Prime Optional Mathematics 10

Let, Slope of diagonal BD be (m2) = m
We have,

Diagonals of a rhombus are bisected to each other at right angle.

So,

m1m2 = –1

or, – 2 x m = –1
3

\ m = 3
2

5. Find the equation of straight line passing through the point (2, –3) which is parallel to the

line joining the points (3, –2) and (–3, 2). B(–3, 2)

Solution: A(3, –2) Q
The given points are A(3, –2) and B(–3, 2 ) (2, –3)

Here, y2 – y1
x2 – x1
Slope of AB(m1) =

= 2+2 P
–3 –3

= 4
–6

= – 2
3

Let, Slope of required line PQ be (m2) = m
By the question,

The lines are parallel,
or, m1 = m2

or, – 2 = m
3

\ m = – 2 .
3

Then,
Equation of straight line PQ passes through a point (2, –3) is,
y – y1 = m(x – x1)

or, y + 3 = – 2 (x – 2)
3

or, 3y + 9 = –2x + 4
` 2x + 3y + 5 = 0 is the required equation.

6. Find the equation of straight lines passes through a

point having x-intercept 3 which makes an angle of 45°

with the line 6x + 5y – 7 = 0 45°
45°
Solution:
6x + 5y – 7 = 0
The given equation of straight line is ,
6x + 5y – 7 = 0 ........................................ (i)
Coefficient of x
Slope of line (i) (m1) = – Coefficient of y

= – 6
5

146 PRIME Opt. Maths Book - X

Let, slope of required line be (m2) = m Steps to find the equation of straight line:
By the question, • SSllooppee ooff rgeivqeunir leidn eli n(me (1)m =2 )f i=n sdupposed
• Finding of point through which
Angle between the two lines is 45°, •
or, q = 45°
or, Tanq = Tan45°
m1 – m2 required line passed.
or, ± 1 + m1 m2 = 1 • Using one-point formula,
y – y1 = m(x – x1)
– 6 –m = 1
5
or, ±
a– 6 km
1 + 5

or, ± –6 – 5m =1
5 – 6m

or, ±(–6 –5m) = 5 –6m

Taking +ve sign Taking –ve sign
– 6 – 5m = 5 – 6m 6 + 5m = 5 – 6m
m = –111
or, m = 11

Now,
Equation of straight line passes through a point (3, 0) [making x- intercept 3] is,
o r , yy –– 0y1 = = m m((xx – – 3 x)1)
xyT a+=k 1–in111g1y (m –x 3=– =3– )1011

Taking m = 11

or, y = 11(x – 3)
or, 11x – y – 33 = 0

\ They are the required equations.

7. Find the equation of straight line passes through a point of intersection of x – 2y = 7 and 3x

+ y = 7 which is perpendicular to the line 4x – 6y + 7 = 0.

Solution: 3x + y = 7
The given equation of straight line is,
4x – 6y + 7 = 0 x
Coefficient of y
Slope of given line (m1) = – Coefficient of

= – 4 x – 2y = 7
–6

= 2
3

Let, the slope of required line be (m2) = m 4x – 6y + 7 = 0
By the question,

The lines are perpendiculars,
or, m1m2 = –1

or, 2 × m = –1
3

\ m = – 3
2

PRIME Opt. Maths Book - X 147

Again, for the intersecting point of other given lines,

x – 2y = 7 " x = 2y + 7 ........................................ (i)

3x + y = 7 " x = 7 – y ........................................ (ii)
3

Solving equation (i) and (ii),

2y + 7 = 7 – y
3

or, 6y + y = – 14
\ y = –2

Substituting the value of ‘y' in equation (i),
x = 2(–2) + 7 = 3
\ Intersecting point is (3, –2).

Then, Equation of straight line passes through a point (3, –2) is,
y – y1 = m(x – x1)

or, y + 2 = – 3 (x – 3)
2

or, 2y + 4 = – 3x + 9
\ 3x + 2y – 5 = 0 is the required equation.

8. Find the equation of perpendicular bisector of the line joining the points A(3, –2) and

B(–3, 6). P
Solution:
The given points are A(3, –2) and B(–3, 6).

Here, y2 – y1
x2 – x1
Slope of given line (m1) =

= 6+2
–3 –3

= 8 A(3, –2) Q(x, y) B(3, 6)
–6

= – 4
3

Let, slope of required line be (m2) = m
By the question,

The lines are perpendicular,
So, m1m2 = –1

or, – 4 x m = – 1
3

\ m = 3 .
4

Again, Using mid - point formula of the straight line AB

Q(x , y) =

= a 3– 3 , –2 + 6 k
2, 2

= (0, 2)

148 PRIME Opt. Maths Book - X

Then, the equation of straight line passes through a point (0, 2).
y – y1 = m(x – x1)

or, y – 2 = 3 (x – 0)
4

or, 3x = 4y – 8
\ 3x – 4y + 8 = 0 is the required equation.

9. Find the equation of sides AB and AD of a square A(2, 4) D
ABCD having opposite vertices A(2, 4) and C(–1, –2). C(–1, –2)

Solution:

The opposite vertices of a square are A(2, 4) and
C(–1, –2).
y2 – y1
Here, Slope of diagonal AC (m1) = x2 – x1

= –2 –4 B
–1 –2

= 2
Let, slope of required lines be (m2) = m
By the question,
Angle between diagonal and sides of a square is 45°,
i.e. q = 45°
or, Tanq = Tan 45°

or, ± m1 – m2 = 1
1 + m1 m2

or, ± 2–m = 1
1+2# m

or, ±(2 – m) = 1 + 2m

Taking +ve sign Taking –ve sign
2 – m =1 + 2m –2 + m = 1 + 2m
or, 1 = 3m –m = 3
1
\ m = 3 m = –3

Now, Equation of straight line passes through a point A(2, 4) is,
o r , yy –– 4y1 = = m m((xx – – 2 x)1)
1
3
Taking m = Taking m = –3

or, y – 4 = 1 (x – 2) y – 4 = –3(x – 2)
3

or, x – 2 = 3y – 12 y – 4 = – 3x + 6
\ x – 3y + 10 = 0 3x + y – 10 = 0
They are the required equations of sides AB & AD.

PRIME Opt. Maths Book - X 149

10. Find the equation of straight line passes C(1, 3)
G
through the centroid of a triangle which is

parallel to the side BC of the triangle having

vertices A(1, 2), B(–2, –3) and C(4, –5). ?
C(4, –5)
Solution:
The vertices of DABC are A(1, 2), B(–2, –3)
and C(4, –5).

Here, = y2 – y1 B
Slope of side BC (m1) x2 – x1 (–2, –3)

= –5 + 3
4+2

= – 2
6

= – 1
3

Let, slope of required line be (m2) = m.
By the question,

The lines are parallel,
So, m1 = m2

or, – 1 = m
3

\ m = – 1 .
3

Again,
Using coordinate of centroid of a triangle,

(x, y) =

=

= (1, – 2)
Now, Equation of straight line passes through the centroid (1, – 2) is,
y – y1 = m(x – x1)

or, y + 2 = 1 (x – 1)
–3

or, 3y + 6 = – x + 1
\ x + 3y + 5 = 0 is the required equation.

Remember :
• Equation of straight line parallel to the line ax + by + c = 0 can

be taken as ax + by + k = 0.
• Equation of straight line perpendicular to the line ax + by + c = 0

can be taken as bx – ay + k = 0.
• Such types of equations also can be taken to find the equation of

straight lines.

150 PRIME Opt. Maths Book - X

Exercise 4.1

1. Answer the following questions.

i) Write down the calculating formula of angle between any two straight lies.

ii) Write down the condition of any two straight lines being perpendicular and parallel.

iii) Write down the equation of straight line perpendicular to ax + by + c = 0.

iv) Write down the equation of straight line parallel to ax + by + c = 0

v) What is the slope of straight line perpendicular to the another straight line which

has slope 2 ?
3

2. Prove that the following lines are parallel to each other.

i) 3x + 4y – 7 = 0 and 6x + 8y – 11 = 0

ii) 4x – 6y + 5 = 0 and 12y – 8x + 7 = 0

iii) 10x – 6y + 11 = 0 and line joining the points (–1, 2) and (2, 3)

iv) (2 – 3)x + ( 3 – 2)y + 8 = 0 and 3x – 3y + 7 = 0

v) y = mx + c and x = 1 y + 2
m

3. Prove that the following lines are perpendicular to each other.
i) 6x – 8y + 13 = 0 and 4x + 3y – 5 = 0
ii) 3x + y – 2 3 = 0 and x – 3y – 3 3
iii) Line joining the points A(2, –3) and B(4, 0) to the line joining the points P(–1, 1),
and Q(5, –3).
iv) y = (1 + 3)x + 3 and x = (–1 – 3)y +7
v) px + qy – r = 0 and qx = py – m

4. Find the acute angle between the following lines :
i) 3x + y – 7 = 0 and x + 2y – 5 = 0
ii) 6x + 5y + 1 = 0and x + 11y – 9 = 0
iii) x – 3y = 7 and y = 3x – 4
iv) (2 – 3)x – y –1 = 0 and (2 + 3)x = y + 3
v) 5y – 6x + 1 = 0 and 5x = 3 – 6y

5. Find the obtuse angle between the following lines :
i) 11x – y – 7 = 0 and 6x + 5y – 3 = 0
ii) 2x + 3y = 5 and 5y = x – 11
iii) y = 1 – 3x and 3x = y + 3 3
iv) 3x – y – 9 = 0 and 3y = x – 11
v) (2 – 3)x = y – 3 3 and y = (2 + 3)x – 5 .

6. Find the value of the constant letter given below.

i ) If the straight lines having equations 3x + 4y – 7 = 0 and 6x + my – 5 = 0 are parallel
to each other, find the value of ‘m'

ii) If the straight lines having equations px – 4y – 11 = 0 and 4x + 12y – 13 = 0 are
perpendicular to each other, find the value of ‘p'

PRIME Opt. Maths Book - X 151

iii) If the straight line having equation 6x – 8y – 17 = 0 is parallel to the line joining the
points (2k + 1, 2) and (–1, 8), find the value of ‘k'

iv) If line joining the points (1, – 3) and (3p, 3) and the straight line having equation 4x
+ 3y – 2 = 0 are orthogonal find the value of ‘p'

v) If line joining the points (3m – 2, 7) and (1, 3) is perpendicular to the line joining the
points (6, 1) and (2, 7) find the value of ‘m'

7. Using slope of straight line.
i) Prove that the vertices A(1, 3), B(–5, 3), C(–5, – 2) and D(1, – 2) are the vertices of a
parallelogram.
ii) Prove that the triangle having vertices P(2, 5), Q(–3, 1) and R(1, – 4) is right angle at
Q.
iii) Prove that the parallelogram is the rhombus having vertices A(2, 4), B(5, – 3), C(–2,
– 6) and D(–5, 1).
iv) Prove that the quadrilateral having vertices P(–3, 3), Q(2, 4), R(3, –1) and S(–2, – 2)
is a square.
v) Prove that vertices A(2, – 2) and B(8, 4), C(5, 7) and (–1, 1) are of the vertices of
rectangle.

8. i) Find the value of ‘m' from the given right angled triangle PQR.
P(1, 4)

(3, –1)Q R(2m, 1)

ii) Find the value of ‘k' from the given square ABCD.

A(3, 3) D(K, 9)

B C(4, 11)
iii) If angle between any two straight lines Px + y – 5 = 0 and x + 2y – 4 = 0 is 45°, find

the value of ‘p'.
iv) If angle between the straight lines x – 3y – 7 = 0 and mx – y – 9 = 0 is 30°, find the

value of ‘m'.
v) Find the slope of a straight line which is parallel to the line joining the points (3, – 2)

and (–5, 4).

9. i) Find the slope of a straight line which is perpendicular to the straight line of equation
4x – 6y + 7 = 0.

ii) Find the slope of a diagonal BD of a rhombus where two opposite vertices are
A(1, – 3) and C(7, 5).

152 PRIME Opt. Maths Book - X

iii) Find the slope of altitude drawn from first vertex of a triangle having vertices
A(3, 2), B(–1, – 4) and C(7, –2).

iv) Find the slope of side AB of a rectangle ABCD where two of the vertices are B(3, – 2)
and C(–1, 4).

v) Find the slope of side AB from the adjoining diagram.

A (3, –1)

B C (1, 3)

10. Find the equation of straight lines under the following conditions.
i) Find the equation of straight lines passes through a point (1, – 3) which is parallel
to the line of equation 4x + 6y – 7 = 0.
ii) Find the equation of straight line passes through a point (–3, 1) which is
perpendicular to the line of equation 6x – 8y + 5 = 0
iii) Find the equation of straight line having x-intercept 2 which is parallel to the line
joining the points (1, 5) and (5, – 1)
iv) Find the equation of straight line having y-intercept –3 which is perpendicular to
the line of equation 8x + 12y + 11 = 0.
v) Find the equation of straight line passes through the mid-point of line joining the
points (3, – 2) and (1, – 4) which is parallel to the line 2x – 4y + 5 = 0.

11. i) Find the equation of straight line passes through a point (2, – 3) which makes an
ii)
iii) angle 45° with the line of equation 6x + 5y – 7 = 0.
iv)
v) Find the equation of straight line passes through a point (3, 1) which makes an

angle 135° with the line x + 2y – 5 = 0.

Find the equation of straight line passes through a point (2, 0) which makes 30°

angle with the line x + 3y + 7 = 0.

Find the equation of straight line passes through a point (0, – 3) Which makes

an angle tan–1 a 1 k with the line 3x – 2y + 13 = 0.
3
Find the equation of sides AB and AD of a square where two opposite vertices

are A(1, 2) and C(3, 1).

12. i) Find the equation of perpendicular bisector of the line joining the points A(1, 4) and
ii) B(5, – 2).
iii. Find the equation of straight line passes through the mid-point of line joining the
iv) points P(3, – 2) and Q(1, 2) which is perpendicular to PQ.
v) Find the equation of diagonal BD of a rhombus ABCD where two opposite vertices
are A(3, 1) and C(–1, 7).
Find the equation of a diagonal of a square where the ends of another diagonal are
(–4, 4) and (6, – 8).
Find the equation of altitude of a triangle drawn from first vertex where the vertices
of the triangle are A(2, – 1), B(– 2, 6) and C(–4, – 3).

PRIME Opt. Maths Book - X 153

13. PRIME more creative questions:

a. i) Find the equation of straight line passes through the intersecting point of the lines
having equations x – 2y = 5and 3x + y = 1 which is parallel to the line 4x – 3y + 7 = 0.

ii) Find the equation of straight line passes through the centroid of a triangle having
vertices A(3, –3), B(–1, 5), C(–5, 7), which is perpendicular to the side BC of DABC.

iii) Find the equation of straight line passes through a point which cuts the line
joining the points (–3, –1) and (3, 5) in the ratio 1 : 2 which is parallel to the line
6x – 4y + 5 = 0.

iv) Find the equation of sides PQ and QR of a square PQRS having a vertex Q(3, 2) and
equation of diagonal QS is 3x –2y + 11 = 0.

v) Find the equation of side BC of a rectangle ABCD where two of the vertices are
A(4, –2) and B(–2, 6).

b. i) Find the equation of sides of an equilateral triangle having vertex (–2, 2) where
equation of its base is X = 0.

ii) Find the equation of sides of right angled isosceles triangle having vertex (–3, –2)
and equation of its base is y = 0.

iii) If the lines (m2 – n2)x – (a + b)y = 0 and (a2 – b2) x + (m + n)y = 0 are orthogonal,

prove that (m – n) (a – b) = 1.

iv) aIfnads3trxai–ghyt=li1nwe haxic+h y = 1 passes through the intersecting point of the lines 2x + y = 4
bis parallel to the line of equation 4x + 2y – 7 = 0, find the value
of ‘a' and ‘b'.

v) If a straight line y = mx + c passes through the intersecting point of 3x – y = 8 and
x + 2y = 5 which is perpendicular to the line 3x + 6y – 11 = 0. Find the value of ‘m'
and ‘c'.

c. i) Find the equation of straight line which makes sum of the intercepts on the axes 6
and parallel to the straight line of equation 4x + 2y – 7 = 0

ii) Find the equation of straight line which makes an area between the axes 6 square
units and perpendicular to the line 3x – 4y + 13 = 0.

iii) Find the foot of the perpendicular drawn from a point (5, –3) to the straight line of
equation 3x – 4y – 2 = 0.

iv) Find the co - ordinate of the foot of the perpendicular drawn from a point (9, –4) to
the straight line of equation 8x – 6y + 4 = 0.

v) If lines mx + ny +c = 0 and mx – ny + p = 0 are orthogonal to each other, prove that
(m + n)(m – n) = 0.

154 PRIME Opt. Maths Book - X

Answer

1. Show to your teacher.

4. i) 45° ii) 45° iii) 30° iv) 60° v) 90°

5. i) 135° ii) 135° iii) 120° iv) 150° v) 120°

6. i) 8 ii) 12 iii) –5 iv) 3 v) 3

8. i) 4 ii) 0, 7 iii) 3, – 1
iv) 3, 0 3

v) – 3
4

9. i) – 3 ii) – 3 iii) –4
2 4

iv) 2 v) ! 1
3 2

10. i) 2x + 3y + 7 = 0 ii) 4x + 3y + 9 = 0 iii) 3x + 2y – 6 = 0
iv) 3x – 2y – 6 = 0 v) x – 2y – 8 = 0

11. i) 11x – y – 25 = 0 or x + 11y + 31 = 0
ii) x – 3y = 0 or 3x + y – 10 = 0
iii)
iv) 3x + y – 2 3 = 0 or y = 0
v) 11x – 3y – 9 = 0 or 7x – 9y – 27 = 0
3x – y – 1 = 0 or x + 3y – 7 = 0

12. i) 2x – 3y – 3 = 0 ii) x – 2y – 2 = 0 iii) 2x – 3y + 10 = 0
iv) 5x – 6y – 17 = 0 v) 2x + 5y + 5 = 0

13.a.i) 4x – 3y – 10 = 0 ii) 2x – y + 5 = 0 iii) 3x – 2y + 5 = 0
v) 3x – 4y + 30 = 0
iv) x – 5y + 7 = 0 and 5x + y + 30 = 0

b. i) x + 3y + 2 3 = 2 and x + 3y + 2 = 2 3 iv) 2 and 4
ii) x – y + 1 = 0 and x + y + 5 = 0
v) 2 and –5

c. i) 2x + y – 4 = 0 ii) 4x + 3y – 12 = 0 or 4x + 3y + 12 = 0
iii) (2, 1) iv) (1, 2)

PRIME Opt. Maths Book - X 155

4.2 Homogeneous equation

The polynomial equation having degree of the variables of each of the terms of the expression
are same is called homogeneous equation.

Examples :
i) x2 – 3xy + 4y2 = 0

Here, degree of the each terms is 2.
ii) x3 – 2x2y + 3xy2 + y3 = 0

Here, degree of the each terms is 3.
iii) ax2 + 2hxy + by2 = 0 is the standard form of the second degree homogeneous

equation passes through the origin.
Where,
a, h and b are the constants.

Corollary 1: Every second degree homogeneous equation represents a pair of straight lines
passes through the origin.
Solution:
Let, the second degree homogeneous equation be,

ax2 + 2hxy + by2 = 0 ............................. (i)
or, by2 + 2hxy + ax2 = 0
Taking b ≠ 0

Dividing both sides by bx2

or, by2 2hxy + ax2 = 0
+ bx2 bx2

bx2 bx2

or, ` y 2 + 2h ` y j + a =0
x b x b
j

'y'
It is the quadratic equation in x .

Hence, it has two roots say m1 and m2.
y
i.e. x = m1 ⇒ y = m1 x

y ⇒ y = m2x
x = m2

Which are the equation of straight lines in slope - intercept form passes through the

origin.

Again, taking b = 0

The equation (i) becomes

ax2 + 2hxy = 0

or, x(ax + 2hy) = 0

Either Or

x=0 ax + 2hy = 0

They also represent the equation of straight lines pass through the origin.

Hence, in both cases the second degree homogenous equation represents a pair of

straight lines passes through the origin.

156 PRIME Opt. Maths Book - X

Corollary 2 : Angle between the line pairs of homogeneous Y
y = m2x
equation.
q y = m1x
Solution: OX

Let, the second degree homogeneous equation be, X' Y'

ax2 + 2hxy + by2 = 0

or, by2 + 2hxy + ax2 = 0

Dividing both sides by bx2.

or, ` y 2 + 2h ` y j + a = 0 .................................... (i)
x b x b
j

'y'
It is the quadratic equation in x .

Hence, it has two roots say m1 and m2.

y ⇒ y – m1x = 0 and
i.e. x = m1

y ⇒ y – m2x = 0.
x = m2

Again, The single equation of the above straight lines is,

(y – m1x) (y – m2x) = 0
or, y2 – (m1 + m2)xy + m1m2x2 = 0
or, Dividing both sides by x2,
y y
or, ` x 2 – ^m1 + m2h ` x j + m1m2 = 0 ................................. (ii)

j

Equating the quadratic equations (i) and (ii), we get,

m1 + m2 = – 2h
b

m1m2 = a
b

Here,

m1 – m2 = ± [ a (a – b)2 = (a + b)2 – 4ab]

=± a– 2h 2 – 4 a
b b
k

4h2 – 4ab
= ± b2

Again, h2 – ab
=± b

For the angle between the any two straight lines 'q'
m1 – m 2
Tanq = ± 1 + m1 m2

2 h2 – ab

=± b

1+ a
b

2 h2 – ab
=± a+b

\q = Tan –1 ;2 h2 – ab E It is the angle between the lines represented
a+ b by second degree homogeneous equation.

PRIME Opt. Maths Book - X 157

i. Condition of the pair of straight lines being perpendicular (orthogonal). Y

i.e. q = 90° C Y' B
or, Tanq = Tan90° Y X

or, 2 h2 – ab = Sin90° X' D
a+b Cos90°
B
or, 2 h2 – ab = 1 A D
a+b 0
X
or, a + b = 0
(i.e. Coeff. of x2 + Coeff. of y2 = 0)

ii. Condition of the pair of straight lines being coincident.

i.e. q = 0° X'

or, Tanq = Tan0° A
C
or, 2 h2 – ab = 0 Y'
a+b 1

or, 2 h2 – ab = 0

or, h2 – ab = 0
Squaring on both sides,
or, h2 – ab = 0

` h2 = ab

Worked out Examples

1. Find the separate equations of the following homogenous equations
i. x2 – 3xy – 18y2 = 0
Solution:
The homogeneous equation is,
x2 – 3xy – 18y2 = 0
or, x2 – (6 – 3)xy – 18y2 = 0
or, x2 – 6xy + 3xy – 18y2 = 0
or, x(x – 6y) + 3y(x – 6y) = 0
or, (x – 6y) (x + 3y) = 0

Either, or,

x – 6y = 0 x + 3y = 0

They are the required separate equations.

ii. ab(x2 – y2) – xy (a2 – b2) = 0
Solution:
The homogeneous equation is,

ab(x2 – y2) – xy(a2 – b2) = 0
or, abx2 – aby2 – a2xy + b2xy = 0
or, abx2 – a2xy + b2xy – aby2 = 0
or, ax(bx – ay) + by(bx – ay) = 0
or, (bx – ay) (ax + by) = 0

158 PRIME Opt. Maths Book - X

Either, or,

ax + by = 0 bx – ay = 0

They are the required separate equations.

iii. x2 – 2xySeca + y2 = 0
Solution:
The homogenous equation is,

x2 – 2xySeca + y2 = 0
or, x2 – 2xySeca + y2 × 1 = 0
or, x2 – 2xySeca + y2 (Sec2a – Tan2a ) = 0
or, x2 – 2xySeca + y2 Sec2a – y2Tan2a = 0
or, (x – ySeca)2 – (yTana)2 = 0
or. (x – ySeca + yTana) (x – ySeca– yTana) = 0

Either, or,
x – y (Seca – Tana) = 0 x – y (Seca + Tana) = 0
They are the required separate equations.

iv. x2 – 4xy + 4y2 – 2x + 4y – 15 = 0
Solution:
The second degree equation is,

x2 – 4xy + 4y2 – 2x + 4y – 15 = 0
or, (x – 2y)2 – 2(x – 2y) – 15 = 0
Let, x – 2y = a
or, a2 – 2a – 15 = 0
or, a2 – (5 – 3) a – 15 = 0
or, a2 – 5a + 3a – 15 = 0
or, a(a – 5) + 3 (a – 5) = 0
or, (a – 5) (a + 3) = 0

Replacing the value of ‘a' or,
or, (x – 2y – 5) (x – 2y + 3) = 0 x – 2y + 3 = 0
Either,
x – 2y – 5 = 0
They are the required separate equations.

v. 3x2 – 5xy – 2y2 + x + 5y – 2 = 0
Solution:
The second degree equation is,
3x2 – 5xy – 2y2 + x – 3y – 2 = 0
or, 3x2 – x(5y – 1) – (2y2 – 5y + 2) = 0

Comparing it with ax2 + bx + c = 0,
a=3
b = – (5y – 1)
c = – (2y2 – 5y + 2).

PRIME Opt. Maths Book - X 159

Then, – b ! b2 – 4ac
x= 2a

=

=

=

Taking +ve sign,

5y – 1 + 7y – 5 Note : This way of factorisation
x= 6
or, x = 2y – 1 (using quadratic equation) can
or, x – 2y + 1 = 0
be used for any kind of second
degree equations to find the

Taking –ve sign, separate equations.

5y – 1 – 7y + 5
x= 6

–y + 2
or, x = 3
or, 3x + y – 2 = 0

` x – 2y + 1 = 0 and 3x + y – 2 = 0 are the required separate equations.

2. Find the single equation of the following pair of straight lines.
i. x – 3y = 0 and 2x – y – 1 = 0.
Solution:
The pair of straight lines are,
x – 3y = 0 .................................................. (i)
2x – y – 1 = 0 .......................................... (ii)
The single equation of them is,
(x – 3y) (2x – y – 1) = 0
or, 2x2 – xy – x – 6xy + 3y2 + 3y = 0
or, 2x2 – 7xy + 3y2 – x + 3y = 0
This is the required single equation.

ii. x – y (Coseca + Cota) = 0 and x – y (Coseca – Cota) = 0
Solution:
The pair of straight lines are,
x – y (Coseca + Cota) = 0 ........................... (i)
x – y (Coseca – Cota) = 0 ............................. (ii)
The single equation of them is,

{ x – y (Coseca + Cota)} {x – y(Coseca – Cota)} = 0
or, x2 – xy(Coseca – Cota) – xy (Coseca + Cota) + y2 (Coseca + Cota) (Coseca – Cota) = 0
or, x2 – 2xy Coseca – Cota + Coseca + Cota ) + y2(Cosec2a – cot2a) = 0
or, x2 – 2xyCoseca + y2 = 0
This is the required single equation.

160 PRIME Opt. Maths Book - X

3. Find the angle between the pair of lines of the following homogeneous equation.
i. 3x2 – 4xy + 3y2 = 0
Solution,
The homogeneous equation is,
3x2 – 4xy + 3y2 = 0
Comparing it with ax2 + 2hxy + by2 = 0,
a= 3
b= 3
2h = – 4 $ h = – 2

Then, for the angle between the line pairs ‘q‘

Tanq = ! h2 – ab
a+b

=

or, Tanq = ! 2 4 – 3
23

or, Tanq = ! 1
3

or, q = Tan–l c! 1 m
3

or, q = 30°, (180° – 30°)

` q = 30° , 150°

ii. x2 – 2xy Tan2a – y2 = 0
Solution:
The given homogeneous equation is,
x2 – 2xy Tan2q – y2 = 0
Comparing it with ax2 + 2hxy + by2 = 0
a=1
b=–1
2h = – 2Tan2a → h = – Tan2a.

Then, For angle between the line pairs ‘q‘

Tanq = !2 h2 – ab
a+b

=

= ! 2Sec2a
0

or, Tanq = Tan90°

` q = 90°

PRIME Opt. Maths Book - X 161

4. Find the value of ‘P' under the following conditions.
i. Line pairs of Px2 + 6xy – 4y2 = 0. are perpendicular to each other
Solution:
The given homogeneous equation is,
Px2 + 6xy – 4y2 = 0
Comparing it with ax2 + 2hxy + by2 = 0,
a=p
b=–4
2h = 6

` h=3

For the line pairs perpendicular,

a +b = 0

or, p + ( –4) = 0

` p=4

ii) Line pairs of 2x2 – 8xy + (p + 1)y2 = 0 are coincident to each other
Solution:
The homogenous equation is,
2x2 – 8xy + (p + 1)y2 = 0
Comparing it with ax2 + 2hxy + by2 = 0
a=2
2h = –8 → h = –4
b = (p + 1)
For the line pairs coincident,

h2 = ab
or, (–4)2 = 2(p + 1)
or, 16 = 2p + 2
or, 14 = 2p

` p=7

iii) Angle between the line pairs of 2x2 – pxy + 3y2 = 0 is 45°.

Solution:

The given homogeneous equation is,

2x2 – pxy + 3y2 = 0

Comparing it with ax2 + 2hxy + by2 = 0,

a=2

b=3 p
2h = –p → h=– 2

For the angle between the line pairs q = 45°
q = 45°

or, Tan q = Tan 45°
2 h2 – ab

or, ± a + b = 1

162 PRIME Opt. Maths Book - X

2 ` p 2 – 6
or, ± 2
j

5 =1

or, ± c 2 p2 – 4 m = 5
2

Squaring on both sides,

or, (p2 – 24) = 25

or, p2 = 49

` p= !7

5. Find the equation of pair of lines passes through the origin which are perpendicular to the

line pairs of 2x2 – xy – 6y2 = 0.

Solution:

The given homogeneous equation is,

2x2 – xy – 6y2 = 0

or, 2x2 – (4 – 3)xy – 6y2 = 0

or, 2x2 – 4xy + 3xy – 6y2 = 0

or, 2x(x – 2y) + 3y(x – 2y) = 0

or, (x – 2y) (2x + 3y) = 0

Either, x – 2y = 0 Or, 2x + 3y = 0

We have,

Equation of straight line passes through the origin and perpendicular to ax + by = 0 is

bx – ay + k = 0.

or, b × 0 – a × 0 + k = 0

\ k=0

So, equation becomes bx – ay = 0

Then, the above equation changes to,

2x + y = 0 and 3x – 2y = 0

The single equation of them is,

(2x + y) (3x – 2y) = 0

or, 6x2 –4xy – 2y2 = 0

or, 6x2 – xy – 2y2 = 0 is the required equation.

6. Find the equations of straight lines pass through a point (2, 1) and perpendicular to the

pair of lines 3x2 + 4xy – 4y2 = 0.

Solution:

The given homogeneous equation is,

3x2 + 4xy – 4y2 = 0

or, 3x2 + (6 – 2)xy – 4y2 = 0

or, 3x2 + 6xy – 2xy – 4y2 = 0

or, 3x(x + 2y) – 2y(x + 2y) = 0

or, (x + 2y) (3x – 2y) = 0

Either, Or,

x + 2y = 0 3x – 2y = 0

PRIME Opt. Maths Book - X 163

Here, we have ,

Equation of straight line passes through a point and perpendicular to ax + by = 0 is

bx – ay + k = 0.

Then, the equation becomes, Note : After finding the separate

2x – y + k1 = 0 .....................................(i) equations of given homogeneous
2x + 3y + k2 = 0 ..............................(ii)
Taking the point (2, 1) for them, equation, try to use the

conditions.
• ax + by + k = 0 for parallel to

2 x 2 – 1 + k1 = 0 ⇒ k1 = – 3 ax + by = 0
2 x 2 + 3 x 1 + k2 = 0 ⇒ k2 = –7 • bx – ay + k = 0 for

perpendicular to ax + by = 0

Then, the equation becomes,

(2x – y – 3 ) (2x + 3y – 7) = 0

or, 4x2 + 6xy – 14x –2xy – 3y2 + 7y – 6x – 9y + 21 = 0

or, 4x2 + 4xy – 3y2 – 22x – 2y + 21 = 0

It is the required equation.

7. If the line pairs of (2p + 1)x2 – 12xy + 4y2 = 0 are coincident to each other, find the value of

‘p'. Also find the equation of straight lines represented by them.

Solution :

The given homogeneous equation is,

(2p + 1)x2 – 12xy + 4y2 = 0

Comparing it with ax2 + 2hxy + by2 = 0,

a = (2p + 1)

b=4

2h = – 12 → h=–6

Now, for the line pairs coincident,

h2 = ab

or, (– 6)2 = (2p + 1 )4

or, 9 = 2p + 1

` p=4

Then, the homogeneous equation becomes,
(2 x 4 + 1) x2 – 12xy + 4y2 = 0

or, 9x2 – (6 + 6) xy + 4y2 = 0
or, 9x2 – 6xy – 6xy + 4y2 = 0
or, 3x(3x – 2y) – 2y(3x – 2y) = 0
or, (3x – 2y) (3x – 2y) = 0

Either, Or,

3x – 2y = 0 3x – 2y = 0

They are the required separate equations.

164 PRIME Opt. Maths Book - X

Exercise 4.2

1. Answer the following questions.
i) Write down the formula to find angle between the line pairs of homogeneous
equation.
ii) Write down the conditions of the line pairs of homogeneous equations being
perpendicular and coincident.
iii) Write down the standard form of second degree homogeneous equation passes
through the origin.
iv) Write down the single equation of x = 0 and x – y = 0
v) Write down the separate equations of homogeneous equation x2 = xy.

2. Find the single equation of the following pair of straight lines :
i) 3x – 2y = 0 and x – 4y = 0
ii) x + 2y – 2 = 0 and 2x – y = 0
iii) mx + ny = 0 and nx – my = 0
iv) x – y (Seca + Tana) = 0 and x – y (Seca – Tana) = 0
v) 3x + y = 0 and x – 3y = 0

3. Find the separate equations of the following homogeneous equations.

i) 2x2 + 5xy – 12y2 = 0 ii) x2 – 7xy + 6y2 – 2x + 12y = 0.

iii) 3x2 – 5xy – 12y2 – 6x – 8y = 0 iv) x2 – 6xy + 9y2 – 3x + 9y – 10 = 0

v) x2 + 4xy + 4y2 – 2x – 4y – 15 = 0

4. Find the separate equation of straight lines represented by the following equations.

i) x2 + 2xy Seca + y2 = 0 ii) X2 – 2xy Cosecq + y2 = 0

iii) x2 – 2xy Tan2a – y2 = 0 y2
v) x2 – y2 – 2xyTana + y2Sec2a = 0 iv) y2Cos2q – 2xy + (1 + x2 Sin2q)x2 = 0

5. Find the separate equations represented by the following equations.

i) mn(x2 – y2) – xy(m2 – n2) = 0 ii) x2 + xy – 2y2 – 3x – 3y + 2 = 0

iii) 2x2 – 7xy + 3y2 + 3x + y – 2 = 0 iv) x2 – y2 – 3x + 3y = 0

v) x2 – y2 – 6y – 9 = 0

6. Find the angle between the line pairs of following homogeneous equations.

i) 3x2 – 7xy + 2y2 = 0 ii) 2x2 – 7xy – 15y2 = 0

iii) 3x2 – 4xy + 3y2 = 0 iv) x2 – 2xy Seca + y2 = 0

v) x2 – 2xy Cot2a – y2 = 0

7. Find the separate equations of the following homogeneous equations. Also find the angle

between the line pairs .

i) x2 – 3y2 = 0 ii) 3x2 – 5xy – 2y2 = 0

iii) x2 + 2xy Coseca + y2 = 0 iv) x2 – 2xy Tan2a – y2 = 0

v) xy(p2 – q2) – pq (x2 – y2) = 0

PRIME Opt. Maths Book - X 165

8. i) Prove that the line pairs of 6x2 – 5xy – 6y2 = 0 are perpendicular to each other.
ii) Prove that the line pairs of 9x2 – 12xy + 4y2 = 0 are coincident to each other.
iii) If the line pairs of 4x2 – 6xy – (k + 1) y2 = 0 are perpendicular to each other, find the
value of ‘k'.
iv) If the line pairs of mx2 – 12xy + 9y2 = 0 are coincident to each other, find the value of
‘m'
v) If angle between the line pairs px2 – 7xy + 2y2 = 0 is 45°, find the value of ‘p'.

9. i) Find the equation of line pairs passes through the origin which are perpendicular to
the line pairs of 3x2 – 7xy + 2y2 = 0

ii) Find the equation of line pairs passes through the origin which are perpendicular to
the line pairs of 6x2 – 5xy – 4y2 = 0.

iii) Find the equation of line pairs passes through a point (1, – 2) which is perpendicular
to the line pairs of 2x2 – 7xy + 6y2 = 0.

iv) Find the equation of line pairs passes through a point (3, 1) which is orthogonal to
the line pairs of 3x2 + xy – 10y2 = 0.

v) Find the equation of line pairs passes through a point (–2, 1) which are parallel to
the line pairs of 4x2 – 4xy – 3y2 = 0.

10. PRIME more creative questions:
i) If the line pairs of 6x2 – 5xy – (3k – 3) y2 = 0? Also find the equation of straight lines
represented by it.
ii) If the line pairs of mx2 – 12xy + 4my2 = 0 are coincident to each other find the positive
value of ‘m'. Also find the separate equation of it.
iii) Find the angle between the line pairs of homogeneous equation (3b2 – a2)x2 – 8abxy
+ (3a2 – b2)y2 = 0
iv) Find the single equation of straight line AB and CD from the given diagram.

Y

B(0, 2) D

X' O A(4, 0)
X

C

Y'
v) Find the single equation of straight lines PQ and RS from the given diagram.

R

(1, 2) Q
x+y–2=0

P
S

166 PRIME Opt. Maths Book - X

Answer

1. Show to your teacher.

2. i) 3x2 – 14xy + 8y2 = 0 ii) 2x2 + 3xy – 2y2 – 4x + 2y = 0
iii) mn(x2 – y2) – xy(m2 – n2) = 0 iv) x2 – 2xySeca + y2 = 0
v) 3x2 – 2xy – 3y2 = 0

3. i) x + 4y = 0 and 2x – 3y = 0 ii) x – 6y = 0 and x – y – 2 = 0
iii) 3x + 4y = 0 and x – 3y – 2 = 0 iv) x – 3y + 2 = 0 and x – 3y – 5 = 0
v) x + 2y + 3 = 0 and x + 2y – 5 = 0

4. i) x + y(seca + Tana) = 0 and x + y (seca – Tana) = 0
ii) x – y(Cosecq + Cotq) = 0 and x – y(Cosecq – Cotq) = 0
iii) x – y(Sec2a + Tan2a) = 0 and x + y (Sec2a + Tan2a) = 0
iv) x – y = 0 and x – y = 0
v) x – yTana = 0 and x – yTana = 0

5. i) mx + ny = 0 and nx – my = 0 ii) x – y – 2 = 0 and x + 2y – 1 = 0
iii) x – 3y + 2 = 0 and 2x – y – 1 = 0 iv) x – y = 0 and x + y – 3 = 0
v) x – y – 3 = 0 and x + y + 3 = 0

6. i) 45° or 135° ii) 45° or 135°
iii) 30° or 150° iv) a or 180° – a
v) 90°

7. i) x + 3y = 0, x – 3y = 0, 60°, 120°
ii) x – 2y = 0, 3x + y = 0, Tan–l(±7)
iii) x + y(Coseca + Cota) = 0, x + y(Coseca – Cota) = 0, 90° – a, 90° + a
iv) x – y(Sec2a + Tan2a) = 0, x + y(Sec2a + Tan2a) = 0, 90°
v) px + qy = 0, qx – py = 0, 90°

8. iii) 3 iv) 4 v) 3

9. i) 2x2 + 7xy + 3y2 = 0 ii) 4x2 – 5xy – 6y2 = 0

iii) 6x2 + 7xy + 2y2 + 2x + y = 0 iv) 10x2 + xy – 3y2 – 61x + 3y + 90 = 0

v) 4x2 – 4xy – 3y2 + 20x – 2y + 21 = 0

10. i) 3, 2x – 3y = 0, 3x + 2y = 0 ii) 3, x – 2y = 0, x – 2y = 0
iii) 60°, 120° iv) x2 + xy – 2y2 – 4x + 4y = 0
v) x2 – y2 – x + 3y – 2 = 0

PRIME Opt. Maths Book - X 167

4.3 Conic Section and Circle

Let us consider a plane and a cone where the plane surface cuts the cone. Depending on the
position of the plane, the intersecting points form some curves as shown in diagrams.

Generating Line
Semi Vertical Angle
Vertex

Axis

cone Plane Surface

Here, Each part of a double cone is called nappe and when a plane surface cuts such nappe the
different forms of curves so introduced are called conic sections.

The part obtained in a cone by intersecting a cone with a plane
surface by making different angles with the axis of the cone is
called conic section. They are of four types called parabola, circle,
ellipse and hyperbola.

• In conic section, the ratio of the distances of a point on conic section from a fixed
point and a fixed straight line is always constant.
• The fixed straight line which is used to construct a conic section and to define
the conic section is called directrix.
• The fixed point about which the conic section is constructed is called focus.
• The constant ratio of distance of any point in conic section from its focus to the
perpendicular distance from that point to the directrix is called ecentricity.

168 PRIME Opt. Maths Book - X

i) Parabola :

The section bounded by the curve so obtained when
intersecting plane is taken parallel to the generating
line of cone is called parabola.

parabola
Note : Any point on a parabola is always equidistant from a fixed point, called the focus and
from a fixed straight line called the directrix.

• In figure PQ = AQ.

A Q
vertex focus

P

parabola

• Value of ecentricity in parabola is always unity ai.e., e = PQ = 1k
• The standard form of equation of parabola is, PA

 y2 = 4ax (for the vertex origin and y = 0 is the axis)
 (y – k)2 = 4a(x – h) (for the vertex (h, k) and y = k is the axis)
 x2 = 4ay (for vertex origin and x = 0 is the axis)
 (x – h)2 = 4a(y – k) (for vertex (h, k) and x = h is the axis)

ii) Circle

The section bounded by the curve so obtained when
intersecting plane is taken perpendicular to the axis of
cone is called circle.

PRIME Opt. Maths Book - X 169

Circle

Circle

• The standard form of equation of circle is
 (x–h)2 + (y – k)2 = r2 for centre (h, k) and radius ‘r'
 x2 + y2 = r2 for centre (0, 0) and radius ‘r'.

iii) Hyperbola :

The section bounded by the curve so obtained
when the intersecting plane is parallel to the
axis of the cone is called hyperbola.

hyperbola
• There are two focus (foci) and two directrix in the section of hyperbola.
• The set of all points where the difference of distance from it's two foci is

always constant.

• Value of ecentricity in hyperbola is greater than unity i.e., e 2 1.

P

focus focus
F2 F1

directrix directrix

• The standard form of equation of hyperbola is,
x2 y2
 a2 – b2 = 1 for vertex origin (0, 0)

 ^x–hh2 ^y–kh2 for vertex (h, k).
a2 – b2

170 PRIME Opt. Maths Book - X

iv) Ellipse :

The section bounded by the curve so obtained when
intersecting plane is taken making a certain angle with
the axis of cone which is greater than the semi - vertical
angle is called elllipse.

Ellipse

Ellipse

• There are two focus (foci) and two directrix in the section of ellipse.
• The set of all points for which the sum of the distance from two foci is always

constant in ellipse.

• Value of ecentricity in ellipse is always less than unity i.e., e 1 1.

P

focus focus
F2 F1

directix P' directix

• The equation of standard form of ellipse is,
y2
 x2 + = 1 for centre at origin (0, 0)
a2
b2

 (x – h)2 + (y – k)2 = 1 for centre at (h, k)
a2 b2

PRIME Opt. Maths Book - X 171

Exercise 4.3

1. Answer the following questions.
i) What do you mean by conic section?
ii) Write down the name of types of conic section.
iii) In what condition parabola is formed? Show in diagram.
iv) Write down the condition of formation of circle as the conic section?
v) Show the formation of ellipse in diagram as the conic section.

2. Which types of conic section are there in the given diagrams ?
i) ii)

iii) iv)

v)

3. Which conic section represents the following equations ?

i) y2 = 4ax ii) ^x – hh2 + ^y – kh2 =1
a2 b2

iii) x2 + y2 = r2 iv) x2 y2 =1
a2 – b2

v) (x – h)2 = 4a(y – k)

172 PRIME Opt. Maths Book - X

4. Define the following terms used in conic section.

i) Nappe ii) Focus

iii) Directrix iv) Ecentricity

v) Vertex

5. Which types of conic section is formed for the followings.
i) When the intersecting plane is parallel to the axis of cone.
ii) When the intersecting plane is perpendicular to the axis of cone
iii) When the intersecting plane makes any angles with the axis of cone
iv) When the intersecting plane is parallel to the generating line of cone
v) When the intersecting plane is parallel to the base of the cone.

6. Write down the general form of the equation of conic section for the followings.
i) Equation of parabola having vertex (h, k) and the axis is y = k
ii) Equation of ellipse having centre (h, k)
iii) Equation of parabola having vertex (0, 0) and the axis is x = 0
iv) Equation of hyperbola having vertex (0, 0)
v) Equation of hyperbola having vertex (h, k)

Answer

1. Show to your teacher.

2. i) Circle ii) Ellipse iii) Parabola
iv) Circle v) Hyperbola iii) Ellipse

3. Discuss with your subject teacher.

4. Discuss with your subject teacher.

5. i) Hyperbola ii) Circle
iv) Parabola v) Circle

6. Discuss with your subject teacher.

PRIME Opt. Maths Book - X 173

4.4 Circle

Circle is the enclosed surface formed during the enclosed path traced out by a locus point
while moving equidistant form a fixed point. A circle is formed when a plane surface taking
perpendicular to the axis of a cone intersects the cone.

• The fixed point is called the centre.
• The constant (equal) distance is called the radius.
• The total length of locus of the moving point is called its
circumference.

1. Equation of circle having centre as origin and radius ‘r'
Solution:
Centre of a circle is 0(0, 0) and any point on its circumference is P(x, y)
Radius = OP = r
Now, using distance formula,

d= P(x, y)
or, d(OP) = r

or, r = x2 + y2 O(0, 0)

or, r2 = x2 + y2

` x2 + y2 = r2 is the equation of circle.

2. Equation of circle having centre (h, k) and radius ‘r' P(x, y)
Solution: r
Let, A(h, k) be the centre of a circle A(h, k)
P(x, y) be any point on its circumference
radius = AP = r.
Now, using distance formula,

d=

or, d(AP) =

or, r =

or, r2 = (x – h)2 + (y – k)2

` Equation of circle having centre (h, k) and radius ‘r' is

(x – h)2 + (y – k)2 = r2

174 PRIME Opt. Maths Book - X

3. Equation of circle having ends of diameter (x1 , y1) & (x2 , y2)
Solution:
P(x, y)

A(x1, y1) B(x2, y2)

Let, A(x1, y1) and B(x2, y2) be the co-ordinate of ends of diameter AB.
P(x, y) be any point on its circumference

Then,
\ APB = 90°

or, AP^BP

or, Slope of AP × Slope of BP = – 1 y2 – y1
y – y1 y – y2 x2 – x1
or, x – x1 × x – x2 =–1 [\ m = ]

or, (y – y1) (y – y2) = –(x – x1) (x – x2)
or, (x – x1) (x – x2) + (y – y1) (y – y2) = 0

\ This is the equation of circle.

4. Second degree equation of circle:
Solution:
Let, the equation of circle having centre (h, k) and radius 'r' be
(x – h)2 + (y – k)2 = r2
or, x2 – 2hx + h2 + y2 – 2ky + k2 = r2
or, x2 + y2 – 2hx – 2ky + (h2 + k2 – r2) = 0
Let, –2h = 2g
– 2k = 2f
h2 + k2 – r2 = c (Any constant)
Then, the equation becomes,
or, x2 + y2 + 2gx + 2fy + c = 0
It is the second degree equation of circle.

5. Centre and radius of circle having equation x2 + y2 + 2gx + 2fy + c = 0.
Solution:
The second degree equation of circle is
x2 + y2 + 2gx + 2fy + x = 0
or, x2 + 2gx + y2 + 2fy = – c
or, x2 + 2.x.g + y2 + y2 + 2.y.f + f2 = g2 + f2 – c
or, (x + g)2 + (y + f)2 = g2 + f2 – c
or, {(x – (–g)}2 + {y – (–f)}2 = _ g2 + f2 – c i2
Comparing it with (x – h)2 + (y – k)2 = r2, we get
Centre (h, k) = (–g, – f)

radius (r) = g2 + f2 – c

PRIME Opt. Maths Book - X 175

Conditions required for the equation of circle.

Equation of circle can be found out by calculating the center of the circle (h, k) and radius of
the circle 'r' by using the formula (x – h)2 + (y – k)2 = r2.

Finding of center and radius of circle in different conditions:
1. When the centre (h, k) and any point (a, b) on its circumference are given.

(a, b)
r

(h, k) Radius (r) =
=

2. When the Centre (h, k) and the circle touches x-axis only.
Y

(h, k)

r Radius (r) = k.
OX

3. When the centre (h, k) and the circle touches y-axis only.
Y

(h, k)
r

O X Radius (r) = h
4. When the radius ‘r' and the circle touches both the axes.

Y

r (h, k) X Radius (r) = h = k.
r Centre = (r, r)

O

176 PRIME Opt. Maths Book - X

5. When the centre (h, k) and the circle touches the straight line ax + by + c = 0.

(h, k) Radius (r) = ax1 + by1 + c
r a2 + b2

ax + by + c = 0 = ah + bk + c
a2 + b2

6. When the centre (h, k) lies in the equation of straight line ax + by + c = 0.

(h, k)

ax + by + c = 0

The equation becomes, ah + bk + c = 0

7. When the equation of circle x2 + y2 + 2gx + 2fy + c = 0 is given.
We have to find the centre and radius of circle.

(h, k) Centre (h, k) = (–g, – f)
r Radius (r) = g2 + f2 – c

8. When the equation of two diameters are given.
D

A

(h, k)

B Centre (h, k) = Intersecting point of two diameters.
C

9. When the equation of a circle and another circle concentric with it is given.

(h, k) r2 Centre (h, k) = Centre of given circle
r1 Radius (r) = will be derived from given condition

10. When the circle passes through more than one points is given.

(x – h)2 + (y – k)2 = r2
Put x & y from the passing points.
Solve equations so obtained to get centre and radius.

PRIME Opt. Maths Book - X 177

Worked out Examples

1. Find the equation of circle having centre (2, – 1) which passes through a point (6, 2).

Solution:

Centre of a circle (h, k) = (2, – 1).

It passes through a point (6, 2). (2, –1) (6, 2)
Here,

Radius(r) =

=

= 16 + 9
= 5 units.
Then,
Equation of circle is,
(x – h)2 + (y – k)2 = r2
or, (x – 2)2 + (y + 1)2 = 52
or, x2 + y2 – 4x + 2y – 20 = 0
It is the required equation of circle.

2. Find the equation of circle having ends of diameter
(3, – 2) and (–3, 6)

Solution:

The co-ordinate of ends of diameter are A(3, – 2) A(3, –2) B(–3, 6)

and B(–3, 6).

Then, equation of circle is, Y
(x – x1) (x – x2) + (y – y1) (y – y2) = 0

or, (x – 3) (x + 3) + (y + 2) (y – 6) = 0
or, x2 – 9 + y2 – 6y + 2y – 12 = 0
or, x2 + y2 – 4y – 21 = 0
It is the required equation of circle.

3. Find the equation of circle having centre (2, 3) which (2, 3)
touches x-axis. r=3
Solution:
Centre of a circle (h1 k) = (2, 3) O X
It touches x-axis, so radius (r) = k = 3

Then, the equation of circle is ,
(x – h)2 + (y – k) = r2

or, (x – 2)2 + (y – 3)2 = 32
or, x2 – 4x + 4 + y2 – 6y + 9 = 9
\ x2 + y2 – 4x – 6y + 4 = 0
It is the required equation of circle.

178 PRIME Opt. Maths Book - X

4. Find the centre and radius of circle having equation 4x2 + 4y2 – 16x + 6y – 71 = 0.
4
Solution:

The given equation of circle is

4x2 + 4y2 – 16x + 6y – 71 =0
4

Dividing both sides by ‘4'
3y
or, x2 + y2 – 4x + 2 = 71
16

or, x2 – 4x + y2 + 3y = 71
2 16

or, x2 – 2.x.2 + 22 + y2 + 2.y. 3 + ` 3 2 = 22 + ` 3 2 + 71
4 4 4 16
j j

or, (x – 2)2 + (y + 3 )2 = 32
4

Comparing it with (x – h)2 + (y – k)2 = r2

We get,

Centre(h, k) = (2, – 3 )
4

Radius (r) = 3 units.

Alternative method:

The given equation of circle is,

4x2 + 4y2 – 16x + 6y – 71 =0
4

or, 4(x2 + y2 – 4x + 3y 71 =0
2 )– 4

or, x2 + y2 – 4x + 3y – 71 =0
2 16

Comparing it with x2 + y2 + 2gx + 2fy + c = 0

We get,
2g = –4 & g = –2

2f = 3 & f= 3
2 4

c=– 71
16

Then,

Centre (h, k) = (–g, –f) = (2, – 3 )
4

Radius(r) = g2 + f2 – c

=

64 + 9 + 71
= 16

= 144
16

= 3 units.

PRIME Opt. Maths Book - X 179

5. Find the equation of circle having two diameters x + 2y = 8 and 3x – y = 3 and the circle

touches the straight line 3x + 4y + 2 = 0. D
Solution:

The given equation of diameters are , A (h, k)
x + 2y = 8 & x = 8 – 2y ................................. (i)

3x – y = 3 & x = y +3 ................................. (ii) B = 0
3 + 4y
2
Solving equation (i) and (ii) +
y +3
C 3x
8 – 2y = 3

or, y + 3 = 24 – 6y

or, 7y = 21

\ y=3

Substituting the value of y in equation (i)
3+3
x= 3 =2

\ Centre (h, k) = intersecting point of two diameters = (2, 3)

Again,

The circle touches the line 3x + 4y + 2 = 0

\ Radius(r) = ax1 + by1 + c
a2 + b2

=

= 4 units.
Then,
Equation of circle is,
(x – h)2 + (y – k)2 = r2
or, (x – 2)2 + (y – 3)2 = 42
or, x2 + y2 – 4x – 6y – 3 = 0
It is the required equation of circle.

6. Find the equation of circle concentric with the circle of equation x2 + y2 – 4x + 6y – 3 = 0

which passes through a point (8, 5).

Solution:

The given equation of circle is , (8, 5)

x2 + y2 – 4x + 6y – 3 = 0

or, x2 – 4x + y2 + 6y = 3

or, x2 – 2.x.2 + 22 + y2 + 2.y.3 + 32 = 22 + 32 + 3 (2, –3)

or, (x – 2)2 + (y + 3)2 = 42

Comparing it with (x – h)2 + (y – k)2 = r2

We get,

Centre (h, k) = (2, –3)

Radius (r) = 4 units.

180 PRIME Opt. Maths Book - X

Again,
For the concentric circle which passes through a point (8, 5).
Centre (h, k) = (2, – 3)

Radius(r) =

=
= 10 units
Then, the equation of circle is,
(x – h)2 + (y – k)2 = r2
or, (x – 2)2 + (y + 3)2 = 102
or, x2 + y2 – 4x + 6y – 87 = 0
It is the required equation of circle.

7. If one end of diameter of a circle of equation 3x2

+ 3y2 + 6x – 12y – 60 = 0 is (2, – 2) find the co-

ordinate of other end. A(2, –2) B(a, b)
Solution:

One end of diameter AB is A(2, – 2) and the other

end supposed to B(a, b).

Where, The equation of circle is,

3x2 + 3y2 + 6x – 12y – 60 = 0

or, 3(x2 + y2 + 2x – 4y – 20) = 0

or, x2 + 2x + y2 – 4y = 20

or, x2 + 2. x. 1. + 12 + y2 – 2. y. 2 + 22 = 12 + 22 + 20

or, (x + 1)2 + (y – 2)2 = 52

Comparing it with (x – h)2 + (y – k)2 = r2
We get,
Centre (h, k) = (–1, 2)
Radius (r) = 5 units.
Again,
We have, centre (–1, 2) is the mid-point of diameter AB.

So, using mid-point formula,
y1 + y2
x= x1 + x2 , y= 2
2

or, –1 = 2+a , 2= –2 + b
2 2

or, 2 + a = – 2, –2 + b = 4

` a = – 4, & b=6

\ Other end of diameter is B(–4, 6)

PRIME Opt. Maths Book - X 181

8. Find the equation of circle passes through the point (1, – 2) (i, –2)
and (7, 6) where centre lies in the line of equation 2x – y = 6.
Solution: 2x – y = 6
Let, The equation of circle be, (x – h)2 + (y – k)2 = r2 (7, 6)
It passes through the points (1, –2) and (7, 6).
Then,
(1 – h)2 + (–2, –k)2 = r2................(i)
(7 – h)2 + (6, –k)2 = r2 ................(ii)

Solving equation (i) and (ii)

(1 – h)2 + (–2, –k)2 = (7 – h)2 + (6 – k)2

or, 1 – 2h+ h2 + 4 + 4k + k2 = 49 – 14h + h2 + 36 – 12k + k2

or, 12h = 16k + 80

or, h= –4k + 20 ................(iii)
3

By the question,

Centre (h, k) lies in the straight line,

2x – y = 6

or, 2h – k = 6

or, 2 a –4k + 20 k –k=6 [From equation (i) ]
3

or, – 8k + 40 – 3k = 18

or, –11k = –22

\ k=2

Putting the value of ‘k' in equation (iii)

h= –4 ^2h + 20 =4
3

Putting the value of (h, k) = (4, 2) in equation (i),
or, (1 –4)2 + (–2 –2)2 = r2
or, r2 = 25
\ r = 5 units

Then, the equation of circle is,
(x – h)2 + (y – K)2 = r2
or, (x – 4)2 + (y – 2)2 = 52
or, x2 + y2 – 8x – 4y – 25 = 0
\ It is the required equation of circle.

9. Find the equation of tangent to the circle of equation x2 + y2 – (4, 8)

8x – 16y – 20 = 0 at point (–4, 2). P B

Solution:

The given equation of circle is, Q
x2 + y2 – 8x – 16y – 20 = 0

or, x2 – 8x + y2 – 16y – 20 = 0 (–4, 2)

or, x2 – 2.x.4 + 42 + y2 – 2.y.8. + 82 = 42 + 82 + 20 A

182 PRIME Opt. Maths Book - X

or, (x – 4)2 + (y – 8)2 = 102

Comparing it with (x – h)2 + (y – k)2 = r2
Centre (h, k) = P(4, 8)
Radius (r) = 10 units = PQ

Again,

Let AB be the tangent to the circle at point Q(–4, 2)

Then PQ^AB.

or, Slope of PQ x Slope of AB = –1

or, y2 – y1 × m = –1
x2 – x1

or, 8–2 × m = –1
4+4

or, m= 4
–3

Then, equation of tangent passes through the point (–4, 2) is,

y –y1 = m(x –x1)

or, y – 2 = – 4 (x + 4)
3

or, 3y – 6 = – 4x – 16

\ 4x + 3y + 10 = 0.

It is the required equation of tangent AB.

10. Prove that the circles having equations x2 + y2 + 4x – 14y + 28 = 0 and x2 + y2 – 20x + 4y + 4 = 0

are touched externally.

Solution:

The given equation of 1st circle is, P r1 r2 Q
x2 + y2 + 4x – 14y + 28 = 0 (–2, 7) (10, –2)
or, x2 + 4x + y2 – 14y = – 28

or, x2 + 2. x. 2 + 22 + y2 – 2. y. 7. + 72 = 22 + 72 – 28

or, (x + 2)2 + (y – 7)2 = 52

Comparing it with (x – h)2 + (y – k)2 = r2, we get
Centre (h, k) = p(–2, 7)
Radius (r1) = 5 units.
Again,
The equation of 2nd circle is,

x2 + y2 – 20x + 4y + 4 = 0
or, x2 – 20x + y2 + 4y = 4
or, x2 – 2.x. 10 + 102 + y2 + 2.y.2 + 22 = 102 + 22 – 4
or, (x – 10)2 + (y + 2)2 = 102
Comparing it with (x – h)2 + (y – k)2 = r2
Centre (h, k) = Q(10, – 2)
Radius (r2) = 10 units.

PRIME Opt. Maths Book - X 183

Again, distance formula for centres P and Q.

d=

or, d(PQ) =

= 225

= 15 units.

Here,

r1 + r2 = 5 + 10
= 15

= PQ

Here, PQ = r1 + r2
\ i.e they are touched externally.

Exercise 4.4

1. Answer the following questions.
i) Write down the equation of circle having centre (h, k) and radius 'r'.
ii) Write down the equation of circle having ends of diameter joining the points (x1, y1)
and (x2, y2).
iii) Write down the second degree equation of circle.
iv) Write down centre and radius of circle of equation x2 + y2 = 9.
v) Write down the centre and radius of circle having equation x2 + y2 + 2gx + 2fy + c = 0.

2. Find the radius of the circle under the following conditions.

i) ii)
B (5, 1)

A (3, 2) T
(1, –2) P

Q
3x + 4y + 3 = 0

A

iii) Y iv) Y

(4, 2) (3, 4)
X
X' O X' O X

184 PRIME Opt. Maths Book - X

v) Y

8 Note : If equation of circle is given
in question, finding the center and
X' O 6 radius of circle makes easy to solve

X the problems.

3. Find the equation of the circle under the following conditions.
i) Having centre (2, – 3) and passes through a point (8, 5)
ii) Having centre (4, 2) and touches a straight line of equation 4x + 3y + 3 = 0.
iii) Having centre (2, –1) and touches y – axis.
iv) Having centre (3, 5) and touches x-axis.
v) Making x-intercept 4 and y-intercept 6 on the axes and passes through the origin.

4. Find the equation of circle under the following conditions.
i) Having ends of diameter (2, – 3) and (6, – 5).
ii) Having ends of diameter (1, 5) and (–7, – 1)
iii) Having centre (3, 2) and touches the straight line of equation 6x + 8y + 6 = 0.
iv) Having centre (–1, 3) and diameter 10.
v) Having centre (3, 4) and passes through origin.

5. Find the equation of circle under the following conditions.
i) Having intercepts 6 and 8 on the axes.
ii) Having radius 4 units and touches both the axes.
iii) Centre lies in the straight line 3x + 2y = 5 and touches both the axes.
iv) Touching x-axis at point (3, 0) and passes through a point (1, 2).
v) Having centre (a, b) and a2 + b2 radius units.

6. Find the centre and radius of the circle from the given equations.

i) x2 + (y + 3)2 = 25 ii) x2 + y2 – 6x – 8y = 0

iii) x2 + y2 + 12x – 10y – 39 = 0 iv) 4x2 + 4y2 – 16x + 24y – 48 = 0

v) 3x2 + 3y2 – 12x + 9y – 6 = 0

7. Find the equation of circle under the given conditions.
i) Having two diameters of equations x + 2y = 7 and 4x – y – 1 = 0 which has radius 5
units.
ii) Having two diameters of equations 3x – y = 7 and 2x + 3y = 12 which passes through
a point. (–5, 8)
iii) Concentric with the circle of equation x2 + y2 – 6x + 4y – 12 = 0 & having radius 10
units.
iv) Concentric with the circle of equation x2 + y2 + 12x – 8y – 48 = 0 which passes
through a point (–2, 1).

PRIME Opt. Maths Book - X 185

v) Passes through the point (4, – 1) and (5, 6) where centre lies in the line x + 2y = 7

8. i) Find the equation of circle passes through the intersecting point of two straight
lines having equations 2x + 3y – 8 = 0 and 4x – 3y + 2 = 0 where centre of the circle
is (4, – 2)

ii) Find the equation of circle passes through the centre of a circle of equation 2x2 +
2y2 – 4x + 8y – 8 = 0. Where centre of the circle is (9, 4).

iii) Find the equation of circle having radius 5 units, passes through a point (4, 2) and
centre lies in first quadrent having equation x – 2y = 5.

iv) Find the equation of circle passes through the point (–3, 6), having radius 10 units
and centre lies in the line x + 2y + 1 = 0

v) Find the equation of circle passes through the point (0, 8) and (6, 0) where origin
point lies in its circumference.

9. Find the equation of circle under the followings.
i) Passes through the points (3, 0), (4, 3), (2, 7).
ii) Passes through the points (5, 3) (6, 2) and (–2, –4)
iii) Passes through the points (–4, 3) ,(–2, 5) and (10, – 11)
iv) Passes through the points (2, 0) and (1, 7) where centre lies in the straight 3x + 4y
– 6 = 0.
v) Passes through the points (–5, –4) and (9, 10) where centre lies in the line x + 2y – 7 = 0.

10. PRIME more creative questions.
a. i) If one end of diameter of circle of equation x2 + y2 – 2x – 6y – 15 = 0 is (5, –1). Find

the co-ordinate of other end.
ii) The center of a circle of equation 2x2 + 2y2 – ax + by – 174 = 0 is (2, –3), find the value

of ‘a' and ‘b'. Also find the radius of circle.
iii) Find the equation of circle having radius 5 units, passes through a point (4, 3) and

center lies on y - axis.
iv) Find the equation of circle which touches x-axis at (7, 0) and makes y-intercept 48

units.
v) The radius of circle of equation x2 + y2 – 4x + ky = 12 = 0 is 5 units, find the positive

value of ‘k'. Also find the centre of the circle.

b. i) Find the equation of straight line which touches the circle of equation x2 + y2 = 25 at
point (3, – 4).

ii) Find the equation of tangent to a circle of equation x2 + y2 – 6x – 14y – 42 = 0 at point
(–3, 1)

iii) Prove that the circle of equation x2 + y2 – 4x + 6y – 12 = 0 touches the straight line of
equation 3x + 4y + 6 = 0.

iv) Prove that the straight line x – y + 9 = 0 cuts the circle of equation x2 + y2 + 6x – 4y –
87 = 0 at two points.

v) Find the length of the chord having equation 3x + 4y + 15 = 0 to a circle of equation
x2 + y2 – 6x + 2y – 15 = 0

186 PRIME Opt. Maths Book - X

c. i) Prove that the circle having equations x2 + y2 + 6x – 4y – 12 = 0 and x2 + y2 – 18x +
14y + 30 = 0 are touched externally.

ii) Prove that the circles having equations x2 + y2 – 4x – 8y – 80 = 0 and x2 + y2 – 4x +
4y – 8 = 0 are touched internally.

iii) Prove that the straight line having equation 3x + 4y = 3 touch at a point to the circle
of equation x2 + y2 – 6x – 2y + 6 = 0.

iv) Prove that the circle having equation x2 + y2 – 2ax – 2ay + a2 = 0 touches both the
axes.

v) Prove that the circle having equation x2 + y2 – 4x – 4y + 4 = 0 touches both the axes.

11. Project work
Collects the equation of circles in different forms in different conditions in a chart paper

and present it in your classroom.

Answer

1. Show to your teacher.

2. i) 5 units ii) 4 units iii) 2 units
iv) 3 units v) 5 units

3. i) x2 + y2 – 4x + 6y – 87 = 0 ii) x2 + y2 – 8x – 4y – 5 = 0
iii) x2 + y2 – 4x + 2y + 1 = 0 iv) x2 + y2 – 6x – 10y + 9 = 0
v) x2 + y2 – 4x – 6y = 0

4. i) x2 + y2 – 8x + 8y + 27 = 0 ii) x2 + y2 + 6x – 4y – 12 = 0
iii) x2 + y2 – 6x – 4y – 3 = 0 iv) x2 + y2 + 2x – 6y – 15 = 0
v) x2 + y2 – 6x – 8y = 0

5. i) x2 + y2 – 6x – 8y = 0 ii) x2 + y2 – 8x – 8y + 16 = 0
iii) x2 + y2 – 6x – 6y + 9 = 0 iv) x2 + y2 – 6x – 4y + 9 = 0
v) x2 + y2 – 2ax – 2by = 0

6. i) (0, –3); 5 units ii) (3, 4); 5 units iii) (–6, 5); 10 units
iv) (2, –3); 5 units v)
(2, – 3 ); 33
2 2 units

7. i) x2 + y2 – 2x – 6y – 15 = 0 ii) x2 + y2 – 6x – 4y – 87 = 0
iii) x2 + y2 – 6x + 4y – 87 = 0 iv) x2 + y2 + 12x – 8y + 27 = 0
v) x2 + y2 – 2x – 6y – 15 = 0

8. i) x2 + y2 – 8x + 4y – 5 = 0 ii) x2 + y2 – 18x – 8y – 3 = 0

iii) x2 + y2 – 4x – 18y + 60 = 0

iv) x2 + y2 – 6x + 4y – 87 = 0 or x2 + y2 + 26x – 12y + 105 = 0

PRIME Opt. Maths Book - X 187

Answer

v) x2 + y2 – 6x – 8y = 0

9. i) x2 + y2 + 2x – 6y – 15 = 0 ii) x2 + y2 – 4x + 2y – 20 = 0
iii) x2 + y2 – 8x + 6y – 75 = 0 iv) x2 + y2 + 4x – 6y – 12 = 0
v) x2 + y2 – 6x – 4y – 87 = 0

10. (–3, 7) ii) a = 8, b = 12, and r = 10 units.
a. i) x2 + y2 – 12y + 11 = 0; x2 + y2 = 25 iv) x2 + y2 – 14x – 50y + 49 = 0
R = 6; (2, 3)
iii)
v)

b. i) 3x – 4y – 25 = 0 ii) x – y + 4 = 0
v) 6 units.

Co-ordinate

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Define the term homogeneous equation.

2. a) If line joining the points (m, 3) and (–3, – 1) is perpendicular to the line of equation
2x + 3y – 7 = 0, find the value of m.

b) Find the angle between the line pairs of equation 3x2 – 4xy + 3y2 = 0

c) Find the equation of circle having centre (2, – 3) which touches the straight line of

equation 3x – 4y – 2 = 0

3. a) Find the equation of perpendicular bisector of the line joining the points A(3, –2)
and B(–5, 4).

b) Find the separate equation of the homogeneous equation between the line pairs of
the given equations.

4. Find the equation of circle having centre lies on the straight line x – 2y = 5, radius 5 units
and the circle passes through a point (5, 1).

188 PRIME Opt. Maths Book - X

Unit 5 Trigonometry

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 1 2 1 – 4 9 15
Weight 1 4 4 –

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

Objectives : At the end of the lesson
• Students are able to solve trigonometrical ratios for multiple and sub-

multiple angle.
• Students are able to use transformation formula in trigonometry &

conditional form in trigonometry.
• Students are able to solve trigonometric equations in different quadrants.
• Students are able to find height & distance using right angled triangle.

Materials
• Trigonometric formula chart.
• Chart of the concept of quadrants.
• Model to find height and distance of an object (clinometer)
• Chart of trigonometric ratios for standard angles.
• Chart of algebraic formula etc.
• Graph paper.

PRIME Opt. Maths Book - X 189

Enjoy the recall: Sin(A – B) = SinA.CosB – CosA.SinB
• Recall from all trigonometric formula of grade IX.
• Compound angles :

Sin(A + B) = SinA.CosB + CosA.SinB

Cos(A + B) = CosA.CosB – SinA.SinB Cos(A – B) = CosA.CosB + SinA.SinB

Tan(A + B) = TanA + TanB Tan(A – B) = TanA – TanB
1 – TanA.TanB 1 + TanA.TanB

Cot(A + B) = CotA.CotB – 1 Cot(A – B) = CotA.CotB + 1
CotB + CotA CotB – CotA

• Recall all the algebraic formula which you learn in junior classes.

5.1 Multiple Angle

Multiple angles are the multiple of an angle A with the constant numerals 2, 3, 4 etc. i.e. 2A,
3A, 4A, ... etc are the multiple angles of A. Trigonometrical ratios can be used for the multiple
angles and formula can be derived by the concept of compound angle as discussed in the
previous chapter of grade IX.

1. Trigonometrical ratios of 2A.

i) Sin2A = Sin(A + A)

= SinA.CosA + CosA.SinA

= 2SinACosA .............................................. (i)

= 2SinA.CosA
1

= 2SinA.CosA
Cos2 A + Sin2 A

2SinA.CosA

= Cos2 A [\ Dividing numerator and dinominator by Cos2A]
Cos2 A + Sin2 A

Cos2 A

= 2TanA .............................................. (ii)
1 + Tan2 A

\ Sin2A = 2SinACosA

Sin2A = 2TanA
1 + Tan2 A

ii) Cos2A = Cos(A + A)

= CosA.CosA – SinA.SinA

= Cos2A – Sin2A .............................................. (i)

= Cos2 A – Sin2 A
Cos2 A + Sin2 A

Cos2 A – Sin2 A

= Cos2 A [\ Dividing numerato and dinominator by Cos2A]
Cos2 A + Sin2 A

Cos2 A

= 1 – Tan2 A .............................................. (ii)
1 + Tan2 A

190 PRIME Opt. Maths Book - X

Also,

Cos2A = Cos2A – Sin2A

= 1 – Sin2A – Sin2A

= 1 – 2Sin2A .............................................. (iii)

= Cos2A – 1 + Cos2A

= 2Cos2A – 1 .............................................. (iv)

\ Cos2A = Cos2A – Sin2A

Cos2A = 2Cos2A – 1

Cos2A = 1 – 2Sin2A

Cos2A = 1 – Tan2 A
1 + Tan2 A

iii) Tan2A = Tan(A + A)
\ Tan2A TanA + TanA
= 1 – TanA.TanA

= 2TanA
1 – Tan2 A

iv) Cot2A = Cot(A + A)
\ Cot2A
= CotA.CotA – 1
CotA + CotA

= Cot2 A – 1
2CotA

2. Trigonometrical ratios of 3A.
i) Sin3A = Sin(2A + A)

= Sin2ACosA + Cos2ASinA
= (2SinACosA)CosA + (1 – 2Sin2A)SinA
= 2SinACos2A + SinA – 2Sin3A
= 2SinA (1 – Sin2A) + SinA – 2Sin3A
= 2SinA – 2Sin3A + SinA – 2Sin3A
= 3SinA – 4Sin3A
\ Sin3A = 3SinA – 4Sin3A

ii) Cos3A = Cos(2A + A)
= Cos2A.CosA – Sin2A.SinA
= (2Cos2A – 1)CosA – (2SinA.CosA)SinA
= 2Cos3A – CosA – 2Sin2A.CosA
= 2Cos3A – CosA – 2CosA(1 – Cos2A)
= 2Cos3A – CosA – 2CosA + 2Cos3A

\ Cos3A = 4Cos3A – 3CosA

iii) Tan3A = Tan(2A + A)
=

=

PRIME Opt. Maths Book - X 191

=

\ Tan3A = 3TanA – Tan3 A
1 – 3Tan2 A

iv) Cot3A = Cot(2A + A)

= Cot2A.CotA – 1
CotA + Cot2A

=

=

\ Cot3A = Cot3 A – 3CotA
3Cot2 A – 1

Important informations to solve the problems:
• Taking coefficient 2 for products and square form like to SinA.CosA, Cos2A and Sin2A.
• Use the following relations as the formula.

• Use the following steps for others too.

i) 2SinA.CosA = Sin2A

ii) 2Sin2A = 1 – Cos2A

iii) 2Cos2A = 1 + Cos2A

iv) 4Cos3A = 3CosA + Cos3A

v) 4Sin3A = 3SinA – Sin3A

vi) Sin2ACos2A = 1 (2SinA.CosA)2 = 1 (1 – Cos22A)
4 4

vii) Sin4A = 1 (2Sin2A) = 1 (1 – Cos2A)2
4 4

viii)1 + Sin2A = Cos2A + Sin2A + 2SinACosA = (CosA + SinA)2

ix) Sin2(120° – q) = 1 (2Sin2(120° – q)
2

= 1 [1 – Cos2(120° – q)]
2

= 1 [1 – Cos(240° – 2q)]
2

x) Cos320° = 1 [4Cos320°]
4

= 1 [3Cos20° + Cos3(20°)]
4

= 1 [3Cos20° + Cos60°]
4

= 1 [3Cos20° + 1 ]
4 2

192 PRIME Opt. Maths Book - X

Worked out Examples

1) If SinA = 4 , find the value of Sin2A and Cos2A.
5

Solution :

SinA = 4
5

\ CosA = 1 – Sin2 A

= 1 – a 4 2
5
k

= 25 – 16
25

= 3
5

Then, = 2SinA.CosA Cos2A = Cos2A – Sin2A
Sin2A

=2× 4 × 3 = a 3 2 – a 4 2
5 5 5 5
k k

= 24 = 9 – 16
25 25

= –7
25

2) If Cos2A = 7 , find the value of CosA.
Solution : 25

Cos2A = 7
25

We have,

Cos2A = 2Cos2A – 1

or, 7 + 1 = 2Cos2A
25

or, 2Cos2A = 32
25

or, Cos2A = 32
25 × 2

\ CosA = ± 4
5

3) If CosA = 1 `a + 1 j , prove that Cos3A = 1 aa3 + 1 k.
2 a 2 a3

Solution: CosA = 1 `a + 1 j
2 a

Cos3A = ?

We have,

Cos3A = 4Cos3A – 3CosA

= 4 & 1 (a + 1 3 –3× 1 `a + 1 j
2 a 2 a
)0

=4× – 3 `a + 1 j
2 a

= – 3 `a + 1 j
2 a

PRIME Opt. Maths Book - X 193

= 1 aa3 + 1 k + 3 `a + 1 j – 3 `a + 1 j
2 a3 2 a 2 a

= 1 aa3 + 1 k
2 a3

4) If Cos30° = 3 prove that: Sin15° = 3 –1
2 22

3 –1
Solution :

22

Let, A = 15°

\ 2A = 30°

As we have,

Cos2A = 1 – 2Sin2A

or, Cos30° = 1 – 2Sin215°

3
or, 2Sin215° = 1 – 2

2– 3
or, 2Sin215° = 2

or, Sin215° = 2– 3 × 2
4 2

4–2 3
or, Sin215° = 8

or, Sin215° =

2

or, Sin215° = e! 3 – 1 o
22

or, Sin15° = 3 – 1 [\ For acute angle Sinq is positive]
22

\ Sin15° = 3 –1 is proved.

22

5) Prove that : Tan ` r + Aj = 1 + Sin2A
4 Cos2A
Solution :
1 + Sin2A
R.H.S. = Cos2A

=

=

= CosA + SinA
CosA – SinA

CosA + SinA

= CosA [\ Dividing each terms by CosA]
CosA – SinA

CosA

= 1 + TanA
1 – TanA

194 PRIME Opt. Maths Book - X

= [ a Tan45° = 1]

= Tan(45° + A)

= Tan( r + A)
4

= L.H.S. is proved.

6) Prove that : Cosec 10 – 3Sec10° = 4
Solution :
L.H.S. = Cosec 10 – 3Sec10°

= Cosec 10 – Cot30°.Sec10°

= 1 – Cos30° . 1
Sin10° Sin30° Cos10°

=

=

= 4Sin20°
Sin2 (10°)

= 4Sin20°
Sin20°

=4
= R.H.S. is proved.

7) Cos2A + Sin2A.Cos2B = Cos2B + Sin2B.Cos2A
Solution :
L.H.S = Cos2A + Sin2A.Cos2B

= 1 [2Cos2A + 2Sin2A.Cos2B]
2

= 1 [1 + Cos2A + (1 – Cos2A)Cos2B]
2

= 1 [1 + Cos2A + Cos2B – Cos2A.Cos2B]
2

= 1 [1 + Cos2B + Cos2A(1 – Cos2B)]
2

= 1 [2Cos2B + Cos2A × 2Sin2B]
2

= Cos2B + Sin2B.Cos2A
= R.H.S. is proved

PRIME Opt. Maths Book - X 195