STRAIGHT LINE FOR 11TH SCIENCE AND ENGINEERING STUDENTS

"Master Straight Lines! Perfect for Class 11 and Engineering Students — A Must-Have for JEE Main & Advanced Success."

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"Straight Lines – Essential Resource for Class 11, Engineering Aspirants, and JEE Main & Advanced Preparation."

1. PAIR OF STRAIGHT LINES
Total Sessions – 02
SESSION – 1
AIM
 Combined equation of two lines
 Def. of homogeneous equation
 Second degree homogeneous equation passing through origin
 Sum and product of the slopes of pair of lines represented by ax
2
+ 2hxy + by
2
= 0.
 Angle between the pair of lines
 Bisectors of the angles between the lines of a homogeneous equation
 Product of perpendicular lengths from a point
 Area of the triangle formed by ax
2
+ 2hxy + by
2
= 0 and lx + my + n = 0.
 Equation of the pair of lines passing through a point, parallel and perpendicular to the given pair of
lines.
EXPLANATION
1] Combined equation of two lines
If a
1
x + b
1
y + c
1
= 0 and a
2
x + b
2
y + c
2
= 0 are the equations of two lines then
(a
1
x + b
1
y + c
1
) (a
2
x + b
2
y + c
2
) = 0 is called combined equation (or) Joint equation of two lines.
* The joint equation of a pair of straight lines is general equation of second degree.
* General equation of second degree is ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0
Ex1: Find the combined equation of two lines x + y – 3 = 0 and 2x – y + 1 = 0
Sol:The combined equation of the given lines
x + y – 3 = 0 and 2x – y + 1 = 0 is
(x + y – 3) (2x – y + 1) = 0.
2x
2
– xy + x + 2xy – y
2
+ y – 6x + 3y – 3 = 0
2x
2
+ xy – y
2
– 5x + 4y – 3 = 0.
2] Homogeneous Equation
An equation in x and y in which the indices x and y are non–negative integers and their sum in each
term is the same is called as homogeneous equation.
E.g.: 2x
2
+ 3y
2
+ 5xy = 0
3x
3
+ 4 x
2
y + xy
2
+ 2y
3
= 0
* The joint equation of a pair of lines through the origin is homogeneous equation of second degree in
x and y.
* A general homogeneous equation of second degree in x and y is ax
2
+ 2hxy + by
2
= 0, where a, b, h
are constants and atleast one of them is not zero.
94

2. eg. 2x
2
– 3xy + y
2
= 0, x
2
+ y
2
= 0 & 5xy = 0
95

3. 3] A homogeneous equation of second degreeax
2
+ 2hxy + by
2
= 0, h
2
ab represents a pair of straight
lines passing through origin.
Proof: ax
2
+ 2hxy + by
2
= 0 ................(i)
Case1: If a = 0, then the equation (i) becomes 2hxy + by
2
= 0
y(2hx + by) = 0, which is a combined or joint equation of two lines y = 0 and 2hx + by = 0 and both
passing through the origin.
Similarly, if b = 0, then the equation (i) represents two lines x = 0 and ax + 2hy = 0, both passing
through the origin.
Case2: If b0, then equation (i) becomes
b(ax
2
+ 2hxy + by
2
) = 0
⟹ b
2
y
2
+ 2bhxy + abx
2
= 0
⟹ (by + hx)
2
– h
2
x
2
+ abx
2
= 0
⟹ (by + hx)
2
– (h
2
– ab)x
2
= 0
⟹(by+hx−√h
2
−abx)(by+hx+√h
2
−abx)=0
⟹by+(h−√h
2
−ab)x=0 and by+(h+√h
2
−ab)x=0
Both the lines pass through origin.
 ax
2
+ 2hxy + by
2
= 0, h2≥ab represents a pair of straight lines passing through origin.
* ax
2
+ 2hxy + by
2
= 0 represents pair of distinct lines if h
2
>ab,
coincident lines if h
2
= ab and imaginary lines if h
2
< ab.
* The above pair of lines by+(h−√h
2
−ab)x=0 and (by+√h
2
−ab)x=0can be rewritten in sloge form
as
y=
(−h+√h
2
−ab)
b
x and y=
−h−√h
2
−ab
b
x
4] If ax
2
+ 2hxy + by
2
= 0 represents the pair of lines y = m
1
x and y = m
2
x then
m
1+m
2=
−2h
b
and m
1m
2=
a
b
Proof: ax
2
+ 2hxy + by
2
 b (y – m
1
x ) (y– m
2
x).
ax2
+ 2hxy + by
2
 by
2
– b (m
1
+ m
2
)xy + m
1
m
2
bx
2
Comparing the co–efficients of x
2
and xy terms,
we get m
1+m
2=
−2h
b
and m
1m
2=
a
b
i.e, m
1
+m
2
=
−2h
b
=
−co−efficientofxy
co−efficientofy
2
m
1m
2=
a
b
=
co−efficientofx
2
Co−efficientofy
2
96

4. 97

5. Ex2: If the slope of one line is twice the slope of the other in the pair of straight lines
ax
2
+ 2hxy + by
2
= 0 then 8h
2
= _____
a) 7ab b) –7ab c) 9ab d) –9ab
Sol: If m
1
and m
2
are the slope of two lines represented by ax
2
+ 2hxy + by
2
= 0.
Then m
1+m
2=
−2h
b
and m
1m
2=
a
b
given that m
2
= 2m
1
⟹3m
1=
−2h
b
⟹m
1=
−2h
3b
and m
1m
2=
a
b
⟹2m
1
2
=
a
b
⟹2(
−2h
3b)
2
=
a
b
8h
2
=9ab⟹
8h
2
9b
2
=
a
b
⟹8h
2
=9ab
5] Angle between the pair of lines
If  is the acute angle between the pair of lines represented by
ax
2
+ 2hxy + by
2
= 0 then tanθ=|
2√h
2
−ab
a+b|
Proof: Let the pair of straight lines represented by
ax
2
+ 2hxy + by
2
= 0 are y = m
1
x & y = m
2
x
We know that m
1
+m
2
=
−2h
b
and m
1
m
2
=
a
b
.
The slopes of the pair of lines are m
1
and m
2
, angle between them is ,
⟹tanθ=|
m
1
−m
2
1+m
1
m
2|
⟹tanθ=|
√(m
1+m
2)
2
−4m
1m
2
1+m
1m
2 |
¿|
√
4h
2
b
2
−
4a
b
1+
a
b|
=|
2√h
2
−ab
a+b|
* Condition for the pair of lines to be coincident, θ=θ⟺h
2
=ab
* Condition for the pair of lines to be perpendicular, θ=
π
2
⟺a+b=0
i. e, co–efficient of x
2
+ co–efficient of y
2
= 0
* Cosθ=
|a+b|
√(a−b)
2
+4h
2
Ex3: Find the acute angle between the pair of lines 2x
2
+ 5xy + 2y
2
= 0.
Sol:2x
2
+ 5xy + 2y
2
= 0.
Comparing with ax
2
+ 2hxy + by
2
= 0
98

6. a = 2, h=
5
2
, b = 2
If  is the acute angle between the pair of lines,
then
tan=|
2√h
2
−ab
a+b|
=|
2√
25
4
−(4)
2+2|
= tan=
3
4
⟹θ=Tan
−1
(
3
4)
Ex4: If the pair of lines given by (x
2
+ y
2
) sin
2
 = (x cos – y sin)
2
are perpendicular to each other,
then  =
a)
π
2
b) 0 c)
π
4
d)
π
3
Sol:
(x
2
+ y
2
) sin
2
 = (xcos – ysin)
2
⟹ x
2
sin
2
+ y
2
sin
2
= x
2
cos
2
+ y
2
sin
2
– 2xy sin cos
⟹ x
2
(cos
2
– sin
2
) – 2xy sin cos = 0
The pair of lines represented by the equation are perpendicular to each other
if co–efficient of x
2
+ co–efficient of y
2
= 0
⟹ cos
2
– sin
2
 = 0cos2 = 0
⟹2α=
π
2
⟹α=
π
4
* The equation of the pair of lines passing through origin and perpendicular to the lines represented by
ax
2
+ 2hxy + by
2
= 0 is bx
2
– 2hxy + ay
2
= 0.
* The quadratic equation whose roots are the slopes of lines represented by
ax
2
+ 2hxy + by
2
= 0 is bm
2
+ 2hm + a = 0 called auxiliary equation.
6] Bisectors of the angles between the lines of a Homogeneous Equation.
The equations of the pair of bisectors of the angles between the pair of lines
ax
2
+ 2hxy + by
2
= 0 is
x
2
−y
2
a−b
=
xy
h
(or) h (x
2
– y
2
) – (a – b) xy = 0
Proof: Let y = m
1
x and y = m
2
x be the pair of lines represented by ax
2
+ 2hxy + by
2
= 0, then
m
1
+m
2
=
−2h
b
and m
1
m
2
=
a
b
.
The equations of the bisectors of the angle between the lines y = m
1
x and y = m
2
x are
y−m
1
x
√1+m
1
2
=±
y−m
2
x
√1+m
2
2
⟹(1+m
2
2
)(y−m
1x)
2
=(1+m
1
2
)(y−m
2x)
2
⟹(1+m
2
2
)(y
2
+m
1
2
x
2
−2xym
1)=(1+m
1
2
)(y
2
+m
2
2
x
2
−2m
2xy)
⟹x
2
(m
1
2
+m
1
2
m
2
2
−m
2
2
−m
1
2
m
2
2
)+y
2
(1+m
2
2
−1−m
1
2
)−¿
2xy(m
1+m
1m
2
2
−m
2−m
1
2
m
2)=0
99

7. ⟹(m
1
2
−m
2
2
)(x
2
−y
2
)−2xy(m
1−m
2)(1−m
1m
2)=0
⟹(m
1+m
2)(x
2
+y
2
)+2xy(2m
1m
2−1)=0(∵m
1≠m
2)
⟹
−2h
b
(x
2
−y
2
)+2xy(
a
b
−1)
=0 ⟹
x
2
−y
2
a−b
=
xy
h
(or) h(x
2
– y
2
) – (a – b) xy = 0
* The equations of the pair of bisectors of the angles between the pair of lines
ax
2
+ 2hxy + by
2
= 0 are always perpendicular to each other.

100

8. Ex5:Find the equation of the bisectors of the angle between the two straight lines
x
2
– xy – 6y
2
= 0.
Sol:x
2
– xy – 6y
2
= 0
Comparing with ax
2
+ 2hxy + by
2
= 0,
we get a =1, b = –6, 2h = –1.
Equations of bisectors is h (x
2
– y
2
) – (a – b) xy = 0
⟹−
1
2
(x
2
−y
2
)−(1+6)xy=0
⟹x
2
−y
2
+14xy=0 ⟹x
2
+14xy−y
2
=0
Ex6: If the pair of straight lines x
2
– 2pxy – y
2
= 0 and x
2
–2qxy – y
2
= 0 be such that each pair bisects
the angle between the other pair then
a) pq = –1 b) p = q c) p = –q d) pq = 1
Sol: Equation of the bisector of the angles between
x
2
– 2pxy – y
2
= 0 is px
2
+ 2xy – py
2
= 0, which is same as
x
2
– 2qxy – y
2
= 0,
⟹
p
1
=
1
−q
=
p
1
⟹pg=−1
7] The product of the perpendiculars from a point (, β) to the pair of lines ax
2
+ 2hxy + by
2
= 0 is
|
aα
2
+2hαβ+bβ
2
√(a−b)
2
+4h
2|
Proof: Let l
1
x + m
1
y = 0 and l
2
x + m
2
y = 0 be the pair of lines represented by
ax
2
+ 2hxy + by
2
= 0
ax
2
+ 2hxy + by
2
⟹l
1
l
2
=a,m
1
m
2
=b,l
1
m
2
+l
2
m
1
=2h
Product of the perpendicular lengths from (α,β) to the lines
l
1
x+m
1
y=0&l
2
x+m
2
y=0
¿|
(l
1α+m
1β)(l
2α+m
2β)
√(l
1
2
+m
1
2
)(l
2
2
+m
2
2
)|
=|
l
1l
2α
2
+(l
1m
2+l
2m
1)αβ+m
1m
2β
2
√l
1
2
l
2
2
+l
1
2
m
2
2
+l
2
2
m
1
2
+m
1
2
m
2
2|
¿
|
aα
2
+2hah+bβ
2
√a
2
+b
2
+(l
1m
2+l
2m
1)
2
−2l
1l
2m
1m
2|
=|
aα
2
+2hαβ+bβ
2
√(a−b)
2
+4h
2|
Ex7: Find the product of perpendicular lengths from the point (1, 2) to the pair of lines
x
2
– 3xy – 3y
2
= 0
Sol: Here(α,β) = (1, 2) & a = 1, b = –3, 2h = –3

101

9. Product of perpendicular lengths ¿|
aα
2
+2hαβ+bβ
2
√(a−b)
2
+4h
2|
¿|
1−6−12
√16+9|
=
17
5

8] The area of the triangle formed by ax
2
+ 2hxy + by
2
= 0 and lx + my + n = 0 is|
n
2
√h
2
−ab
am
2
−2hlm+bl
2|
Proof: Let l
1
x + m
1
y = 0 and l
2
x + m
2
y = 0 be the pair of lines represented by
ax
2
+ 2hxy + by
2
= 0.
The equation of the given line is lx + my + n = 0 ––––––(3)
Solving equations (1) and (3), we get the point of intersection
A(
mn
1
ml
1
−lm
1
,
−nl
1
ml
1
−lm
1)
Solving equations (2) and (3), we get the point of intersection,
B(
mn
2
nl
2
−lm
2
,
−nl
2
ml
2
−lm
2)
or (OAB)=
1
2
|x
1
y
2
−x
2
y
1|
¿
1
2|
n
2
(l
1m
2−l
2m
1)
(ml
1
−lm
1)(ml
2
−lm
2)|
=
1
2|
n
2
√(l
1m
2+l
2m
1)
2
−4l
1l
2m
1m
2
m
2
l
1l
2−(l
1m
2+l
2m
1)ml+m
1m
2l
2|
¿
1
2|
n
2
√4h
2
−4ab
am
2
−2hlm+bl
2|
=|
n
2
√h
2
−ab
am
2
−2hlm+bl
2|
Ex8: The area of the triangle formed by the lines x
2
+ 4xy + y
2
= 0, x + y = 1 is :
a) √3 b) 2 c) 1 d)
√3
2
Sol:Area of triangle formed by the given lines ¿|
n
2
√h
2
−ab
am
2
−2hlm+bl
2|
a = 1, b = 1, 2h = 4, l = 1, m = 1, n = –1
¿|
1√4−1
1−4+1|
=
√3
2
9] The equation of the pair of lines through the origin and making an angle with the line
lx + my + n = 0 is (lx + my)
2
– tan
2
(mx – ly)
2
= 0 and the area of the triangle is
n
2
tanα(l
2
+m
2
)
10]The equations of the pair of lines passing through and parallel to the pair of lines
ax
2
+ 2hxy + by
2
= 0 is a (x –)
2
+ 2h (x –)(y –β) + b(y – β)
2
= 0
11]The equations of the pair of lines passing through (α,β) and perpendicular to the pair of lines
ax
2
+ 2hxy + by
2
= 0 is b(x –)
2
– 2h (x –)(y –β) + a(y – β)
2
= 0

102

10. 12]The equation to the pair of bisectors of the coordinate axes is x
2
– y
2
= 0.
13
]If one of the line in ax
2
+ 2hxy + by
2
= 0 bisects the angle between the coordinate axes then
(a + b)
2
= 4h
2
14]A pair of lines is equally inclined to a line L = 0 if L = 0 is parallel to one of the angle bisectors.
15]If the pair of lines ax
2
+ 2hxy + by
2
= 0 equally inclined to the co–ordinates axes then
h = 0 and ab < 0.
16]Two pairs of lines are equally inclined to each other if and only if two pairs of lines have same pair
of angular bisectors.

103

11. Ex9: If a, h, b are in AP. then the triangle area formed by the pair of lines ax
2
+ 2hxy + by
2
= 0 and the
line x – y = –2 in square units is:
a)|
a+b
a−b|
b)|
a
2
+b
2
a−b|
c) |
a−b
a+b|
d) |
a
2
+b
2
a+b|
Sol:a, h, b are in AP 2h = a + b
Given equation of line is x – y = –2, Here l = 1, m = –1, n = 2
Area of triangle ¿|
n
2
√h
2
−ab
am
2
−2hlm+bl
2|
¿|
2√
(a+b)
2
4
−ab
a+2h+b|=|
(a−b)
(2a+2b)|
=|
a−b
a+b|
Ex10: Find the equation to the pair of lines passing through (–1, 2) and parallel to 2x
2
– xy – y
2
= 0
Sol: Equation to the pair of lines passing through and parallel to ax
2
+ 2hxy + by
2
= 0 is
a (x –)
2
+ 2h (x –)(y –β) + b(y –β)
2
= 0
2 (x + 1)
2
–1 (x + 1)(y – 2) – (y – 2)
2
= 0
2(x
2
+ 2x + 1) – (xy – 2x + y –2 ) – (y
2
+ 4 – 4y ) = 0
2x
2
– xy – y
2
+ 6x + 3y = 0
Ex11:Find the equation to the pair of lines passing through the point (1, –1) and perpendicular to the pair
of lines x
2
– xy + 2y
2
= 0
Sol: Equation to the pair of lines passing through (α,β)and perpendicular to the lines
ax
2
+ 2hxy + by
2
= 0 is
b(x –)
2
– 2h (x –) (y –β) + a (y –β)
2
= 0
⟹ 2(x – 1)
2
+ (x – 1) (y + 1) + (y +1)
2
= 0
⟹ 2(x
2
+ 1– 2x) + xy + x – y – 1 + y
2
+ 1 + 2y = 0
⟹ 2x
2
+ xy + y
2
– 3x + y + 2 = 0
* The triangle formed by the lines ax
2
+ 2hxy + by
2
= 0 and lx + my + n = 0 isosceles if
h (l
2
– m
2
) = (a – b)lm .
Ex12: If x
2
– y
2
= 0, kx + 2y = 2 form an isosceles triangle, then k =
a) 1 b) 2 c) 3 d) 0
Sol:The triangle formed by given lines is isosceles if h (l
2
– m
2
) = (a – b) lm
Here h = 0, a = 1, l = k, m = 2, n = –2
0 (k
2
– 4) = (2) (k) (2) ⟹ k = 0.

104

12. 105

13. CLASS EXERCISE
1] Find the combined equation of the two lines x – 2y = 0 and x – 3y = 0
2] Find the pair of straight lines represented by the homogeneous equation x
2
+ 2xy cot – y
2
= 0
3] Find the acute angle between the pair of lines x
2
– 7xy + 12y
2
= 0.
4] The equation 3x
2
+ 10xy – 8y
2
= 0 represents
a) real and distinct lines b) coincident lines
c) imaginary lines d) parallel lines
5] Find the equation of the bisectors of the angle between the two straight lines
2x
2
– 3xy + y
2
= 0.
6] Find the equation to the pair of lines passing through the origin and perpendicular to 5x
2
+3xy=0
7] The product of the perpendiculars from (3, –1) to the pair of lines 3x
2
– 5xy – 9y
2
= 0.
8] If the area of the triangle formed by the lines 3x
2
– 2xy – 8y
2
= 0. and the line 2x + y – k = 0 is 5 sq.
units then k = ____.
9] The triangle formed by the pair of lines x
2
– 4y
2
= 0 and the line x – a = 0 is always
a) equilateral b) isosceles c) Right angled d) scalene
10]If 2x
2
– 5xy – 2y
2
= 0 represents two sides of a triangle whose centroid is (1, 1) then the equation of
the third side
a) x + y – 3 = 0b) x – y – 3 = 0 c) x + y + 3 = 0 d) none
11]Find the equation to the pair of lines passing through (1, 2) and parallel to the co–ordinate axes is __.
12]The pair of lines 2x
2
+ 3xy + 5y
2
= 0, 4x
2
+ 21xy + 25y
2
= 0 are
a) mutually perpendicular
b) equally inclined to the axes
c) equally inclined to each other
d) each pair bisects the angle between the other.
HOME EXERXISE
1] Find the combined equation of the two lines 2x + 3y = 0 & 3x – 2y = 0
2] Find the pair of straight lines represented by the homogeneous equation 2x
2
+ xy – y
2
= 0
3] Find the acute angle between the pair of lines 2x
2
+ 5xy + 3y
2
= 0
4] The difference of slopes of the lines represented by y
2
– 2xy sec
2
+ (3 + tan
2
) (–1 + tan
2
) x
2
= 0
5] Find the equation of the bisectors of the angle between the two straight lines x
2
+ 3xy – 9y
2
= 0

106

14. 6] Find the equation to the pair of lines passing through the origin and perpendicular to
x
2
– xy – 2y
2
= 0
7] The product of the perpendiculars from (–1, 2) to the pair of lines 2x
2
– 5xy + 2y
2
= 0
a) 4 b) 3 c) 8 d)
5
2
8] Find the area of the triangle formed by the lines 3x
2
– 4y + y
2
= 0 and 2x – y = 6 is
9] Find the centroid of the triangle formed by the lines 12x
2
– 20xy + 7y
2
= 0 and 2x – 3y + 4 = 0
10]The adjacent sides of a parallelogram are 2x
2
– 5xy + 3y
2
= 0 and one diagonal is
x + y + 2 = 0. Find the vertices and equation to the other diagonal .
11]The equation of the pair of the lines through (1, –1) and perpendicular to the pair of lines
x
2
– xy – 2y
2
= 0.
12]Show that the orthocentre of the triangle formed by the pair of lines ax
2
+ 2hxy + by
2
= 0 and the
line px + qy = 1 is the point (λp,λq)then λ=
a+b
aq
2
−2hpq+bp
2.
13]Show that the lines (3x – 4y)
2
– 3(3y + 4x )
2
= 0 and 3x – 4y + 5 = 0 form an equilateral triangle
and find its area.

107

15. SESSION – 2
AIM
 Condition for the second degree non–homogeneous equation to represent a pair of straight lines.
 Angle between the pair of lines
 Point of intersection of the pair of lines.
 Product of perpendicular distances from a point.
 Distance between pair of parallel lines
 Equations of the bisectors of the angles between pair of straight lines.
 Intercepts on axes.
 Condition for a pair of lines to intersect axes.
 Homogenization
EXPLANATION
1] The general equation of second degree non–homogeneous equation
ax
2
+ 2hxy + by
2
+2gx + 2fy + c = 0 represents a pair of straight lines then
i] = abc + 2fgh – af
2
– bg
2
– ch
2
= 0 (or) |
ahg
hbf
gfc|
= 0
ii]h
2≥ab, g
2≥ ac, f
2≥bc .
Proof: Let l
1
x + m
1
y + n
1
= 0 and l
2
x + m
2
y + n
2
= 0 be the pair of lines.
ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c≡ (l
1
x + m
1
y + n
1
)(l
2
x + m
2
y + n
2
)
⟹l
1
l
2
=a,m
1
m
2
=b,n
1
n
2
=c,l
1
m
2
+l
2
m
1
=2h,l
1
n
2
+l
2
n
1
=2g,m
1
n
2
+m
2
n
1
=2f
iii] Consider (2h) (2g) (2f) = (l
1
m
2
+ l
2
m
1
) (l
1
n
2
+ l
2
n
1
) (m
1
n
2
+ m
2
n
1
)
8fgh=l
1l
2(m
1
2
n
2
2
+m
2
2
n
1
2
)+n
1n
2(l
1
2
m
2
2
+l
2
2
m
1
2
)+m
1m
2(l
1
2
n
2
2
+l
2
2
n
1
2
)
+2l
1
l
2
m
1
m
2
n
1
n
2
= a (4f
2
– 2bc) + c (4h
2
– 2ab) +b (4g
2
– 2bc) + 2abc
= 4{af
2
+bg
2
+ch
2
−abc}
⟹abc + 2fgh – af
2
– bg
2
– ch
2
= 0 (or) |
ahg
hbf
gfc|
= 0
iv]Consider h
2
−ab=
(l
1
m
2
+l
2
m
1)
2
−4l
1
l
2
m
1
m
2
4
¿
(l
1
m
2
−l
2
m
1)
2
4
≥0⟹h
'
≥ab.
Similarlyg
2
≥ac andf
2
≥bc
2] ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 represents a pair of lines then the equation
ax
2
+2hxy+by
2
=0 represents a pair of lines parallel to them through the origin.

108

16. 3] The acute angle () between the pair of lines ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 is same as the
acute angle between the pair of lines ax
2
+ 2hxy + by
2
= 0.
Tanθ=|
2√h
2
−ab
a+b|
.
4] The point of intersection of the pair of lines represented by ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 , is
(
hf−bg
ab−h
2
,
gh−af
ab−h
2)
,h
2
≠ab (or)
(
bc−f
2
hf−bg
,
fg−ch
hf−bg)
hf≠bg (or) (
ch−fg
af−gh
,
g
2
−ac
af−gh)
af≠gh
Ex1: Find the pair of lines represented by the equation x
2
– y
2
– x + 3y – 2 = 0.
Sol:x
2
– y
2
– x + 3y – 2 (x – y + l) (x + y + m)
⟹x
2
– y
2
– x + 3y – 2 ≡ x
2
– y
2
+ x (m + l) + y (l – m) + lm
⟹ l + m = –1, l – m = 3 , lm = –2
 The required pair of lines are x – y + 1 = 0 and x + y – 2 = 0.
Ex2: Find the value of k such that x
2
– y
2
+ 2x + k = 0 represents a pair of lines.
Sol:x
2
– y
2
+ 2x + k = 0
Comparing with ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0
a = 1, b = –1, h = 0, g = 1, f = 0, c = k.
The given equation represents a pair of lines
 = abc + 2fgh – af
2
– bg
2
– ch
2
= 0
⟹ – k + 0 – 0 + 1 – 0 = 0 ⟹ k = 1
Ex3: The angle between the pair of lines 2 (x + 2)
2
+ 3(x + 2) (y – 2) – 2(y – 2)
2
= 0 is
a)
π
4
b)
π
3
c)
π
6
d)
π
2
Sol:2(x + 2)
2
+ 3(x + 2)(y – 2) – 2(y –2)
2
= 0 is a pair of lines parallel to 2x
2
+ 3xy – 2y
2
= 0 and
passing through (–2, 2).
The angle between the pair of lines 2x
2
+ 3xy – 2y
2
= 0 is
π
2
.(∵a + b = 2 – 2 = 0)
Ex4:Find the angle between the pair of lines 2x
2
+ 5xy + 2y
2
+ 3x + 3y + 1 = 0.

109

17. Sol:Tanθ=|
2√h
2
−ab
a+b|

¿|
2√
25
4
−4
4|
(∵a=2,b=2,h=
5
2
)
¿
3
4
⟹θ=Tan
−1
(
3
4)
Ex5: If the lines x
2
+ 2 xy – 35y
2
– 4x + 44y – 12 = 0 and 5x +y – 8 = 0 are concurrent, then
 = _____.
a) 0 b) 1 c) –1 d) 2
Sol:x
2
+ 2 xy – 35y
2
– 4x + 44y – 12 = 0.
a = 1, b = –35, h = 1, g = –2 , f = 22, c = –12.
Point of intersection ¿(
hf−bg
ab−h
2
,
gh−af
ab−h
2)
¿(
22−70
−35−1
,
−2−22
−35−1)
¿(
−48
−36
,
−24
−36)
=(
4
3
,
2
3)
.
As the lines are concurrent, the point of intersection lies on the line 5x +y – 8 = 0
5(
4
3)
+(
2
3)
=8 ⟹
2
3
=8−
20
3
=
4
3
⟹=2
5] The product of the perpendiculars from (α,β) to the pair of lines
ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 is |
aα
2
+2hαβ
2
+2gα+2fβ+c
√(a−b)
2
+4h
2 |
6] If ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 represents a pair of parallel lines then
h
2
= ab and bg
2
= af
2
.
Also the distance between the two parallel lines is 2√
g
2
−ac
α(a+b)
(¿)2
√
f
2
−bc
b(α+b)
Proof: Let lx + my + n
1
= 0 and lx + my + n
2
= 0 be the parallel lines.
Then ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c (lx + my + n
1
)(lx + my + n
2
)
⟹ l
2
= a , m
2
= b
2
, n
1
n
2
= c , 2lm = 2h &
l(n
1
+ n
2
) = 2g , m (n
1
+ n
2
) = 2f.
⟹l
2
m
2
=h
2
⟹ h
2
= ab &
l(n
1
+n
2)
m(n
1
+n
2)
=
2g
2f
⟹
l
2
m
2
=
g
2
f
2
⟹af
2
=bg
2
And abc + 2fgh – af
2
– bg
2
– ch
2
= 0
⟹ 2fgh = 2af
2
(or) 2fgh = 2bg
2

110

18. ⟹ gh = af (or ) bg = fh ⟹
a
h
=
h
b
=
g
f
Distance between the parallel lines
¿|
n
1
−n
2
√l
2
+m
2|
=|
√(n
1+n
2)
2
−4n
1n
2
√a+b |
¿|√
4g
2
l
2
−4c
√a+b|=|
2√
g
2
−ac
α(a+b)|
=2|
f
2
−bc
b(α+b)|
7] The equations of the bisectors of the angles between the lines represented by
ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 are h{(x−α)
2
−(y−β)
2
}=(a−b)(x−α)(y−β), where (α,β) is
the point of intersection of the lines.
8] Length of the intercept made by the pair of lines represented by
ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 on
i) x–axis is
2√g
2
−c
|a|
ii) y–axis is
2√f
2
−c
|b|
9] If the pair of lines ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 intersect on
i) x–axis then g
2
= ac and 2fgh = af
2
+ch
2
ii) y–axis then f2
= bc and 2fgh = bg
2
+ ch
2
Ex6: The product of the perpendicular distances from (1, 1) on the pair of straight lines
12x
2
+ 25xy + 12y
2
+ 10x + 11y + 2 = 0.
Sol: Product of perpendicular lengths from (α,β) on the pair of straight lines
¿|
aα
2
+2hαβ+bβ
2
+2gα+2fβ+c
√(a−b)
2
+4h
2 |
=
12+25+12+10+11+2
√
(12−12)
2
+(
25
2)
2
=
72
25
Ex7:If the distance between the pair of parallel lines x
2
+ 2xy + y
2
– 8ax – 8ay – 9a
2
= 0 is 25√2 then a
= _____
a) 1 b) 2 c) 3 d) 5
Sol: Distance between parallel lines ¿2
√
g
2
−ac
a(a+b)
=25√2
⟹2
√
16a
2
+9a
2
1(1+1)
=25√2⟹5a=25⟹a=5
Ex8:Find the intercept made by the pair of lines 6x
2
– 7xy – 3y
2
– 24x – 3y + 18 = 0 on x–axis and y–
axis.

111

19. Sol: Intercept on x–axis ¿
2√g
2
−ac
|a|
=
2√144−108
|6|
=
2×6
6
=2
Intercept on y–axis = 2√f
2
−bc
|b|
=
2√
9
4
+54
|−3|
=
15
3
=5
Ex9: If the equation ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 represents a pair of straight lines then show
that
g
2
−ac
fg−ch
=
fg−ch
f
2
−bc
Sol: abc + 2fgh – af
2
– bg
2
– ch
2
= 0
2fgh – ch
2
= af
2
+ bg
2
– abc
2fghc – c2
h
2
= acf
2
+ bg
2
c– abc
2
(fg – ch)
2
– f
2
g
2
+ acf
2
+ bg
2
c – abc
2
= 0 ⟹ (fg – ch)
2
– f
2
(g
2
– ac) + bc (g
2
–ac) = 0
⟹ (fg – ch)
2
= (g
2
– ac) (f
2
–bc) ⟹
g
2
−ac
fg−ch
=
fg−ch
f
2
−bc
Ex10: Area of the triangle formed by the lines x + y = 3 and angle bisectors of the pair of straight lines x
2
–
y
2
+ 2y = 1 is:
a) 2 sq.units b) 4 sq.units c) 6 sq.units d) 8 sq.units
Sol:x
2
– y
2
+ 2y = 1
x
2
−(y
2
−2y+1)=0
⟹ x
2
– (y –1)
2
= 0 ⟹ (x−y+1)(x+y−1)=0
⟹ x – y + 1 = 0 ––––––––––(i)
x + y –1 = 0 ––––––––––(ii)
Equations of bisectors of equations (i) & (ii) are
x−y+1
√2
=±
x+y−1
√2
⟹ x – y + 1 = x + y –1 and x – y + 1 = – x – y +1
⟹ y = 1 and 2x = 0 ⟹ x = 0
 Area of the triangle formed with the lines x + y = 3, y = 1 and x = 0 is
1
2
×2×2=2sq. units.
10]Homogenization:

112

20. The equation of the pair of lines, joining origin to the points of intersection of the curve
S≡ax
2
+2hxy+by
2
+2gx+2fy+C=0 and the line L≡lx+my+n=0 is
ax
2
+2hxy+by
2
+2(gx+fy)(
lx+my
−n)
+C
'
(
lx+my
−n)
2
=0.
Ex 11:
Show that the pair of lines joining origin to the point of intersection of the line x−y−√2=0 with the
curve x
2
−xy+y
2
+3x+3y−2=0 are at right angle.
Sol:Given line x−y−√2=0⟹
x−y
√2
=0 ….(1)
Homogenizing curve x
2
−xy+y
2
+3x+3y−2=0 ….(2) with (1)
x
2
−xy+y
2
+3(x+y)(
x−y
√2)
−2(
x−y
√2)
2
=0
⟹√2(x
2
−xy+y
2
)+3(x
2
−y
2
)−√2(x
2
−2xy+y
2
)=0
⟹3x
2
+√2.xy−3y
2
=0 …. (3)
(3) represents the pair of lines joining origin to the points of intersection of (1) & (2)
In (3) Cottx
2
+cotty
2
=3+(−3)=0
∴ The pair of lines in (3) are perpendicular.
Ex 12:
Find the angle between the pair of lines joining the origin to the points of intersection of the line
x+y−2=0 and the curve x
2
−2xy+y
2
−x+y+1=0
Sol:Equation to the given line can be rearranged as
x+y
2
=1
Homogenizing the equation
x
2
−2xy+y
2
−x+y+1=0 with that of (1)
we get x
2
−2xy+y
2
−(x−y)
(x+y)
2
+1(
x+y
2)
2
=0
⟹3x
2
−6xy+7y
2
=0
∴ The angle θ between the pair of lines (2) is
Cosθ=
3+7
√(3−7)
2
+(−6)
2
=
10
√52
=
5
√13
⟹θ=Cos
−1
(
5
√13)
CLASS EXERCISE
1] The value of k such that 3x
2
+ 11xy + 10y
2
+ 7x + 13y + k = 0 represents a pair of straight lines is

113

21. a) 1 b) 2 c) 3 d) 4
2] If ax
2
+ 5xy – 6y
2
– 10x + 11y + c = 0 represents a pair of perpendicular lines then c = ____
a) 2 b) –2 c) 4 d) –4
3] Find the angle between the pair of lines 12x
2
– 10xy + 2y
2
+ 14x – 5y + 2 = 0
4] Find the point of intersection of the pair of lines x
2
– y
2
+ 2x + 2y = 0
5] If the pair of lines ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 intersect on the y–axis, then
a) 2fgh = bg
2
+ ch
2
b)bg
2
ch
2
c) abc = 2fgh d) none
6] The product of perpendicular distances from the origin on the pair of straight lines
12x
2
+ 25xy + 12y
2
+ 10x + 11y +2 = 0
a)
1
25
b)
2
25
c)
3
25
d)
4
25
7] Find the distance between the parallel lines 9x
2
– 6xy + y
2
+ 18x – 6y + 8 = 0
8] Find the intercept made by the pair of lines 2x
2
+ 4xy – 6y
2
+ 3x + y + 1 = 0 on x –axis
9] Find the values of k and l so that x
2
+ 6xy + ky
2
+ lx + 12y – 5 = 0 represents a pair of parallel lines.
Find also the parallel line midway between them
10]If the lines joining the origin to the points of intersection of the line x+2y=k with the curve
2x
2
−2xy+3y
2
+2x−y−1=0 are at right angles, find k.
11]Prove that an angle between the pair of lines joining points of intersection of the line lx+my=1
with the curve x
2
+y
2
=a
2
is 2cos
−1
[
1
a√l
2
+m
2]
and hence find the condition that the two lines
may be coincident (c) perpendicular.
12]Find the angle between the lines drawn from the origin to the points of intersection of
3x−y+1=0 and x
2
+2xy+y
2
+2x+2y−5=0.
HOME EXERCISE
1] Two lines 9x
2
+ y
2
+ 6xy – 4 = 0 are
a) Parallel and coincident b) coincident only
c) Parallel but not coincident d) perpendicular
2] If x
2
+ 6xy + 9y
2
+ 4x + 12y + 3 = 0 represents a pair of straight lines, then  = ______
a) –3 b) –4 c) 1 d) 5
3] If ax
2
+ 6xy + by
2
– 10x + 10y – 6 = 0 represents a pair of perpendicular straight lines then |a| =
____
a) 2 b) 4 c) 1 d) 3
4] Find the point of intersection of the pair of lines 12x
2
– 10xy + 2y
2
+ 14x – 5y + 2 = 0
5] If xy + x + y + 1 = 0, x + ay – 3 = 0 are concurrent, then a = _____
a) 2 b) 3 c) –4 d) 4

114

22. 6] The product of perpendicular distances from the (–2, 3) on the pair of straight lines
x
2
– y
2
+ 2x + 1 = 0.
a) 3 b)4 c) 5 d) 6
7] Find the distance between the parallel lines x2+2√2xy+2y2+4x+4√2y+1=0.
8] Find the intercept made by the pair of lines 6x
2
– 5xy – 6y
2
+ x + 5y – 1 = 0 on x–axis and y–axis.
9] If ax
2
+ 2hxy + by
2
+ 29x + 2fy + c = 0 represents a pair of lines then show that the square of the
distance from the origin to their point of intersection is
c(a+b)−f
2
−g
2
ab−h
2
and If the lines are
perpendicular then the square of the distance is
f
2
+g
2
h
2
+b
2
.
10]If the pair of lines ax
2
+ 2hxy + by
2
+ 2gx + 2fy + c = 0 are equidistant from the origin then prove
that d
4
– g
4
= c (bf
2
– ag
2
).
11]The chords of the curve 3x
2
−y
2
−2x+4y=0 subtend a right angle at the origin. Prove that all such
chords are concurrent at a point.
12]Show that the lines drawn from the origin to the points of intersection of the line 6x−y+8=0 with
the curve 3x
2
+4xy−4y
2
−11x+2y+6=0 are equally inclined to the coordinate axes.
13]Find the equation of the lines joining origin to the point of intersection of line x+y=1 and
(a) x
2
+y
2
=a
2
(b) y
2
=x. Also find the angle between them.

115

23. KEY
SESSION – 1
CLASS EXERCISE
1) x
2
−5xy+6y
2
=0 2) x+y(cotα−cosecα)=0;x+y(cotα+cosecα)=0
3) cos
−1
(
13
√170)
4) a 5) 3x
2
+2xy−3y
2
=0 6) 3xy−5y
2
=0
7)
−33
13
8) k=5 9) b 10) a 11) xy−2x−y+2=0
12) c
HOME EXERCISE
1) 6x
2
+5xy−6y
2
=0 2) 2x−y=0;x+y=0 3) cos
−1
(
5
√26)
4) 4
5) 3x
2
−20xy−3y
2
=0 6) 2x
2
−xy−y
2
=0 7) a 8) 36 sq. units
9) (
8
3
,
8
3)
10) (−1,−1)(
−11
5
,−
9
5)(
−6
5
,−
4
5)
;9x−11y=0
11) 2x
2
−xy−y
2
−5x−y+2=0 12) 13) Area ¿
1
√3
sq.units
SESSION – 2
CLASS EXERCISE
1) d 2) d 3) co
−1
(
7√2
10)
4) (−1,1) 5) a
6) b 7)
4
3
8)
1
2
,
5
6
9) k=9,l=4;x+3y+2=0

HOME EXERCISE
1) c 2) c 3) b 4) (
3
2
,
5
2)
5) c 6) b
7) 2 8)
5
6
,
1
6

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