Vedanta Optional Math Teacher's guide

its slope is, m1=– Coefficient of x
Coefficient of y

=– 1 =1
–1
y2-y1
slope of line joining points (4,6) and (10,12), m2= x2-x1
=1
= 12–6 = 6
10–4 6

Here, m1 =m2, hence the line (i) and the line joining points (4,6) and (10,12) are parallel.

7, Show that lines joining the points (7, –5) and (3,4) is perpendicular to the line 4x–2y +7=0.

Solution
Slope of line joining points (7,–5) and (3,4) is given by, m= y2-y1

= 4+5 = 9 =– 9 x2-x1
3-7 –4 4
Slope of the line 4x –9y+7=7 is m2 =– 4 = 4
now, m1. m2 -9 9
= 9 . 4 =–1
-4 9
Since the product of slopes is –1, the lines are perpendicular to each other.

8. Find the value of k is that the lines represented by kx +3y+5 =0 and 4x=3y+10=0 are
perpendicular to each other.

Solution
Here given equations of lines are

kx +3y+5 =0 ...........(i)

and 4x=3y+10=0.............(ii)

slope of line (i), m1=– Coefficient of x
Coefficient of y
k
=– 3

Slope of line, m2=– 4 = 4
-3 3
As the two lines are perpendicular to each other

oomrr,,1–.km=k32=9.–413 =–1
4

9. Find the equation of the straight line that passes through the point (4,5) and perpendicular
6
to the line having slope 5 .

Solution
Let the required line be

y – y1=m(x – x1)
It passes through the point (4,5)

y – 5 = m(x – 4).........(i)

Vedanta Optional Mathematics Teacher's Guide ~ 10 201

sAlsoptheeoltingievseanrelinpeeripse(nmd1i)c=ula65r to each other

m.m1 = –1 =–1
or, m. 6
5

m= – 5
6

put the value of m is equation (i), we get

y – 5=– 5 (x – 4)
6
or, 6y – 30=–5x +20

5x + 6y =50 which is the required equation.

10. (a) Find the equation of straight line which passes through the point (2,3) and parallel
to the line x – 2y – 2=0 .

Solution

Given equations of straight line is x – 2y – 2=0

its slope is (m1) = – Coefficient of x
Coefficient of y

=– 1 = 1
-2 2
we find the equation of straight line which is parallel to given line. So its slope (m)=m1= 1
2

Now, equation of required line passing through (2,3) and with slope m is given by

y– yy1–=3m=(x–12x1()x–2)
or,

or, 2y – 6 =x–2

or, x – 2y+4=0

x–2y+4=0 is the required equation.

(b) Find the equation if the straight line which passes through the point of intersection of
the straight line 3x+4y=7 and 5x–2y=3 and perpendicular to the line 2x+3y=5.

Solution

For the point of intersection of the straight lines

3x+4y=7 and 5x–2y=3,

from 1st equation, and from 2nd equation

3x+4y=7 5x–2y=3
or, x= 3+2y
or, x = 7-4y
3 5
from both

7-4y = 3+2y
3 5

or, 35–20y=9 + 6y

202 Vedanta Optional Mathematics Teacher's Guide ~ 10

y=1

now, x=3+25×1 = 1
Let (x,y)=(1,1) be (x1,y1) and the slope of required line be m,

m ×slope of equation (2x+3y=5)=–1 (pepr to each other)

or, m× -2 =–1
3
3
m= 2

now, eqn. of st. lines is,

y–y1 =m(x–x1)

or, y – 1= 3 (x –1)
2

or, 2y – 2=3x–3

or, 3x–2y–1 =0

3x–2y–1=0 is the required equation.

(c) Find the equation if the straight line which passes through the point of intersection of
2x – 3y +1=0 and x+2y=3 and parallel to the line 4x+3y=12.

Solution
For the point of intersection of the straight lines

2x – 3y +1=0 and x+2y=3

from 1st equation, and from 2nd equation

2x – 3y =–1 x=3–2y

or, x = 3y-1
2
from both

3y-1 = 3–2y
2
or, 3y–1=6 – 4y

y=1

Also, x=3–2×1=1

Let (x,y)=(1,1) be (x1,y1)
for slope (m), two lines passing through (1,1) and 4x+3y=12 are parallel.

Now, eqn. of straight lines is,

y–y1 =– m1=(x–-34x1()x –1)
or, y

or, 3y – 3=–4x+4

4x+3y=7 is the required equation.

(d) Find the equation of the straight line that divides the line joining the points p(–1,–4) and
Vedanta Optional Mathematics Teacher's Guide ~ 10 203

Q(7,1) in the ratio of 3:2 and perpendicular to it.

Solution

The given two points are P(–1,–4) and Q(7,1)

Slope of PQ (m1)= y2-y1 = 1+4 = 5
x2-x1 7+1 8

Let R(x,y) be the point of PQ which divides at in 3:2 ratio. 3 R2

m1x2 +m2x1 m1y2 +m2y1 P(–1,–4) Q(7,1)
(x,y)= m1 +m2 m1 +m2
,

3.7+2.(-1) 3.1+2.(-4)
, 3+2
= 3+2

19 ,-1 19 ,-1
=5 5

Since the required line is perpendicular to PQ and passes through , its slope (m2) is

given by m1.m2=–1 8
5
or, 5 .m2=–1 or, m2=–
8

Now, required equation is

y–y1=m(x–x1)

y +1=– 8 19
5 x- 5

or, 40x+25y=127
40x+25y=127 is the required equation.

(e) Find the equation of the straight line which passes through the centroid of ∆ABC with
vertices A(4,5), B(–4,–5) and C(1,2) and parallel to 7x +5y=35.

A(4,5)
Solution

For centroid of ∆ABC,

G(x,y)= x1 +x2+x3 , y1+y2 +y2
3 3

4-4+1 5-5+2 G
=3 ,3

= 1 2
,
33
B(–4,–5) C(1,2)
7x+5y=35
Let m be the slope of required line which is
perpendicular to 7x +5y=35 and passes through G

204 Vedanta Optional Mathematics Teacher's Guide ~ 10

12
3,3
m =slope of eqn. (7x +5y=35) [ parallel to each other]

m2=– 7
5
Now, required equation is

y–y1=m(x–x1)

y– 2 =– 7 1
3 5 x- 3

or, 75xy–+513y0==–1377x+ 7
or, 3

21x+15y=17 is the required equation.

11. If the angle between the lines (a2–b2)x–(p+q)y=0 and (p2–q2)x+(a+b)y=0 is 90˚, prove
that (a–b)(p–q)=1.

Solution
Given equations of the lines are

(a2–b2)x–(p+q)y=0 and ...........(i)

and (p2–q2)x+(a+b)y=0.............(ii)

From equation (i),slope(m1) = – Coefficient of x =– (a2-b2) = a2-b2
Coefficient of y –(p+q) p+q

From equation (ii), slope (m2)==– (p2-q2)
a+b
Since the lines (i) and (ii) are perpendicular to each other

m1.m2=–1 p2-q2
a+b
or, a2-b2 . - =–1
p+q

or, (a+b)(a-b)(p-q)(p+q) =1
(p+q)(a+b)

(a–b)(p–q) = 1 proved

12.(a) From the point p(–2,4), perpendicular PQ is drawn to the line AB:7x–4y+15=0. Find
the equation of PQ.

Solution P(–2,4)
Equation of AB is

7x–4y+15=0

Its Slope is (m1)= 7 = 7
(-4) 4

PQ is perpendicular to AB, A QB

Vedanta Optional Mathematics Teacher's Guide ~ 10 205

So slope of PQ (m2)=– 4 ( m1.m2=–1)
7

Now the equaton of line PQ is given by

yy––y14==m–7(4x–(xx1+) 2)

4x+7y=20 is the required equation.

12(b). Find the equation of the altitude PM drawn from the vertex P to QR in ∆PQR
p(2,3),Q(–4,1) and R(2,0).
P(2,3)

Solution

Here PM is perpendicular to QR.

Slope of line QR (m2)= y2–y1 = 0–1
x2–x1 2+4
1 –1
= –6 = 6

Since they PM is perpendicular to QR perpendicular
to Q16R×, mm12×=m+21=–1, where m2 is the slope of PM
+

m2=6 Q(–4,1) M R(2,0)
Equation of PM is

y–y1=m(x–x1)
y – 3=6(x–2)

y – 3=6x–12

or, 6x–y–9=0

6x–y=9 is the required equation.

13. Find the equation of the line perpendicular bisector to the join of the following two
points.
N

(a) P(–3,5) and Q(–6,7)

Solution
Given points are P(–3,5) and Q(–6,7)

Mid point PQ=M x1 +x2 , y1+y2 P(–3,5) M Q(–6,7)
2 2

-3-6 5+7
=M 2 ,2

-9
=M 2 , 6

206 Vedanta Optional Mathematics Teacher's Guide ~ 10

Slope of PQ (m1)= y2-y1
x2-x1

= 7-5 =– 2
-6+3 3
Let MN be the perpendicular bisector of PQ with slope(m2).

othrm,e–n2 =, m23 321.–mm2 =2 =–1–1
Now, equation of MN is

y–y1=m(x–x1), (⸪ m=m2 = 3 )
9 2

or, y – 6= 3 x +2
2

or, 4y – 24=6x +27

or, 6x–4y+51=0

6x–4y +51=0 is the required equation.

(b) P(5,6) and Q(7,10)

Solution

We have to find the equation of perpendicular bisector of S
MN R(6,8) N(7,10)

Mid point PQ=M x1 +x2 , y1+y2
2 2

5+7 6+10 M(5,6)
=M 2 ,2

=(6,8)

Slope of MN is:

m1 = y2–y1
x2–x1

= 10–6
7–5

= 4
2

=2

Let RS be the perpendicular of MN

Since the lines are perpendicular, slope of MN=m2 is given by

m1 × 32m2.m=2 –1
or, – =–1

Vedanta Optional Mathematics Teacher's Guide ~ 10 207

or, 2×m2=–1
–1
m2 = 2

Now, equation of st. line RS is:

yor–,yy1=–m8=(x––x211)(,x–6)
x+2y =22 is the required equation.

(c) P(2,4) and Q(–2,–4)

Solution M
Let N be the mid point of RS

Then the coordinates of S

2–2 4–4
are 2 ,2 =(0,0)
(m1)= y2–y1
Slope of PQ R(2,4) N S(–2,–4)
x2–x1
= –4–4
–2–2
=2

Let MN be the perpendicular bisector of RS, then we get slope of MN is (m2)=– 1 ,
( m1 .m2 = –1 ) 2

Now, equation of RS is

oyr–,yy1=–m0=(x––x121)(,x–w0)here m=m2 =– 1
2

or, 2y=–x

x+2y =0 is the required equation.

14 (a) In rhombus PQRS P(2,4) and R(8,10) are the opposite vertices. Find the equation of
diagonal QS.

Solution
In rhombus PQRS the diagonals bisect each other at right angles. Let M be the mid point of

the diagonals. Then the coordinates of M are 2+8 4+10 =(5,7)
2 ,2 P(2,4)
y2–y1 10–4 S
Slope of PR (m1)= x2–x1 = 8–2 =1

Since PR QS, slope of QS (m2)=– 1 ( m1 .m2 = –1 )
Now, equation of QS is

y–y1=m(x–x1), M
or, y – 7=–1(x–5)

or, y–7 =–x+5

x+y =12 is the required equation. Q R(8,10)

208 Vedanta Optional Mathematics Teacher's Guide ~ 10

14(b) M(5,1) and P(–3,3) are two opposite vertices of square MNPQ. Find the equation of
diagonal QN.

Solution M(5,1) Q
MNPQ is a square with opposite vertices M(5,1) and P(–3,3).
In a square the diagonals bisect each other. Mid point of MP is

5–3 1+3
2 , 2 =(1,2)

y2–y1 (1,2)
x2–x1
Slope of MP (m2)= = 3–1 =– 1
–3–5 4

MP is perpendicular to NQ. Slope of QN (m2)=4 N P(–3,3)

Hence equation of QN passing through (1,2) and wiht slope 4 is given by,

y–y1=m(x–x1),
or, y – 2=4(x–2)

or, y – 2 =4x–8

4x – y =6 is the required equation.

15(a) Determine the equation of straight lines through (1,–4) that make an angle of 45˚ with
the straight line 2x +3y+7=0.

Solution P(1,–4)

Let MN be the given line with equation 2x+3y+7=0 45˚ 45˚
M 2x+3y+7=0 N
Its slope is (m1) = – Coefficient of x
2 Coefficient of y
3
=–

Let PM and PN be two lines passing through P(1,–4) which

make 45° with MN.

Let m2 be the slope of PM or PN
Then required equation are given by

y – y1=m(x–x1)
or, y+4=m(x–1)............(i)

Angle between the lines θ=45°
Now, tanθ =± (m1–m2)

1 +m1m2

2
– 3 –m2
tan45º =± –2
1+ 3 m2

or, 1= ± (2+3m2)
3–2m2

Vedanta Optional Mathematics Teacher's Guide ~ 10 209

or, 3–2m2 =±(2+3m2)
Taking positive sign, we get

3–2m2 = 2+3m2

or, –5m2 = –1
m2 = 1
5
Taking negative sign, we get

3–2m2 = 2+3m2

m2= –5 1
5
When =m2=51 ,then from equation(i),we get
y +4
(x–1)

or, 5y+20 = (x–1)

x – 5y =21

When m2 =–5, then, from equation(ii),we get
y +4 =–5(x–1)

or, 5x +y =1

Hence the required equations are x – 5y =21 and 5x +y =1.

(b) Find the equation of the straight lines passing through the point (3,2) and making angle
of 45˚ with the line x–2y–3=0.

Solution

Let MN be the line with equation x–2y–3=0..........(i)

slope is (m1) = – Coefficient of x
Coefficient of y
1
=– –2 P(3,2)

= 1
2

Let MP and NP be two lines passing through P(3,2) which

make 45° with MN. and the slope m. 45˚ 45˚
M N
Let the slope of MP or NP be m.

Then required equation are given by

y – y1=m(x–x1)
or, y–2=m(x–3)............(i)

Angle between the lines θ=45°
Now, tanθ =± (m1–m2)
1 +m1m2 , where m2 = m

1
m– 2
tan45º =±

1
1+ 2 m

210 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, 1= ± (1–2m)
2+m

or, 2+m=±(1–2m)

Taking positive sign,

2+m=1–2m

or, 3m=-1

m= – 1
3
Taking negative sign, we get,

2+m=–1+2m

m= 3

When m= – 1 , then from equation(i),we get
3

y –2 = – 1 (x–3)
3
or, 3y–6 =–x+3

⸫ x +3y=9

When m=3, then from equation(i),we get

y –2 =3(x–3)

or, 3x–y =7

Hence the required equations are x + 3y =9 and 3x–y =7.

(c) Find the equation of two lines passing through the point (1,–4) and making an angle of
45˚ with the lines 2x–7y+5=0.

Solution P(1,–4)
Let MN be the line with equation 2x–7y+5=0

Its slope is (m1) = – Coefficient of x
Coefficient of y
2
= 7 45˚ 45˚
M 2x–7y+5=0
Let required lines be MP and NP passing through (1,–4) and N
the slope m.

Let the slope of MP or NP be m.

Then required equation is given by

y – y1=m(x–x1)
or, y+4=m(x–1)............(i)

Angle between the lines θ=45°
Now, tanθ =± (m1–m2)

1 +m1m2
2
7 –m
tan45º =±

3
1+ 7 m

Vedanta Optional Mathematics Teacher's Guide ~ 10 211

or, 1= ± (2–7m)
Taking pos7i+tiv2emsign, we get

7+2m=2–7m

m= – 5
9
Taking negative sign, we get,

7+2m=–2+7m

m= 9
5
5
When m= – 9 , from equation(i),we get

y +4 = – 5 (x–1)
9

or, 9y+36 =–5x+5

5x + 9y +31=0

When m= 9 , from equation(i),we get
5
9
y +4 = 5 (x–1)

or, 9x–5y =29
Hence the required equations are 5x + 9y +31 and 9x–5y =29.

(d) Find the eqn. of straight line passing through the point (3,–2) and making an angle of 60˚
with the line 3 x+y–1=0.
P(3,–2)
Solution

Let the given line be QR with equation 3 x+y–1=0.

Its slope is (m1) = – Coefficient of x
Coefficient of y
=– 3

Let QP and RP be two lines passing through P(3,–2) making
angle 60° with QR.
60˚ 60˚
Let the slope of QP or RP be m.

Then the required equation is Q 3x+y–1=0 R

y – y1=m(x–x1)
or, y+2=m(x–3)............(i)

Now, tanθ =± m1–m2
1 +m1m2

tan60º =± (– 5 –m)
1 +(– 3 )m

or, 3 =± ( 3 +m)
1– 3m

212 Vedanta Optional Mathematics Teacher's Guide ~ 10

Taking negative sign, we get
3 –3m=– 3 –m

or, 2m=-2 3
m= 3

Taking positive sign, we get,
3 –3m= 3 +m

∴ m=0
When m= 3 , then from equation(i),
y +2 = 3 (x–3)
or, y+2 = 3 x–3 3
or, 3 x-y=3 3 +2
When m = 0, then from equation (i), y + 2 = 0
Hence the required equations are y+2=0 and 3 x-y=3 3 +2.

(e) Determine the value of m so that 3x–my–8=0 will make an angle of 45° with the line
3x+5y–17=0.

Solution P
:Let MN and MP be the given lines with angle 45° between them.
45˚
Equation MN : 3x–5y–8=0. 3x+5y–17=0 3x–my–8=0

slope of MN (m1) =– 3
–m
3
= m

Equation MN : 3x+5y–17=0

slope of MP(m2)= – 3 M N
5
33
m+5
Now, tanθ =± 3 3
1+ m – 5

33
m+5
or, tan45°=± 3
3
15 +13+mm. – 5
or, 1 =±
5m–9

or, 5m–9=±(15+3m)

Taking positive sign, we get
5m–9=15+3m
or, 2m=24

m = 12

Vedanta Optional Mathematics Teacher's Guide ~ 10 213

Taking negative sign, we get,

5m–9=–15–3m

or, 8m=–6

⸫ m=12 or – 3 .
4

16(a). Find the equation of the sides of an equilateral triangle whose vertex is (1,2) and base
is y =0.
y

Solution A(1,2)

Let ABC be an equilateral triangle with vertex
A(1,2) with base y=0
there . ABC= BCA= CAB=60° 60˚

slope of BA=tan60°= 3

slope at CA=tan120° =– 3

Now, for equation of BA 60˚ 60˚ 120˚
y=0 C
let, (x1,y1)=(1,2) x' B 0 x
(y–2)= 3 (x–1)

or, 3 x –y+2– 3 =0

⸫ 3 x–y+2– 3 =0

Again, for equation of CA

(y – 2)=– 3 (x – 1)

or, y –2 =– 3 x– 3

⸫ 3 x + y+ 3 –2=0

Hence the required equations are 3 x–y+2– 3 =0 and 3 x + y+ 3 –2=0.

(b). Find the equation of the sides of right y

angled isoceles triangle whose vertex is
at (–2,–3) and base is x =0.

Solution: Let P(–2,–3) be the vertex of right R

angled isoceles triangle PQR with P=90° 135˚

For side PR, its slope (m1)=tan45° x' 45˚ x
Equation of PR is O

y–y1=m(x–x1) P(–2,–3) x=0
or, y +3=1(x+2)

x – y=1

Again, for equation of side QP, slope Q
(m)=tan135°=–1 y'

Equation of QP is y–y1=m(x–x1)
or, y +3=–1(x+2)

or, x+y+5=0

⸫ Hence the required equation are x – y=1 and x+y+5=0.

214 Vedanta Optional Mathematics Teacher's Guide ~ 10

(c). Find the equation of the sides of right angled isoceles triangle whose vertex is at (2,4)
and equation of base x =0.

Solution
Let ∆MNP be an isoceles right angled triangle with p=90°

For side NP, slope (m1)=tan45°=1 y
Now, equation of NP is M
x=0
y–y1=m(x–x1), P(2,4)
or, y – 4=1(x–2) N x

x–y+2=0

For side MP, slope (m2)=tan135° = –1 x'
Equation of MP is y–y1=m(x–x1)
or, y – 4=–1(x–2)

x+y=6 y

Hence the required equation are x +y+2=0 and x+y=6.

Questions for practice

1. The line px+3y+5=0 is perpendicular to the line joining the points (4,3) and (6,–3),
find the values of P.

2. Find the equation of a straight line passing through the point of intersection of the
straight lines x–y=7 and x+y=15 and parallel to the line 4x+3y=10.

3. Find the equation of the line segment PQ which passes through the point (3,4) and
the mid point of line segment joining M(–4,–5) and N(7,8).

4. If the line x + y = 1 passes through the point of intersection of the lines x+y=3
a b

and 2x–3y=1 and is parallel to the line x–y=6, then find the values of a and b.

5. Find the equation of perpendicular bisector of line segment joining M(–4,–5) and
N(8,9).

Pair of straight lines

Estimated Teaching periods : 8

1. Objectives

S. N. Level Objectives

1. Knowledge(K) To define equation of pair of lines.
To define homogeneous equation.
To tell general equation of second degree.
To tell formula to find angle between two lines represented by
ax2+2hxy+by2=0
To state conditions for coincidence and perpendicularity.

Vedanta Optional Mathematics Teacher's Guide ~ 10 215

To find a single equation of a pair of lines.

2. Understanding(U) To find pair of lines of given homogeneous equation in x and y.
To identify given homogeneous equation represents a pair of
perpendicular or coincident lines.

3. Application(A) To find angle between two lines represented by general
equation of second degree.
To find separate equation of two lines represented by general
eqn. of second degree.

4. Higher Ability To derive formula to find angle between two lines represented
(HA) by ax2+2hxy+by2=0 and find condition for coincidence and
perpendicularity
To show ax2+2hxy+by2=0 represents a pair of lines through origin.

2. Required teaching materials :
Chart paper with figure of angle between two lines represented by ax2+2hxy+by2=0.

3. Teaching Learning Activities:
– Take two equations like x+2y=0 and x–2y=0 and multiply them to get x2–4y2=0,
discuss the conclusion.
– Find two separate equations represented by x2+8xy+12y2=0, discuss the conclusion.
– Define general equation of second degree in x and y .
– Define a homogeneous equation with examples.
– Show that ax2+2hxy+by2=0. represents a pair of lines through the origin.
– Discuss how to find the angle between the lines represented by ax2+2hxy+by2=0. ?
– Discuss the conditions for perpendicularity and coincidence of the lines represented
by ax2+2hxy+by2=0.
– Do some problems from exercise to guide the students.

Note:
i) The equation ax2+2hxy+by2=0 always represents a pair of lines through the origin
in the form of y=m1x and y=m2x.

2. The angle between the lines represented by ax2+2hxy+by2=0 is

=tan-1 ±2 h2–ab
a+b


3. The ax2+2hxy+by2=0 represents a pair of lines, then

i) Condition for coincidence of the lines h2=ab

ii) Condition for perpendicularity/orthogonality a+b=0

4. The roots of quadratic equation ax2+bx+c=0 are

–b± b2–4ac
x= 2a

216 Vedanta Optional Mathematics Teacher's Guide ~ 10

Some solved problems

1. Find the single equation represented by the pair of straight lines.
3x+4y=0 and 2x–3y=0

Solution
Given equations of lines are
3x+4y=0 and 2x–3y=0
the single equation represented by above equations is
(3x+4y)(2x–3y)=0
or, 6x2–9xy+8xy–12y2=0
or, 6x2–xy–12y2=0

6x2–xy–12y2=0 is the required equation.

2. Find the separate equations of of straight lines represented by the following equations.
x2+ y2 –2xy+2x–2y =0

Solution
Here,x2+ y2 –2xy+2x–2y =0
or, (x2–2xy+y2 )+2(x–y) =0
or, (x–y)2 +2(x–y) =0
or, (x–y)(x–y+2) =0
Hence x–y =0 and x–y+2 =0 are the required equations.

3. Show that 6x2–5xy–6y2=0 represents a pair of lines.

Solution
Here, 6x2–5xy–6y2=0
or, 6x2–9xy+4xy–6y2=0
or, 3x(2x–3y)+2y(2x–3y)=0
or, (2x–3y)(3x–2y)=0
Either 2x–3y=0 or 3x–2y=0 ,
each of which are equation of straight lines.

4. Determine the lines represented by each of the given equations.
(a) x2+2xy+y2–2x–2y–15=0

Solution
Here, x2+2xy+y2–2x–2y–15=0
or, (x+y)2–2(x+y)–15=0
or, (x2+y)2–5(x+y)+3(x+y)–15=0
or, (x+y) (x+y–5)+3(x+y–5)=0
or, (x+y–5) (x+y+3)=0
Either, or, (x+y–5)=0 or (x+y+3)=0

Vedanta Optional Mathematics Teacher's Guide ~ 10 217

Each of which represents a straight line
Hence the given equation represents a pair of lines.

(b) 2x2+xy–3y2+2y–8=0

Solution
Here, 2x2+xy–3y2+2y–8=0

or, 2x2+xy+(–3y2+10y–8)=0

which is in the form of ax2+bx+c=0,

where, a=2, b=y and c=-3y2+10y–8=0

By using formula,
–b± b2–4ac

x= 2a

–y± y2–4.2.(-3y2+10y–8)
= 2.2

–y± y2+24y2–80y+64
= 2.2

–y± (5y–8)2
=4
Taking positive sign, we get
4x=–y+5y–8
or, 4x–4y+8=0
or, x–y+2=0
Taking negative sign, we get
4x=–y–5y+8
or, 4x+6y–8=0
or, 2x+3y–4=0
Hence the required equation are x–y+2=0 and 2x+3y =4

(c) x2+9y2+6xy+4x+12y–5=0

Solution
Here, x2+6xy + 9y2+4x+12y–5=0
or, x2+2.x.3y + (3y)2+4(x+3y)–5=0
or, (x+3y)2+4(x+3y)–5=0
or, (x+3y)2+5(x+3y)–(x+3y)–5=0
or, (x+3y)(x+3y+5)–1(x+3y+5)=0
or, (x+3y+5)(x+3y–1)=0
Hence the required equation are x+3y+5=0 and (x+3y–1)=0

218 Vedanta Optional Mathematics Teacher's Guide ~ 10

5. Prove that two lines represented by the following equations are perpendicular to each other.
9x2–13xy–9y2+2x–3y+7=0

Solution
Here, 9x2–13xy–9y2+2x–3y+7=0
Comparing at with ax2–2hxy+by2+2gx+2fy+c=0
we get, a=9, h=– 13 , b =-9, g=1, f= – 3 , c=7

22
condition for perpendicularity
a+b=0
Here, a+b=9–9=0

a+b=0
Hence the lines represented by given equation are perpendicular to each other.

6. Show that the given equations represents a pair of coincident lines.
9x2–6xy+y2 = 0

Solution
Here, 9x2–6xy+y2 = 0

Comparing it with ax2–2hxy+by2=0 we get,

a=9, b =1, g=1, h=–3

condition for coincident,

h2=ab

ie. (–3)2 =9.1

9=9(true) proved.

7. Find the value of λ when the lines represented by each of the following equations are
coincident.

(10λ–1)x2+(5λ+3)xy+(λ–1)y2=0

Solution: Comparing given equation

(10λ–1)x2+(5λ+3)xy+(λ–1)y2=0 with

ax2+2hxy+by2+2gx+2fy+c=0 , we get,

a= 10λ–1, h= 5λ+3 ,b= λ–1
2
Now, condition for coincident

h2=ab

or, 5λ+3 2
2
= (10λ–1). (λ–1)

or, 25λ2+30λ+9 = 10λ2–10λ–λ+1
4

or, 25λ2+30λ+9= 40λ2–44λ +4
or, 15λ2–74λ–5=0
or, 15λ2–75λ+λ–5=0
or, 15λ(λ–5) +1(λ–5)=0

Vedanta Optional Mathematics Teacher's Guide ~ 10 219

or, (λ–5)(15λ+1)=0

Either λ–5=0 or λ=5 1
15
or, 15λ+1=0 or, λ= –
1
λ=5,– 15

8. Find the value of λ when the give equation represents a pair of lines perpendicular to
each other.

(3λ+4)x2–48xy–λ2y2=0

Solution
Comparing the given equation with
ax2+2hxy+by2=0, we get
a= 3λ+4, h=–24 ,b=–λ2
Condition for perpendicularity, a+b=0

or, 3λ+4–λ2= 0
or,λ2–3λ–4= 0
or,λ2– 4λ+λ – 4=0
or, λ (λ– 4)+1(λ – 4)=0
or, (λ– 4)(λ +1)=0
Either , λ– 4=0 or λ=4
or, λ+1=0 or λ=–1

λ=4,–1

9. Find the angle between the following pair of lines.
(a) x2+9xy+14y2=0

Solution
Here, x2+9xy+14y2=0

comparing it with ax2+2hxy+by2=0, we get

a=1, h= 9 , b=14
2
Let θ be the angle between them.
±2 h2–ab
tanθ=
a+b

±2 81 –14
4
= 1+14

81–56
= ±2 2×15

=± 5
15

220 Vedanta Optional Mathematics Teacher's Guide ~ 10

=± 1
takin3g
positive sign, tanθ= 1
3
θ=71.57°
1
taking negative sign, tanθ= – 3
θ=108.43°

(b) 2x2+7xy+3y2=0

Solution
Comparing it with ax2+2hxy+by2=0, we get

a=2, h= 7 , b=3
2
Let θ be the angle between the pair of lines, we get,
±2 h2–ab
tanθ=
a+b

±2 49 –2.3
4
= 2+3
5
= ±2 2.5

=±1

taking positive sign, we get,

tanθ=1
or, tanθ=tan45°

θ=45°

taking negative sign, we get
tanθ= –1

or, tanθ=tan135°
θ=135°

Hence, θ=45° ,135°.

(c) x2–7xy+y2=0

Solution

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

a=1, h=– 7 , b=1
2
Let θ be the angle between the pair of lines, we get,
±2 h2–ab
tanθ=
a+b

±2 49 –1
4
= 1+1

Vedanta Optional Mathematics Teacher's Guide ~ 10 221

= ± 3 5
2
taking positive sign, we get,

tanθ= 3 5
2
θ=73.4°

taking negative sign, we get
35

tanθ= – 2
or, tanθ=tan135°

θ=106.6°
Hence, θ=73.4° ,106.6°.

(d) 9x2–13xy–9y2+2x–3y +7=0

Solution

comparing it with ax2+2hxy+by2+2gx+2fy+c=0, we get

a=9, h=– 13 , b=–9, g=1, f=–– 3 , c=7
2 2
Let θ be the angle between the pair of lines, we get,
±2 h2–ab
tanθ=
a+b
169
±2 4 +81

= 9–9
=∞

=tan90°

θ=90°

(e) x2+6xy+9y2+4x+12y –5=0

Solution
comparing it with ax2+2hxy+by2+2gx+2fy+c=0, we get

a=1, h=3, b=9, g=2, f= 6, c=–5

Let θ be the angle between the pair of lines,
±2 h2–ab
tanθ=
a+b

±2 9–9
= 1+9
=0
or, tanθ=tan0°

θ=0°

10. Find the angle between the following pair of lines.
(a) x2–2xycot –y2=0

222 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution
comparing given equation with ax2+2hxy+by2=0, we get

a=1, h=–cot , b=–1

Let θ be the angle between the pair of lines,

tanθ= ±2 h2–ab

a+b

±2 cot2 +1
= 1–1

=∞

or, tanθ=tan90°
θ=90°

(b) x2+2xycosec +y2=0

Solution
Comparing given equation with ax2+2hxy+by2=0, we get

a=1, h=cosec , b=1

Let θ be the angle between the pair of lines,

tanθ= ±2 h2–ab

a+b

±2 cosec2 –1
= 1+1

=±cot
or, tanθ=tan(90°± )

θ=90°±

(c) y2+2xycot –x2=0

Solution
The given equation can be written as

y2+2xycot –x2=0

Comparing given equation with ax2+2hxy+by2=0, we get

a=1, h=–cot , b=–1

Let θ be the angle between the pair of lines,
±2 h2–ab
tanθ=
a+b

±2 cot2 +1
= 1–1

=∞
θ=90°

Vedanta Optional Mathematics Teacher's Guide ~ 10 223

11. Find the separate equation of two lines represented by the following equations.
(a) x2+2xycot –y2=0

Solution
Here, x2+2xycot –y2=0
or, x2 – 2xycot – y2(cosec2 –cot2 )=0
or, x2 + 2xycot + y2cot2 – y2cosec2 =0
or, (x+ycot )2 – (ycosec )2=0
or, (x+ycot + ycosec )(x+ycot – ycosec )=0
Hence the required separate equations are
x + y(cot + cosec )=0
and x – y(cot – cosec )=0

(b) x2 + 2xysec + y2=0

Solution
Here, x2 + 2xysec + y2=0
or, x2 + 2xysec + y2(sec2 –tan2 )=0
or, x2 + 2xysec + y2sec2 – y2tan2 =0
or, (x+ysec )2 – (ytan )2=0
or, (x+ysec + ytan )(x+ysec – ytan )=0
Hence the required separate equations are
x + y(sec + tan )=0
and x + y(sec – tan )=0

(c) x2+2xycosec + y2=0

Solution
Here, x2+2xycosec + y2=0
or, x2 + 2xycosec + y2(cosec2 –cot2 )=0
or, (x2 + 2xycosec + y2cosec ) – y2cot2 =0
or, (x+ycosec )2 – (ycot )2=0
or, (x+ycosec + ycot )(x+ycosec – ycot )=0
Hence the required separate equations are
x+y(cosec + cot ) = 0
and x+y(cosec – cot )=0

(d) x2 – 2xycot2 –y2=0

Solution
Here, x2 – 2xycot2 –y2=0
or, x2 – 2xycot2 – y2(cosec22 –cot22 )=0
or, x2 – 2xycot2 + y2cot22 – y2cosec22 =0
or, (x – ycot2 )2 – (ycosec2 )2=0

224 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, (x – ycot2 + ycosec2 )(x – ycot2 – ycosec2 )=0
or, {x – y(cot2 – cosec2 )}{x – y(cot2 + ycosec2 )}=0
Hence the required separate equations are
x – y(cot2 – cosec2 )=0
and x – y(cot2 + cosec2 )=0

14(a). Find the equation of two lines represented 2x2 + 7xy + 3y2=0. Find the point of
intersection . Also find the angle between them.

Solution
Here, 2x2 + 7xy + 3y2=0

or, 2x2 + 6xy + xy + 3y2=0

or, 2x(x + 3y) + y(x+3y) =0

or, (x + 3y)(2x + y)=0

Hence the equation of two lines are

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

2x + y = 0...............(ii)

solving equations (i) and (ii) , we get x=0 and y = 0

(0,0) is the point of intersection of lines (i) and (ii)

To find the angle between the lines.

comparing the given equation 2x2 + 7xy + 3y2=0

we get, a= 2, h = 7 and b =3
2
Let θ be the angle between the lines

tanθ = ±2 h2–ab

a+b

±2 49 –2.3
4
= 1+1
5
2. 2

=± 5

= ±1
taking positive sign, we get,
tanθ= 1
= tan45°

θ=45°

taking negative sign, we get
ttanθ= –1
= tan135°

θ=45°
θ= 45° or 135°

Vedanta Optional Mathematics Teacher's Guide ~ 10 225

14(b). Find the separate equation of two lines represented x2 – 5xy + 4y2=0. Also find the
angle between them and their point of intersection.

Solution
Here, x2 – 5xy + 4y2=0

or, x2 – 5xy + 4y2=0

or, x(x – 4y) – y(x – 4y) =0

or, (x – 4y)(x – y)=0

Hence the equation of two lines are

x – 4y = 0.................(i)

x – y = 0...............(ii)

solving equations (i) and (ii) , we get

(x, y) = (0,0)

comparing the given equation x2 – 5xy + 4y2=0, we get,

a= 1, h =– 5 and b = 4
2
Let θ be the angle between the lines

tanθ = ±2 h2–ab

a+b

±2 25 –1.4
4
= 1+4
5
=± 2 3
5
θ=tan–1 ± ,

θ=31°,149°

(c) Find the separate equations of two lines represented by the equation x2– 2xycosec +
y2=0 . Also find the angle between them.

Solution
Here, x2– 2xycosec + y2=0
or, x2 – 2xycosec + y2(cosec2 –cot2 )=0
or, (x2 – 2xycosec + y2cosec ) – y2cot2 =0
or, (x – ycosec )2 – (ycot )2=0
or, (x – ycosec + ycot )(x – ycosec – ycot )=0
Either
x – y(cosec – cot ) =0...............(i)
x – y(cosec + ycot )=0...............(ii)
Which are the required equation of straight lines

226 Vedanta Optional Mathematics Teacher's Guide ~ 10

Comparing the given equation with ax2 +2hxy + by2=0, we get,

a= 1, h =–cosec , b = 1
Let θ be the angle between the lines

tanθ = ±2 h2–ab

a+b
±2 cosec2 – 1
= 1+1
= ±cot

= ±tan(90˚ ± )

= tan(90˚ ± )

θ = 90˚ ±

15(a). Find the pair of lines parallel to the lines x2 – 3xy + 2y2 = 0 and passing through
the origin.

Solution
Here, x2 – 3xy + 2y2 = 0
or, x2 – 2xy – xy + 2y2 = 0
or, x(x – 2y) –y(x – 2y) =0
or, (x – 2y) (x – y) = 0
x – 2y = 0..........(i)
and x – y = 0 ..........(ii) are the required equations.
Equations of the lines parallel to (i) and (ii) and passing through the origin are,
x – 2y = 0 and x – y = 0

(b). Find the pair of lines parallel to the 2x2 – 7xy + 5y2 = 0 and passing through the point (1,2).

Solution
Here, 2x2 – 7xy + 5y2

or, x2 – 5xy – 2xy + 5y2 = 0

or, x(2x – 5y) –y(2x – 5y) =0

or, (2x – 5y) (x – y) = 0

2x – 5y = 0..........(i)

and x – y = 0 ..........(ii)

Equations of the lines parallel to (i) and (ii)

are, 2x – 5y + k1 = 0 ...........(iii)
and x – y + k2 = 0 ................(iv)
The lines (i) and (ii) passes through the point (1,2), we get,

2 × 1 – 5 × 2 + k1 = 0 ⇒ k1 =8
and 1 – 2 + k2 = 0 ⇒ k2 =1
put the values of k1 and k2 in eqn. (iii) and (iv), we get,
2x – y + 8 = 0 and x – y + 1 = 0.

Vedanta Optional Mathematics Teacher's Guide ~ 10 227

16(a). Find the pair of lines perpendicular to the lines x2 – 5xy + 4y2 = 0 and passing
through the origin.

Solution
Here, x2 – 5xy + 4y2 = 0
or, x2 – 4xy – xy + 4y2 = 0
or, x(x – 4y) –y(x – 4y) =0
or, (x – 4y) (x – y) = 0
Either, x – 4y = 0..........(i)
or, x – y = 0 ..........(ii)
From equations (i), equation of the lines perpendicular to it and passing through the origin
are,
4x + y = 0
Also, from equation (ii), equation of line perpendicular to it is
x+y=0

Required pair of lines are
4x + y = 0 and x + y = 0.

(b). Find the equation of two straight lines which pass through the point (2,3) and
perpendicular to the lines x2 – 6xy + 8y2 = 0.

Solution
Here, x2 – 6xy + 8y2 = 0
or, x2 – 4xy – 2xy + 8y2 = 0
or, x(x – 4y) – 2y(x – 4y) =0
or, (x – 4y) (x – 2y) = 0
Either x – 4y = 0..........(i)
or, x – 2y = 0 ..........(ii)
Equations of the lines perpendicular to above lines are,
4x + y + k1 = 0 ...........(iii)
and 2x – y + k2 = 0 ................(iv)
Both of above lines passes through the point (2,3), we get,

4 × 2 + 3 + k1 = 0 →k1 = –11
and 2 + 2×3 + k2 = 0 →k2 = –7
put the values of k1 and k2 in eqn. (iii) and (iv), we get,
4x + y –11 = 0 and 2x+y – 7 = 0 which are the required equation of the lines.

17 (a). Find the two separate equations when the lines represented by kx2 + 8xy – 3y2 = 0
are perpendicular to each other .

Solution
Here, kx2 + 8xy – 3y2 = 0 ...................(i)
Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = k, b = –3, h = 4

228 Vedanta Optional Mathematics Teacher's Guide ~ 10

Conditions for perpendicularity is
a+b=0
or, k – 3 = 0 or, k ⇒ 3
put the value of k in eqn.(i), we get
3x2 + 8xy – 3y2 = 0
or, 3x2 + 9xy – xy – 3y2 = 0
or, 3x(x + 3y) – y(x + 3y) =0
or, (x + 3y) (3x – y) = 0

x + 3y = 0 and 3x – y = 0 are the required equation of the lines.

(b). Find the two separate equations when the lines represented by 6x2 + 5xy – ky2 = 0 are
perpendicular.

Solution

Here, 6x2 + 5xy – ky2 = 0 ...................(i)

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

a = 6, b = –k, h = 5
2
conditions for perpendicularity is

a+b=0

or, 6 – k = 0
k=6

put the value of k in eqn.(i), we get
6x2 + 5xy – 6y2 = 0
or, 6x2 + 9xy – 4xy – 6y2 = 0
or, 3x(2x + 3y) – 2y(2x + 3y) =0

or, (2x + 3y) (2x – 2y) = 0
2x + 3y = 0 and 2x – 2y = 0 are the required equation of the lines.

18(a). Show that the pair of lines 3x2 – 2xy – y2 = 0 are parallel to the lines 3x2 – 2xy – y2
–5x + y + 2 =0

Solution
Here, 3x2 – 2xy – y2 = 0
or, 3x2 – 3xy + xy – y2 = 0
or, 3x(x – y) + y(x – y) =0
or, (x – y) (3x + y) = 0
Either, x – y = 0 ...........(i)
3x + y = 0 ..........(ii)
Also, 3x2 – 2xy – y2 –5x + y + 2 =0
3x2 – (2y + 5)x + (y – y2 + 2) =0
which is in the form of
ax2 + bx + c = 0

Vedanta Optional Mathematics Teacher's Guide ~ 10 229

where a = 3, b = – (2y + 5), c = y – y2 + 2
Comparing it with ax2 + 2hxy + by2 = 0, we get,

–b± b2–4ac
x= 2a

(2y + 5)± (2y + 5)2 – 4 . 3.(y – y2 + 2)
= 2.3

(2y + 5)± 4y2 +20y +25 –12y +12y2 – 12)
= 6

2y± 16y2 +8y + 1
=6

2y± (4y + 1)2
=6
= 2y ± (4y + 1)

Taking positive sign, we get

6x = 2y+4y + 1

or, 6x – 6y – 1=0 ........... (iii)

or, x–y+2=0

Taking negative sign, we get

6x = 2y – 4y – 1

or, 6x + 2y + 1=0 ........... (iv)

From eqn.(i), slope (m1) = 1

From equation (iii), slope (m3) = 1 m1 = m3

The lines (i) and (iii) are parallel to each other.

Again, from equation (ii), slope(m2 ) = –3
from equation (iv), slope (m4) = –
6 =–3
2
m2 = m4

The lines (ii) and (iv) are parallel to each other .

proved

18(b). Show that the pair of lines 4x2 – 9y2 = 0 and 9x2 – 4y2 = 0 are perpendicular to each
other.

Solution
Here, 4x2 – 9y2 = 0
or, (2x – 3y) (2x + 3y) = 0
Either, 2x – 3y = 0 ...........(i)
or, 2x + 3y = 0 ..........(ii)
Again ,9x2 – 4y2 = 0

230 Vedanta Optional Mathematics Teacher's Guide ~ 10

(3x + 2y) (3x – 2y) = 0

Either, 3x + 2y = 0 ...........(iii)

or, 3x – 2y = 0 ..........(iv)

From equation (i), slope (m1) = 2 3
3 2

From equation (iii), slope (m3) = –

2 – 3 = –1
m1 . m3 = 3 2

The lines (i) and (iii) are perpendicular to each other.

Again, from equation (ii), slope(m2 ) =– 2
3

from equation (iv), slope (m4) = 3
2

Now, m2 . m4 = – 2 3 = –1
3 2

The lines (ii) and (iv) are perpendicular to each other . proved

Questions for practice

1. Find the single equation representing the pair of lines y = x and y = –x.

2. Find the two separate equations represented by

i) 2x2 + 7xy + 3y2 = 0 ii) x2 + 2xysecθ + y2 = 0
3. Determine the two straight lines represented by 6x2 – xy – 12y2 –8x + 29y –14 =0

4. Find the equation to the straight lines through the origin and at right angles to the
lines x2 – 5xy + 4y2 = 0

5. Find the value of k when the pair of lines represented by (k + 2)x2 +8xy + 4y2 = 0
are coincident.

6. Show that the angle between a pair of lines represented by 2x2 – 7xy + 3y2 + 2x –
6y=0 is 60°.

7. Find the two separate equations of straight lines passing through the point (1,0) and
parallel to the lines represented by the equation

x2 + 3xy + 2y2 = 0

8. Find the equation of two lines which pass through the point (3, –1) and perpendicular
to the pair x2 – xy – 2y2 = 0.

9. Find the value of p so that the two lines represented by the equation (p + 1)x2 – 12xy
+ 9y2 = 0 are coincident.

10. Prove that the lines represented by x2 – 7xy + 12y2 = 0 are perpendicular to the lines
represented by 12x2 + 7xy + y2 = 0

Vedanta Optional Mathematics Teacher's Guide ~ 10 231

Conic Section and Circle

Estimated Teaching periods : 10 Hours

1. Teaching Objectives

S,N. Level Objectives

1. Knowledge(K) To define terms vertex axis, generator of a cone.
To define ellipse, parabola, hyperbola.

Understanding To identify types of conic sections from given figures.
(U)
2. – To derive equation of circle x2 +y2 = a2 and
(x –h)2 + (y – k)2 = r2 by using distance formula.

To use equation of circles (x – h)2 + (y – k)2 = r2
3. Application (A) (x – x1)(x – x2) + (y – y1)(y – y2) = 0

to find equation of circles.

4. Higher Ability To solve verbal problems related to circle
(HA) – eqn of circle passing through three or more given points.
To show given four points concyclic.

2. Required teaching materials
– diagrams of conic sections.

– graph papers.

3. Teaching strategies
– Discuss different types of conic sections
– Circle, parabola, ellipse, hyperbola by using plane figures of intersection of a plane
and a cone.
– Discuss to derive the following equations of circles
x2 + y2 = a2
– (x – h)2 +(y – k)2 = r2 , (x – h)2 +(y – h)2 = h2 , (x – k)2 +(y – k)2 = k2
– (x – x1)(x – x2) + (y – y1)(y – y2) = 0
– x2 + y2 +2gx + 2fy + c = 0

To each of formula, illustrated examples are to be given.

Notes :
Equations of circles in different forms.

i) Equation of circle with centre at the origin 0(0,0) and radius r : x2 + y2 = a2
ii) Equation of circle with centre at (h, k) and radius r; (x – h)2 + (y – k)2 = r2
iii) Equation of circles touching x – axis ie. r = k,
(x – k)2 +(y – k)2 = r2

iv) Equation of a circle touching y–axis, ie. r = h:

(x – h)2 +(y – k)2 = r2

v) Equation of a circle touching both axis : h = k = r

(x – h)2 +(y – h)2 = h2 ,

232 Vedanta Optional Mathematics Teacher's Guide ~ 10

vi) Equation of a circle in a diameter form
(x – x1)(x – x2) + (y – y1)(y – y2) = 0
vii) General Equation of circle : x2 + y2 +2gx + 2fy + c = 0
where center (h,k) =(–g, –f)
radius (r) = g2 + f2 – c
If circle x2 + y2 +2gx + 2fy + c = 0 passes through origin, then c = 0, equation of circle is
x2 + y2 +2gx + 2fy = 0.
viii) The point of intersection of two diameters is the center of the circle.

Some solved problems

1. Write radius and centre of circle x2 + y2 + 2gx + 2fy + c = 0

Solution
Equation of circle is x2 + y2 + 2gx + 2fy + c = 0
Radius (r) = g2 + f2 – c
centre (h,k) =(–g, –f)

2. Find the equation of circle with centre (–2, –3) and radius 6 units.

Solution
Here, centre (h,k)=(–2, –3)
radius (r) = 6 units
Equations of circle is (x – h)2 +(y – k)2 = r2
ie. (x + 2)2 +(y + 3)2 = 62
or, x2 + 2.x.2 + 22 +y2 +2.y.3 +32 =36
or, x2 + y2 + 4x +6y =36

3. Find the equation of circle whose end of diameters are (4,5) and (–2, –3).

Solution
Here, centre ( x1, x2) = (4,5) ,(y1 , y2) = (–2, –3)
Equations of the required circle is
(x – x1)(x – x2) + (y – y1)(y – y2) = 0
ie. (x – 4)(x + 2) + (y – 5)(y + 3) = 0
or, x2 – 2x – 8 + y2 – 2y – 15 = 0
ie. x2 + y2 – 2x –2y =23

x2 + y2 – 2x –2y = 23 is the required equation.

4. Find the centre and radius of the circles.

(a) x2 + y2 + 6x + 4y –12 = 0

Vedanta Optional Mathematics Teacher's Guide ~ 10 233

Solution
Here, x2 + y2 + 6x + 4y –12 = 0
comparing it with x2 + y2 + 2gx + 2fy + c = 0
g = 3, f = 2, c = –12
centre (–g, –f) = (–3, –2)
Radius (r) = g2 + f2 – c

= 9 + 4 – 12

=5
= 5 units.

(b) 9x2 + 9y2 – 36x + 6y = 107

Solution
Here, 9x2 + 9y2 – 36x + 6y = 107

dividing both sides by 9, we get,

x2 + y2 – 4x + 2y = 107
3 9
comparing it with x2 + y2 + 2gx + 2fy + c = 0

we get, g = –2, f = 1 ,c=– 107
3 9

centre (–g, –f) = 2,– 1
3

Radius (r) = g2 + f2 – c

= 22 + – 1 2 107
3 9
+

= 4 + 1 + 107
9 9

36 + 1 + 107
=9

144
=9
= 16
=4

(c) x2 + y2 – 2axcosθ – 2aysinθ = 0

Solution
Here, x2 + y2 – 2ax cosθ – 2ay sinθ = 0
comparing it with x2 + y2 + 2gx + 2fy + c = 0
g = –acosθ , h =–asinθ, we get

234 Vedanta Optional Mathematics Teacher's Guide ~ 10

centre (–g,–f) = (asinθ, asinθ)
Radius (r) = g2 + f2 – c

= (–acosθ)2+(–asinθ)2

= a2cos2θ + a2sin2θ

= a2(sin2θ +cos2θ)
= a units.

5. Find the equation of circle whose centre is (4,5) and touches x – axis.

Solution x' y' C(4,5) x
Here centre (h,k) = (4,5) (4,0)
The required circle touches x – axis at (4,0) and radius (r) O
= k=5 y
Now, the equation of the circle is, (x – h)2 +(y – k)2 = k2
ie. (x – 4)2 +(y – 5)2 = 52
x2 – 8x + 16 + y2 –10y +25 = 25
x2 + y2 –8x – 10y + 16 = 0 is the required equation.

6. Find the equation of a circle whose centre is (4,–1) and passing through (–2,–3) .

Solution P(–2,–3)
Distance between the centre (4,–1) and point (–2,–3) is the radius of the circle.

radius (r) = distance between C(4,–1) and P(–2,–3)

= (–2–4)2+(–3+1)2

= 36 + 4 C(4,–1)

= 40
=2 10 units.
Now, the equation of the circle is, (x – h)2 +(y – k)2 = r2
ie. (x – 4)2 +(y + 1)2 = 40
x2 – 8x + 16 + y2 + 2y + 1 = 40
x2 + y2 –8x + 2y= 23 is the required equation of the circle.

7. Find the equation of a circle whose centre is the p oint of intersection of x + 2y – 1 = 0
and 2x – y – 7 = 0 and passing through the point (6,4) .

Solution
Given equation of lines are,

Vedanta Optional Mathematics Teacher's Guide ~ 10 235

x + 2y – 1 = 0 .........(i)
2x – y – 7 = 0 ..........(ii)
Solving equations (i) and (ii) we get, (x, y) = (3, –1).
The point of intersection of the lines (i) and (ii) is the centre of the circle.

centre (h, k) = (3, –1)
The point (6, 4) is on the circumference of the circle.
radius (r) = distance between (3, –1) and a point on circumference.
= (x2 – x1)2+(y2 – y1)2

= (6 – 3)2+(4 + 1)2 x–2y–1=0

= 32 + 52 (3,–1) r (6,4)
= 9 + 25
2x–y–7=0
= 34 units. (4,0)
Hence the equation of circle is,
(x – h)2 +(y – k)2 = r2
ie. (x – 3)2 +(y + 1)2 = 34

x2 + y2 – 6x + 2y= 24 is the required equation of the circle.

7. Determine the points of intersections of a straight line and the circle x + y = 3, x2 +
y2 – 2x – 3 = 0. Also find the length of the intercepts(chord).

Solution x+y=3 Q
Given equation of lines are, P
x + y = 3........(i)(a line)
x2 + y2 – 2x – 3 = 0 ..........(ii)(a circle)
Solving equations (i) and (ii) we get, the point of intersection of
the line and the circle.
From equation (i), y = 3 – x ..........(iii)
put the value of y in equation (ii), we get
x2 + (3 – x)2 –2x –3 = 0
or, x2 + 9 – 6x +x2 –2x –3 = 0
or, 2x2 – 8x + 6 = 0
or, x2 – 4x + 3 = 0
or, x2 – 3x –x + 3 = 0

236 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, (x – 3)(x – 1) =0
x = 1, 3

From equation(i), when x = 1, then y = 2
when x = 3, then y = 0
Hence the required points of intersection are (1,2) and (3,0)
Also, the length of chord, PQ = (3 – 1)2+(0 – 2)2
= 4+4

=8
= 2 2 units.

9. Find the centre and radius of circle passing through the points P(2, –1), Q(2,3) and R(4,1).
Also find the equation of the circle.

Solution
Let the equation of required circle be
(x – h)2 +(y – k)2 = r2 ...............(i)
where centre = c(h,k), radius = r
The circle (i) passes through the points P(2, –1), Q(2,3) and R(4,1).
Now, CP =CQ =CR
ie. CP2 = CQ2 =CR2
Taking CP2 = CQ2
ie. (2 – h)2 +(–1 – k)2 = (2 – h)2 +(3 – k)2
ie. h2 + k2 – 4h + 2k +5 = 4 + h2 –4h + 9 –6k + k2
or, 8k = 8

k=1

again,

Taking CP2 = CR2

ie. (2 – h)2 +(1 + k)2 = (4 – h)2 +(1 – k)2

ie. h2 + k2 – 4h + 2k +5 = h2 +k2 – 8h – 2k + 17

put k = 1,

4h + 4.1 = 12

or, 4h = 8

h=2

centre (h,k) = (2, 1)

Radius (r) = (x – h)2+(y – k)2 ( (x,y)=(2,–1),(h,k)=(2,1)

= (2 – 2)2+(–1 – 1)2
= 2 units.

Equation of the required circle is, (from (i))

(x – 2)2 +(y – 1)2 = 4

Vedanta Optional Mathematics Teacher's Guide ~ 10 237

or, x2 + y2 – 4x – 2y +5 = 4
x2 + y2 – 4x – 2y +1 = 0 is the required equation of the circle.

Note:
Alternatively above question can be solved by taking equation x2 + y2 +2gx + 2fy + c =0
putting the points and solving three obtained equations for g,f and c.

10. Find the equation of the circle passing through the points (5,7), (6,6) and (2,–2).

Solution
Let the required equation of the circle be
x2 + y2 +2gx + 2fy + c = 0 ...........(i)
The circle passes through the points (5,7), (6,6) and (2,–2), we get
52 + 72 +2. g. 5 + 2 .f .7 + c = 0
or, 10g + 14f + 74 + c = 0 .................... (ii)
Similarly,
12g + 12f +72 +c =0 .............(iii)
and 4g – 4f + 8 +c = 0 .............(iv)
Subtracting (iii) from (ii), we get,
–2g + 2f + 2 = 0
or, g – f = 1 ...........(v)
Again subtracting (iv) from (iii)
8g + 16f + 64 =0
or, g + 2f + = –8 ................(vi)
Solving equation (v) and (vi), we get,
f = –3 and g =–2

centre = (–g, –f) = (2,3)
Put the values of g and f in equation (ii),
we get, 10×(–2) + 14 × (–3) + 74 + c =0
or, – 20 – 42 + 74 + c =0

c = –12
put the values of g, f and c in eqn(i), we get
x2 + y2 – 4x – 6y – 12 = 0
which is the required equation.

11. Find the equation of the circle passing through the points (3,2) and (5,4) and centre lies
on the line 3x – 2y = 1.

Solution
Let c (h,k) be the centre of the circle.
It lies on the line 3x – 2y =1.
ie. 3h – 2k = 1 ...........(1)
Let A(3,2) and B (5,4) be the points on the circumference of the circle. Then

238 Vedanta Optional Mathematics Teacher's Guide ~ 10

CA = CB
or, CA2 = CB2
(3 – h)2 +(2 – k)2 = (5 – h)2 +(4 – k)2
or, 9 – 6h + h2 + 4 – 4k + k2 = 25 – 10h +h2 + 16 – 8k + k2
or, 13 + 4h + 4k = 41
or, 4h + 4k = 28

h + k = 7 ............(ii)
solving equation (i) and (ii), we get
h = 3 and k = 4

centre = (h,k) =(3,4)
Now, equation of the circle is (x – h)2 +(y – k)2 = r2
(x – 3)2 +(y – 4)2 = r2
where, r = (3 – 3)2+(2 – 4)2 =2
or, x2 – 6x + 9 + y2 – 8y + 16 = 4
or, x2 – 6x + 9 + y2 – 8y + 16 = 4

x2 + y2 – 6x – 8y + 21 = 0 is the required equation of the circle.

12. Find the equation of the circle which touches the x –axis at (4, 0) and cuts off an intercepts
of 6 units from the y – axis positively.

Solution

Let c(h,k) be the centre of the required circle. Let A(4,0) be the point on x –axis and DE = 6
units from y – axis.

The point of (4,0) on the line 3x – 2y =1. y'
ie. 3h – 2k = 1 ...........(1) E
Let A(3,2) and B (5,4) be the points on the

circumference of the circle. Then OX . 3 C(h,k)
Draw CB perpendicular on y – axis and CA
CA = k = radius = r B4
h = BC = 4 untis = OA, BD = 3 I
Now, CD = K = BC2+BD2
= 42+32 3

D

= 25 = 5 x' O A(4,0) x
centre = (h,k) =(4, 5) y

Now, equation of the required circle is given by

(x – h)2 +(y – k)2 = k2 , r = k = 5

(x – 4)2 +(y – 5)2 = 52
or, x2 – 8x + 9 + y2 – 105 + 16 = 4

⸫ x2+y2–8x–105+16=0

Vedanta Optional Mathematics Teacher's Guide ~ 10 239

13. Prove that the points A(2, –4), B(3, –1), C(3, –3) and D(0, 0) are concyclic.

Solution
Let P(h,k) be the centre of the circle and A(2, –4), B(3, –1), C(3, –3) and D(0, 0) be three
points on the circumference of the circle.
Then
PA = PB = PC
or, PA2 =PB2 = PC2
Taking PA2 = PB2
ie. (h – 2)2 +(k + 4)2 = (h – 3)2 +(k + 1)2
or, h2 –4h + 4 + k2 + 8k + 16 =h2 –6h + 9 + k2 +2k +1
or, 2h + 6k = –10
or, h + 3k = –5 ...............(i)
Taking PB2 = PC2
ie. (h – 3)2 +(k + 1)2 = (h – 3)2 +(k + 3)2
or, k2 + 2k + 1 = k2 +6k +9
or, – 4k = 8

k = –2
Put the value of k in equation (i), we get
h + 3(–2) = –5
or, h = 1

(h,k) =(1, –2)
radius(r) =PA = (1 – 2)2+(–2 + 4)2
= 1 +4

=5
Now, equation of required circle is
(x – h)2 +(y – k)2 = r2
(x – 1)2 +(y + 2)2 = 5

x2 + y2 – 2x + 4y = 0 ........(ii)
put the point D(0,0) in equation (ii), we get,
0 = 0 (true)

D(0,0) also lies on the circle.
Hence the points A, B, C and D are concyclic.

14. Show that the two circles x2 + y2 = 36 and x2 + y2 – 12x – 16y + 84 = 0 touch externally.

Solution
Let the given circles be p1 and p2 as shown in the given figure.
x2 + y2 = 36 ............(i)
x2 + y2 – 12x – 16y + 84 = 0 ..........(ii)
From equation (i), radius = r1 = 6, centre c1 = (0,0)

240 Vedanta Optional Mathematics Teacher's Guide ~ 10

Again, from circle (ii), P1 P2
radius (r2) = g2 + f2 – c , centre = c2=(6,8)

= (–6)2 + (–8)2 – 84

= 36 + 64 –84 C1 C2

= 16 = 4 units

Distance between c1 and c2 = 62 + 82
= 100 = 10 units

sum of the radii = r1 + r2 = 6 + 4 = 10
Since the distance between the centres of two circles is equal to the sum of radii of the
circles. So given two circles touch externally. proved

15. Show that two circles touch internally. x2 + y2 = 81 and x2 + y2 – 6x – 8y + 9 = 0 .

Solution
Given equations of circles are
x2 + y2 = 81 ............(i)
x2 + y2 – 6x – 8y + 9 = 0 ..........(ii)
From equation (i), radius = r1 = 9 imoyd, centre (h,k) = (0,0)
Again, from equation (ii),
radius = g2 + f2 – c

= (–3)2 + (–4)2

= 9 + 16

= 25 r1=9 C1

= 5 units 4

radius (r2) =5 32 + 42 = 9 + 16 r2=5C2
centre = c2 =(3, 4) = 25 = 5
Difference of raddi = r1 – r2 = 9 – 5 = 4

Distance between the centres of the circles =

Since the distance between the centres of two circles is equal to the difference of the of
radii of the circles, the two circles touch internally. proved

Conic Section

16. Text book Q. N. 3(paper 184)
Solution
(a) (b) (c)Ellipse (d) Hyperbola

Vedanta Optional Mathematics Teacher's Guide ~ 10 241

17. Find the equation of tangent to the circles x2 + y2 = 25 at (3, 4).

Solution
Let PT be the tangent to the circle

x2 + y2 = 25............(i)

at the point A(3, 4).

Now, radius of circle = 5 O
A(3,4)
centre = O(0,0)

Slope OA (m2)= y2–y1 = 4–0 4
x2–x1 3–0 =– 3

Let slope of tangent PT be m2 P T
By plane geometry, we know that OA is perpendicular to PT

m1. m2 = –1
3
ie. mc = – 4

Now, equation of tangent PT is given by

y – y1 =m(x – x1)3
where m= m32 =– 4
ie. y – 4 =– 4 (x – 3)
or, 4y – 16 = –3x + 9

3x + 4y = 25

Note: Equation of tangent to the circles x2 + y2 =r2 is xx1 + yy1 = r2

Questions for practice

1. Find the equation of the circle with centre (2,3) and radius 5.
2. Find the centre and radius of the circle whose equation is x2 + y2 + 4x – 6y + 4=0
3. Find the equation of the circle whose centre is the point of intersection of x + 2y – 1 =0

and 2x – y –7 = 0 and it passes through (3,1).
4. Find the equation of the circle which touches positive axes of x and y and whose radius is 6 units.
5. Find the equation of the circle of the ends of a diameter are (3,2) and (7, –2).
6 (a). Find the equation of the circle passing through the points (1,0),(2,–2) and (3,1).
(b). Find the equation of the circle through the points (1,2),(3,1) and (–3,–1).
7. Show that the points (3,3),(3,–3) and (–3,3) and (–3,–3) are concyclic.
8. Find the equation of the circle passing through the points (1, –5) and concentric with the circles,

x2 + y2 – 4x – 8y – 81 = 0 (2,–2).
9. If y = x + 2 is the equation of a chord of the circle x2 + y2 + 2x = 0 . Find the equation of the circle

of which this chord as an a diameter. Also find the length of the chord.
10. Find the equation of the tangent to the circle x2 + y2 =100 at (6,8).
11. Find the point of intersection of the line x + y = 3 and the circle x2 + y2 – 2x – 3 =0. Also find

the length hord.(Ans: (1,2), (3,0), 2 2 )
12. Show that the two circle touch externally. x2 + y2 – 4x – 6y – 3 = 0 and x2 + y2 – 22x – 4y + 109 = 0

242 Vedanta Optional Mathematics Teacher's Guide ~ 10

UNIT

nine Trigonometry

Multiple Angles

1. Objectives Estimated Periods : 7
S,N. Level
Objectives
(i) Knowledge(K) To define multiple angles of angle A as 2A, 3A ect.
To tell the formulae of trigonometric ratios of multiple angles.
(ii) Understanding(U) To explain to derive the formulae of multiple angles by using
compound angle formulae.
(ii) Application(A) To solve problems of trigonometric identities of multiple angles.
To interpreter sin2A, cos2A, tan2A geometrically.
(iv) Higher Ability To solve trigonometric identities of difficult questions of
(HA) multiple angles.

2. Teaching Materials
Formula chart of compound angles and multiple angles.

3. Learning stategies
– Review the formula of trigonometric ratios of compound angles studied in class 9.
– Define multiple angles of A as 2A, 3A, ... ... ...
– Show how to derive formulae of multiple angles of trigonometric ratios.
eg. sin2A = 2 sinA . cos A, sin3A = 3 sinA – 4sin3A
– Discuss how to solve trigonometric identition wiht examples.

4 List of formulae 2 tanA 2 cotA
+ tan2A + cot2A
1. sin2A = 2 sinA . cosA = 1 = 1
2. cos2A =
3. tan2A = cos2A – sin2A = 2 cos2A – 1 = 1 – 2 sin2A = 1 – tan2A = = cot2A – 1
1 + tan2A cot2A + 1
2 tanA

1 – tan2A
cot2A – 1
4. cot2A = 2 cotA

5.sin3A = 3 sinA – 4 sin3A

⇒ 4 sin3A = 3 sinA – sin3A

6. cos3A = 4 cos3A – 3 cosA

⇒ 4 cos3A = 3 cosA + cos3A

7. tan3A = 3 tanA – tan3A
1–3 tan2A

8. cot3A = cot3A – 3 cotA
3 cot2A –1

Vedanta Optional Mathematics Teacher's Guide ~ 10 243

Some solved problems

1. If cosθ = 1  p + 1 , then show that:
2 p

i) cos2θ = 1  p2 + 1
2  p2 

ii) cos3θ = –21  p3 + 1 
 p3 

Solution

i) Here, cos2θ = 2 cos2θ – 1

=2 1  p + 1 2 –1
4  p 

= 1 p2 + 2 . p . 1 + 1 – 1
2 p p2 

= 1 p2 + 2 + 1 – 1
2 p2 

= 1  p2 + 2 + 1 – 2
2  p2

= 1 p2 + 1
2 p2 

ii) cos3θ = 4 cos3θ – 3 cosθ

= cosθ(4 cos2θ – 3)

= 1 p + p1  4 . 1  p + 1 2 – 3
2 4 p 

= 1 p + p1   p2 + 2 . p. 1 + 1 – 3
2  p p2 

= 1 p + 1  p2 + 2 + 1 – 3
2 p p2

= 1 p + 1  p2 – 1 + 1 
2 p p2

= 1 p + p1  p2 – p . 1 + 1
2 p p2 

= 1 p3 + 1
2 p3 

244 Vedanta Optional Mathematics Teacher's Guide ~ 10

2. a) If cosθ = 5 6 , Show that cos2θ = 11
b. 2 25.

If tanθ = 5 , show that tan2θ = 120
12 119

Solution

i) Here, cosθ = 5 6
2

LHS = cosθ = 2cos2θ – 1

= 2  6 2 – 1

5 2

= 2 36 – 1
50

= 36 – 1
25

= 36 – 25
25

11
= 25 = RHS proved

ii) Here, tanθ = 5
12

LHS = tan2θ = 2 tanθ
1 – tan2θ

2 . 5
12
=
25
1 – 144

5 144
= 6 × 144 – 25

120
= 119 = RHS proved

3. Prove that following.

a. 1 cos2θ = 1 – tanθ
+ sin2θ 1 + tanθ

Solution

LHS = 1 cos2θ
+ sin2θ

= sin2θ cos2θ – sin2θ cosθ
+ cos2θ + 2 sinθ

Vedanta Optional Mathematics Teacher's Guide ~ 10 245

= (cosθ + sinθ) (cosθ – sinθ)
(cosθ + sinθ)2

= cosθ – sinθ
cosθ + sinθ

(dividing numerator and denominator by cosθ)

= 1 – tanθ = RHS proved
1 + tanθ

Alternate Method
cos2θ
LHS = 1 + sin2θ

1 – tan2θ

= 1 + tan2θ
+
1 2 tanθ
1 + tan2θ

= 1 – tanθ = RHS proved
1 + tanθ

b. cosθ – cosθ = tan2θ
cosθ – sinθ cosθ + sinθ

Solution

LHS = cosθ – cosθ
cosθ – sinθ cosθ + sinθ

= cosθ(cosθ + sinθ) – cosθ(cosθ – sinθ)
(cosθ – sinθ) (cosθ + sinθ)

= cos2θ + sinθ . cosθ – cos2θ + sinθ . cosθ
cos2θ – sin2θ

= 2 sinθ . cosθ
cos2θ

= sin2θ = tan2θ = RHS proved
cos2θ

1 + sin2A sinA + cosA
c. cos2A = cosA – sinA

Solution

LHS = 1 + sin2A
cos2A

= sin2A + cos2A + 2 sinA . cosA
cos2A – sin2A

= (cosA (cosA + sinA)2 sinA)
+ sinA) (cosA –

= sinA + cosA = RHS proved
cosA – sinA

246 Vedanta Optional Mathematics Teacher's Guide ~ 10

d. sin5θ – cos5θ = 4 cos2θ
sinθ cosθ

Solution

LHS = sin5θ – cos5θ
sinθ cosθ

= cosθ . sin5θ – sinθ .cos5θ
sinθ . cosθ

= sin(5θ – θ)
sinθ . sinθ

= sin4θ = sin(2θ)
sinθ . cosθ sinθ . cosθ

= 2sin2θ . cos2θ
sinθ . cosθ

= 4 sinθ . cosθ . cos2θ
sinθ . cosθ

= 4 cos2θ = RHS proved

e. 1 cos2θ = tan π – θ
+ sin2θ 4

Solution

LHS = 1 cos2θ
+ sin2θ

1 – tan2θ

= 1 + tan2θ
+
1 2 tanθ
1 + tan2θ

= 1 – tan2θ × 1 + 1 + tan2θ tanθ
1 + tan2θ tan2θ + 2

= (1 – tanθ) (1 + tanθ)
(1 + tanθ)2

= 1 – tanθ
1 + tanθ

tan π – tanθ
4
=
π
1 + tan 4 . tanθ

= tanπ4 – θ = RHS proved

4. Prove that following.

Vedanta Optional Mathematics Teacher's Guide ~ 10 247

a. 1 sin2θ – cosθ = cotθ
– sinθ – cos2θ

Solution

LHS = 1 sin2θ – cosθ
– sinθ – cos2θ

= 2 sinθ . cosθ – cosθ
1 – sinθ – 1 + 2 sin2θ

= cosθ(2 sinθ – 1)
sinθ(2 sinθ – 1)

= cotθ = RHS proved

b. 1 + sin2θ – cos2θ = tanθ
1 + sin2θ + cos2θ

Solution

LHS = 1 + sin2θ – cos2θ
1 + sin2θ + cos2θ

= (1 – cos2θ) + sin2θ
(1 + cos2θ) + sin2θ

= 2 sin2θ + 2 sinθ . cosθ
2 cos2θ + 2 sinθ . cosθ

= 2 sinθ(sinθ + cosθ)
2 cosθ(sinθ + cosθ)

= sinθ
cosθ

= tanθ = RHS proved

c. sinθ + cosθ + cosθ – sinθ = 2 sec2θ
cosθ – sinθ cosθ + sinθ

Solution

LHS = sinθ + cosθ + cosθ – sinθ
cosθ – sinθ cosθ + sinθ

= (sinθ + cosθ)2 + (cosθ – sinθ)2
(cosθ – sinθ) (cosθ + sinθ)

= sin2θ + 2 sinθ . cosθ + cos2θ + cos2θ – 2 sinθ . cosθ + sin2θ
cos2θ – sin2θ

= 2(sin2θ + cos2θ)
cos2θ – sin2θ

= 2
cos2θ

= 2 sec2θ = RHS proved

248 Vedanta Optional Mathematics Teacher's Guide ~ 10

d. (1 + sin2θ + cos2θ)2 = 4 cos2θ(1 + sin2θ)

Solution

LHS = (1 + sin2θ + cos2θ)2

= (1 + 2 sinθ . cosθ + 2 cos2θ – 1)2

= [2 cosθ(sinθ + cosθ)]2

= 4 cos2θ(sinθ + cosθ)2

= 4 cos2θ(sin2θ + 2 sinθ . cosθ + cos2θ)

= 4 cos2θ(1 + sin2θ) = RHS proved

e. 1 – 1 cotθ = cotθ
tan2θ – tanθ cot2θ –

Solution

LHS = 1 tanθ – 1 cotθ
tan2θ – cot2θ –

= 1 – 1
tan2θ –
tanθ 1 – 1
tan2θ tanθ

= 1 tanθ – tan2θ . tanθ
tan2θ – tanθ – tan2θ

= 1 + tan2θ . tanθ
tan2θ – tanθ

1 + 2 tanθ tanθ
1 – tan2θ
=
2 tanθ
1 – tan2θ – tanθ

= 2 1– tan2θ + 2tan2θ
tanθ – tanθ + 1 – tan3θ

= 1 + tan2θ
tanθ + tan3θ

= 1+ tan2θ
tanθ(1 + tan2θ)

= 1
tanθ

= cotθ = RHS proved

5. Prove that:

a. 1 – sin2θ =  cotθ – 1 2
1 + sin2θ cotθ + 1 

Vedanta Optional Mathematics Teacher's Guide ~ 10 249

Solution

LHS = 1 – sin2θ
1 + sin2θ

= 1 + 1 2 tanθ
1 + 1 + tan2θ

2 tanθ
+ tan2θ

= 1 + tan2θ – 2 tanθ
1 + tan2θ + 2 tanθ

=  1 + tanθ 2
1 – tanθ 

1 – 1 2
cotθ
= 1
1 + cotθ

=  cotθ – 1 2 = RHS proved
 cotθ + 1 

1 + tan2 π – θ
4
b. = cosec2θ
tan2 π θ
1 – 4 –

Solution

LHS = 1 + tan2 π – θ
1 – 4 – θ

tan2 π
1 4

= π
4
1 – tan2 – θ

1 + tan2 π – θ
4

= 1

cos  π – θ 
4

= 1

cos  π – 2θ
 2

= 1
sin2θ

= cosec2θ

= RHS proved

250 Vedanta Optional Mathematics Teacher's Guide ~ 10