y – 1) = (5, 3), find the value of x and y.

Solution:

The equal ordered pairs are:
(x + 2, y – 1) = (5, 3)
By equating the corresponding elements.
x+ 2=5 and y – 1 = 3
or, x = 5 – 2 and y = 3 + 1
∴ x=3 and y = 4

\ x=3
y=4

PRIME Opt. Maths Book - VI 3

Exercise 1.1

1. Answer the following questions:
i. What is ordered pair?
ii. What do you mean by antecedent?
iii. What is consequence?
2
iv. Are the ordered pairs (3, 1) and a6 – 3, 2 k equal ?

v. Are the ordered pairs (2, 3) and (5 – 3, 6 ) equal?
2

2. Write down the ordered pairs for the followings:
i. Ordered pairs of first two odd numbers with first two even
numbers.
ii. Ordered pairs of first 3 english alphabets with their position
in counting numbers.
iii. Ordered pairs of all vowel alphabets with the position of
counting numbers.
iv. Ordered pairs of 5 countries with respect to their capitals.
v. Ordered pairs of 3 temples with respect to their location
(district).

3. If the following ordered pairs are equal, find the value of ‘x’ and y.
i. (x, y – 2) = (2, 3) ii. (x + 2, y) = (5, 4)
iii. (x – 2, y – 1) = (2, 1) iv. (x + 3, y + 2) = (5, 7)
v. (x – 1, y + 2) = (3 – x, 6 – y)

1. Show to your teacher. Answer

2. i) (1, 2) and (3, 4) ii) (a, 1), (b, 2) and (c, 3)

iii) (a, 1), (e, 5), (i, 9), (o, 15), (u, 21)

iv) Do yourself v) Do yourself

3. i) x = 2, y = 5 ii) x = 3, y = 4 iii) x = 4, y = 2
iv) x = 2, y = 5 v) x = 2, y = 2

4 PRIME Opt. Maths Book - VI

1.2 Cartesian product:

Let us consider A = {1, 2} and B = {3, 4} are any two sets. Then the all
possible ordered pairs from the set A to the set B can be taken as (1,
3), (1, 4), (2, 3), (2, 4). The set of these all possible ordered pairs is
called Cartesian product A × B written as A × B ={(1, 3), (1, 4), (2, 3),
(2, 4)}

The set of all possible ordered pairs (x, y)
from the non empty sets A to B is called
Cartesian product A × B.

Example:
If A = {a, b}, B = {p, q}

The Cartesian product of A and B is
A × B = {a, b} × {p, q}

= {(a, p), (a, q), (b, p), (b, q)}

The Cartesian product of B and A is,
B × A = {p, q} × {a, b)

= {(p, a), (p, b), (q, a), (q, b)}

Arrow diagram: B×A
A×B pa
qb
ap
bq

Relation:

Let us consider the two sets,
A = {Ram, siva},
B = {sita, parvati, Durga}
Then, Cartesian product of A and B is,
A × B = {(Ram, Sita), (Ram, Parvati), (Ram, Durga) (Siva, Sita)
(Siva, Parvati), (Siva, Durga)}

PRIME Opt. Maths Book - VI 5

But the relation of husband and wife can be taken from the set as,
{(Ram, Sita), (Siva, Parvati)} only. It is called the relation from the set
A to the set B which is the sub- set of Cartesian product A × B.

The sub-set (x, y) of Cartesian product A × B

where there are the related order pairs from
the set A to B is called relation.

In the above example,
Set A = {Ram, Siva} is called domain.
Set B = {Sita, Parvati, Durga} is called co-domain
Range = The sub-set of co-domain where the elements are related to
the domain is called range.
Arrow diagram:

Ram Sita

Siva Parvati

Durga

Worked out Examples

1. If A = {x, y}, find the Cartesian product A × A. Also show in arrow
diagram.

Solution:
Set A = {x, y}
Then, A × A = {x, y} × {x, y}

= {(x, x), (x, y), (y, x), (y, y)}

Arrow diagram: A×A
xx

yy

6 PRIME Opt. Maths Book - VI

2. If A = {a, b, c}, B = {2, 3}, find A × B, Also show in arrow diagram.

Solution:
A = {a, b, c}
B = {2, 3}

Then,
A × B = {a, b, c} × {2, 3}

= {(a, 2), (a, 3), (b, 2), (b, 3), (c, 2), (c, 3)}

Arrow diagram: A×B

a2
b
c 3

3. If A={2, 3, 4} ,B={4, 5}, find the relation A to B where there is the
relation more than by 1. Also show in arrow diagram.

Solution:
A= {2, 3, 4}
B={4, 5}
Then, A × B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}
Taking the relation more than by 1.
R= { (3, 4), (4, 5) }

Arrow diagram: R

24
3
45

Exercise 1.2

1. i. What is Cartesian product ?
ii. What do you mean by relation?
iii. Is {(3, 6),(2,1)} a relation from A={1, 2, 3, } to B={3, 4} ?
iv. Is {(2,6), (3,7)} a relation from P={2, 3, 4} to Q ={5, 6, 7} ?
V. Define the terms domain and range.

PRIME Opt. Maths Book - VI 7

2. Write down the Cartesian product from the followings.
i. A × B Where A = {a, b}, B = {m, n}
ii. P × Q Where P = {1, 2}, Q = {a, b, c}
iii. B × A Where A = {a, b}, B = {m, n}
iv. Q × P Where P = {1, 2}, Q = {a, b, c}
v. A ×B Where A = {1, 2, 3}, B = {4, 5, 6}

3. Show all the Cartesian product of Q. No.2 in arrow diagram ?

4. Which of the following are the relation from the set A= {3, 4, 5}
and B= {6, 7, 8} ? Give reason.
i. RRRRR14523 = {(1, 3), (2,4), (3, 5)}
ii. = {(3,6), (4, 7), (5 8)}
iii. = {(3, 7), (4, 8)}
iv. = {(3, 6), (5, 8)}
v. = {(4, 6), (5, 8), (6, 7)}

5. Find the domain and range from the following relations. Also
show in arrow diagram.
i. RRRRR12345 = {(1, 2), (3, 4)}
ii. = {(a, 1), (b, 2), (c, 3)}
iii. = {(p, 1), (q, 1), (r, 1)}
iv. = {a, p), (b, a), (c, r)}
v. = {(2, 4), (3, a), (4, b)}

6. If A = {2, 3, 4}, B = {4, 5, 6}, find the following relations from the
set A to the set B. Also show in arrow diagram.
i. First component is lesser than second.
ii. First component is less then second by 1.
iii. First component is equal to second.
iv. Sum of the components is 8.
v. First component is double the second.

8 PRIME Opt. Maths Book - VI

Answer

1. Show to your teacher.
2. i. A × B = {(a, m), (a, n), (b, m), (b, n)}

ii. P × Q = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, C)}
iii. B × A = {(m, a), (m, b), (n, a), (n, b)}
iv. Q × P = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
v. A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4) (3, 5), (3, 6)}

3. i. A×B ii. P×Q iii. B×A

am 1 a m a
2 b b
bn c n

iv. Q×P v. A×B

a 1 1 4
b 2 2
c 3 5
6

4. i. No relation ii. Relation iii. Relation
iv. Relation because it belongs to A × B.
v. No relation because it does not belong to A × B.

5. i. Do main = {1, 2}, range = {2, 4}, R2
ii. Domain = {a, b, c}, range = {1, 2, 3}

R1 a 1
12 b 2
34 c 3

(In this way for remaining question also.)

PRIME Opt. Maths Book - VI 9

6. i. R1 = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6)}

R1

24

35

46

ii. R2 = {(3, 4), (4, 5)}

R2

24

35
46

iii. R3 = {(4, 4)}

R3

24

35
46

iv. R4 = {(2, 6), (3, 5), (4, 4)}

R4

24

35
46

v. R5 = {(2, 4), (3, 6)}

R5

24

35
46

10 PRIME Opt. Maths Book - VI

1.3 Polynomials

Let us consider the algebric expression,
x2 + 2x + 3 It is polynomial.
3 2
x2 + x + x2 It is not polynomial.

x4 + 3x3 + 2x2 + 5x – 1 It is polynomial.
x2 + 3x–1 + 2x–2 It is not polynomial.

The algebric expression which has positive
integers as the power (exponent) of the
variables is called polynomial.

• Polynomials should be written in descending order of the
exponents of the variables which is called standard form of
a Polynomial.

Example: p(x) = x4 +3x3 – 2x2 + 5x – 2

Variable:

The algebric symbols used in the expression is called variable.
Examples: x, y, z, u and v etc.

Constant:

The mathematical numerical value used in the terms of the expression
is called constant.

Examples: 2, 3, 4, –3, a, b and c etc.

Term:

One part of the expression including constant and variable is called
term.

Examples: 2x², 3x4, 2, 5x³etc.

PRIME Opt. Maths Book - VI 11

Index of the variable (Exponent)

The power of the variables used in the algebric term is called index
(Exponent).

Examples: 3x5 " 5 is index of variable x.

Algebric expression:

The combination of the algebric terms with +ve or –ve sign is called
expression.

Examples: 2x3 + 3x2 – 5x + 2.

Coefficient:

A constant (real number) used as the multiplier of the variable in an
algebric term is called coefficient.
Examples: 2x3 2 is the coefficient.

Degree of the Polynomials:

The sum of the powers of the variables used in an algebric term of the
polynomial is called degree.
The maximum degree of the terms have to be taken as degree of the
polynomials.

Example: x2y3 + x3y – 3x2 + 2

1st term (degree) = 2 + 3 = 5
2nd term (degree) = 3 + 1 = 4
3rd term (degree) = 2
4th term (degree) = 0
\ Degree of the polynomial is 5.

Types of Polynomial:
• According to number of terms :
It has only one term.
i) Monomial
p(x) = 3x
It has two terms.
ii) Binomial
p(x) = ax + b

12 PRIME Opt. Maths Book - VI

iii) Trinomial It has three terms.
p(x) = ax2 + bx +c

iv) Polynomial It has more then three terms.
P(x) = ax4 + bx3 + cx2 + dx +e.

• According to degree : zero degree .

i) Constant polynomial
P(x) = 5
First degree.
ii) Linear polynomial
P(x) = ax + b
iii) Quadratic polynomial Second degree.
P(x) = ax2 + bx + c
third degree.
iv) Cubic polynomial
P(x) = ax3 + bx2 + cx + d
Fourth degree.
v) Biquadratic polynomial
P(x) = ax4 + bx3 + cx2 + dx + e

Equal Polynomials:

Any two polynomials which are similar or equal in all respect i.e. same
degree, no. of terms, variables as well as coefficients are called equal
polynomials.

Example: p(x) = x2 + 3x – 2
q(x) = x2 + 3x – 2

Hence, p(x) = q(x).

Addition and Subtraction of Polynomials.

Addition and subtraction of algrbric expressions can be done by
adding or subtracting the like term of the two or more polynomials.

Example: p(x) = 3x2 + 2x + 5
q(x) = x2 – 3x + 2

then,
p(x) + p(x) = (3x2 + 2x + 5) + (x2 – 3x + 2)

= 3x2 + 2x + 5 + x2 – 3x + 2

PRIME Opt. Maths Book - VI 13

= 3x2 + x2 + 2x – 3x + 5 + 2
= 4x2 – x + 7

p(x) – q(x) = (3x2 + 2x + 5) – (x2 – 3x + 2)
= 3x2 + 2x + 5 – x2 + 3x – 2
= 3x2 – x2 + 2x + 3x + 5 – 2
= 2x2+ 5x + 3

Product of the polynomials.

The product of the polynomials is the multiplication of each terms of
a polynomial by each and every terms of the another polynomial. Also
there is the using of addition and subtraction rule to find the product
of the polynomials.

Examples:
p(x) = x² + 2x –3
p(x) = x + 2

Then, = (x² + 2x – 3)(x + 2)
p(x).q(x) = x(x² +2x – 3) + 2(x² +2x – 3)
= x³ + 2x² – 3x + 2x² +4x – 6
= x³ + 4x2 + x – 6

Worked out Examples

1. Find the degree of the polynomial x²y³ + 3x²y – 5x²y². Also write
down its type.
Solution :
Polynomial,p(x) = x²y³ + 3x²y – 5x²y²
degree of 1st term = 2 + 3 = 5
degree of 2nd term = 2 + 1 = 3
degree of 3rd term = 2 + 2 = 4

\ Degree of the polynomial is 5.
It is trinomial according to number of terms.

14 PRIME Opt. Maths Book - VI

2. If p(x) = x3 + 2x2 – 3x + 2 and q(x) = 5 + x2 – 2x + x3, find the sum
of the polynomials.
Solution :
p(x) = x3 + 2x2 – 3x + 2
q(x) = 5 + x2 – 2x + x3
Then,
Sum = p(x) + q(x)
= (x3 + 2x2 – 3x + 2) + (5 + x2 – 2x + x3)
= x3 + 2x2 – 3x + 2 + 5 + x2 – 2x + x3
= (x3 + x3) + (2x2 + x2) – 3x – 2x +2 +5
= 2x3 + 3x2 – 5x + 7

3. Find the difference of p(x) = 3x³ + 2x² – 5 + 3x and q(x) = 5x² – 2
– x – x³.
Solution :
p(x) = 3x3 + 2x2 – 5 + 3x
q(x) = 5x2 – 2 – x – x3.
Difference = p(x) – q(x)
= (3x3 + 2x2 – 5 + 3x) – (5x2 – 2 – x – x3)
= 3x³ + 2x2 – 5 + 3x – 5x2 + 2 + x + x3
= (3x3 + x3 ) + (2x2 – 5x2) + 3x + x – 5 + 2
= 4x3 – 3x2 + 4x – 3.

4. Find the product of 2x² – 3x – 2 and 2x + 3.
Solution:
p(x) = 2x² – 3x – 2
q(x) = 2x + 3
Then,
product = p(x). q(x)
= (2x² – 3x – 2)(2x + 3)
= 2x(2x² – 3x – 2) + 3(2x² – 3x – 2)
= 4x³ –6x² – 4x + 6x² –9x – 6
= 4x³ – 13x – 6

PRIME Opt. Maths Book - VI 15

Exercise : 1.3

1. Answer the following questions.
i. What is algebraic term?
ii. What do you mean by algebraic expression?
iii. What is Polynomial?
iv. Write down coefficient, index and variable of the term 3x4.
v. Write down the polynomial 3x – 2 + 2x³ – x² in standard
form.
2. Which of the following are the polynomials? Give reasons.
i. x3 – 3x2 + 2x + 5
ii. x4 + x2 –3x + 2 – x3
3
iii. x2 + 3x – x
iv. 1 + 4x2
2x –
3x 2

v. 3x² + 2 x – 3x + 2
vi. 4x + 3x² – 7 + x³ + 2x4

3. Write down the degree of the following polynomials. Also write
down the types of the polynomials.
i. p(x) = 3x –2
ii. q(x) = 5
iii. f(x) = 3x³ – 4x2 + 2x – 2
iv. g(x) = x2 – 3x + 2
v. h(x) = x4 – 3x3 + 2x2 + 3x – 5
vi. p(x) = 2x³y – 3xy² + 5x²y4

4. Add the following polynomials.
i. f(x) = x2 + 3x – 2, g(x) = x2 – x – 1
ii. p(x) = 3x – 2, q(x) = 4 – 2x + x2
iii. 2 + 3x2 – x and x2 – 5 + 3x
iv. p(x) = x3 – 2x2 + 3x – 2, q(x) = x3 – x2 + 2x – 5
v. f(x) = 2x3 – 3x2 + 3x –1, g(x) = 3 – 5x + 2x2 –x3

16 PRIME Opt. Maths Book - VI

5. Subtract the following polynomials.
i. f(x) = 2x2 + 2x – 3, g(x) = x2 – x – 1
ii. p(x) = 5 –2x + 3x2, q(x) = 3x – 2 – x2
iii. 3 + 2x – x2 and 5x – 3 – 2x2
iv. p(x) = 2x3 –2x + x2 – 1 and q(x) = x3 – 3 – x + 2x2
v. f(x) = 3x2 + 1 – 3x – x3 and g(x) =3x2 – 2x3 – 2x –1

6. Prime more creative questions.
i. Subtract x³ + 3x² – 3x + 1 from the sum of 2x³ – x² + 5x – 3
and 3x² – 6x – 2.
ii. Find the sum of x³ + 2x² – 3x – 2 and x³ – x² + 5x + 7.
iii. Find the product of 3x² – 2x + 1 and x – 2.
iv. Multiply p(x) = 3x² + 3x – 2 by q(x) = 2x2 – 3x + 2.
v. If p(x) = x² – 2x + 3 and q(x) = 3x – 2, find the value of
p(x).q(x).
vi. Multiply the polynomials 2x³ + 2x² – 4x + 5 and 2x² – 3x + 1.

Answer

1. Show to your teacher. ii. x2 + x + 2
2. Show to your teacher. iv. 2x3 – 3x2 + 5x – 7
3. Show to your teacher.
4. i. 2x2 + 2x – 3

iii. 4x2 + 2x – 3
v. x3 – x² – 2x + 2

5. i. x2 + 3x – 2 ii. 4x2 – 5x + 7
iii. x2 – 3x + 6 iv. x3 – x2 – x + 2
v. x3 – x + 2

6. i. x³ – x² + 2x – 6 ii. 2x³ + x² + 2x + 5
iii. 3x³ – 8x² + 5x – 2 iv. 6x4 – 3x3 – 7x2 + 12x – 4
v. 3x³ – 8x² + 13x – 6 vi. 4x5 – 12x³ + 25x² – 19x + 5

PRIME Opt. Maths Book - VI 17

1.4 Surds

Let us discuss some of the roots:
4 = 2² = 2(not surd)
3 = 1.732....................... (surd)
9 = 3² = 3 (not surd)

3 2 = no exact value (surd)
3 8 = 3 2³ = 2 (not surd)

In the above examples.
Some roots has fixed value but some of them has no fixed value. The
roots which has no fixed value are taken as surd.

The root of rational number which has not
the exact rational number is called surd.

Pure surd and mixed surds:

Let us observe the following examples.
2 is pure surd: $ pure square root.
3 is pure surd: $ pure square root.

3 4 is pure surd: $ pure cube root.
2 3 is mixed surd: $ Having coefficient more than one.
53 2 is mixed surd: $ Having coefficient more than one.

Like and unlike surds

Let us observe the examples.
2, 3, 5 , 2 3, 3 7 are like surds $ Having same degree of root.
2, 3 4 , 4 5, 6 12 are unlike surds $ Having different degree of root.

3 5, 3 10, 23 2 are like surds.
23 4 , 5 20 6 7 , 12 100 are unlike surds.

18 PRIME Opt. Maths Book - VI

Worked out Examples

1. Convert the pure surd 12 into mixed surd.
Solution:
12 = 2 × 2 × 3
= 22 × 3
= 2 3 (It is mixed surd)

2. Convert the mixed surd 23 3 into pure surd.
Solution:
23 3 = 3 23 × 3
= 3 8×3
= 3 24 . ( It is pure surd)

3. Evaluate: 22 + 32 + 42 + 7
Solution:
22 + 32 + 42 + 7 = 4 + 9 + 16 + 7
= 36
= 62
=6

Exercise : 1.4

1. Answer the following questions.
i. What is surd?
ii. Write down one example of each of pure surd and mixed
surd.
iii. Write down one example of each of like surd and unlike surd.
iv. Is 2 3 a pure surd ? Why ?
v. Is 12 a mixed surd? Why ?

2. Convert the following pure surds into mixed surds.

i. 20 ii. 98 iii. 3 54

iv 3 40 v. 4 80

PRIME Opt. Maths Book - VI 19

3. Convert the following mixed surds into pure surds:

i. 2 3 ii. 23 5

iii. 43 2 iv. 3 7

v. 33 2

4. Evaluate the followings: ii. 2 + 32 + 52
i. 32 + 7 iv. 3 42 + 52 – 32 – 5
iii. 3 3²–2² + 3
v. 52 + 42 + 32 + 22 – 5

5. More creative questions.
i. Which is greater between 3 3 and 12 ?
ii. Convert into pure surds and write down greater surd
between 2 12 and 5 3 .
iii. Evaluate 3 2³ + 3² + 4²– 6 .
iv. Write down in ascending order of the surds2 3, 5 2 and
3 5.
v. Find the product of 2 3 and 3 2 . Also convert in pure surd.

Answer

1. Show to your teacher.

2. i. 2 5 ii. 7 2 iii. 33 2 iv. 23 5 v. 24 5

3. i. 12 ii. 3 40 iii. 3 128 iv. 63 v. 3 54

4. i. 4 ii. 6 iii. 2 iv. 3 v. 7

5. i. 3 3 ii. 5 3 iii. 3 iv. 5 2 , 3 5 and 2 3
v. 216, 6 6

20 PRIME Opt. Maths Book - VI

Algebra

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Write down antecedent and consequence from the ordered

pair (2, 3)

2. a. If (x + 2, 4) and (4, 6 – y) are equal ordered pairs, �ind the value
of ‘x’ and ‘y’.

b. Evaluate : 3² + 5 ² – 18
c. Write down the degree of the polynomial 2x – 5 + x4 – 3x³ + x².

Also write down the type of polynomial according to degree.

3. a. If A = {1, 2, 3}, B = {4, 5}, �ind the cartesian products A × B and
B × A.

b. Convert 2 3 and 3 2 in pure surds. Also write down which
are is greater.

4. Subtract the polynomial x³ – 2x² + 3x – 2 from the sum of 2x³ –
x² – x + 3 and x³ + 4x² – x + 3.

PRIME Opt. Maths Book - VI 21

Unit Test - 2
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. What do you mean by polynomial?

2. a. Convert 23 3 into pure surd.
b. If A × B = {(1, a), (2, a), (2, b)}, find the sets A and B.
c. If ordered pairs (x, y + 2) and (2, 4) are equal, find the value of
x and y.

3. a. Find the product of (x + 2) and (x² – 3x + 5)
b. Simplify : 12 + 27 – 48

4. If A = {a, b, c}, B = {1, 2, 3}, find A × B and show in arrow diagram.

22 PRIME Opt. Maths Book - VI

2 Matrices

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
1– 2 6 10
No. of Questions – 1 4–

Weight –2

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

Objectives : At the end of the lesson
• Students are able to represents the numbers in matrix form.
• Students are able to know the concept of order of matrix and types

of matrices.
• Students are able to operate the matrix addition, subtraction,

transpose and multiplication by scalar.
• Students are able to know the properties of addition of matrices.

Materials Required:
• Chart paper.
• List of price chart of goods of a market.
• List of properties of matrix addition
• List of types of matrices

PRIME Opt. Maths Book - VI 23

2.1 Introduction

The arrangement of the elements in different rows and columns can
be taken as follows by defining the rows and columns as given in
example. This type of arrangement of elements is called matrix. The
cost per kg of vegetables in Kalimati vegetables market in a particular
day is as follows.

Shop A Beans Tomato Pumpkin
Shop B 15 25 20
Shop C 20 30 15
18 24 12

Above cost of the vegetables can be represented using square bracket
as,

Here, Rows represents the shops and column represents the type of
vegetables. This type of presentation of numbers is called a matrix.

The rectangular array of the numbers in different
rows and columns which are taken into a square
bracket or round bracket is called a matrix.

• A matrix is denoted by capital letter A, B, C, ...
• The members used in a matrix are called elements which are

denoted by small letters a, b, c, d, e, f, ........................
• The horizontal arrangement of the number is a matrix are

called rows.
• The vertical arrangement of the number in a matrix are called

columns.
• The no. of rows (m) and no. of columns (n) can be written as

24 PRIME Opt. Maths Book - VI

m × n which is called the order of the matrix. We read m cross
n or m by n.
• An element of a matrix is represented by the no. of rows and
no. of columns. It lies as lower subscript e.g. the element lies
in 2nd row and 3rd column is denoted by a23.

A=

Here, a11 = element in first row and first column.
a23 = element in second row and third column.

Order of the matrix A = 3 by 3 written as 3 × 3.

Types of matrices

1. Row matrix
The matrix having only one row is called row matrix.
Ex : A = [a11 a12 a13]1 × 3 B = [2 4]1 × 2

2. Column matrix: only
The matrix having one column isBca=lleSSSSSSSSSRTSS1352dWXVWWWWWWWWWW4c×o1 lumn matrix.
Ex : A = TSRSSSSSSSaaa132111WWWWXWVWWW3×1

3. Null matrix:

The matrix having all the elements zero is called null matrix. It is
denoted by ‘O’.

Ex : O = <0 0 0F
0 0 0 2×3

4. Rectangular matrix:
The matrix having unequal number of rows and columns is called
rectangular matrix.

Ex : A =

PRIME Opt. Maths Book - VI 25

5. Square matrix:
The matrix having equal number of rows and columns is called
square matrix.
C = RSSSSSSTSS174 369WWVXWWWWWW3×3
Ex : A = [2]1×1 B = <1 2F 2
3 4 2×2 5
8

6. Equality of the matrices

Any two matrices having same order and same corresponding
elements are called equal matrices.

Ex : A = <2 3F , B= <2 3F
4 1 4 1

Here, A = B

7. General element of a matrix:
The element of the matrix ijsisdnenoootfecdobluymainj ass.
where i is no. of rows and the general element

Ex : If aaaaa2112ij1122=====22222i – j – 1 = 1
Then, × 1 – 2 = 0
× 1 – 1 = 3
× 2 – 2 = 2
× 2

\ 2 × 2 matrix is, A = <a11 a12F = <1 0F
a21 a22 3 2

Worked out Examples

1. Represent the cost of articles given below in a matrix.

Shop A pen pencil
Shop B 15 10
12 8

Solution:

The cost of pen and pencil of shop A and B respectively are as

following where,
Rows " Represents the shops.
Columns " Represents the articles.

26 PRIME Opt. Maths Book - VI

Matrix is,

A = <15 10F
12 8

2. If a matrix A = <2 3 1F , find the order of the matrix. Also write
4 –2 0
down the elements a21 and a13.
Solution:
The given matrix is,

A = <2 3 1F
4 –2 0

No. of rows = 2
No. of columns = 3
` Order of the matrix = 2×3
= element in second row and 1st column = 4
a21 = element in 1st row and 3rd column = 1
a13

3. If <x y 2 F and <2 2F are equal matrices, find the value of x and y.
3 + 2 3 5

Solution:

The equal matrices are: <x y 2 2F = <2 2F
3 + 3 5

By equating the corresponding elements,
x = 2 and y + 2 = 5
x = 2 and y = 5 – 2

` x = 2 and y = 3
` x =2

y =3

4. If aij = 3i + j, find a11, a12, a21 and a22. Also write down 2 × 2 matrix.
Solution:
3aij×=13+i +1 j
Here, 3×1+2 =
3×2+1 =
a11 = 3×2+2 = 4 ` 2 × 2 matrix = <4 5F
a12 = = 5 7 8
a21 = 7
a22 = 8

PRIME Opt. Maths Book - VI 27

Exercise : 2.1

1. Write down the following informations in matrix form. Also write
down the order of the matrices so formed.
i. Cost per kilogram of fruits are given below.

Shop - 1 Mango Guava
Shop - 2 Rs. 6 Rs. 8
Rs. 5 Rs. 9

ii. The numbers of boys and girls of different schools.

School - A Boys Girls
School - B 260 320
School - C 420 300
510 420

iii. The expenditure of three students in different days of school
tiffin.

student - A student - B student - C
Day - 1 Rs. 40 Rs. 60 Rs. 50
Day - 2 Rs. 45 Rs. 40 Rs. 35

2. Write down the types of matrices for the followings. SSSSRTSSSS132WWWWWWWWXV

i. < 1 2 4F ii. 00 iii.
–3 1 2 00

iv. < 2 3F v. 63 2 1@
–1 2
SSSSSSSRST–242 115WWVWXWWWWW
vi. 3
0
–3

3. Write down the order of the matrices given in Q No. 2.

28 PRIME Opt. Maths Book - VI

4. If mWatrriixteAd=owSSSSSTRSSS364n –o122rd––e053rWWWWWVXWWWo,ffimndattrhixe followings.
i. A.
ii. Write down the element in second row and third column.
iii. FFWiinnriddtetthhdeeowesulnemmthoeefnaet 1lae1ijm+weahn2e1tr+ae3a2i3.=2. 3, j = 1.
iv.
v.

5. Find 2 × 2 matrix for the followings. iii. aij = 3i – 2j
i. aij = 2i + j ii. aij = 2i + 3j

6. Find x and y from the following equal matrices.

i. <2 xF and <2 1F
y 4 3 4

ii. A = =3 x + 1G and B = <3 3F
1 y + 2 1 5

iii. P= =2 x - 2G and Q = <2 1F
1 y+3 1 4

iv. < 3 1 5F and =3 y - 2G
x - 2 2 2

v. A = <3 1F and B = =x - 2 y + 4G
0 2 0 2

1. Show to your teacher. Answer

2. Show to your teacher.
3. i. 2 × 3 ii. 2 × 2 iii. 3 × 1 iv. 2 × 2 v. 1 × 3 vi. 3 × 3
4. i. 3 × 3 ii. – 5 iii. 2 iv 9 v. 6

5. i. <3 4F ii. <5 8 F iii. <1 –1F
56 7 10 42

6. i. x = 1, y = 3 ii. x = 2, y = 3 iii. x = 3, y = 1
iv. x = 3, y = 7 v. x = 5, y = – 3

PRIME Opt. Maths Book - VI 29

2.2 Operation on matrices

The simplification of two or more matrices in a single matrix by
using any kind of mathematical operations indicates the operation on
matrices. Which are addition, subtraction, multiplication with scalar,
transpose etc. Here we are discussing some of them in grade VI.

2.2.1. Addition of matrices

Marks obtained by three students Madhav, Pralika and Pranisha in
two monthly test examinations in optional maths are as follows of two
months.

Ashadh 1st 2nd Shrawan 1st 2nd
Madhab 60 Madhab 70 75
Pralika 85 75 Pralika 80
Pranisha 90 Pranisha 95
95 85 80 95

Total marks obtained by them in two tests in two months as,

Madhab 1st 2nd
Pralika
Pranisha 60 + 70 = 130 75 + 75 = 150
85 + 95 = 180 90 + 80 + 170
95 + 80 = 175 85 + 95 = 180

TRTSSSSSSSS689h055is789in550fWWWWWVWWXWo+rmSSSSTSSSSR789a005tio789n505VWXWWWWWWWca=n be expressed in m= aSSSTSSSSRS111tr837i050x f111o875rm000WXWWWWWWWV as,

The sum of any two matrices having same order is called a
new single matrix obtained by adding the corresponding
elements of the matrices. The single matrix so formed has
the order same as the given matrices.

30 PRIME Opt. Maths Book - VI

eg. If A = <–21 53F, B = <–34 –21F

Then,

A+B= <2 3F + <3 2F
–1 5 –4 –1

==2 + 3 3 + 2G
–1–4 5 – 1

= <5 5F
–5 4

2.2. 2. Difference of the matrices.

The difference of any two matrices having same
order is called a new single matrix obtained by
subtracting the corresponding elements of the
matrices. The single matrix so formed has the
order same as the given matrices.

eg. If P = <52 3 12F, Q = <–21 1 53F
–3 1

Then,

P–Q = <2 3 1F – <–1 1 5F
5 –3 2 2 1 3

=

= <3 2 –4F
3 –4 –1

PRIME Opt. Maths Book - VI 31

2.2.3. Transpose of the matrix.

Let us taking an example where cost of apple of a shop of three
days are given.

1st shop 2nd shop

Sunday 150 140
Monday 160 145
Tuesday 170 150
It can be written in such a way by changing the information of

rows and column as,

1st Shop Sunday Monday Tuesday
2nd Shop 150 160 170
140 145 150

ASu=chSSSSSTSRSS111e756x000am111445p050leWWWWWWXVWWs can be written in matrix form as,
After changing rows and columns =

It is called transpose of the matrix A and denoted by AT.

The new matrix obtained by interchanging the
rows and columns of a matrix is called transpose
of the matrix.
• Transpose of A is denoted by AT or A` or A.

Note : Order of the transpose matrix is different than matrix A for the
rectangular matrix but same for the square matrix.

eg. If A = <32 4 95F then.
6

AT = RSSSSSSSTS534 629WWWWWWWWXV

32 PRIME Opt. Maths Book - VI

2.2.4. Multiplication of a matrix with a scalar.

Let us consider the cost of apple given above becomes double in the
next week which can be written as

(old cost) (new cost)

The new matrix formed by multiplying each element
of a given matrix by a given scalar quantity is called
the multiplication of a matrix with the scalar.
i.e. If A = <ac dbF; then, KA = <kkac kkdbF

Eg. If A = <13 ––21F, �ind 3A.

Solution : A = <13 ––12F
3A = 3<13 ––12F = <39 ––63F

Worked out Examples

1. If A = :13 2 D and B = :12 1 D , �ind A + B.
0 3
Solution:
:13 2
A = 0 D

B = :12 1 D
3
then,
:13 2 :12 1
A + B = 0 D + 3 D

= :13 + 2 2 + 1 D
+ 1 0 + 3
3 3
= : 4 3 D

PRIME Opt. Maths Book - VI 33

2. SIfoPlu=tio831n: 4 B and Q= :12 –1 D , �ind 2P – Q.
2 –3

P = 831 4 B
2

Q = :12 –1 D
–3
then,
= 2831 4 :12 –1
2P – Q 2 B – –3 D

= :62 - 2 8 + 1 D
- 1 4 + 3

= 8 41 9 B
7

3. SIfoAlu=tio:2n3: –1 D and B = 8 3 1 B , �ind (A + B)T
1 –2 4

A = :23 –1 D , B = 8 –23 1 B
1 4

then,
(A + B)T = (<32 –11F – <–32 14F2

= : 4 +3 - 1+ 1 D
3 -2 1+ 4

= <17 50FT

= <70 15F

4. If A + B = <32 12F, B = =10 -32G, �ind the matrix A.
Solution:

A + B = <32 12F , B = =1 - 2G
0 3
now,
A = (A + B) – B

= <32 12F – B

34 PRIME Opt. Maths Book - VI

A = <3 2F – =1 - 2G
2 1 0 3

= =3 - 1 2 + 2G
2 - 0 1 - 3

= <2 -42F
2

Exercise : 2.2

1. Add the following matrices.

i. A= <2 1F and B = <–1 3F
1 0 2 1

ii. P = <3 –2F and Q = <1 –4F
–1 4 –2 –1

iii. M= <3 –2F and N = <4 3F
1 1 2 4

iv. A= <1 3 –2F and B = <2 1 2F
1 2 4 3 1 –3

v. P= <3 2F and Q = <–1 3F
1 4 2 1

2. Subtract the pair of matrices given below.

i. P= <3 2F and Q = <–1 3F
1 4 2 1

ii. A= <1 3 –2F and B = <2 1 2F
1 2 4 3 1 –3

iii. M= <3 –2F and N = <4 3F
1 1 2 4

iv. P = <3 –2F and Q = <1 –4F
–1 4 –2 –1

v. A= <2 1F and B = <–1 3F
1 0 2 1

PRIME Opt. Maths Book - VI 35

3. Find the transpose of the following matrices.

i. A = [2 3 4] ii. B = <5F
4
iv. Q = STRSSSSSSS153 624WVXWWWWWWW
iii. P = <2 3 1F
5 2 3

v. M = <3 –2F
–4 1

4. If A = <3 1F , B = <2 –2F , find the following matrices.
2 4 1 –3

i. 2A + B ii. 3A + 2B

iii. AT + BT iv. (A + B)T

v. (2A – B)T

5. Prime more creative questions:

i. If A + B = <1 4F and A = <2 1F , find the matrix B.
3 1 1 0

ii. If B = <3 2F and A –B = <1 15F, find the matrix A.
1 –1 1

iii. Imf aPtr=ix.<–13 2F and Q = <–2 –4F , prove that 2P + Q is null
–2 6 4

iv. Imf aAtr=ix<.–44 2F and B = <–2 1F , prove that A – 2B is a null
–2 2 –1

v. If A + B = <2 5F and A – B = <4 1F , find the matrices A and
B. 3 5 –1 3

36 PRIME Opt. Maths Book - VI

Answer

1. i. <1 4F ii. < 4 –6F iii. <7 1F
3 1 –3 3 35

iv. <3 4 0F v. <2 5F
431 35

2. i. < 4 –1F ii. <–1 2 –4F iii. <–1 –5F
–1 3 –2 1 7 –1 –3

iv. <2 2F v. < 3 –2F
15 –1 –1

3. Show to your teacher.

4. i. <8 0F ii. <13 –1F iii. < 5 3F
55 86 –1 1

iv. < 5 3F v. <4 3 F
–1 1 4 11

5. i. <–1 3F ii. <4 3F
21 24

v. <3 34F , <–1 2F
1 2 1

PRIME Opt. Maths Book - VI 37

Matrices

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. What do you mean by null matrix?

2. a. If A = <2 ––31F, find the transpose of the matrix A.
1

b. If A = <2 1F , find the matrix 3A.
–1 3

c. If A = <2 1F and B = <1 10F, find the matrix A – B.
3 4 2

3. a. If =x + 4 3 G = <1 3F , find the value of x and y.
2 – 1 2 5
y

b. If A = SSSSSSTRSSo–302rde524r o––1f31tVXWWWWWWWWh, efinmdatrix
i.
ii. sum of a32 and a22. A.

4. If A = <2 1F and B = < 1 2 F, find (A + B)T and AT + BT.
3 4 –1 –2

38 PRIME Opt. Maths Book - VI

3 Co-ordinate Geometry

2. Co-ordinate Geometry

2.1 Graph and quadrants
2.2 Distance formula
2.3 Mid-point formula

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
1– 3 7 16
No. of Questions 1 1 4–

Weight 12

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

Objectives : At the end of the lesson
• Students are able to know rectangular Co-ordinate axis.
• Students are able to know run and rise.
• Students are able to use distance formula.
• Students are able to understand mid-point formula.

Materials Required:
• Chart paper
• Number line system.
• Chart of rectangular Co-ordinate axis.
• Concept of sign in quadrants in chart.

PRIME Opt. Maths Book - VI 39

3.1 Rectangular co-ordinate axis:

The horizontal number line is taken as X X` and called x-axis where
the vertical number line is taken as yy` and called y-axis. They are
intersected at ‘O’ at right angle where the intersected point ‘O’ is called
the origin.

Y

4

2nd Quadrent 3 1st Quadrent
2

1
0
X’ –4 –3 –2 –1 1 2 3 4 X

–1

3rd Quadrent –2 4th Quadrent

–3

–4

Y’
Here,
The region $ XOY is called first quadrant.
The region $ X`OY is second quadrant.
The region $ X`OY` is third quadrant.
The region $ XOY` is fourth quadrant.
The point ‘O’ $ Origin point.

Taking a point A(2, 3) in the co-ordinate axis.
In the point A(2, 3).
2 $ Along x - axis towards right side (+ve). Y

3 $ Along y - axis towards up (+ve). 4 A(2, 3)

(2, 3) $ The point on 1st quadrant. 3

2

Here, 1
OM = 2 0
AM = 3 X’ –4 –3 –2 –1 1 2 3 4 X

–1 M

Y’

40 PRIME Opt. Maths Book - VI

Taking a point P(x, y).
x is taken as value of x - component (abscissa) and
y is taken as value of y - component (ordinate).
Then the point (x, y) is called the co-ordinate of the point P(x, y)

which is the pair of x - component and y - component . It is shown in
diagram according to rectangular co-ordinate axis.

Y

P(x, y)

X’ 0 X

M

Y’

Here,
OM = x (abscissa)
PM = y (ordinate)

The pair of abscissa and ordinate of a point
in round bracket is called co-ordinate of the
point which is p(x, y).

Sign used in quadrant for a point (x, y).
1st quadrant $ x (positive), y (positive).(+, +)
2nd quadrant $ x (negative), y (positive). (–, +)
3rd quadrant $ x (negative), y (negative). (–, –)
4th quadrant $ x (positive), y (negative).(+, –)

PRIME Opt. Maths Book - VI 41

Y

(–, +) (+, +)

X’ 0 X

(–, –) (+, –)

Y’

Note :
Co-ordinate of origin point = (0, 0)
Co-ordinate of point on x- axis= (x, 0)
Co-ordinate of point on y-axis = (0, y)
Co-ordinate of any point = (x, y)

Any two points in graph
Let us taking A (1, 2) and B (5, 5).
Y

6 B(5, 6)

5

4

3 C(5, 3)
A(1, 3)

2

1 0
1
X’ –7 –6 –5 –4 –3 –2 –1 2 3 4 5 6 7 X

–1 M N

–2

Here, OM = 1 Y’
AM = 3, ON = 5
BN = 6

42 PRIME Opt. Maths Book - VI

Horizontal displacement of A and B is run.
i.e. Run = AC = MN = ON – OM = 5 –1 = 4.

Vertical displacement of A and B is rise.
i.e. Rise = BC = BN –CN = BN – AM = 6 – 3 = 3

Distance between A and B is AB,
Where,
AB = ^ACh2 + ^BCh2
= ^4h2 + ^3h2
= 16 + 9
= 25
= 5 units.

Distance between any two point
can be expressed as,
d = (run) ² + (rise) ²

Distance between any two points:

Y B (x2, y2)
C
) ‘d’

(x , y 1

1

A

OM NX

Let, ‘d’ be the distance between any two points A(x1, y1) and B(x2, y2).
Where,
d= AB
Run = AC =MN = ON – OM = x2 – x1
Rise = BC =BN – CN = BN –AM = y2 – y1

PRIME Opt. Maths Book - VI 43

Using Pythagorean theorem,
In right angled T ABC,
AB2 = AC2 + BC2
or, d2 = (run)2 + (rise)2
d = run)2 + (rise)2
` d = (x2 –x1)2 + (y2 –y1)2
It is the distance between any two points A & B.

Worked out Examples

1. Plot the points A (1, 2), B(3, 5) C(–3, 4) and D(–5, 0) in graph and
write down the name of the shape formed by joining the points
one after another.
Solution:
The given points are A (1, 2), B (3, 5), C (–3, 4) and D (–5, 0).
Y

6

C(–3, 4) 5 B(3, 5)
4

3

2 A(1, 2)

X’ D(–5, 0) –4 –3 –2 1 1 2 3 4 5 6 7 X
–7 –6 –5 0

–1

–1

–2

Y’
Quadrilateral is formed by joining the points one after another.

2. Find the distance between any two points where run = 3 units
and rise = 4 units.
Solution:
For the straight line,
Run = 3 units
Rise = 4 units.

44 PRIME Opt. Maths Book - VI

Then,
Distance(d) = ^Runh2 + ^Riseh2

= 32 + 42
= 9 + 16
= 25
= 5 units.

3. Find the distance between the points A(1, 2) and B(4, 6).
Solution:
The given points are:
A(1, 2) = (x1 , y1)
B(4, 6) = (x2 , y2)
Then,
Distance between the points A and B is,
d = ^Runh2 + ^Riseh2
= ^x2 –x1h2 + ^y2 –y1h2
= ^4 – 1h2 + ^6 – 2h2
= 9 + 16

= 25 = 5 units.
4. Find the distance between the points A(3, 4) and the origin, B(0,

5) and the origin, Also prove that AO = BO.
Solution:
For the two points A(3, 4) and origin.
A(3, 4) = (x1 , y2)
O(0 , 0) = (x2 , y2)
Distance d(AO) = ^Runh2 + ^Riseh2

= ^x2 – x1h2 + ^y2 – y1h2

= ^0 – 3h2 + ^0 – 4h2

= 9 + 16

= 25
= 5 units.

PRIME Opt. Maths Book - VI 45