Shubharambha OPT Mathematics 10 Final for CTP 2077

(c) Graph of y = ax2 + bx + c

The given quadratic function is y = ax2 + bx + c

or, y =a x2 + ab x + c
a

= a x2 + 2.x. b + b 2 b 2 c
2a 2a 2a a
– +

b 2 b2 c
2a 4a2 a
= a x + – +

= a x+ b 2 b2 + c
2a 4a
–

= a x+ b 2 b2
2a 4a
+c–

y = a x + b 2 4ac – b2
2a 4a
+

When x = – b , y = 4ac – b2
2a 4a

∴ Vertex of the parabola of y = ax2 + bx + c is – b , 4ac – b2 .
2a 4a

Here, the equation of line of symmetry is x = – 2ba.

The equation of tangent is y = 4ac – b2
4a

WORKED OUT EXAMPLES

1. Draw he graph of :
(a) y = x2 + 3x + 2

Solution: Here, y = x2 + 3x + 2

or, y = x2 + 2.x. 3 + 3 2 3 2
2 2 2
– +2

= x + 3 2 9 + 4
2 4
–

= x+ 3 2 9 – 16
2 4
–

+ 3 2 –7
2 4
=x –

y = x + 3 2 7 –4 –3 –2 –1 0 1 2 3 4 5
2 4
+

Infinity Optional Mathematics Book - 10 101

∴ Vertex of parabola = –23, 7 = (–1.5, 1.75)
4

x –1.5 –2 –1 –3 0 –4 1
y 1.75 0 0 2 2 6 6
(x, y) (–1.5, 1.75) (–2, 0) (–1, 0) (–3, 2) (0, 2) (–4, 6) (1, 6)

Y

X' y = x2 + 3x + 2 X
O



(b) y = –x2 + 6x – 5 Y'

Solution : Here, y = – x2 + 6x – 5

= –(x2 – 6x + 5)

= – (x2 – 2.x.3 + 32 – 32 + 5)

= – {(x – 3)2 – 9 + 5} –3 –2 –1 0 1 2 3 4 5 6

y = –(x – 3)2 + 4

When x = 3, then y = 4 then vertex of parabola (3, 4)

x3425160

y 4 3 3 0 0 –5 –5

(x, y) (3, 4) (4, 3) (2, 3) (5, 0) (1, 0) (6, –5) (0, –5)

Y
A

X' O X

B

Y'
102 Infinity Optional Mathematics Book - 10

Here, the equation of line of symmetry AB is x = 3. Y y = x3
Graph of cubic function: X

A function of three degree of the single

variable is called cubic function. For

example, y = x3, y = x3 – 5x, y = (x – 2)3,

y = 5x3 – 3x2 – 2x + 5 etc are the cubic

function of variable x. The general form of

cubic function is written as y = ax3 + bx2 + O

X'

cx + d where a, b, c and d are real constant

and a ≠ 0.

Draw the graph of y = x3

Can you draw the graph of y = – 1 x3 ?
2

Try yourself.

Y'

Solving of quadratic equation by graphical method
The process of finding the value of the variable involved in the equation is known
as solving of the equation. Quadratic equations can be solved by different methods
like factorization method, completing whole square, using formula and by graphical
method. We know that the nature of the graph of quadratic equation is called
parabola. The point on x-axis where the parabola touches or intersect x-axis are
called roots or values of the variable of quadratic equation.

WORKED OUT EXAMPLES

1. Solve: x2 – 4x – 5 = 0 –3 –2 –1 0 1 2 3 4 5
Solution: Here, x2 – 4x – 5 = 0
Let y = x2 – 4x – 5
Now, y = x2 – 2 × x × 2 + 22 – 22 – 5
= (x – 2)2 – 4 – 5
y = (x – 2)2 – 9
When x = 2, then y = – 9
∴ The vertex of parabola = (2, –9)

Infinity Optional Mathematics Book - 10 103

x 2 3 1 4 0 5 –1

y –9 –8 –8 –5 –5 0 0

(x, y) (2, –9) (3, –8) (1, –8) (4, –5) (0, –5) (5, 0) (–1, 0)


Y

B A X
X' (–1, 0) O (5,0)

Y'
Here, the parabola cuts x-axis at A(5, 0) and B( –1, 0). Hence the required

solutions are x = 5 and x = –1.

2. Solve x2 + 2x – 3 = 0 graphically (By parabola and straight line)

Solution: Here, x2 + 2x – 3 = 0
or, x2 = 3 – 2x
Let y = x2 = 3 – 2x
Then y = x2 ......... (i) [It gives parabola]

y = 3 – 2x ........ (ii) [It gives straight line]

From equation (i)
y = x2

x 0 1 –1 2 –2 3 –3

y0114499

(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4) (3, 9) (–3, 9)
From equation (ii)
y = 3 – 2x

x120

y 1 –1 3

(x, y) (1, 1) (2, –1) (0, 3)

104 Infinity Optional Mathematics Book - 10

Y

B(–3, 9)

A(1, 1)
X' X

O

Y'
Here, the straight line y = 3 – 2x cuts the parabola y = x2 at two points

A(1, 1) and B(–3, 9).

Hence, x = – 3 or 1.

3. Find the equation of the given parabola.

Solution : Let the equation of parabola be y = ax2 + bx + c ........ (1)

When the parabola passes through (2, – 9), then we get

–9 = a(2)2 + b(2) + c Y

or, –9 = 4a + 2b + c ...........(2)

When the parabola passes through

(4, –5), then

–5 = a(4)2 + b(4) + c X' C DX
or, –5 = 16a + 4b + c .......... (3) (–1, 0) O

When the parabola passes through

(–1, 0), then

0 = a(–1)2 + b(–1) + c B(4,–5)

or, 0 = a – b + c ........... (4)

Now, subtracting equation (3) from

equation (2) A (2,–9)
– 9 – (–5) = 4a + 2b – 16a – 4b Y'

Infinity Optional Mathematics Book - 10 105

or, –4 = –12a –2b
or, 2 = 6a + b ........... (5)
Again subtracting equation (4) from equation (3) we get
–5 – 0 = 16a + 4b – a + b
or, –5 = 15a + 5b
or, –1 = 3a + b ........... (6)
Again, subtracting equation (6) from equation (5), we get
2 + 1 = 6a + b – 3a – b
or, 3 = 3a
∴ a = 1.
Substituting the value of a in equation (6), we get
–1 = 3 × 1 + b
or, –1 – 3 = b
∴ b = – 4
Again, substituting the value of a and b in equation (3), we get
–5 = 16 × + 4x (– 4) + c
or, –5 = 0 + c
∴ c = –5
Hence, the equation of parabola is y = ax2 + bx + c = x2 – 4x – 5.

Note: When the vertex of parabola is at origin, the equation of parabola is
(i) y = ax2.

(ii) When the vertex of parabola is on y-axis, then the equation of parabola
is y = ax2 + c.

4. Solve the given quadratic equation and linear equation. –3
11
y = x2 + 2 and 4x – y = 1 (–3, 11)
Solution: Here, the given quadratic equation is
y = x2 + 2 (y = ax2 + c, c = 2, vertex of parabola is (0, 2).

x 0 1 – 1 2 –2 3
y 2 3 3 6 6 11
(x, y) (0, 2) (1, 3) (–1, 3) (2, 6) (–2, 6) (3, 11)

The given linear equation is x01 2
4x – y = 1 y –1 3 7
or, y = 4x – 1 (x, y) (0, –1) (1, 3) (2, 7)

106 Infinity Optional Mathematics Book - 10

Y
B(3, 11)

A(1, 3)

X' X
O

Y'



From the graph, a straight line cuts parabola at A(1, 3) and B(3, 11).

∴ (x, y) = (1, 3) or (3, 11)

Exercise 1.15

Section 'A'

1. Draw the graph of the following quadratic functions.
(b) y = 12x2
(a) y = –x2 (c) y = –3x2

2. Draw the graph of the following quadratic functions.

(a) y = x2 + 3 (b) y = –x2 + 5

(c) y = –2x2 – 1 (d) y = 2x2 – 5

3. Draw the graph of the following quadratic functions.

(a) y = x2 + 4x + 4 (b) y = x2 – 6x + 9

(c) y = x2 – 3x + 2 (d) y = –x2 + 2x + 8

(e) y = x2 – 3x (f) y = –(x – 2)2

4. Draw the graph of the following cubic equations.

(a) y = x3 (b) y = –x3 (c) y = x3 – 2

Infinity Optional Mathematics Book - 10 107

Section 'B'

5. Solve the following quadratic equations graphically. Also, find the

equation of line of symmetry.

(a) x2 + 2x – 3 = 0 (b) x2 – 4x – 5 = 0

(c) x2 + 7x + 12 = 0 (d) x2 – 6x + 9 = 0

(e) x2 + 4x + 4 = 0 (f) 2x2 – 5x – 3 = 0

6. Solve graphically.

(a) y = x2 and y = x + 6 (b) y = x2 + 1 and y = 3x + 1

(c) y = 4x2 + 8x + 5 and x + y = 3 (d) y = x2 + 2x – 8 and y = –5

7. Find the equation of the parabola given below.


(a) Y (b) Y

A(2, 4) A(–2, 2) X

X' O
X' O X B(0, –2)

Y'

(c) Y' (d) Y
Y
A(–4, 4) C(0, 4) X'
X
A(3, 1)

O X
C(1, –3)
X'
B(–2, 0) O

B(5, –3)

Y' Y'
108 Infinity Optional Mathematics Book - 10

UNIT

2 CONTINUITY

Review

Discuss the following questions.
(a) Show the sets of natural numbers (N), whole numbers (W), integers (Z) and

rational numbers (Q) in a number line.
(b) Draw a Venn diagram to represent the relation among N, W, Z and Q.
(c) What are real numbers?
(d) Give examples of discrete and continuous data.

Introduction of continuity

Let's consider the examples given below to understand continuity.
1. A rabbit is running on a road.


In this case, the rabbit is jumping from one point to the other. There is a gap
between two consecutive points. So, it does not represent continuity.

2. A tortorize is crawling on the road.


In this case, there is no gap between any consecutive points. That means the
tortorize does not leave any points while crawling on the road. So it represents
continuity.
3. A set of integers

– 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5
The set of integers is presented in a number line (shown above). It is clear that
there is a gap between two consecutive integers. So, the set of integers does not
represent continuity.

Infinity Optional Mathematics Book - 10 109

4. A set of all real numbers.

– 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5
A set of real numbers is presented in a number line (shown above). It shows
continuity because there is a real number for every point of the number line.

Continuity represents a line that you could draw without lifting your pen from the
paper.

Exercise 2.1

Section 'A'
1. Find out which of the following represent continuity.
(a) A rabbit is running. (b) A tortorize is crawing.
(c) A snake is crawing. (d) Sizes of shoes.
(e) Heights of a plant. (f) Age of person.
2. (a) What is the last number of a set of natural numbers?
(b) What is the last number of a set of whole numbers?

Section 'B'
3. Show the following numbers in number line.

(a) Natural numbers from 1 to 8.
(b) Natural numbers from 10 to 15.
(c) Whole numbers from 100 to 150.
(d) Integers from – 3 to + 3.
(e) Integers from – 4 to + 5.
(f) Real numbers from – 5 to + 5.
4. (a) The height of a plant was 10 mm on saturday. The height of the plant

increases 2 mm every day. Show it in a number line and also find the
height of the plant on the next friday.
(b) A shopkeeper deposited Rs. 1000 in a finance company in first day and
deposited Rs. 100 every day. Show it in a number line and the find the
sum collected in 12th day.

110 Infinity Optional Mathematics Book - 10

Investigation of discontinuities

Let discuss the discontinuity in a number line with the help of following examples

(a) – ∞ – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5∞

In the above number line, the real numbers are discontinuous at – 2 and 3. That
means the real numbers are continuous from –∞ to –2 which can be written as
(– ∞, – 2), from – 2 to 3 which an be written as (– 2, 3) and from 3 to + ∞ which can
be written as (3, + ∞)

(b) 0 1 2 3 4 5∞
–∞ –5 –4 –3 –2 –1

In the above number line, the real numbers are discontinuous at 3. That means real
numbers are continuous from – ∞ to – 2 (included) which can be written as (– ∞, – 2],
from – 2 (included) to 3 which can be written as [–2, 3) and from 3 to +∞ which can
be written as (3, + ∞)

(c) 0 1 2 3 4 5∞
–∞ –5 –4 –3 –2 –1

In the above number line, the real numbers are discontinuous at – 2. That means
real numbers are continuous from – ∞ to – 2 which can be written as (– ∞, – 2),
from – 2 to 3 (included) which can be written as (– 2, 3] and from 3 (included) to +
∞ which can be written as [3, + ∞)

(d) 0 1 2 3 4 5∞
–∞ –5 –4 –3 –2 –1

In the above number line, the real numbers are continuous from – ∞ to – ∞. That
means the real numbers are continuous from – ∞ to – 2 (included) which can be
written as (– ∞, – 2], from – 2 (included) to 3 (included) which can be written as
[– 2, 3] and from 3 (included) to + ∞ which can be written as [3, + ∞)

Now, we discuss the discontinuity in the graph. If any one of the three conditions
for a function to be continuous fails, then the function is said to be discontinuous
at that point.

Infinity Optional Mathematics Book - 10 111

(a) Jump discontinuity Y
In this type of discontinuity, the limit
of function f(x) when x tends to 4 from 6
the right is 4 and the limit of the 5
function f(x) when x tends to 4 from 4
the left is 6. So the limit of the function 3
f(x) when x tends to 4 from the left and 2
right are not equal. 1

O 1 23 4 5 6 7 X

Y

(b) Infinite discontinuity 6

In this type of discontinuity, the 5

value of f(x) at x = 4 is not defined.

4

So, the limit of the function f(x) as x

tends to '4' is also not defined. 3

2

1

O 1 23 4 5 6 7X
1 2 34 5 6 X
Y

(c) Point discontinuity 7
6
In this type of discontinuity, the 5
4
limiting value of f(x) as x tends to 3
2
'4' is 4 and the functional value is 1

6. i.e. the limit of f(x) at x tends to

4 is not equal to functional value

f(4) at x = 4.

Symbolically, lim f(x) ≠ f(4).
x→4

O

If there is break or gap or hole in the curve of a function at a certain point then the
curve is said to be discontinuous at that point.

112 Infinity Optional Mathematics Book - 10

WORKED OUT EXAMPLES

1. From the following curves, find out (i) the points where the curve is
discontinuous (ii) the points where the curve is continuous and (iii)
the points where the curve is not defined.

(a) (b)

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

Y' Y'

Solution:
a. (i) There is a hole at x = 3 and jump at x = 4. So the given curve is

discontinuous at x = 3 and x = 4.
(ii) The given curve is continuous in the intervals [0, 3), (3, 4] and (4, 7].
(iii) The given curve is not defined at x = 3.
b. (i) There is a gap is the given curve at x = 2. So the given curve is

discontinuous at x = 2.
(ii) The given curve is continuous in the intervals [– 3, 2) and (2, 6].
(iii) The given curve is not defined at x = 2.

Infinity Optional Mathematics Book - 10 113

Exercise 2.2

Section 'A'
1. (a) Define a continuous curve.
(b) What do you mean a curve fails continuity?

2. Identify continuous or discontinuous from the following curves.


(a) Y (b) Y

XO X

O (d) Y

(c) Y

XO X

O (f)

(e) Y Y

X O X
O
114 Infinity Optional Mathematics Book - 10

Section 'B'

3. From the graphs given below, find out (i) the points where the curve
is discontinuous (ii) the points where the curve is continuous (iii) the
points where the curve is not defined.

(a) Y (b) Y

6 6

55

44

33

22

11

O 1 2345 X X
O 12 3 4 5 6

(c) Y (d) Y
5

5 4
4 3
3
2 2
1 1
O 1 2 34 56X
X –5 –4 –3 –2 –1 O X'
–1 1 2 34 5

–2

–3
–4

–5
Y'

(e) Y (f) Y

44

33

22

11

X'– 4 – 3 – 2 – 1 O 1 2 3 4 5 X X –5 –4 –3 –2 –1 O X'
–1 –1 1 2 34 5

–2 –2

–3 –3
Y' –4

Y'

Infinity Optional Mathematics Book - 10 115

4. The following curves drawn from – 5 to + 5. Write down the points
where the curves are continuous and discontinuous.

(a) Y
4

3

2
1

X' – 5 – 4 – 3 – 2 – 1 O 1 23 4 X
5

–1

–2

–3

Y'
(b) Y

4

3

2
1

X' – 5 – 4 – 3 – 2 – 1 O 1 23 4 X
5

–1

–2

–3

Y'

116 Infinity Optional Mathematics Book - 10

(c) Y

3

2
1

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

–2

–3
–4

Y'
(d) Y

4
3

2

1

X' –4 –3 –2 –1 O X
–5 –1 1 2 34 5

–2

–3
–4

Infinity Optional Mathematics Book - 10 117

(e) Y
4
3

2
1

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

–3

Y'
(f) Y

4 X
3 1 2 34 5
2
1
X' –4 –3 –2 –1 O
–1
–2

–3
–4

Y'

118 Infinity Optional Mathematics Book - 10

Notational Representation of continuity

Consider a function f : R → R defined by f(x) = x x 2 . Can the values of f(2), f(4), f(8),
–
f(12) be determined? The value of f(2) can not be determined. So, the function is not

defined at x = 2. The values of f(8), f(4) and f(12) are 2, 4 and 6 respectively called
3 5
the values of the function f(x).

Left Hand Limit
Consider a function f(x) = x + 2. Find the functional values of f(x) at x = 2.9, 2.99,
2.999, 2.9999 ............. i.e. x approaches to 3 from the left. Also find the functional
value of f(x) at x = 3. Look at the table given below.

x 2.9 2.99 2.999 2.9999 3

f(x) 4.9 4.99 4.999 4.9999 5

What is the value of f(x) at x = 2.9, 2.99, 2.999 and 2.9999 round off to the whole

number? Is it equal to f(3)? From the above example, it is concluded that when x

approaches to 3 from the left, functional values of f(x) approaches to 5. So, 5 called

left hand limit of the function f(x).

x approaches to 3 from the left i.e. x = 2.9, 2.99, 2.999, 2.9999 ............. is denoted

by x → 3 – 0 or x → 3–. The left hand limit of f(x) when x approaches to 3 from the

left is denoted by x lim – f(x) or lim f(x).
→3 x → 3 – 0

When x tends to a from the left i.e. x → a– or x → a – 0 then f(x) also tends to f(a).

The value f(a) is called left hand limit of f(x). It is denoted by x lim f(x) = f(a) or
→ a–
lim f(x) = f(a).
x →a– 0

Right Hand Limit

Consider a function f(x) = x + 2. Find the functional values of f(x) at x = 3.1, 3.01,

3.001, 3.0001, ............. that is x approaches to 3 from the right. Also find the

functional value of f(x) at x = 3. Look at the table given below.

x 3.1 3.01 3.001 3.0001 3

f(x) 5.1 5.01 5.001 5.0001 5

What is the values of f(x) at x = 3.1, 3.01, 3.001 and 3.0001 round off to the whole

number? Is it equal to f(3)? From the above example, it is concluded that when x

approaches to 3 from the right then the functional values of f(x) approaches to 5.

So, 5 is called right hand limit of the function f(x).

x approaches to 3 from the right i.e. x = 3.1, 3.01, 3.001, 3.0001 ............. is denoted

Infinity Optional Mathematics Book - 10 119

by x → 3 + 0 or x → 3+. The right hand limit of f(x) when x approaches to 3 from

the right is denoted by lim f(x) or lim f(x)
x → 3+
x→3+0

When x tends 'a' from the right i.e. x → a+ or x → a + 0 then f(x) also tends to f(a).

The value f(a) is called right hand limit of f(x). It is denoted as lim f(x) = f(a)
x → a+
or, lim f(x) = f(a)

x→a+0

Limit of a function

Consider a function f(x) = x + 2. Then left hand limit of f(x) as x tends to 3 from

the left is x lim f(x) = 5 and the right hand of f(x) as x tends to 3 from the right
→ 3–

is x lim f(x) = 5. Since lim f(x) = lim f(x) = 5, the limit of the function f(x)
→ 3+ x → 3– x → 3+

is defined at x = 3 and its value also equals to 5 which is denoted as lim f(x) = 5
x→3


The limit of a function f(x) exists at x = a if the left hand limit of f(x) at x = a and

right hand limit of f(x) at x = a are equal.

Symbolically, x lim f(x) = x lim f(x) = lim f(x)
→ a+ → a– x→a

Continuity of a function
Consider a function f(x) = x + 2. We have already discussed the left hand

limit of f(x) at x = 3 i.e. lim f(x) = 5 and the right hand limit of f(x) at x = 3
x → 3–
x l→im3+f(x) = 5. Since
i.e. x lim f(x) = lim f(x) = 5, the limit of the function f(x)
→ 3– x → 3+
lim
exists at x = 3 i.e. x → 3 f(x) = 5.

Lets find the functional value of f(x) at x = 3. So, f(3) = 2 + 3 = 5
Here, the limiting value and functional value of f(x) at x = 3 are equal

i.e. lim f(x) = f(3). So the function f(x) is continuous at x = 3.
x→3

A function is said to be continuous at a point if the limiting value and the
functional value of the function are equal at that point.

Alternatively, a function f(x) is said to be continuous at x = a if lim f(x) = f(a)
x→a

120 Infinity Optional Mathematics Book - 10

WORKED OUT EXAMPLES

1. If f:R → R denoted by f(x) = x + 5 then show that the function f(x) is
continuous at x = 1.

Solution: Here, f(x) = x + 5.
Functional value of f(x) at x = 1.
f(1) = 1 + 5 = 6.
Left hand limit of f(x) at x = 1.

x 0.9 0.99 0.999 0.9999

f(x) 5.9 5.99 5.999 5.9999

As the value of x approaches to 1 from the left, the value of f(x) approaches to

6. So, the left hand limit of f(x) at x = 1 is 6. i.e. x lim f(x) = 6
→ 1–

Right hand limit of f(x) at x = 1.
x 1.1 1.01 1.001 1.0001
f(x) 6.1 6.01 6.001 6.0001

As the value of x approaches to 1 from the right, the value of f(x) approaches

to 6. So, the right hand limit of f(x) at x = 1 is 6. i.e. lim f(x) = 6
x → 1+

Since f(1) = lim f(x) = l→im1+f(x) =6, the function f(x) is continuous
x → 1– x

at x=1

2. Test whether a function f(x)= 2x + 1 for 1≤x≤2
x + 4 for x > 2

is continuous or not at x = 2.

Solution: Here,

f(x) = 2x + 1 for 1≤x≤2
x + 4 for x<2

Functional value of f(x) at x = 2.

For x = 2, f(x) = 2x + 1

∴ f(2) = 2 × 2 + 1 = 5

Left hand limit of f(x) at x = 2.

In left hand limit, x approaches to 2 from the left i.e. x < 2. So, f(x) = 2x + 1


Infinity Optional Mathematics Book - 10 121

x 1.9 1.99 1.999 1.9999
f(x) 4.8 4.98 4.998 4.9998

As the value of x approaches to 2 from the left, the value of f(x) approaches to
5. So, the left hand limit of f(x) at x = 2 is 5. i.e. x l→im2–f(x) = 5

Right hand limit of f(x) at x = 2.
In the right hand limit, x approaches to 2 from the right. i.e. x > 2.
So, f(x) = x + 4

x 2.1 2.01 2.001 2.0001
f(x) 6.1 6.01 6.001 6.0001

As the value of x approaches to 2 from the right, the value of f(x) approaches

to 6. So, the right hand limit of f(x) at x = 2 is 6. i.e. x l→im2+f(x) = 6
Since f(2) = x l→im2–f(x) ≠ x l→im2+f(x), the function f(x) is not continuous at x = 2.

Exercise 2.3

Section 'A'
1. (a) What does x → 2– mean?
(b) What does x → 3 + 0 mean?
(c) Write the following notation in a sentence.

(i) x l→ima+ f(x) (ii) lim f(x)
x → a–

(iii) x l→im2+ f(x) (iv) lim f(x)
x→5

(d) Write the following sentences in a notation.

(i) Left hand limit of f(x) at x = a

(ii) Right hand limit of f(x) at x = a

(iii) Left hand limit of f(x) at x = 5

(iv) Right hand limit of f(x) at x = 3

2. (a) Under what condition, the limit of a function f(x) exists at x = a?

(b) Write the condition that the function f(x) is countinuous at x = a.

Section 'B'

3. Find the left hand limit of the following functions at x = 3.

(Taking x = 2.9, 2.99, 2.999, 2.9999)

122 Infinity Optional Mathematics Book - 10

(a) f(x) = x + 1 (b) f(x) = 2x – 1 (c) f(x) = x2 – 9
x–3

4. Find the right hand limit of the following functions at x = 2.

(Taking x = 2.01, 2.001, 2.0001, 2.00001) x2 – 4
x–2
(a) f(x) = x + 3 (b) f(x) = 3x + 1 (c) f(x) =

Section 'C'

5. (a) Given that a function f : R → R defined by f(x) = x + 2. Then

(i) Find the left hand limit of f(x) at x = 1

(ii) Find the right hand limit of f(x) at x = 1

(iii) Find the value of f(1).

(iv) Does the limit of f(x) at x = 1 exist?

(v) Is the function f(x) continuous at x = 1?

(b) If f(x) = 3x + 2 is a real valued function then

(i) Find lim f(x)
x → 2–

(ii) Find lim f(x)
x → 2+

(iii) Find f(2)

(iv) Does lim f(x) exist?
x→2

(v) Is the function f(x) continuous at x = 2?

6. (a) Show that the function f(x) = x + 3 is continuous at x = 4.
x2 – 25
(b) Show that the function g(x) = x–5 is not continuous at x = 5.

7. (a) State whether the function f(x) = 4x – 1 for 0≤ x < 1 is
x + 6 for 1≤ x < 2
continuous or not at x = 1.

(b) Is the function f(x) = x + 3 for 1≤x≤4 is continuous at x = 4?
2x – 1 for 4<x≤6

Infinity Optional Mathematics Book - 10 123

UNIT

3 MATRICES

Review

Discuss the following questions in the group.

1. Define and give the examples of each of the following matrices.

Square matrix, Diagonal matrix, Scalar matrix, Unit matrix, Triangular

matrix, Symmetric matrix, Equal matrix, Sigular matrix, non-singular matrix

and Transpose of a matrix.

2. Construct a matrix A of order 2 × 3 whose elententt aij = 3i + j.
3. Find the value of x and y when

9 x – 5 2 = 4 3
10 7 7 y 3 4

4. If A = 3 7 –3 2 0
1 64 and B = 3 – 1 then find A + BT

5 7
5. Matrix A has X rows and (x + 5) columns. Matrix B has Y row and (11 – y)

columns. If AB and BA both exist, then find the value of x and y.

6. Find the product of (i) 31 57
24 60

(ii) – 2 –5 –5 01
4 6 –2 03
56

(iii) 0 – 1 –2 4 –2
03

1 – 1 A2 15
– 1 1
7. If A = , verify that = 2A

8. If A = 1 2 , verify that M2 – 2M = 5I. Where I is a unit matrix.
3 1

9. If A = 2 1 and B = –3 , find the matrix C when AC = B.
–4 5 2

10. Which matrix post multiplies 1 1 , to get matrix 19 –3 ?
3 4 75 – 13

11. Which matrix pre-multiplies 2 4 , to get a matrix 14 20 ?
6 8

124 Infinity Optional Mathematics Book - 10

3.1 Determinant of a Matrix

Let A = a c be a square matrix of order 2 × 2.
b d

The determinant of the matrix A denoted by det. A or A is a number ad – bc.

∴ The determinant is a pure number associated with a square matrix.
a c
i.e. For matrix A = b d

Determinant of A = A c
a d
= b

= ad – bc

Note: (i) If A = [b], then A = b
(ii) Determinant of only square matrix can be calculated.

WORKED OUT EXAMPLES

1. If A = 5 7 , then find A.
1 2
Solution:

Here, A = 5 7
1 2

Now, A= 5 7
1 2

=5×2–7×1

= 10 – 7 = 3

2. Evaluate: –7 7
–1 6
Solution:

Here, –– 71 67 = –7 × 6 – (–1) × 7 = – 42 + 7
= – 35

3. Find the value of k when k – 3 = 11
4 k 5k 6
Solution:

Here, k4 – k3 = 1 1
5k 6

or, k2 + 12 = 6 – 5k
or, k2 + 5k + 6 = 0
or, k2 + 3k + 2k + 6 = 0

Infinity Optional Mathematics Book - 10 125

or, (k + 3) (k + 2) = 0
Either k + 3 = 0
∴ k = – 3
or, k + 2 = 0
∴ k = – 2
Hence, k = – 3, – 2

Singular and non-singular matrices

A square matrix whose determinant is zero is called singular matrix.
3 6
Let, A = 4 8 then

A = 3 6 = 24 – 24 = 0
4 8

Since A = 0, then matrix A is a singular matrix.

A square matrix whose determinant is not zero is called non-singular matrix.

Let, B = 7 – 1 then
4 3

B = 7 1 = – 21 – 4 = – 25
4 –3

Since B ≠ 0, the matrix B is a non-singular matrix.

WORKED OUT EXAMPLES

1. Find the value of k when the matrix k + 3 5 is a singular.
Solution: 2 k

Here, the matrix k + 3 5 is a singular matrix.
2k

So, its determinant is zero.

i.e. 2k + 3 5 =0
k

or, k2 + 3k – 10 = 0

or, k2 + 5k – 2k – 10 = 0

or, k(k + 5) – 2(k + 5) = 0

or, (k + 5) (k – 2) = 0

Either k + 5 = 0 or, k – 2 = 0

126 Infinity Optional Mathematics Book - 10

∴ k = – 5 ∴ k = 2

Hence, k = – 5, 2

2. Find the value of x when the determinant of the matrix x cos2θ is 1.
– 1 sinθ

Solution:
By question, x cos2θ = 1

– 1 sin θ

or, x sinθ + cos2θ = 1

or, xsinθ = 1 – cos2θ

or, x = sin2 θ
sin θ

∴ x = sinθ

Adjoint of 2 × 2 Matrix

A new matrix obtained by interchanging the diagonal elements and sign changed
of the remaining elements of the square matrix of order 2 ×2 is called adjoint of the
square matrix. The adjoint of square matrix A is denoted by adjoint of A or Adj.A.

Let A = 3 6 , then adjoint of a matrix A = Adjoint of A = –2 –6
7 –2 –7 3

Inverse of a Matrix
Let us consider two non singular matrices

A= 7 5 and B = 3 –5
4 3 –4 7

Now, AB = 7 5 3 –5
4 3 –4 7

= 7 5 3 –5
4 3 –4 7

21 – 20 – 35 + 35
= 12 – 12 – 20 + 21

= 1 0 =I
0 1

Again, BA = 3 –5 7 5
–4 7 4 3

21 – 20 – 15 – 15
= – 28 + 28 – 20 + 21

= 1 0 =I
0 1

Infinity Optional Mathematics Book - 10 127

10
Here, AB = BA = I, where I = 0 1 , the unit matrix. In this case, the matrices A
and B are said to be inverse of each other.

Thus, two square matrices A and B are inverse to each other if their product is
identity (or unit) matrix. i.e. AB = BA = I

The inverse of a matrix A is denoted by A–1
By definition,
A.A–1 = A–1.A = I

Process of finding inverse of a matrix a b
c d
Let us consider a non-singular matrix A =

Then |A|= a b = ad – bc ( ≠ 0 )
c d

Adjoint of A = d –b p q
–c a r s

Let inverse of the matrix A = A–1 =

Then, by the definition of inverse matrix.

AA–1 = I p q 1
ab r s 0 0
or, cd = 1

or, ap + br aq + bs = 1 0
cp + dr cq + ds 0 1

Now, equating the corresponding elements, we have
ap + br = 1 … (i)
cp + dr = 0 … (ii)
aq + bs = 0 … (iii)
cq + ds = 1 … (iv)

Solving (i) and (ii), we get p = ad d bc and r = –c
– ad – bc

Solving (iii) and (iv), we get q = –b and s = ad a bc
ad – bc –

d –b

∴ A-1 = p q = ad – bc ad – bc
r s –c a

ad – bc ad – bc

= ad 1 bc d –b
– –c a

128 Infinity Optional Mathematics Book - 10

A–1 = 1 adjoint of A = Adjoint of A A
|A| Determinant of

Hence, inverse of a non singular matrix is obtained by dividing the adjoint of matrix
A by its determinant.

Note: The inverse of a singular matrix does not exist as the determinant of that
matrix is zero.

WORKED OUT EXAMPLES

1. Find the inverse of the following matrices.

(a) A= 3 6
2 4
Solution:
3 6
Here, A = 2 4

Now,|A| = 3 6 = 12 – 12 = 0
2 4
Since |A| = 0, A–1 does not exist.

(b) B= 7 8
4 1
Solution:
7 8
Here, B = 4 1

∴NoBw–,1|eBx|is=ts.74 8 = 7 – 32 = – 25 ≠ 0
1

Again adjoint of B = 1 –8
–4 7

Now, B–1 = 1 adjoint of B –1 8
|B| 25

= 1 × 1 –8 = 25 –7
– 25 –4 7 4

25 25

2. If the inverse of the matrix x 2x – 9 is the matrix 35 , find the
values of x and y. – y 3 y x

Solution:

Here, x 2x – 9 3 5 = 1 0
–y 3 y x 0 1

Infinity Optional Mathematics Book - 10 129

or, 3 –x + 2xy – 9y 5x + 2x2 – 9x = 1 0
3y + 3y – 5y + 3x 0 1

or, 3x + 2xy – 9y 2x2 – 4x = 1 0
0 3x – 5y 0 1

Now, equating the corresponding elements

3x – 5y = 1…(i) 2x2 – 4x = 0 …(ii)

From equation (ii), 2x2 – 4x = 0

2x2 – 4x = 0

or, 2x (x – 2) = 0

Either 2x = 0

∴ x = 0

or, x – 2 = 0

∴ x = 2

When x = 0 then

3x – 5y = 1 (from equation (i) )

3 × 0 – 5y = 1
∴
y = –1
5

When x = 2 then

3 × 2 – 5y = 1

or, – 5y = 1 – 6

∴ y = 1

Hence, x = 0, 2 and y = –51, 1 0 – 1
– 1 0
3. Show that the inverse of the matrix is the matrix itself.
Solution:
0 –1
Here, the given matrix = –1 0

Now, –0 1 –1 0 –1
0 –1 0

= 0 + 1 – 0 – 0
– 0 – 0 + 1 + 0

= 1 0
0 1

Since the product of the matrix and itself is unit matrix, the inverse of the
matrix is the matrix itself.

130 Infinity Optional Mathematics Book - 10

3.2 Solving Linear Simultaneous Equations in
Two Variables Using Matrix Method

Let us consider the two linear simultaneous equations of two variables as

a1x + b1y = c1 …………. (i)

a2x + b2y = c2 …………. (ii) a1 b1 x = c1
The above two equations can be written in matrix form as a2 b2 y c2

which is in the form of AX = B ………. (iii)
a1 b1 x c1
where, A = a2 b2 y = c2

If |A|= a1 b1 ≠ 0, then A–1 exists.
a2 b2

Now, pre-multiplying equation (iii) on both sides by A–1, we get

A–1 (AX) = A–1 B

or, (A–1 A) X = A–1 B [∴ A–1 A = I]

or, I X = A–1 B [∴ IX = X]

∴ X = A–1 B which gives the solution of the equations.

WORKED OUT EXAMPLES

1. Solve the following equations by matrix method.

(a) 3x + 5y = 21 and 2x + 3y = 13.

Solution:

Here, the given equations are

3x + 5y = 21 ……… (i)

2x + 3y = 13 ……… (ii)

The above equations is written in matrix form as

35 x = 21
23 y 13

Which is in the form of AX = B where

A= 3 5 ,x= x and B = 21
2 3 y 13

Now, |A| = 3 5 = 9 – 10 = – 1 ≠ 0
2 3
∴ A–1 exists.

Adjoint of A = 3 –5
–2 3

Infinity Optional Mathematics Book - 10 131

A–1 = 1 adjoint of A
|A|

= 1 3 –5 = –3 5
–1 –2 3 2 –3

Now, X = A–1B –3 5 21
2 –3 13
or, X =

or, X = – 63 + 65
42 – 39

or, yx = 2
3

Equating the corresponding elements of the equal matrices, we get
x = 2 and y = 3
Hence, x = 2 and y = 3

(b) 3y + 4x = 2xy and 18y – 4x = 5xy

Solution:

Here, the given equations are 3y + 4x = 2xy

3x + 4 = 2 ……….. (i) [Dividing both sides by xy]
y

and 18y – 4x = 5xy

or, x18 – 4 = 5 …… (ii) [Dividing both sides by xy]
y

The above equations is written in matrix form as

1

3 4 x = 2
18 –4 1 5

y

i.e. AX = B, where

1

A = 3 4 ,X= x and B = 2
18 –4 1 5

3 4 y
18 –4
Now, |A|= = – 12 – 72 = – 84 ≠ 0

∴ A–1 exists.

Adjoin of A = –4 –4
– 18 3

A–1 = 1 adjoint of A
|A|

= – 1 –4 –4
84 – 18 3

132 Infinity Optional Mathematics Book - 10

Now, X = A–1B
1 –4 –4 2
or, X = – 84 – 18 3 5

or, X = – 1 – 28
84 – 21

– 28 × –1
84
or, X =
–1
– 21 × 84

11

x3
or, =

11

y4

Equating the corresponding elements of equal matrices, we get
1 1 1 1
x = 3 and y = 4

∴ x = 3 and y = 4

3.3 Cramer's rule

This is the method of solving simultaneous linear equations with the help of

determinant. This method was first used by Swis mathematician Gabriel Cramer

and hence this method received the name as Crame's rule.

Lets consider two linear equations

a1x + b1y = c1 ............ (i)
a2x + b2y = c2 ............ (ii)
Multiplying equation (i) by b2 and equation (ii) by b1 and subtracting, we get

a1b2x + b1b2y = b2c1
–a 2b1x +– b1b 2 y = –b1c2

(a1b2 – a2b1­) x = b2c1 – b1c2­

∴ x = b2c1 – ab21bc21
a1b2 –

Again, multiplying equation (i) by a2 and equation (ii) by a1 and subtracting, we get

a1a2x + a2b1y = a2c1
–a 1a2x +– a1b 2 y = –a1c2
(a2b1 – a1b2­) y = a2c1 – a1c2­

Infinity Optional Mathematics Book - 10 133

∴ y = a2c1 – a1c2 = a1c2 – a2c1
a2b1 – a1b2 a1b2 – a2b1

Using the determinant notations, we have

a1 b1 x= c1 b1 and a1 c1 y= a1 c1
a2 b2 c2 b2 a2 b2 a2 c2

c1 b1 a1 c1
c2 b2 D1 a2 c2 D2
∴ x= a1 b1 = D and y = a1 b1 = D

a2 b2 a2 b2

Where, D1 = c1 b1 , D2 = a1 c1 and D= a1 b1
c2 b2 a2 c2 a2 b2

WORKED OUT EXAMPLES

1. Solve by Cramer's rule : 2x + 3y = 12 and 3x –2y = 5

Solution: Here, the given eqations are 2x + by = 12 ........ (i) and 3x – 2y = 5 ....... (ii)

Now, compairing the above equations with a1x + b1y = c1 and a2x + b2y = c2
We get,

a1= 2, b1= 3, c1= 12, a2= 3, b2= – 2, c2= 5

Here, D = a1 b1 = 2 3 = – 4 – 9 = – 13
a2 b2 3 –2

D1 = c1 b1 12 3
c2 b2 = 5 – 2 = – 24 – 15 = – 39

D2 = a1 c1 = 2 12 = 10 – 36 = – 26
a2 c2 3 5

Now, x = D1 = – 39 = 3
D – 13

y = D2 = – 26 = 2
D – 13

Hence, x = 3 and y = 2.

134 Infinity Optional Mathematics Book - 10

Exercise 3.1

Section 'A'

1. Calculate the determinant of the following matrices.
(a) 53 4 (b) 41 23 (c) 2– 3 67
8

(d) 0– 1– 5 (e) –– 84 36 (f) –– 12 –3
1 6

(g) – ab –b (h) – csoinsθθ sinθ (i) tanA secA
a cosθ secA tanA

2. Find the value of:

(a) 45 12 (b) –– 7 – 3 (c) 03
4m – 1 –5 6
(d) 2aa + 1 1 – 4a x –1 x
1– 2a (e) x2 + 1 x2 + x + 1

3. Prove the followings

(a) xy2 y2 = (x – y) (x2 + xy + y2) (b) xz – +yy xy + z = x2 – z2
x – y

(c) ab bc = c (a + b) (a – b) (d) a – 1 a2 –a+1 = – 2
ac a + 1 a2 +a+1

(e) xy xx + z = (x – y) (x + y +z)
+ z

4. Find the value of x in each of the followings.

(a) 4x 2 = 12 (b) 2x x = 0 (c) x x =5
6 3 4 x 3 2x

(d) 2x 1 = 5x (e) x x– 1 x–2 =0
1 3x x–3

5. (a) Show that the matrix 10 6 is a singular matrix.
5 3

(b) Show that the matrix –1 7 is a non-singular matrix.
2 9

(c) Find the value of x when a matrix x x + 716 is a singular.

– 16 9x

(d) For what value of x, a matrix 7x x is a singular.
3x –1

Infinity Optional Mathematics Book - 10 135

Section 'B'

6. (a) If A = 2 4 , find the determinant of 5A + 3I.
7 8

(b) If A = 3 4 and B = 2 1 , find the determinant of 3A – 4B.
–2 4 –1 3

(c) If M = 1 –1 and N = –3 7 then find |MN|.
–1 1 4 6

(d) If A = 2 0 and B = 4 –6 , find the determinant of 5A – 1 B + 2I
–1 3 –6 –4 2

where I is a unit matrix.

(e) If A = 2 3 and B = 2 4 , find |A2 – 3B – I|.
0 5 6 8

(f) If P = 3 2 and Q = 5 0 , then find the determinant of PQ.
4 6 1 3

(g) If A = 1 2 , then show that |A2 – 2A – 5I| = 0 where I is a unit matrix
3 1

of order 2 × 2.

(h) Construct a matrix A of order 2 × 2 whose elements aij = 3i – 5j. Also
determine |A|.

7. (a) If M = 3k –2 and the value of its determinant is 71, find the value of
4 3

k.

(b) If B = 5 1 and |B| = 14, find the value of x.
–4 x

(c) If A = –4 6 , B = 4 8 and the determinant of A + B – 2I is – 10,
4 8 y 6

Find the value of y.

(d) If A = 1 2 ,B= 2 1 and |AB| = 24, find x.
3 x 3 2

(e) If A = 2 x ,B= 1 3 and |A2 – B2| = –16, find the value of x.
0 2 2 4

8. Show that |PQ| = |P| |Q|, when

(a) P = 3 1 and Q = –1 3 (b) P = –3 –4 and Q = 5 7
5 2 0 2 –1 –2 3 6

(c) P= 5 –3 and Q = 4 2
2 –4 –3 5

136 Infinity Optional Mathematics Book - 10

9. Show that the following two matrixes are inverse to each other.

(a) 53 1 and 2 –1 (b) 52 83 and 8 –3
2 –5 3 –5 2

(c) 5– 7 3 and –4 –3
–4 7 5

10. Find the values of x and y when the following two matries are inverse
to each other.

(a) A= x 1 and 2 –1 (b) A= 1 2 and 5 x
5 2 –5 x y 5 –2 1

(c) P= 2x 7 and 9 y (d) M= 6 4 and 3y –2
5 9 –5 4 x 3 –2 3

(e) A= x 2x – 9 and 3 5
–y 3 y x

11. (a) In what condition, the matrix has no inverse? Show that 2 6 has no
inverse. 3 9

(b) Show that the matrix 3 5 has its inverse.
2 7

(c) For what value of x, the matrix 6 3 has no inverse?
x 4

(d) For what value of y, the matrix y–3 y–1 does not have its inverse?
y 2y – 2

12. Show that the inverse of each of the following matrixes is the matrices.

(a) 0 1 (b) 0 – 1 (c) 1 1 (d) 1 0
10 –1 0 0 –1 01

13. Find the inverse of the following matrices.

(a) 97 4 (b) 43 –– 54 (c) 53 21
5

(d) –– 2 0 (e) –– 32 –– 85
4 3

14. (a) A= 2 3 and B = 3 8 then find (AB)– 1.
3 4 2 5

(b) If A = 5 7 and B = 3 4 then show that A–1B–1 = (BA)– 1. .
3 4 7 9

15. Find the value of x and y by using inverse matrix method.

(a) 21 –3 x = 1 (b) 87 54 xy = 2
1 y 2 1

Infinity Optional Mathematics Book - 10 137

(c) 32 1 x = 5 (d) 21 – 11 yx = 6
–2 y 8 3

(e) 2– 3 – 1 x = 1
1 y 2
Section 'C'

16. Solve the following system of equations by matrix method if possible.

(a) 4x – 2y = 7 and 2x – y = 6 (b) 2x + 3y = 3 and 2x + 3y = 11

(c) 3x – 2y = 5 and x + y = 5 (d) x = 2y – 1 and y = 2x

(e) 2y = 3x – 2 and 5x – 3y = 5

(f) 4 (x – 1) + 5 (y + 2) = 10 and 5 (x – 1) – 3(y + 2) = – 6

(g) 52 x – y = 5 and 4 x – y = – 2 (h) x + y = 1 and x – y = 1
3 6 4 2 4

(i) 2x + 4 = – y = 40 – 3x (j) 2x +3 y = 3x –3 5y =1
5 4

(k) y4 – x = x + y = 1 (l) 3x + 4 = 7 and x + 1 = 3
2 y y

(m) 4x + 2y = 8 and 2 + 3y = 10 (n) 3x –1 = 5x – 3
x 5 8

(0) 8y – 3 = 4y – 3 (p) x8 – 9y = 1 and x10 + y6 = 7
7 3

(q) 5x + 7y = 31xy and 7x + 5y = 29xy

(r) 7y – 2x = 22xy and 5x + 3y = 27xy

(s) 3xy – 10y = 6 and 5xy + 3 = 21y

(t) 2 – 4xy = –11x and 6 + xy = 6x

17. Solve the following by Cramer's rule.

(a) 3x – 2y = 5 and x + y = 5 (b) x = 2y – 1 and y = 2x

(c) 6x + y = 1 and x – y = 1 (d) y – x = x + y = 1
4 2 4 4 2

(e) 4 (x – 1) + 5 (y + 2) = 10 and 5 (x – 1) – 3(y + 2) = – 6

138 Infinity Optional Mathematics Book - 10

UNIT

4 CO-ORDINATE GEOMETRY

Review

Let us discuss on the following questions.

(i) What is the distance between two points A(x1, y1) and B(x2, y2)?
(ii) What do you mean by locus of a moving point?

(iii) When a point P(x, y) divides the line joining the points A(x1, y1) and B(x2, y2)
internally in the ratio of m1:m2 then what are the coordinates of internal point
P(x, y)?

(a) When P(x, y) divides the line joining the points A(x1, y1) and B(x2, y2)
externally in the ratio of m1: m2 then what are the coordinates of P(x, y)?

(b) If P(x, y) bisect the line AB, what are the coordinates of P(x, y)?

(iv) What is the slope or gradient of a line?

(v) Is 2x + 3y – 5 = 0 a first degree equation? Can you write it into three standard

forms? x y
a b
(vi) In a equation + = 1 what does 'a' and b represent?

(vii) What is the equation of straight line joining the two points P(x1, y1) and

Q(x2, y2).

4.1 Angle between two lines

Two intersecting lines intersect at two angles, which are supplementary to each

other. For example, if angle between two intersecting lines is 30°, then the other

angle will be (180° – 30°) = 150°. Where, 30º is called an acute angle and 150° is
Y
called obtuse angle.

Angle between two straight lines when their BD
slopes are given:

Let, AB and CD be two straight lines having

equations y = m1x + c1 and y = m2x + c2 respectively. Pa
AB and CD are intersecting at P. If lines AB and θ

CD intersect X-axis at A and C, then ∠XAB = θ1 X' θ2 θ1 X
and ∠XCD = θ2 then slope of AB = m1 = tanθ1 and OC A
slope of CD = m2 = tanθ2. Again, if ∠APC = θ be
the angle between the lines AB and CD, then Y'

Infinity Optional Mathematics Book - 10 139

θ1 = θ + θ2 [ Exterior angle of triangle is equal to the sum of two interior
opposite angles]

or, θ = θ1 – θ2
Taking, trigonometric ratio tan on both sides, we get

tanθ = tan(θ1 – θ2)

or, tanθ = 1ta+ntθa1n–θ1t.atannθθ22

or, tanθ = m1 – m2 ............. (i) [ tanq1 = m1 and tanq2 = m2 ]
1+ m1m2

Let, ∠APD = α then

θ + α = 180°

or, α = 180° – θ

Taking, tan on both sides, we get

tanα = tan (180° – θ)

tanα = –tanθ

or, tanα = – m1 – m2 ............. (ii)
1 + m1m2

Since, 'θ' be the angle between two lines AB and CD then combining equation
(i) and (ii) we get

tanθ = ± m1 – m2
1 + m1m2

or, θ = tan-1 ± m1 – m2
1+ m1m2

Also, cotθ = 1 = 1
tanθ m1 – m2
± 1 + m1m2

cotθ = ± 1 + m1.m2
m1 – m2

Note: The positive value of tanθ or cotθ gives the acute angle between the two lines.

Where as the negative value gives the obtuse angle.

Condition for parallelism: AB
We know that,

tanθ = ± m1 – m2 C D
1+ m1m2

Let, two lines AB and CD are parallel then θ = 0°

140 Infinity Optional Mathematics Book - 10

Now, tan0° = ± m1 – m2
1 + m1m2

or, 0=± m1 – m2
1 + m1m2

or, 01 = ± m1 – m2
1 + m1m2

or, m1 – m2 = 0
∴ m1 = m2
Hence, two lines will be parallel to each other if m1 = m2 i.e. slopes are equal

Condition for perpendicularity:
When two lines AB and CD are perpendicular, then angle between them θ =

90°

We have, tanθ = ± m1 – m2 C
1 + m1m2 AD B

or, tan90° = ± m1 – m2
1 + m1m2

or, csoins9900°° = ± m1 – m2
1 + m1m2

or, 1 = ± m1 – m2
0 1 + m1m2

or, 1 + m1m2 = 0

or, m1m2 = –1

Hence, the two lines will be perpendicular to each other if m1m2 = – 1 i..e
product of their slopes is –1.

Angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
The given equation of the lines are
a1x + b1y + c1 = 0 .......... (i)
and a2x + b2y + c2 = 0 .......... (ii)

Slope of equation (i) (m1) = – Coefficinet of x = – a1
Coefficient of y b1

Slope of equation (ii) (m2) = – Coefficinet of x = – a2
Coefficient of y b2

If θ be the angle between two lines (i) and (ii) then

Infinity Optional Mathematics Book - 10 141

m1 – m2 – a1 + a2
1 + m1m2 b1 b2
tanθ = ± =±
– a1 – a2
= ± 1+ b1 . b2

– a1 b2 + a2 b1

b1 b2
b1b2 + a1 a2

b1 b2

tanθ = ± a2b1 + a1b2
a1a2 + b1b2

or, θ = tan-1 ± a2b1 – a1b2
a1a2 + b1b2

Condition for parallelism

We have,

tanθ = ± a2b1 – a1b2
a1a2 + b1b2

The two line will be parallel if θ = 0° i.e. tan0° = 0

or, 0 = ± a2b1 – a1b2
a1a2 + b1b2

or, a2b1 – a1b2 = 0

or, a2b1 = a1b2

or, a1 = b1 is the required condition for parallel.
a2 b2

Condition for perpendicularity:
We have,

tanθ = ± a2b1 – a1b2
a1a2 + b1b2

The two lines will the perpendicular to each other if θ = 90°

142 Infinity Optional Mathematics Book - 10

or, tan90° = sin90° = 1
cos90° 0

or, tan90° = ± a2b1 – ba11bb22
a1a2 +

or, 01 = ± a2b1 – ba11bb22
a1a2 +

or, a1a2 + b1b2 = 0 is the required condition for perpendicular.

Equation of a line parallel to the line ax + by + c = 0

The given equation of line is ax + by + c = 0 .......... (i)

Slope of the line (i) (m1) = – Coefficinet of x = – a
Coefficient of y b

Let, the equation of a line parallel to (i) be y = mx + c .......... (ii)

Slope of line (ii) (m2) = m

Since, the line representing equation (i) and (ii) are parallel so m1 = m2

or, – ba = m

From equation (ii), y = – a x + c
b

or, y= –ax + bc
b

or, ax + by – bc = 0

Let, –bc = k (constant) then the equation is ax + by + k = 0

Hence, the equation of the line parallel to ax + by + c = 0 is ax + by + k = 0
Equation of any line perpendicular to the line ax + by + c = 0

The given equation of line is ax + by + c = 0 .......... (i)

Slope of equation (i) (m1) = – a
b

Let, the equation of a line perpendicular to (i) be y = mx + c ........... (ii)

Infinity Optional Mathematics Book - 10 143

Slope of the the (ii) (m2) = m
Since, the lines representing by equation (i) and (ii) are perpendicular so

m1m2 = –1

or, – ab . m = –1

or, m = b
a

Substituting m= b in equation (ii), we get
a

y = b x + c
a

or, y = bx + ac
a

or, bx – ay + ac = 0
Let, ac = k (constant) then, the above equation is bx – ay + k = 0
Hence, the equation of the line perpendicular to ax + by + c = 0 is bx – ay + k = 0

Note: (i) We choose only the constant term in the given equation to get the
equation of any line parallel to the given line.

(ii) The equation of any line perpendicular to the given line, we interchange
the coefficient of x and y, sign of one of the variables x and y and the
constant term.

WORKED OUT EXAMPLES

1. If A (–2, 1), B(2, 3) and C(–2, –4) are three points, find the angle between

the straight lines AB and BC.

Solution: Here, A(–2, 1)

A(–2, 1), B(2, 3) and C(–2, –4) are three points.

To find: Angle between AB and BC. C(–2,–4)

θ

Let the slope of line AB and BC be m1 and m2 B(2, 3)
respectively. Then

Slope of AB (m1) = y2 – y1 = 3–1 = 2 = 1
x2 – x1 2+2 4 2

144 Infinity Optional Mathematics Book - 10

and, slope of BC (m2) = y2 – y1 = –4 – 3 = –7 = 7
x2 – x1 –2 – 2 –4 4

Let, θ be the angle between AB and BC. Then by formula,

tanθ = ± m1 – m2
1 + m1m2

1 – 7 2–7
2 4 4
or, tanθ = ± =±
8+7
1 + 1 . 7 8
2 4

=± – 5 × 8 =± –2 =± 2
4 15 3 3

or, θ = tan-1 ± 2 which is the required angle.
3

2. Find the acute angle and obtuse angle between the lines 7x – 4y = 0

and 3x – 11y + 5 = 0 3x – 11y + 5 = 0

Solution: Here,

The given equation of lines are 7x – 4y = 0 .............. (i)

and 3x – 11y + 5 = 0 ............. (ii)

To find: Acute angle and obtuse angle.

Coefficient of x 7 7 7x – 4y = 0
Coefficient of y –4 4
Slope of line (i) (m1) = – = – =

Slope of line (ii) (m2) = – Coefficient of x = – 3 = 3
Coefficient of y –11 11

Let, θ be the angle between line (i) and (ii) then by formula,

m1 – m2 7 – 3 77 – 12 65
1 + m1m2 4 11 44 65
tanθ = ± =± =± =± = ± (1)
7 3 44 + 21
1 + 4 . 11 44

For acute angle, taking (+ve) sign

tanθ = 1

or, tanθ = tan45°

θ = 45°

For obtuse angle, taking (–ve) sign

tanθ = – 1 = tan(180° – 45°)

or, tanθ = tan135°

Infinity Optional Mathematics Book - 10 145

θ = 135°
Hence, the required acute angle and obtuse angles are 45° and 135° respectively.
3. Show that the lines 3x – 3y + 5 = 0 and 3y – 3x + 7 3 = 0 are parallel.

Solution: Here,
The given equation of lines are

3x – 3y + 15 = 0 ......... (i) and 3y – 3x + 7 3 = 0 .......... (ii)

To show : Given lines are parallel

Slope of line (i) (m1) = – Coefficient of x = – 3 = 1
Coefficient of y –3 3

Slope of line (ii) (m2) = – Coefficient of x = – (– 3) = 1
Coefficient of y 3 3

Since, slope of line (i) (m1) = Slope of line (ii) (m2)
Hence, two lines are parallel.

4. If the straight lines 2x + 3y + 6 = 0 and ax – 5y + 20 = 0 are perpendicular
to each other find the value of a.

Solution : Here, ax – 5y + 20 = 0

The given equation of lines are

2x + 3y + 6 = 0 ............. (i)

and ax – 5y + 20 = 0 ............. (ii)

To find: The value of 'a' Coefficient of x 2
of line (i) (m1) = – Coefficient of y – 3
Now, Slope =

2x + 3y + 6 = 0

Slope of line (ii) (m2) = – a = a
–5 5

Since, the lines representing by equation (i) and (ii) are perpendicular so

m1.m2 = –1

or, – 23. a
5 = – 1

or, a = 15
2

Hence, the value of a is 15 .
2
5. Find the equation of a line parallel to 3x – 2y = 4 and passing through

the midpoint of the line segments joining the points (2, – 4) and (2, 4).

Solution: Here,
The given equation of line is 3x – 2y = 4 ............. (i)

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Let, a line segment AB with the points A(2, –4) and B(2, 4)
To find: The equation of line parallel to the given line.
The midpoint of the line AB joining the points A(2, –4) and B(2, 4) is

2 + 2, 4 – 4 = (2, 0)
2 2

Slope of line (i) (m1) = – Coefficient of x = –3 = 3
Coefficient of y –2 2

Since, the lines are parallel, so slope of EF = Slope of GH
3
Let, slope of EF = m = 2 EG

The equation of line passing through the 3x – 2y = 4
3
point (2, 0) and having slope m = 2 is A(2, –4) B(2, 4)
y– 0y1==23m(x(x––2x)1) FH

or, y–

or, 2y = 3x – 6
or, 3x – 2y = 6 is the required equation line.

6. Find the equation of a straight line passing through the point (2, –4)
and perpendicular to the line 5x + 7y + 12 = 0.

Solution: Here,

The given equation of line is 5x + 7y + 12 = 0 ......... (i) (2, – 4)

Passing point (2, –4)

To find: The equation of straight line

Slope of line (i) (m1) = – Coefficient of x = – 5
Coefficient of y 7

The equation of straight line passing through the 5x + 7y + 12 = 0

point (2, –4) is

y – y1 = m(x – x1)
or, y + 4 = m(x – 2) ........... (ii)

Slope of line (ii) (m2) = m
Since, the lines representing by equation (i) and (ii) are perpendicular so

m1.m2 = – 1

or, –75 .m = –1

or, m = 7
5

Infinity Optional Mathematics Book - 10 147

Substituting m = 7 in equation (ii), we get
5

y + 4 = 75(x – 2)

or, 5y + 20 = 7x – 14

or, 7x – 5y = 34 is the required equation of line.

7. Find the equation of the straight lines which passes through the point

(2, 3) and making angle 45° with the line x – 3y – 2 = 0 A
Solution: Here, (2, 3)

The slope of given line x – 3y – 2 = 0 is 45° 45° C
B x – 3y – 2 = 0
–Coefficient of x 1 1
Slope (m1) = Coefficient of y = – –3 = 3

Let, the slope of required lines AB and AC be (m2) = m, then
1
tan45° = ± m1 – m2 =± 3 – m
1 + m1m2
1 + 1 .m
3
or, 1 = ± 1 – 3m
3+m

or, 3 + m = ± (1 – 3m)

Now, taking (+ve) sign, we get

3 + m = 1 – 3m

or, 4m = 1 – 3

m = – 2 = – 1
4 2
Again, taking (–ve) sign, we get

3 + m = –(1 – 3m)

or, 3 + m = – 1 + 3m

or, –3m + m = –1 – 3

or, –2m = – 4

or, m = 2

148 Infinity Optional Mathematics Book - 10

The equation of straight line passing through the point (2, 3) is

y – y1 = m(x – x1)

y – 3 = m(x – 2) ............ (i)

Case I : When 3m==––1212(xth–e2n) from equation (i) we get
y–

or, 2y – 6 = –x + 2

or, x + 2y = 8

Case II: When m = 2 then from equation (i), we get

y – 3 = 2 (x – 2)

or, y – 3 = 2x – 4

or, 2x – y = 1

Hence, the required equation of straight lines are

x + 2y = 8 and 2x – y = 1

8. Find the equation of the perpendicular bisector of the line joining

the points (7, 2) and (3, –6).

Solution: Here, let, CD is the perpendicular bisector of the line AB and M is the

midpoint of AB.

By midpoint formula, the coordinates of M are C
x1 + x2 y1 + y2 M
x = 2 and y = 2

7 + 3 2 – 6
x = 2 y = 2

A(7, 2) B(3, –6)
x = 5 y = –2

∴ M(5, –2)

Slope of line AB (m1) = y2 – y1 = –6 – 2 = – 8 = 2 D
x2 – x1 3–7 – 4

Let the slope of line CD (m2) = m

Since, CD is perpendicular to AB so m1 × m2 = – 1

or, 2 × m = – 1

emqu=at–ion12 of straight line passing through (5, –2) and having slope – 1 is
The 2

y – y1 = m(x – x1)

Infinity Optional Mathematics Book - 10 149

or, y + 2 = – 1 (x – 5)
2

or, 2y + 4 = –x + 5

or, x + 2y = 5 – 4

or, x + 2y = 1 is the required equation of the perpendicular bisector of given

line.

Exercise 4.1

Section 'A'

1. (a) Write the formula to find the angle between y = m1x + c1 and y = m2x +
c2.

(b) Write the condition of parallel and perpendicular of two straight lines.

(c) Find the slope of line having equation –3x – 8y = 1

(d) Find the midpoint and the slope of line joining the points (2, 3) and (4,
7).

2. (a) Obtain the slope of line which is parallel to the line 3x – 2y + 10 = 0.

(b) Find the slope of line which is perpendicular to the line 7x + 8y – 20 = 0

(c) If 'a' and 'b' be the slopes of two straight lines than find the angle between
two lines.
Section 'B'

3. Find the acute angles between the following pair of straight lines.
(a) x – y – 5 = 0 and x – 7y + 7 = 0
(b) 3x – y + 6 = 0 and y + 3 = 0
(c) 2y + 7 = 0 and y – 3x + 1 = 0

4. Find the obtuse angle between the following pair of straight lines.
(a) 3x – 4y = 10 and x – 2y + 7 = 0
(b) y – (2 + 3)x = 5 and y – (2 – 3) x = 10
(c) 2y + 3y + 4 = 0 and 3x + 2y – 1 = 0

5. Find the angle between the following pair of straight lines.
(a) x – 3y = – 2 3and x + 3y + 3 3 = 0
(b) 3x + 4y + 1 = 0 and 7x + y + 2 = 0
(c) x + 2y – 1 = 0 and 3x – y + 2 = 0

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