DGT MH –CET 12th MATHEMATICS Study Material 1

Pairs of Straight Lines 70

04 Pairs of Straight Lines

The two factors equated to zero will give the

Syllabus : Combined Equation of Two Lines  separate equations of lines.
Separate Equation ofTwo Lines Homogeneous Example 2
Equation of Second Degree General Equation The separate equation of lines represented by the
of Second Degree. Removal of First Degree equation
Terms
x2 – 6xy + 8y2 = 0 is
a. x2 + y2 = 0 b. x+5y =0
We know that a linear equation in two variables
c. x–2y = 0 d. 5x + 2y = 0
ax + by + c = 0, a, b, c  r and a, b, c not all zero
represents a straight line. Sol (c) Separate equation of lines represented by the
equation
Now, consider separate equation of two lines
u1  a1x + bty + c1 = 0 and u2= a2x + b2y + c2 = 0. x2 – 6xy + 8y2 = 0
We know, u1+ u2= 0 represents family of lines. i.e. (x – 4y) (x – 2y) = 0
Then, ifwe combine these two equations i.e.ut  x– 4y = 0 and x – 2y = 0
.u2 = 0, this represents a pair of straight lines. or x2 – 6xy + 8y2 = 0
Now, we will study combined and separate By Shri Dharacharya method,
equations of two lines.
6y  ( 6y)2  4.8y 2
Combined Equation of Two Lines x=
2
In order to find the joint (combined) equation of
two lines, make RHS of two lines equal to zero  x = 3y + y 9 8
and then multiply the two equations.
x = 3y + y
• Example 1  x = 4y or x = 2y
Find the joint equation of lines y = x and y = - x.  x – 4 y = 0 or x– 2y = 0
a. x2 + y2 = 0 b. –x2 + y2 = 0
Homogeneous Equation of Second Degree
b. x2 – y2 = 0 d. None of these
An equation of the form ax2 + 2hxy + by2 = 0 in
Sol(c) The given lines can be written as which the sum of powers of x and y in every
x – y = 0 and x + y = 0 term is the same (here 2) is called homogeneous
 Joint equation of lines is equation of degree 2.
(x – y) (x + y) = 0 or x2 – y2 = 0 e.g. 2x2 – xy – y2 = 0
Wrong process Since, lines are y = x and and 6x2 + 5xy – 4y2 = 0
y = – x, then joint equation is are homogeneous equations of second degree.
y2 = – X2 Homogeneous equation of degree 2 represents two
 x2 + y2 = 0 straight lines through origin, if
This process is wrong, since RHS of two equations
h 2  ab > 0
are not equal to zero.
Separate Equation of Two Lines Nature of Lines
In order to find the separate equation of two lines The lines represented by ax2 + 2hxy + by2 = 0 are
when their joint equation is given, first of all make i. real and distinct, if h2 – ab > 0
RHS equal to zero and then resolve LHS into
ii. coincident, if h2 – ab = 0
two linear factors or use Shri Dharacharya method.
iii. imaginary, if h2 – ab < 0

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 DGT MH –CET 12th MATHEMATICS Study Material 2

Pairs of Straight Lines 71

• Example 3 3 3
The lines represented by the equation mlm2 = 
 2 2
2x2 – p xy + 2y2 = 0 be real, then p lies in the
interval Acute Angle between the Lines
Represented by ax2 +2hxy +by2 = 0
a. ( ,  4)  (4,  ) b. [-4,4]
The angle  between the pair of lines represented
b. (–4, 4] d. None of the above by ax2 + 2hxy + by2 = 0
Sol (c) We have. 2x – pxy + 2y2 = 0
2

It represents two real and distinct straight lines. if  2 (h  ab) 

2

h2 – ab > 0. given by   tan 1  

 ab 
2
 p 
   – 2.2 > 0
 12   2 h 2  ab 
and  = sin–1  (a  b)2  4h 2 
p4  
40
4 The angle between the· lines represented by
 2
p – 16 > 0 ax2 + 2hxy + by2 = 0 is the same as the angle
between the lines represented by
 p < – 4 or p > 4
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Sum and Product of Slopes
Example 5
When lines represented by
The angle between the pair of straight lines
ax2 + 2hxy + by2 = 0
represented by the equation
pass through the origin. let their equations be
12x2 – 10xy + 2y2 + 11x – 5y + 2 = 0 is
y =m1x and y = m2x
then (y–m1x) and (y – m2x) 2 3
a. tan–1 b. tan–1
must be factors of ax2 + 2hxy + by2 = 0, then 7 7
ax2 + 2hxy + by2 = b(y – m1x) (y– m2x) 1
c. tan–1 d. None of these
(making coefficient of i equal on both sides) 7
Now. comparing both sides, then we get Sol (c) We have, a = 12, h = –5 and b =2
2h = – b(m1 + m2) and a = bm1m2 Let  be the angle between the lines, then
2h a 2 h 2  ab 2 25  24
m1 + m 2 = – and m1 m2 = 1
b b tan  = = 12  2 =
ab 7
Example 4
The sum and product of slopes of the pair of 1
 = tan–1
straight lines represented by the equation 7

3x 2  5xy  2y 2 = 0 is Corollary 1
Condition for the lines to be perpendicular
3    3 The lines are perpendicular if the angle between
a.  and  b.  and
2 2 2 2

them is
5  3 2
c. and d. none of these
2 2 
2 2 i.e. =
Sol (b) We have 3 x –5xy – 2y = 0 2
Let m1 and m2 be the slopes of the lines. 
 cot  = cot
 5 2
m1 + m2= –  and 
2 2

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Pairs of Straight Lines 72

 cot  = 0 • The parallel lines will be coincident only as both
pass through a point.
ab
 =0 • The general equation
2 (h 2  ab) ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
 a+b=0 will represent two parallel lines, if g2 – ac > 0 and
2
i.e. Coefficient of x + Coefficient of l = 0 a h g
  and the distance between them is
Hence, the lines represented by ax2 + 2hxy + h b f
by2 = 0 are perpendicular, iff a + b = 0
i.e. Coefficient of x2 + Coefficient of x2 + g 2  ac f 2  bc
coefficient of y2 = 0 2 or 2
a(a  b) b(a  b)
Pair of any two perpendicular lines through
the origin
General Equation of Second· Degree
 Lines represented by The equation ax2 + 2hxy + by2 + 2gx+ 2fy + c = 0
is the general equation of second degree and
ax2 + 2hxy + by2 = 0 represents a conics (pair of straight lines, circle,
be perpendicular, then a + b = 0 or b = – a parabola, ellipse, hyperbola). It contains six
Hence, the equation becomes constants a, b, c, f, g and h.
ax2 + 2hxy – ay = 0 i.e. a = coefficient of x z, b = coefficient of y2
c = constant term, g = half the coefficient of x,
 2h 
or x2 +   xy – y2 = 0 I = half the coefficient of y,
 a 
h = half the coefficient of xy.
or x2 + pxy – y2 = 0
Condition for Straight Lines
where, p is any constant.
The necessary and sufficient condition for
Corollary 2 ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Condition for the lines through the origin to represent a pair of straight lines is that
Pair of lines perpendicular to the lines represented abc + 2 fgh – af 2 – bg – ch2 = 0
by
ax2 + 2hxy + by2 = 0 a h g
and through origin is bx2 – 2hxy + ay2 = 0 h b f
or =0
Aid to memory For perpendicular pairs g f c
interchange the coefficients of x2 and y2 and
change the sign of xy. Some Useful Results
If y = m1x + c1 and y = m2x + c2 are two lines.
Corollary 3
Then, ax2 + 2hxy + by2 + 2gx + 2fy + c
Condition for the lines to be coincident
= b(y – m1x – c1) (y – m2x – c2)
The lines are coincident, if the angle between them
is 0° (orj ) = b[y2 – (m1 + m2) xy + m1m2x2
i.e.  = 0 (or ) + (m1c2 + m2c1) x – (c1 + c2) + c1c2
 tan  = 0 On equating coefficients, we get
2h
2 (h 2  ab) m1 + m2 = –
 =0 b
ab
b
 h2 – ab = 0 m1m2 =
a
 h2 = ab
Hence, the lines represented by ax2 + 2h xy + 2g
m1 c2 + m2c1 =
by2 = 0 are coincident, if f h2=ab, then b
ax2 + 2hxy + by2 is a perfect square.

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Pairs of Straight Lines 73

Example 6 Alternate Method
Equation 12 × 2 – 10xy + 2y + 11x –5y + = 0
2
2

represents a pair of straight lines, then  is

13  13 
Here, x ×15 + 2 ×
2
× 13 × 4– 8 ×  
a. 2 b. 3
2 2
b. –2 d. None of these – 2 × (13)2 –15 × (4)2 = 0
Sol (a) Comparing the given equation with the equation and h2 = (4)2 = 16 = 8x2 = ab
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 we get  The given equation
8×2 + 8xy + 2y2 + 26x + 13y + 15 = 0 ...(i)
11 5
a = 12,h = – 5 b = 2, g = ,f= c= represents two parallel straight lines.
2 2
If the given equation represents a pair of straight Point of Intersection of the Lines
lines, then Represented by
abc + 2fgh – fgh – bg2 – ch2 = 0 ax2 +2hxy+by2 +2gx + 2fy + c = 0
The point of intersection of the lines represented
 5  11 by
 12×2 ×  + 2 ×      ( 5) –12 × 25 = 0
 2 2 ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is
121   f 2  bc 
–2× –  × 25 = 0  g 2  ca  
  2 , 
  h  ab   2
4
  h  ab  
A = 2, also h2 – ab = 25 – 24 =  > 0
So, the given equation will represents a pair of
 bg  hf af  gh 
straight lines, if  = 2 or  2 , 2 
 h  ab h  ab 
Example 7
The equation 8x2 + 8xy+2y2+26x + 13y + 15 =0
Example 8
represents a pair of straight lines which is If the pair of lines
a. parallel b. perpendicular ax2 + 2hxy + by2 + 2gx + 2.fy + c = O
b. imaginary d. None of these intersect on the Y-axis, then
Sol (a) The given equation is a. 2f = g2 + h2 b. 2fgh = bg2 + ch2
8x2 + 8xy + 2y2 + 26x + 13y + 15= 0 b. 2g = bf2 + ch2 d. None of these
Writing Eq. (i) as quadratic equation in x, we get Sol (b) We have,
8x2 + 2x + 2y2 + 13y + 15 = 0 abe + 2fgh – af2 – bg2 – ch2 = 0 ...(i)
Let the point of intersection on Y-axis be (0,).
(4y  13)  4(4y  13) 2 (2y 2  13  15 Then, from
 x
16  bg  hf af  gh 
 2 , 2  , we ave
(4y  13)  4(4y  13)  8(2y  13y  15)
2 2  h  ab h  ab 
 x=
8 fh = bg
The equation of the Y-axis is x = O.
(4y  13)  7
 x= Solving this and the equation of the pair of lines,
8 we get
 8x = – 4y – 13 + 7, i.e. 2x + 2y + 3 = 0 by2 + 2fy + c = 0
and 8x = – 4y –13 – 7, i.e, 2x + y + 5 = 0 which must have equal roots
i.e. the given Eq. (i) represents two straight lines f2 = bc
2x + y + 5 = O and 4x + 2y + 3 = 0 Eq. (ii) can be written as f.gh = bg2
i.e. 2x + y + – = 0 Putting f2 = bc, fgh = bg2 in Eq. (i), we get
both lines are parallel. bg2 = ch2
2fgh = bg2 + ch2

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Pairs of Straight Lines 74

2 2
Bisectors of the Angle between the a. x + y + 4xy – 6x – 4y + l = 0
Lines Given by a Homogeneous b. x2 + y2 – 4xy – 6x + 6y – 4 = 0
Equation b. x2 + y2 – 4xy + 6x – 6y + 6 = 0
The joint equation of bisectors of the angles d. None of the above
between the lines represented by the equation Sol (c) We have the equation of pair of straight lines
ax2 + 2hxy + by2 = 0 6x2 + 5xy – 4y + 7x + 13y – 3 = 0
x 2  h 2 xy Its points of intersection of the pair of lines is the
is point of intersection of
ab h
12x + 5y + 7 = 0 and 5x – 8y + 13 = 0.
Example 9 On solving, we get x = –1 and y = 1.
Find the equation of bisectors of the angle between Hence, equation of pair of bisector is
the lines represented by3x2 – 5xy + 4y2 =0.
a. 9x2 + 6y2 – 2x = 0 b. 3x2 + 2xy – y2 = 0 (x  1) 2  (y  1) 2 2 (x  1)(y  1)

b. 5x2+xy+4y2 = 0 d. Sx2+ 2xy– sy2 = 0 10 5
Sol (d) Given equation is 3x2 – 5xy + 4y2 = 0 ...(i)  x2 + y2 – 4xy + 6x – 6y + 6 = 0
Comparing it with the equation Removal of First Degree Terms
ax2 + 2hxy + by2 = 0 = (ii) Let point of intersection of lines represented by
5 ax2 + 2hxy + by + 2gx + 2fy + c = 0 ...(i)
then a = 3,h = – , b = 4 is (, )
2
Hence, equation of bisectors of the angle between bg  fh af  gh
Here, (, ) = ,
the pair of the line (i) is h 2  ab h 2  ab
x 2  y2 xy x 2  y 2 2xy For removal of first degree terms, shift the origin
   to (,)
3  4 5/ 2 1 5
i.e. replacing x by (X + ) and y by (Y + ) in line
 5x2 – 2xy – 5y2 = 0
(i).
x2 – y2 = 2xy
Alternate Method Direct equation after removal
Bisectors of the Angle between the Lines of first degree terms is
Given by General Equation aX2 + 2hXY + by2 + (g + f + c) = 0
The pair of bisectors of the lines represented by
bg  fh af  gh
ax2 + 2hxy + by + 2gx + 2fy + c = 0 where, a = and  = 2
h 2  ab h  ab
(x  a) 2 (x  ) 2 (x  ) (y  ) Example 11
is 
(a  b) h
Find the new equation of curve
where, (, )be the point of intersection of the 12x2 + 7xy – 12y2 – 17x – 31y –7 = 0
pair of straight lines represented by Eq. (i).
after removing the first degree terms.
If ax2+2hxy + by2 + 2gx + 2fy + c = 0
a. 6X2 + 5XY – 10Y2 = 0
represents two straight lines, then the equation of
b. 12X2 + 7XY – 12Y2 = 0
lines through the origin and parallel to them is
c. 12y2 – 5XY + 12X2 = 0
ax 2  2hxy  by 2  0 d. 10X2 + 5XY – 10Y2 = 0
Sol (b) Let  12x2 + 7xy –12y2 –17x – 31y –7 = 0
Example 10
The equation of pair of bisectors of the lines 
 = 24x + 7y – 17 = 0
represented by the equation x
6x2 + 5xy – 4y2 + 7x + 13y – 3 = 0 is

and  7x – 24y – 31 = 0
x

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Pairs of Straight Lines 75

Their point of intersection is (x,y) (1, -1) y  3x
Here,  =1, = – 1  =1 ..(ii)
2
Shift the origin to (1, –1) and replacing x = X + 1
Making Eq. (i) homogeneous with the help of
and y = Y – 1 in Eq. (ii), then
Eq. (i), then the required equation is
12 (X + 1)2 + 7 (X + 1)(Y – 1) –12(Y –1)2  y  3x   y  3x 
x2 + 2xy + 3y2 + 4x   + 8y   –11
1
– 17(X + 1) – 31(Y–1) –7 = 0  2   2 
12X2 + 7XY – 12y2 = 0  y  3x 
2

Equation of the Line Joining the Origin to   =0

 2 
the Points of Intersection of a Given Line
and a Given Curve
or x2 + 2xy + 3y2 + 2xy – 6x2 + 4y2 – 12xy –
The combined equation of the straight lines joining
the origin to the points of intersection of a second 11
(y – 3X)2 = 0
degree curve 4
ax2 + 2hxy + by2 + 2gx +2fy + c =0
11 2
and a straight line lx + my + n = 0 is  – 5x2 – 8xy + 7y2 – (y – 6xy + 9x2) = 0
4
 2x  my   20x2 – 32xy + 28y2 – 11y2 + 66xy – 99x2 = 0
ax2 + 2hxy + by2 + 2gx  
 n   119x2 – 34xy – 17y2 = 0
2
or 7x2 – 2xy – y2 = 0 ...(iii)
 lx  my   lx  my  This is the equation of lines joining the origin to
+ 2 fy   +c   =0
 n   n  the points of intersection of Eqs. (i) and (ii).
Example 12 Comparing Eq. (iii) with ax2 + 2h.xy + by2 = 0,
The angle between the lines joining the origin to we get a = 7, h = –1, b = –1
the points of intersection of the straight line If  is the acute angle between pair of lines of Eq.
y =3x + 2 with the curve (iii), then
x2 + 2xy + 3y2 + 4x + 8y – 11 = 0 is
2 h 2  ab 2 (1  7) 2 8 4 2
2 2  tan  =  = =
 ab 7 1 6 6

a. tan–1  3 
 b.
2
2 2
=
c. d. None of these 3
Sol (a) Equation of curve is
2 2 
x2 + 2xy + 3y2 + 4x + 8y –11 = 0 
  = tan  3 

–1
and line y = 3x + 2 ..(i)

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Pairs of Straight Lines 76

Practice Exercise
Exercise 1
(Topical Problems)

Combined Equation of Two Lines and 7. The angle between the pair of lines represented
Homogeneous Equation of Second Degree by 2x2 – 7xy + 3y2 = a is
1. The combined equation of lines 2x + y + 3 = a a. 60° b. 45°
and x – y + 4 = a is
7
a. x2 + 2y2 + 10x – 12y + 5 = a b. tan–1   d. 30°
6
b. x2 + y2 + 2x + 3y + 5 = a
b. 2x2 – xy – y2 + 11x + y – 12 = a 8. If e is the acute angle between the lines given by
x2 – 2pxy + y2 = 0, then
d. None of the above
a. cos  = P b. tan  = P
2. Separate equation of lines represented by the
b. see  = p d. cot  = p
equation 6x2 + 5xy – 4y2 = a is
9. If the pairs of straight lines ax2 + 2hxy – ay2 = a
a. x – 2y = 0 and x + 3y = 0 and bx2 + 2gxy – by2 = a be such that each bisects
b. 2x – y = a and 3x + 4y = 0 the angles between the other, then
b. x + y = 0 and x – 4y = 0 a. hg + ab = 0 b. ah + bg = 0
2
d. None of the above b. h – ab = 0 d. ag + bh = 0
3. The two straight lines given by 10. If the equation x2 + 4xy + 5y2 = a represents
x2(tan2  + cosy2) – 2xy tan  + y2 sin2  = a two lines inclined at an angle n, then A is equal to
a. 5/4 b. 4/5
make with the axis of x angles such that the
difference of their tangents is b. – 45 d. None of these
11. The product of the perpendiculars drawn from
a. 4 b. 3
the point (1, 2) to the pair of lines x2 + 4xy + y2 =
b. 2 d. None of these a is
4. If the slopes of the lines given by a. 9/4 b. 3/4
ax2 + 2hxy + by2 = a are in the ratio 3 : 1, then h2 b. 9/16 d. None of these
is equal to 12. The equation of straight lines through the point
ab 4ab (x1, y1) and parallel to the lines given by
a. b. ax2 + 2hxy + by2 = 0 is
3 3
a. a(y – y1)2 + 2h(x – x1) (y – y1)+ b(x – x1)2 = 0
4a b. a(y – y1)2– 2h(x – x1)(y – y1)+ b(x – x1)2 = 0
b. d. None of these
3b c. b(y – y1)2+ 2h(x – x1) (y – y1)+ a(x – x1)2 = 0
5. The equation to the pair of lines perpendicular to d. None of the above
the pair of lines 3x2 – 4xy + y2 = 0, is
13. The combined equation of the pair of lines through
a. x2 + 4xy + 3y2 = 0 b. x2 – 4xy – 3y2 = 0 the point (1, 0) and perpendicular to the lines
b. x2 + 4xy + y2 = 0 d. None of these represented by 2x2 – xy – y2 = a is
6. If the pairs of lines ax2 + 2hxy + by2 = 0 and a. 2x2 – xy – y2 – x + y –1 = 0
a' x2 + 2h' xy + b' y2 = a have one line in common, b. 2y2 + xy – x2 + 2x – Y – 1 = 0
then (ab' – a' b)2 is equal to b. 2y2 + xy – x2 – x – y + 2 = 0
a. (h' b – hb') (ha' – h' a) d. None of the above
b. 4(h' b – hb') (ha' – h' a) 14. The combined equation of the lines L1, and L2 is
b. 2(h' b – hb') (ha' – h' a) 2x2 + 6xy + y2 = 0 and that of the lines L3 and L4
d. 4(h' b + hb') (ha' + h' a) is 4x2 + 18xy + y2 = 0. If the angle between
L1 and L4 be , then the angle between L2 and
L3 will be

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 22. The equation of two straight lines through the point

a. – b. 2 (x1' y1) and perpendicular to the lines given by
2
ax1 + 2hxy + by2 = 0 is
 a. b(x – x1)2 – 2h(x – x1) (y – y1) + a(y – y1)2 = 0
b. +  d. 
4 b. b(x – x1)2 + 2h(x – x1) (y – y1)+ a(y – y1)2 = 0
15. If the pair of lines represented by c. a(x – x,)2 – 2h(x – x1)(y – y1) + b(y – y1)2 = 0
ax2 + 2hxy + by2 = 0, b 0, are such that the sum d. None of the above
of the slopes of the lines is three times the product 23. The triangle formed by the lines whose combined
of their slopes, then equation is (y2 – 4xy – x2) (x + y – 1) = 0, is
a. 3b + 2h = 0 b. 2a+3h=O a. equilateral b. right angled
b. 3a + 2h = 0 d. None of these b. isosceles d. obtuse angled
16. If the sum of slopes of the lines given by 24. The combined equation of the pair of lines through
4x2 + 2kxy – 7y2 = 0 is equal to the product of the origin and perpendicular to the pair of lines
slopes, then k is equal to given by ax2 + 2hxy + by2 = 0, is
a. ax2 – 2hxy + by2 = 0
a. –4 b. 4
b. bx2 + 2hxy + ay2 = 0
b. -2 d. 2
c. bx2 – 2hxy + ay2 = 0
17. The combined equation of the images of pair of
d. bx2 + 2hxy – ay2 = 0
lines given by ax2 + 2hxy + by2 = 0 in the line
25. If the slope of one of the lines represented by
mirror y = 0, is
ax2 + 2hxy + by2 = 0 be the square of the other,
a. ax2 – 2hxy + by2 = 0
a  b 8h 2
b. bx2 – 2hxy + ay2 = 0 then 
h ab
b. bx2 + 2hxy + ay2 = 0
a. 3 b. 4
d. None of the above
c. 5 d. 6
18. The angle between the lines represented by
26. Let a and b be non-zero and real numbers. Then,
x2 + 2xy – sec + y2 = 0 is
the equation (ax2 + by2 + c) (x2 – 5xy + 6y2) = 0
a. 4 b. 2 represents
b.  d. None of these a. four straight lines, when c = 0 and a, b are of
19. The equation 3x + 2hxy + 3y2 = 0 represents a
2
the same sign
pair of straight lines passing through the origin. b. two straight lines and a circle, when a = band
The two lines are c is of sign opposite to that of a
a. real and distinct, if h2 > 3 b. two straight lines and a hyperbola, when a and
b. real and distinct, if h2 > 9 bare of the same sign and c is of sign opposite to
that of a
b. real and coincident, if h2 = 3
d. a circle and an ellipse, when a and b are of the
d. real and coincident, if h2 > 3
same sign and c is of sign opposite to that of a
20. The set of values of h for which the equation 27. The slopes of lines represented by
4x2 + hxy – 3y2 = 0 represents a pair of real and
x2 + 2hxy + 2y2 = 0 are in the ratio 1 : 2, then h
distinct lines, is
equals
a. R b. (–3, 4)
1 3
b. (3, 4) d. (4, ) a. + b. +
2 2
21. One bisector of the angle between the lines given
by a (x – 1)2 + 2h(x–1) y+by2=0 is 2x + y – 2 = 0. c. + 1 d. + 3
The equation of the other bisector is 28. If one of the lines given by 6x2 – xy + 4cy2 = 0 is
a. x – 2y + 1= 0 b. x – 2y – 2 = 0 3x + 4Y = 0, then c equals
b. x – 2y – 1= 0 d. None of these a. 1 b. –1
b. 3 d. –3

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2 2
29. The equation of pair of lines joining origin to the 36. The lines represents by ax + 2hxy + by = 0 are
points of intersection of x2 + y2 = 9 and x + y = 3, perpendicular to each other, if
is a. h2 = a + b b. a + b = 0
a. x2 + (3 – x) 2 = 9 b. xy = 0 2
c. h = ab d. h = 0
b. (3 + y)2 + y2 = 9 d. (x – y)2 = 9 37. If the pair of straight lines given by
30. The centroid of the triangle formed by the pair of Ax2 + 2Hxy + By2 = 0 (H2 > AB) forms an
straight lines 12x2 – 20xy + 7y2 = 0 and the line equilateral triangle with line ax + by + e = 0, then
2x – 3y + 4 = 0, is
(A + 3B) (3A B) is equal to
 7 7  8 8 a. H2 b. – H2
a.   ,  b.   , 
 3 3  3 3 c. 2H2 d. 4H2
General Equation of Second Degree
8 8 4 4 38. The equation
c.  ,  d.  , 
 3 3  3 3
ax2 + 2 ab xy + by2 + 2gx + 2fy + c = 0
2 2
x y 2xy represents a pair of parallel straight lines, if
31.   = 0 represents pair of straight lines
a b h a. ag2 = bf2 b. a2g = b2f
such that slope of one line is twice the other. Then,
c. bg2 = af2 d. b2g = a2f
ab : h2 is
39. If the lines joining the origin to the points of
a. 9 : 8 b. 8 : 9
intersection of the line y = mx + 2 and the curve
c. 1 : 2 d. 2 : 1 x2 + y2 = 1are at right angles, then
32. The equation 4x – 24xy + 11y2 = 0 represents
2
a. m2 = 1 b. m2 = 3
a. two paraliellines 2
c. m = 7 d. 2m2 = 1
b. two perpendicular lines 40. If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents
b. two lines through the origin parallel straight lines, then
d. a circle a. hf = bg b. h2 = be
33. The distance between the pair of parallel lines b. a2f = b2g d. None of these
given by x2 –1005x + 2006 = 0 is 41. The equation
a. 1001 b. 1000 8x2 + 8xy + 2y2 + 26x + 13 y + 15 = 0
c. 1005 d. 2006 represents a pair of straight lines. The distance
34. The area (in sq unit) of the triangle formed by between them is
x + y + 1= 0 and the pair of straight lines
c c
x2 – 3xy + 2y2 = 0 is a. b.
(a  b)  4h
2 2
(a  b) 2  4h 2
7 5
a. b. c
12 12
c. d. None of these
(a  b) 2  4h 2
1 1
c. d. 42. If the equation ax2 + 2hxy+by2+2gx+2fy + c = 0
12 6
represents two straight lines, then the product of
35. If the lines px2 – qxy – y2 = 0 makes the angles
the perpendicular from the origin on these straight
 and with X-axis, then the value of tan ( +)
lines, is
is
q q c c
a. b. a. b.
1 p 1 p (a  b)  4h
2 2
(a  b) 2  4h 2

p p c
c. d. c. d. None of these
1 q 1 q (a  b) 2  4h 2

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43. The straight lines joining the origin to the points 50. If the lines
of intersection of the line kx + hy = 2hk with the x2 + 2xy – 35y2 – 4x + 44y – 12 = 0
curve (x – h)2 + (y – k)2 = e2 are at right angles, if
and 5x+y – 8 = 0 are concurrent, then the value
a. h2 + k2 = c2 b. h2 + k2 = 2c2 of  is
b. h2 – k2 = e2 d. None of these a. 0 b. 1
44. The equation ax + by2 + cx + cy = 0, c  0
2
b. –1 d. 2
represents a pair of straight lines, if
51. In order to eliminate the first degree terms from
a. a + b = 0 b. b + c = 0 the equation 2x2 + 4xy + 5y2 – 4x – 22y + 7 = 0,
c. b + c = 0 d. None of these the point to which origin is to be shifted, is
45. If the equation, 12x +txy–py2 –18x + qy + 6 = 0
2
a. (1,– 3) b. (2, 3)
represents a pair of perpendicular straight lines, b. (–2,3) d. (1,3)
then 2 2
52. If ax – y + 4x – y = 0 represents a pair of lines,
a. p = 12, q = – 1 b. P = 12, q = 1 then a is equal to
c. P = –12, q = 1 d. P = 1,q = 12 a. –16 b. 16
46. Observe the following columns: b. 4 d. – 4
Column I Column" 53. The lines represented by the equation
A. If the pair of lines represented by the P.  +  = 8
x2 – y2 – x + 3y – 2 = 0 are
equation 12x2–10xy+21+11x–5y +2 = 0 be
3x – y + A= 0 and 4x – 2y + = 0, then a. x + y – 1= 0, x – y + 2 = 0
B. If the pair of lines represented by the Q.  +  = 9 b. x – y – 2 = 0, x + y + 1= 0
equation 6x2+17xy+12y2+22x+31y+20= 0 b. x + y + 2 = 0, x – y – 1= 0
be 2x+3y + = 0 and 3x + 4y +  = 0, then d. x – Y + 1= 0, x + y – 2 = 0
C. If the pair of lines represented by the
54. The value of 'A, for which the equation
equation 8x2+8xy+21+26x+13y+15= 0 be R. +  = 3
x2–y2–x + y – 2 = 0 represents a pair of straight
2x + y + A= 0 and 4x + 2y +  = 0, then
lines, are
S. 2 + 2 = 41
T. 2 + 2 = 34 a. 5/2 b. + 5
Codes b. + 3 d. 2/5
A B C A B C 55. The value of 'p' for which the equation
a. R Q,S P,T b. Q.S P T x2 + pxy + y2 – 5x – 7y + 6 = 0 represents a pair
of straight lines, is
b. Q R S d. P.Q Q T
a. 5/2 b. 5
47. The pairs of straight lines x – 3xy + 2y2 = 0 and
2

x2 – 3xy + 2y2 + x – 2 = 0 form a c. 2 d. 2/5

a. square but not rhombus 56. The equation of the pair of straight lines
perpendicular to the pair
b. rhombus
2x2 + 3xy + 2y2 + 10x + 5y = 0 and passing
b. parallelogram
through the origin, is
d. rectangle but not a square
a. 2x2 + 5xy + 2y2 = 0
48. If 3x2+ xy – y2– 3x + 6y + k = 0 represents a pair
b. 2x2 – 3xy + 2y2 = 0
of lines, then k is equal to
b. 2x2 + 3xy + y2 = 0
a. 0 b. 9
d. 2x2 – 5xy + 2y2 = 0
b. 1 d. – 9
57. If the equation kx2 – 2xy – y2 – 2x + 2y = 0
49. The value of 'A such that
represents a pair of lines, then k is equal to
x2 – 10xy + 12y2 + 5x – 16y–3 = 0 represents a
a. 2 b. – 5
pair of straight lines, is
c. – 2 d. 3
a. 1 b. –1
c. 2 d. –2

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Exercise 2
(Miscellaneous Problems)
1. The lines represented by 8. The point of intersection of the pair of straight
x2 + 2xy + 2y2 = 0 and the lines represented by lines given by
(1+ ) x2 – 8xy + y2 = 0 are equally inclined, then 6x2 + 5xy – 4y2 + 7x + 13 y – 3 = 0, is
a. is any real number b.  > 2 a. (1, 1) b. (1,– 1)
c. = + 2 d. < –2 c. (–1, 1) d. (–1, –1)
2. The equation (x3 – 3xy2) + y3 – 3x2y = 0 9. If first degree terms and constant terms are to be
represents three straight lines passing through the removed from the equation
origin such that 12x2 + txy –12y2 –17x–31y –7 = 0,
a. they are equally inclined to one another then the origin must be shifted at the point
b. two of which are at right angles a. (1,–1) b. (–1, 1)
c. two of which are coincident b. (–1, –1) d. None of these
d. None of the above 10. The straight lines represented by
3. The equation x3 + x2y – xy2 – y3 = 0 represents (y – mx)2 = a2(1 + m2) and (y – nx)2 = a2(1 + n2)
three straight lines passing through the origin such form a
that
a. rectangle b. trapezium
a. two of them are coincident and two of them
b. rhombus d. None of these
are perpendicular
11. A pair of perpendicular straight lines is drawn
b. two of them are coincident but no two are
through the origin forming with the line
perpendicular
2x + 3y = 6 an isosceles triangle right angled at
c. two of them are perpendicular but no two are the origin. The equation to the line pair is
coincident
a. 5x2 – 24xy–5y2=0 b. 5x2 – 26xy – 5y2 = 0
d. None of the above
b. 5x2+24xy – 5y2=0 d. 5x2 + 26xy – 5y2 = 0
4. The straight lines represented by
12. If the origin is shifted to the point (ab / (a – b), (0)
x2 + mxy – 2y2 + 3y – 1 = 0 meet at the point without rotation, then the equation
a. (1/3,-2/3) b. (–1/3, –2/3) (a – b) (x2 + y2) – 2abx = 0 becomes
c. (1/3, 2/3) d. None of these a. (a – b) (x2 + y2) – (a + b) XY + ab x =a2
5. The angle between the straight lines joining the b. (a + b)(X2 + y2) = 2ab
origin to the point of intersection of
c. (X2 + y2) = (a2 + b2)
3x2 + 5xy – 3y2 + 2x + 3y = 0 and 3x–2y = 1 is
d. (a – b)2 (X2 + Y2) = a2b2
  13. If one of the lines of my2 + (1– m2 ) xy – mx2 = 0
a. b.
3 4 is a bisector of the angle between the lines xy = 0,
then m is
 
c. d. a. 3 b. 2
6 2
6. All chords of the curve 3x2– y2 – 2x + 4y = 0 b. –1/ 2 d. –1
which subtend a right angle at the origin always 14. The slopes of the lines represented by
asses through the point x2 + 2hxy + 2y2 = 0 are in the ratio 1 : 2, then
a. (1, 2) b. (–1, 2) h equals
b. (1,–2) d. (–1, –2) 3
7. If the equation x4 + bx3y + cx2y2 + dxy + ey4 = 0 a. + –1 b. +
2
represents two pairs of perpendicular lines, then
b. + 1 d. + 3
a. b + d =1 and e= –1 b. b + d = 0 and e = –1
b. b+d = 0 and e = 1 d. None of these

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2 2 2 2
15. The equation x y –9y +6x y + 54 y = 0 represents 22. The combined equation of the lines 'l1' 'l2 is ?
a. a pair of straight lines and a circle 2x2 + 6xy + y2 = 0 and that of the lines m1, m2 is
4x2 + 18xy + y2 = 0. If the angle between '1 and
b. a pair of straight lines and a parabola
m2 be , then the angle between l2 and m1 will be
c. a set of four straight lines forming a square
a.  / 2 –  b. 2a
d. None of the above
c. / 4 +  d. 
16. The equation x – y = 4 and x2 + 4xy + y2 = 0
23. The equations a x + 2h(a + b)xy + b2y2 = 0
2 2
represent the sides of
and ax2 + 2hxy + by2 = 0 represent
a. an equilateral triangle b. a right angled triangle
a. two pairs of perpendicular straight lines
b. an isosceles triangle d. None of these
b. two pairs of parallel straight lines
17. The distance between the two lines represented
by equation c. two pairs of straight lines which are equally
inclined to each other
9x2 – 24xy + 16 y2 – 12x + 16 Y – 12 = 0 is
d. None of the above
a. 8/5 units b. 8/5 units
24. The lines a2x2 + bcy?=a (b + c)xy will be oincident,
b. 11/S units d. None of these
if
18. If the pairs of lines x + 2xy + ay2 = 0 and
2
a a = 0 or b = c
ax2 + 2xy + y2 = 0 have exactly one line in b. c = 0 or a = b
common, then the joint equation of the other two
b. a = b or a = c
lines is given by
d. a = b + c
a. 3x2 + 8xy – 3y2 = 0
25. If the slope of one of the lines represented by the
b. 3x2 + 10xy + 3y2 = 0
equation ax2 + 2hxy + by2 = 0 is A times that of
b. y2 + 2xy – 3x2 = 0 the other, then
d. x2 + 2xy – 3y2 = 0 a. 4 h = ab (1 + ) b. h = ab(1 +)2
19. The angle between the pair of lines whose c. 4h2 = ab(1 + )2 d. None of these
equation is
26. The equation y2 – x2 + 2x – 1= 0 represents
4x2 + 10xy + my? + 5x + 10y = 0 is
a. a pair of straight lines
a. tan–1(3/8)
b. a circle
b. tan–1(3/ 4)
b. a parabola
c. tan–1 2 25  4m / m  4 , m  R d. an ellipse
d. None of the above 27. The pair of straight lines passing through the point
(1, 2) and perpendicular to the pair of straight
20. The pair of lines represented by
lines 3x2 – 8xy + 5y2 = 0, is
3ax2 + 5xy + (a2 – 2) y2 = 0 are perpendicular to
a. (5x + 3y + 11) (x + Y + 3) = 0
each other for
b. (5x + 3y –11) (x + Y – 3) = 0
a. two values of a b. a
b. (3x + 5y –11)(x + Y + 3) = 0
b. for one value of a d. for no value of a
d. (3x – 5y + 11)(x + Y – 3) = 0
21. If the equation of the pair of straight lines passing
through the point (1, 1), one making an angle 8 28. If two sides of a triangle are represented by
with the positive direction of X-axis and the other x2 – 7xy + 6y2 = 0 and centroid is (1, 0), then the
making the same angle with the positive direction equation of third side is
of Y-axis, is a. 2x + 7y + 3 = 0 b. 2x + 7y – 3 = 0
x2 – (a + 2) xy + y2 + a(x + y – 1) = 0, a– 2, then b. 2x – 7y + 3 = 0 d. 2x – 7y – 3 = 0
the value of sin 2  is 29. If the angle between the lines represented by the
a. a–2 b. a + 2 equation y2 + kxy – x2 tan2 A = 0 is 2A, then k is
c. 2 / (a+2) d. 2/a equal to
a. 0 b. 1
c. 2 d. tan A

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30. If the lines represented by the equation
4
2x2 – 3xy + y2 = 0 make angles  and  with c. d. 10
10
X-axis, then cot  + cot2  is equal to
37. Two lines are given by (x – 2y)2 + k(x – 2y) = 0.
3 The value of k, so that the distance between them
a. 0 b.
2 is 3, is

7 5 1 2
c. d. a. b. +
4 4 5 5
31. The line x – 2y = 0 will be a bisector of the angle c. + 3. 5 d. None of these
between the lines represented by the equation
38. The locus of the point P(x, y) satisfying the relation
x2 – 2hxy – 2y2 = 0, if h is equal to
1 (x  3) 2  (y  1) 2 + (x  3) 2  (y  1) 2 = 6,
a. b. 2 is
2
a. point
1
c. –2 d.  b. pair of coincident straight lines
2
c. circle
32. If r (1– m2) + m(p – q) = 0, then a bisector of the
d. ellipse
angle between the lines represented by the
equation px2 – 2rxy + qy2 = 0, is 39. The equation of pair of straight lines joining the
point of intersection of the curve x2 + y2 = 4 and
a. y = x b. y = – x
y – x = 2 to the origin, is
c. y = mx d. ym = x
a. x2 + y2 =(y – x)2
33. The equation of the perpendiculars drawn from
b. x2 + y2 = 4(y –x)2
the origin to the lines represented by the equation
2x2 –10xy +12y2 +5x –16y –3 = 0,is b. x2 + y2 + (y – x)2 = 0
a. 6x2 + 5xy + y2 = 0 b. 6y2 + 5xy + x2 = 0 d. x2 + y2 + 4 (y – x)2 = 0
b. 6x2– 5xy + y2 = 0 d. None of these 40. The lines joining the origin to the points of
intersection of the line 3x – 2y = 1and the curve
34. The equation 2x2 + 4xy – py2 + 4x + qy + 1= 0
3x2 + 5xy – 3y2 + 2x + 3y= 0, are
will represent two mutually perpendicular straight
lines, if a. parallel to each other
a. P = 1and q = 2 or 6 b. perpendicular to each other
b. P = 2 and q = 0 or 8 c. inclined at 45° to each other
c. P = 2 and q = 0 or 6 d. None of the above
d. P = – 2 and q = – 2 or 8 41. The pair of straight lines joining the origin to the
35. The acute angle formed between the lines joining points of intersection of the line y = 2 2 x + c
the origin to the points of intersection of the curves and the circle x2 + y2 = 2 are at right angles, if
x2 + y2 – 2x –1 = 0 and x + y = 1,is a. c2 – 4 = 0 b. c2 – 8 = 0
 1 c. c2 – 9 = 0 d. c2 –10 = 0
a. tan–1    b. tan–1(2) 42. The value of h for which the equation
 2
3x2 + 2hxy – 3y2 – 40x + 30y – 75 = 0 represents
1 a pair of straight lines, are
b. tan–1   d. 600
2 a. 4, 4 b. 4, 6
36. The distance between the parallel lines b. 4, – 4 d. 0, 4
9x2 – 6xy + y2 + 18x – 6y + 8 = 0, is 43 Which of the following second degree equation
represented a pair of straight lines?
1 2
a. b. a. x2 – xy – y2 = 1 b. –x2 + xy – y2 = 1
10 10
c. 4x –4xy+ y = 4 d. x2 + y2 = 4
2 2

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44. The equation of one of the lines represented by 52. If in the general quadratic equation
the equation pq(x2 – y2) + (p2 – q2)xy = 0, is f(x, y) = 0,  = 0 and h2 = ab, then the equation
a. px + qy = 0 b. px – qy = 0 represents
2 2
c. p x + q y = 0 d. q2x – p2y = 0 a. two parallel straight lines
45. If the slope of one line of the pair of lines b. two perpendicular straight lines
represented by ax2 + 4xy + y2 = 0 is 3 times the c. two coincident lines
slope of the other line, then a is d. None of the above
a. 1 b. 2 53 Difference of slopes of the lines represented by
c. 3 d. 4 equation
x2(sec2 – sin2) – 2xy tan + y2 sin2 = 0, is
46. If the acute angles between the pairs of lines
a. 4 b. 3
3x2 –7xy + 4y2 = 0 and 6x2 – 5xy + y2 = 0 be 1
b. 2 d. None of these
and 2 respectively, then
54. The lines (lx + my)2 – 3 (mx – ly)2 = 0 and
a.  = 2 b. 2 = 2
lx + my + n = 0 form
c. 21= 22 d. None of these a. an isosceles triangle
47. The lines represented by the equation b. a right angled triangle
x2 + 2 3 xy + 3y2 – 3x – 3 3 y – 4 = 0, are b. an equilateral triangle
d. None of the above
a. perpendicular to each other
55. If the lines (p – q)x2 + 2(p + q)xy + (q – p)y2 = 0
b. parallel
are mutually perpendicular, then
c. inclined at 45° to each other a. p = q
d. None of the above b. p = 0
48. If the bisectors of the angles between the pairs of b. q = 0
lines given by the equation ax2 + 2hxy + by2 = 0 d. p and q may have any value
and ax 2 + 2hxy + by 2 + A(x 2+ y2) = 0 are 56. If 6x2 + 11xy –10y2 + x + 31y + k = 0 represents
coincident, then A is a pair of straight lines, then k is equal to
a. a b. b a. –15 b. 6
c. h d. any real number b. –10 d. – 4
49. The orthocentre of the triangle formed by the lines 57. The pair of lines passing through the origin and
xy = 0 and x + y = 1,is parallel to the lines represented by the equation
2x2 – xy – 6y2 + 7x + 21y –15 = 0, is
1 1
a. (0,0) b.  ,  a. 2x2 – xy – 6y2 = 0 b. 6x2 – xy + 2y2 = 0
2 2
c. 6x2 – xy – 2y2 = 0 d. 2x2 + xy – 6y2 = 0
58. The equation xy + a2 = a (x + y) represents
1 1 1 1
c.  ,  d.  ,  a. a parabola
 3 3 4 4
b. a pair of straight lines
50. If the equation 2x2 – 2hxy + 2y2 = 0 represents b. an ellipse
two coincident straight lines passing through the
d. two parallel straight lines
origin, then h is equal to
59. If the equation ax2 + by2 + cx + cy = 0 represents
a. + 6 b. 6 a pair of straight lines, then
a. a (b+c) = 0 b. c(a + b) = 0
b. – 6 d. + 2
b. b(c + a) = 0 d. a + b + c = 0
51. The equation of one of the lines represented by 60. If the equation x2 + 2y2 – 5xy + 5x – 7y + 3 = 0
the equation x2 – 2xy cot 8 – y2 = 0, is represents two straight lines, then the value of 
a. x – y cot  = 0 will be
b. x + y tan  = 0 a. 3 b. 2
c. y sin  + x(cos  + 1) = 0 c. 8 d. – 8
d. x cos  + y (sin  + 1) = 0

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Pairs of Straight Lines 84

61. If the angle between the two lines represented by 68. The equation of the lines passing through the
2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0, is tan–1 origin and having slopes 3 and – 2, is
m, then m is equal to a. 3y2 + 8xy – 3x2 = 0
1 b. 3x2 + 8xy _3y2 = 0
a. b. 1
5 b. 3y2 – 8xy + 3x2 = 0
d. 3x2 + 8xy + 3y2 = 0
7
c. d. 7 69. The condition of representing the coincident lines
5
by the general quadratic equation f(x, y) = 0, is
62. The point of intersection of the lines represented a.  = 0,h2 = ab
by the equation 2x2+3y2+7xy + 8x + 14y + 8 = 0,
is b.  = 0 and a + b = 0
a. (0, 2) b. (1, 2) c.  = 0, h2 = ab, g2 = ac, f2 = bc
c. (– 2, 0) d. (– 2, 1) d. h2 = ab, g2 = ac and f2 = be
70. If the sum of the slopes of the lines given by
63. The equation of second degree x2 + 2 2 xy + x2 – 2cxy – 7y2 = 0 is four times their product,
2y2 + 4x + 4 2 y + 1= 0 represented a pair of then c has the value
straight lines, the distance between them is a. – 2 b. –1
b. 2 d. 1
4
a. 4 b. 71. If the angle between the pair of straight lines
3
represented by the equation x2 – 3xy + Ay2 +
c. 2 d. 2 3 3x – 5y + 2 = 0 is tan (1/3), where A is non-
64. The lines joining the points of intersection of line negative real number, then A is equal to
x + y = 1and curve x2 + y2 – 2y + A = 0 to the a. 2 b. 0
origin are perpendicular, then the value of A will c. 3 d. 1
be 72. If the equation ax + 2hxy + by2 = 0 has the one
2

1 1 line as the bisector of angle between the coordinate

a. . b. – axes, then
2 2
a. (a – b)2 = h2 b. (a + b)2 = h2
1 b. (a–b)2 = 4h2 d. (a+b)2 = 4h2
c. d. 0
2 73. If the lines ax2 + 2hxy + by2 = 0 represents the
65. The equation of the line joining origin to the points adjacent sides of a parallelogram, then the equation
of intersection of the curve x2 + y2 = a2 and of second diagonal if one is lx + my = 1, will be
x2 + y2 – ax – ay = 0, is a. (am + hl) x = (bl + hm) y
a. x2 – y2 = 0 b. xy – x2 = 0 b. (am – hl) x = (bl–hm)y
b. xy = 0 d. y2 + xy = 0 b. (am – hl)x = (bl + hm) y
66. If the slope of one of the lines represented by d. None of these
ax2 + 2hxy + by2 = 0 is the square of the other,
74. The equation of second degree
then
8. a2b + ab2 – 6 abh + 8h3 = 0 x2 + 2 2 xy + 2y2 + 4x + 4 2 y + 1= 0
b. a2b + a b2 + 6 a bh + 8h3 = 0 represents a pair of straight lines. The distance
c. a2b + ab2 – 3 abh + 8h3 = 0 between is
d. a2b + ab2 – 6abh – 8h3 = 0 a. 2 – 3 b. 2 5
67. If 4 ab = 3h2, then the ratio of slopes of the lines
b. 2 d. 0
represented by the equation a x2+2 hxy+b y2 = 0,
will be 75. The point of intersection of lines represented by
the equation 3x2 + 8xy – 3y2 + 29x – 3y + 18 = 0
a 2 :1 b. 3:1 is
c. 2 : 1 d. 1 : 3

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Pairs of Straight Lines 85

 3 5  3 5  1
a.  ,  b.  ,  a. – b. – 2
2 2  2 2  2
c. (–3, –5) d. (3,5) c. + 1 d. 2
76. All chords of the curve 3x2 – y2 – 2x + 4 y = 0 81. The distance between the pair of parallel lines
which subtend a right angle at the origin, pass x2 + 2xy + y2 – 8ax – 8ay – 9a2 = 0 is
through the fix point a. 2 5 b. 10
a. (1,2) b. (1,–2)
c. 10 a d. 5 2
c. (–1, 2) d. (–1,–2)
77. The pair of lines joining origin to the points of 82. If the bisectors of angles represented by
intersection of the two curves ax2 + 2hxy + by2 ax2 + 2hxy + by2 = 0 and a'x2 + 2h' xy + b' y2 = 0
+ 2gx = 0 are same, then
and a' x2 + 2h' xy + b' y2 + 2g' x = 0 will be at a. (a–b)h'=(a'– b')h b. (a–b) h = (a'– b')h'
right angles, if b. (a+b)h'=(a'–b')h d. (a–b)h'=(a'+b')h
a. (a' + b')g'=(a + b)g
b. (a + b)g'=(a' + b')g 83. If pairs of straight lines x2 – 2pxy – y2 = 0 and
c. h2 – ab = h'2 – a' b' x2 – 2qxy – y2 = 0 be such that each pair bisects
the angle between the other pair, then
d. a + b + h2 = a' + b' + h'2
a. pq = 1 b. pq = –1
78. The angle between the pair of straight lines formed
by joining the points of intersection of x2 + y2 = 4 b. pq = 2 d. pq = –2
and y = 3x + c to the origin is a right angle. Then, 84. The angle between the pair of straight lines
c2 is equal to y2 sin2  – xy sin2  + x2 (cos2  –1) = 0 is
a. 20 b. 13 a.  / 3 b. / 4
b. 115 d. 5 b. /6 d. n/2
79. The angle between lines joining origin and 85. The distance between the pair of parallel lines
intersection points of line 2x + Y = 1and curve x2 + 4xy + 4y2 + 3x + 6y – 4 = 0 is
3x2 + 4 yx – 4x + 1= 0 is
2
a. n/2 b. n/3 a. 5 b.
5
c. n/4 d. n/6
80. If one of the lines my2 + (1– m2) xy – mx2 = 0 is 1 5 5
a bisector of the angle between the lines xy = 0, c. d. e.
5 2 2
then m is/are

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Pairs of Straight Lines 86

MHT-CET Corner
1. If one of the lines of the pair ax2 + 2hxy + by2 = 0 6. If S is the angle between the lines
bisects the angle between positive direction of the ax2 + 2hxy + by 2 = 0, then angle between
axes, then a, band h satisfy the relation x2 + 2xy sec a + y2 = 0 is
a. a + b = 2 | h | b a + b = – 2h a. a b. 2s
c. – s d. 3s The equation
c. a – b = | h | d. (a – b)2 = 4h2
7. The equation
2. If a pair of lines x2 – 2pxy – y2 = 0 and x2 – 2qxy 12x2 + txy + ay2 + 13x – y + 3 = 0 represents a
– y2 = 0 is such that each pair bisects the angle pair of perpendicular lines. Then, the value of a is
between the other pair, then
7
a. pq = –1 b. pq = 1 a. b. –19
2
1 1 1 1 c. –12 d. 12
c.  0 d.  =0
p q p q 8. If x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 bisect
3. The angle between the lines angles between each other, then
x2 – xy – 6y2 – 7x + 31y –18 = 0 is a. p + q = 1 b. pq = 1
c. pq + 1= 0 d. p2 + pq + q2 = 0
  9. The angle between the lines represented by the
a. b.
4 6 equation 2x2 + 3xy – 5y2 = 0, is
   
c. d. a. b.
2 3 3 2
4. The equation of the lines passing through the 12 7
1 c. tan–1 d. tan–1
5 3
origin and having slopes 3 and –
3 2 2
10. If the equation 4x + hxy + y = 0 coincident
a. 3y2 + 8xy – 3x = 0 b. 3x + 8xy + 3y2 = 0
2 2
lines, then h is equal to
c. 3y2 – 8xy – 3x2 = 0 d. 3x2 + 8xy – 3y2 = 0 a. 1 b. 3
5. Joint equation of pair of lines through (3, – 2) and c. 2 d. 4
parallel to x2 – 4xy + 3y2 = 0 is 11. The sum of slopes of lines 3x2 + 5xy – 2y2 = 0 is
a. x2 + 3y2 – 4xy – 14x + 24y + 45 = 0 5 5
a. – b.
b. x2 + 3y2 + 4xy – 14x + 24y + 45 = 0 3 2
c. x2 + 3y2 + 4xy – 14x + 24y – 45 = 0 5 2
c. – d. –
d. x2 + 3y2 + 4xy – 14x – 24y – 45 = 0 2 3

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Pairs of Straight Lines 87

Answers
Exercise 1
1. (c) 2. (b) 3. (c) 4. (b) 5. (a) 6. (b) 7. (b) 8. (c) 9. (a) 10. (b)
11. (d) 12. (c) 13. (b) 14. (d) 15. (b) 16. (c) 17. (a) 18. (c) 19. (b) 20. (a)
21. (c) 22. (a) 23. (b) 24. (c) 25. (d) 26. (b) 27. (b) 28. (d) 29. (b) 30. (c)
31. (a) 32. (c) 33. (a) 34. (c) 35. (a) 36. (b) 37. (d) 38. (c) 39. (c) 40. (a)
41. (b) 42. (b) 43. (a) 44. (a) 45. (c) 46. (a) 47. (c) 48. (d) 49. (c) 50. (d)
51. (c) 52. (b) 53. (d) 54. (c) 55. (a) 56. (b) 57. (d)
Exercise 2
1. (c) 2. (a) 3. (a) 4. (c) 5. (d) 6. (c) 7. (b) 8. (c) 9. (a) 10. (c)
11. (a) 12. (d) 13. (d) 14. (b) 15. (c) 16. (a) 17. (a) 18. (b) 19. (b) 20. (a)
21. (c) 22. (d) 23. (c) 24. (a) 25. (c) 26. (a) 27. (b) 28. (d) 29. (a) 30. (d)
31. (c) 32. (c) 33. (a) 34. (c) 35. (b) 36. (b) 37. (c) 38. (b) 39. (a) 40. (b)
41. (c) 42. (a) 43. (c) 44. (b) 45. (c) 46. (a) 47. (b) 48. (d) 49. (a) 50. (d)
51. (c) 52. (a) 53. (c) 54. (c) 55. (d) 56. (a) 57. (a) 58. (d) 59. (c) 60. (b)
61. (a) 62. (c) 63. (c) 64. (d) 65. (b) 66. (a) 67. (d) 68. (b) 69. (c) 70. (c)
71. (a) 72. (d) 73. (b) 74. (c) 75. (b) 76. (b) 77. (b) 78. (a) 79. (a) 80. (c)
81. (d) 82. (a) 83. (b) 84. (d) 85. (a)
MHT-CET Corner
1. (b) 2. (a) 3. (a) 4. (d) 5. (a) 6. (a) 7. (c) 8. (c) 9. (d) 10. (d)
11. (b)

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 DGT MH –CET 12th MATHEMATICS Study Material 19

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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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DGT Group - Tuitions (Feed Concepts) XIth – XIIth | JEE | CET | NEET | Call : 9920154035 / 8169861448
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