GROUP (A) – CLASS WORK PROBLEMS
Q -3) Find the joint joint equatio equation n of the lines lines passing through the point (–1, 2) 2 ) and perpendicular
Q -1) Find the combined combined equation equation of the bisector bisector of the angles between the co-ordinate axes. Ans. Let lines L 1and L 2be the bisector of the coordinate axes θ1
=
π
4
and
θ2
θ 1 and θ 2be
their inclinations
3x – – 4y – – 53 = 0. Ans. Let lines L 1and L 2 be the lines passing through the point (–1, 2) and perpendicular to the lines x + + 2y + + 37 = 0 and 3 x – – 4y – – 53 =
π
= – 4
0 respectively. Slopes of the lines x + 2y +
L 1 bisects first and 3 rd quadrant L 2 bisects 2nd and 4 th quadrant L 2
to the lines x + + 2y + + 37 = 0 and
Y
1 37 = 0 and 3x 3 x – – 4y – – 53 = 0 are – and and 2
–3 3 = respectively. –4 4
L 1 π
4
π
X
– 4
Slopes of the lines L 1and L 2 are 2 and
∴
–4 3
respectively. Since the lines L 1and L 2 pass through the point (–1, 2), their equations
Let slopes be m 1 and m 2
– π = 1 m 2 = tan = –1 4 4 Since the lines are passing through origin
m 1 = tan
arey arey – – 2 = 2 (x ( x + + 1) and y – – 2 =
π
their equations are y = m1 x and y = m2 x
∴
y – – 2 = 2x 2 x + + 2 and 3y – – 6 = – 4x 4 x – – 4
∴
2x – – y + + 4 = 0 4x + + 3y – – 2 = 0
∴
their joint equation is (2x (2x – – y + + 4) (4x (4 x + + 3y – – 2) = 0
( x – y ) ( x + y ) = 0 ⇒ x 2 – y 2 = 0
8x 2 + 6xy – xy – 4x – x – 4xy – xy – 3y 2 + 2y + y + 16x 16x + + 12y 12y
∴
x – y = 0 and x + y = 0 Combined equation is
–4 (x (x + + 1) 3
– 8 = 0 8x 2 + 2xy – – 3y 2 + 12x 12x + + 14y 14y – – 8 = 0
∴
Q -4) Find the combined equation equation of lines
Q -2) -2) Find the joint joint equation equation of lines lines which pass pass
through the origin, one of which whi ch is parallel
through the point (1, 2) and are parallel to
and other is perpendicular to the line x + +
the the lines co-ordinate axes.
2y +1857 +1857 = 0.
Ans. Equation of the coordinates axes are x = 0 and y = y = 0 ∴
the equat e quations ions of the lines passing through
(1, 2) and parallel parallel to t o the coordinate axes axe s are x = = 1 and y = = 2. i.e., x – – 1 = 0 and y – – 2 = 0 ∴
their joint equation is (x ( x – – 1) (y (y – – 2) = 0
∴
xy – – 2x – – y + + 2 = 0
Ans. The given equation of the line is is x + + 2y + + 1857 = 0 Let slopes of one line be m 1 and other line be m 2 The slope line which is to the line x + + 2y + + 1857 = 0 is ∴
m 1 =
∴ The
–1 – 1 2
–1 2
slope of line which is
line m 2 = 2
⊥ to
the given
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∴ The
is y = =
equation of line to the general line
through the origin having the inclinations
–1 x 2
600 and 1200 Ans. Slope of the line having inclination
x + + 2y = = 0
The equation of line y = =
⊥ to
the general line is
2x i.e. i.e. 2x – – y = = 0
∴ The
Q -4) Find the joint equation equation of lines passing
θ is
tan θ. Inclinations of the given lines are 60 0 and
combined equation of line is ( x + + 2y )
120 1200
(2x – – y ) = 0 ∴
2x 2 + 4xy – – xy – – 2y 2 = 0
their slopes are are m 2 =
2x 2 + 3xy – – 2y 2 = 0
m 1 =
tan 60 0 =
3 and
tan 1200 = tan(1800 – 600)
= – tan 600 = – 3 . GROUP (A) – HOME WORK PROBLEMS
Since the lines pass through the origin, their equations are
Q -1) -1) Find the joint equati equation on of lines and x + + y = = 0
– y = = x –
0
(Ans. x 2 – y 2 = 0)
Ans. The joint equation of the lines x – y = = 0 and x + + y = =
0 is
(x – y ) (x + y ) = 0 ∴ x 2 – y 2
=0
Q -2) -2) Find the joint joint equation equation of lines lines
+ y = = x +
3
and 2x – – y – – 1 = 0 (Ans. x 2 + 3xy + y 2 – 7x – 4y + + 3 = 0) Ans. The joint equation of the lines
x
+ y = = 3 and
2x + + y – 1 = 0 is ∴ 2x 2
+ x y – – x + 2x y + + y 2 – y – 6x – 3y + 3 = 0
∴ 2x 2
+ 3x y + + y 2 – 7x – 4y + 3 = 0.
through the origin having slopes 3 and 2 Ans. We know that the equation of the line passing through the origin and having slope m is is y = = mx .
3x – – y = 0 and 2 x – – y = 0 respectively. ∴
their joint equations equations is
(3x – – y )(2 )(2x – – y ) = 0 ∴ 6x 2 –
3x y – 2x y + + y 2 = 0
∴ 6x 2 –
5x y + + y 2 = 0.
3x – y = 0 and
3x + y = 0
∴
the joint equation of these lines is
(
3x – y ) ( 3x + y ) = 0
∴ 3x 2 – y 2
= 0.
Q -5) -5) Find the joint joint equation equation of lines lines which pass pass through the point (3,2) and are parallel to the lines x = = 2 and y = = 3 ∴
the equations of the line s passing through through
(3, 2) and parallel to the lines
x = =
2 & y = = 3
∴ The
joint equation is
(x – – 3)(y – – 2) = 0 xy –
2x – 3y + + 6 = 0
2x + 3y – xy – – 6 = 0. Q -6) -6) Find the joint equation equation of lines passing
Equation of the lines passing through the
i.e., their equations are
3x
are x = = 3 & y = = 2.
Q -3) -3) Find the joint equation equation of lines passing
and y = = 2x respectively. respectively.
i.e.,
= – 3x and y =
Ans. Equations of the lines are x = = 2 and y = = 3
(x + y – – 3) (2x + y – 1) = 0
origin and having slope 3 and 2 are
y = =
y = =
3x
through the origin having slopes 1 + 3 and 1 – 3 Ans. The equation of the line passing passing through the origin and having slope
m is is y = = mx
Equations of the lines passing through the origin and having slopes 1 + 3 and 1 – 3 are y = =
(1 +
3 ) x and y = = 1 –
(
3 ) x
respectively i.e., The equation are
(1 +
3 ) x – – y = = 0 and (1 – 3 ) x – – y = = 0
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∴
(1 – 3 ) x – y (1 + 3 ) x – y = 0
( x –
3x – y
)(x +
x2 +
3x 2 – xy – 3x 2 – 3x 2 + 3xy
i.e., x – – 3 = 0
)
Hence, the equations of the required lines
3x – y = 0
– xy –
are
3xy + y 2 = 0
–2x 2 – 2xy + y2 = 0 2x 2 + 2xy – y 2 = 0 Q -7) Find the combine combined d equation equation of lines lines which are parallel to the Y -axis - axis and at the distance of 9 units from the Y -axis -axis Ans. Equations of the lines, which are parallel to the Y -axis -axis and at a distance of 9 units from it are x = = 9 and x = = – 9 i.e., x – – 9 = 0 and x + + 9 = 0
x – – 2y + + 1 = 0 and x – – 3 = 0 ∴
(x – – 2y + + 1)(x 1)(x – – 3) = 0 ∴
x 2 – 2xy 2xy + x – 3x 3x – 6y 6y – 3 = 0
∴
x 2 – 2xy 2xy – – 2x 2x + 6y 6y – 3 = 0.
Q -9) Find the combined equation equation of lines through the origin, which wh ich are perpendicular to the lines x + + 2y = = 9 and 3x + + y = = 8 Ans. Let L 1 and L 2 be the lines passing through the origin and perpendicular to the lines x + + 2y = = 9 and 3x 3 x + + y = = 8 respectively. slope of given line are
∴
′
X 9
∴
Y x = = 9
their combined equation is
(x – – 9)(x 9)(x + + 9) = 0 ∴
x 2 – 81= 81= 0.
Q -8) -8) Find the joint equation equation of lines passing through the point (3,2), one of which is parallel to the lines x – – 2y = = 29 and other one is perpendicular to the line y = = 3 Ans. Let L 1 be the line passes through (3,2) and parallel to the line x – x – 2y 2y = = 29 whose slope is
origin, their equations are y = 2x and y = y =
1 x 3
i.e., 2x 2x – – y = = 0 and x – – 3y = = 0 ∴
their combined equation is
(2x (2x – – y )(x )(x – – 3y ) = 0 ∴
2x 2 – 6xy 6xy – – xy + 3y 3y 2 = 0
∴
2x 2 – 7xy 7xy + 3y 3y 2 = 0.
Q-10)Find Q-10)Find combined equation equation of lines through the origin, which are parallel to lines + 3y = = 27 and 2x = = y – – 3 x + the origin and parallel to the lines
1 ∴ slope of the line L 1 is 2
equation of the line L 1 is y – – 2 =
Since the lines L 1 and L 2 pass through the
Ans. Let L 1 and L 2 be the lines passing through
–1 1 = . –2 2
∴
slope of the line L 1 and L 2 are 2 and 1 respectively. 3
X 9 ′
x = = –9
their joint equation is
1 3 – and – = – 3 respectively. 2 1
Y
O
equation of the line L 2 is x = = 3,
1 (x – – 3) 2
∴
2y – 4 = x – – 3
∴
x – 2y 2y + + 1 = 0
Let L 2 be the line passes through (3,2) and perpendicular to the line y = = 3. equation of the line L is L is of the form
x + + 3y = = 27 and 2x 2 x = = y – – 3 respective respectively. ly. Slopes of the lines x + + 3y = = 27 and 2x 2 x = = y – – 3 are ∴
–1 –2 and 3 –1
slope of the lines L 1 and L 2 are
equations are y=
–1 x & y = = 2x 3
–1 & 2 their 3
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∴ The separate equations are
(2x (2x – – y )(3y )(3y + + x ) = 0 6xy + 2x 2x 2 – 3y 3y 2 + xy = 0
(
)
(
)
y = 2 + 3 x and y = 2 – 3 x
2x 2 + 7xy 7xy – – 3y 3y 2 = 0.
Q-11) Q-11) Find the joint equation of lines which pass
Q -3) -3) Find the condit condition ion that the the two lines lines of
through (4, – 3) and are parallel to the lines
+ c )xy may be coincident a 2x 2 + bcy 2 = a (b +
= 1 and y = 5 x =
and perpendicular
Ans. Equations of the lines are x = = 1 and y = = 5 ∴
the equations of the lines line s passing through
(4, – 3) and parallel parallel to the lines x = = 1 and y = = 5 are x = = 4 & y = = – 3 i.e., x – – 4 = 0 & y + + 3 = 0
Ans. Given equation is xy …(I) ⇒ a 2 x 2 + bcy 2 = a (b + c ) xy 2 2 Compare with Ax + 2Hxy + By = 0
We get A = a 2 ; B = bc bc ; H =
The joint equation is
– a (b + c ) 2
i) Equati Equation on will will repres represent ent a pair pair of
(x – – 4)(y 4)(y + + 3) = 0 xy + 3x 3x – 4y 4y – 12 = 0
coincident lines if H 2 – AB = 0
3x – 4y 4y + xy – xy – 12 12 = 0
2
– a 2 2 (b + c ) – a .bc = 0
GROUP (B) – CLASS WORK PROBLEMS
a 2 2 (b + c ) – a 2bc = 0 4
Q -1) -1) Find val values ues of of h , if lines given by 3x 2 + 2hxy + + 3y 2 = 0 are real?
a 2 (b 2 + 2bc + c 2 ) – 4a 2bc = 0
Ans. Comparing the equation
a 2 (b 2 + 2bc + c 2 – 4bc ) = 0
3x 2 + 2hxy + hxy + 3y 2 = 0 with ax 2 + 2hxy + hxy + by 2 = 0, we get, a = 3, h = h, b = 3.
a 2 (b 2 – 2bc + c 2 ) = 0
The lines represented represented by 3 x 2 + 2hxy + hxy + 3y 2 = 0
2
are
a 2 (b – c ) = 0
real, if h 2 – ab ≥ 0 i.e., if h 2 – 3(3)
≥
a (b – c ) = 0
0
i.e., if h 2 ≥ 9
Either a = 0 or b – c = 0
i.e., if h ≥ 3 or h ≤ – 3
Either a = 0 or b = c
Hence, the values of h are all real numbers
Equation represents two coincident
greater than or equal to 3 or less than or
lines if a = 0 or b = c
equal to – 3, i.e., (–
∞,
3] ∪ [3, ∞] = R – – (– 3,3).
ii) Equation (i) will represent a pair of ⊥ lines if A + B = 0
Q -2) -2) Find separate separate equat equation ion of the lines lines represented by x 2 – 4xy + + y 2 = 0 Ans. Join equation x 2 – 4xy + y 2 = 0
⇒ a 2 + bc = 0 Q -4) Find the joint equation equation of a pair of of lines
i.e., x 2 – 4xy + y 2 = 0
through the origin and perpendicular to
y 2 = m ∴ m – 4m +1 = 0 ∵ x
pair of lines 2x 2 – 3xy – 9y 2 = 0
2
m =
4 ± ( 4 ) – 4 (1) (1) 2
=
4 ± 12 =2± 3 2
∴
y = (2 ± 3 ) x
=
4 ± 16 – 4 2
i s, Ans. Joint equation of lines is,
2x 2 – 3xy – 9y 2 = 0 So, Separate equation is, 2 –
y ∵ x = m
2
9m + 3m – 2 = 0 9
2
6
3
3y 9y 2 – 2 =0 x x
2
0
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3m + 2 = 0 and 3m – 1 = 0
m =
∵
–2 1 m = 3 3
⇒ k 2 = 36 ⇒ k = ± 6
y –2 y 1 = = x 3 x 3
Q -6) Find Find k , if one of the lines given by
2x + 3y = 0
…(i)
– x + 3y = 0
…(ii)
–2 Slope of equation (i) = 3 1 Slope of equation (ii) = 3 So, the slopes of
⊥ r
Lines L 1 and is,
Ans. Equation of pair of lines, ... (i)
4x 2 + kxy – y 2 = 0 Equation of lines is, 2x + y = 0 y = –2x
... (ii)
Solving (i) & (ii) 2
4x 2 – 2kx 2 – 4x 2 = 0
Now, Lines L 1 and L 2 pass through the origin ∴ 3x
4x 2 + kxy – y 2 = 0 is 2x + y = 0
4x 2 + kx (–2x ) – ( –2x ) = 0
–3 m 1 = m 2 = 3 2
y –3 = x 2
2
k 2 – k =1 ⇒ = 1, 36 6
+ 2y = 0
Dividing by x 2
–2kx 2 = 0 –2k = 0 k = 0
y = 3 = –3x + y = 0 x Joint equation of perpendicul ar lines,
( 3x + 2y ) ( 3x – y ) = 0 2
Q -7) -7) If one of the the lines lines given given by ax 2 + 2hxy + by 2 = 0 is perpendicular to
the line x + my + n = 0 then show that
2
9x – 3xy + 6xy – 2y = 0 2
2
al 2 + 2h m + bm bm 2 = 0
Q -5) Find Find k if if the slope of one of the lines given by 5x 2 + kxy + y 2 = 0 is 5 times the slope of the other Ans. Given eq’n is 5x 2 + kxy + y 2 = 0 Comparing given eqn .with
Slope of lx + my + n = 0 is
– l l m
∴ slope
of
r
⊥
to
lx + my + n =0 = 0 is
ax 2 +2hxy +2hxy + + by 2 = 0, ∴ Eqn.of
a =5 =5 ,2h ,2h = = k , b = = 1 ∵ m1 + m 2 =
m1 m 2 =
a b
–2h
=
b
=
... (i)
Ans. al 2 + 2h m + bm 2 = 0
9x + 3xy – 2y = 0
– k 1
5
Passing through origin,
1
... (i)
m1m 2 = 5
... (ii)
Bu But m1 = 5m 2
... (iii)
Solving (i) & (iii) & Solving (i) & (ii) ∴ 5m2
+ m2 = –k a k an nd 5m2 .m 2 = 5
∴ 6m 2
= – k and k and 5m 22 = 5
line
lx + my + n =0 = 0
y =
m1 + m2 = – k
m l
m
... (ii)
x
Solving (i) & (ii) 2
m m ax + 2hx + + b x = 0 x l 2
ax 2 +
∴ a +
2hx l 2hm
x 2 +
+
bm 2 =0 l 2
bm 2
=0
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Q -8) Find Find ‘α’if the sum of the squares of slopes of the lines repersented by αx 2 –
= the distance of P (x ( x , y ) from the line (x + + y = = 0)
3xy + y 2 = 0 is 5. ∴
Comparing the equation Ans. Comparing 2
1+ 4 ∴
we get, a = a, 2 a, 2h h = = – 3, b = = 1. Let m 1 and m 2 be the slopes of the lines 2
2
represented by ax – 3xy + y = 0
and m1 + m 2 = – ∴ ∴
a a = = a b 1
2
(m 1 + m 2) = (3) 2 1
2 2
2 1 2 1
2 2 2 2
∴
m + m
2
=
(x + y)
2
2
2
+ 2y ) = 5 ( x + y )
2
( x 2 + 4xy + 4y 2 ) = 5 ( x 2 + 2xy + y 2 )
∴ 2x
2
+ 8xy + 8y 2 = 5x 2 + 10xy + 5y 2
∴ 3x
2
+ 2xy – 3y 2 = 0.
lines which bisect the angle between the
m + m + 2m 1m 2 = 9 m + m + 2a = 9
1 +1
5
∴ 2(x
This is the required requi red joint joi nt equation equat ion of the
2
∴
( x + 2y )
∴2
2h – ( –3) = =3 m1 + m 2 = – 1 b
x +y
=
2
+ by 2 = 0, ax – 3xy + y = 0 with ax 2 + 2hxy +
∴
x + 2y
lines represented by x 2 + 3xy – 2y 2 = 0
= 9 – 2a 2a
Q-10)Show
But m 12 + m 22 = 5
that
the
lines
given
by
3x 2 – 8xy – 3y 2 = 0 and x + 2y = = 3 contain
∴
9 – 2a 2a = 5
∴
2a = 4
the sides of an isosceles right angled
∴
a = 2.
triangle. Ans. Given equation 3x 2 – 8xy – 3y 2 = 0
Q-9) Q-9) Find the joint equa equation tion of of lines lines which which bisect bisect angles between lines represented by
3x ( x – 3y ) + y ( x – 3y ) = 0
x 2 + 3xy + 2y 2 = 0
( x – 3y ) ( 3x + y ) = 0
Ans. x 2 + 3xy + 2y 2 = 0 ∴
3x 2 – 9xy + xy – 3y 2 = 0
⇒ equations are
x 2 + 2xy + xy + 2y 2 = 0
( x – 3y ) = 0 & ( 3x + y ) = 0
x (x + + 2y ) + y (x + + 2y ) = 0 (x + + 2y ) (x (x + + y ) = 0 ∴ (x ∴ seperate equations ∴
of
the
lines
represented by x 2 + 2xy + xy + 2y 2 = 0 are x + + 2y = = 0 and x + + y = = 0.
Slopes of these lines are m1 =
1 & m 2 = – 3 3
Product of Slopes = Angle isector
m1m 2 =
Y P(x,y)
∴ Lines
are
⊥
1 3
×
– 3 = –1
given by 3x 2 – 8xy – 3y 2 = 0
to to each other.
Slopes are m 1 =
…. (i)
1 and m 2 = –3 3
X
Slope of the line x + 2y = 3 is m = Let
θ1
–1 2
be the angle between
Let P ( P (x , y ) be any point on one angle bisector.
x + 2y = 3 & x – 3y = 0
Since the points on the angle bisectors are are
From (i), (ii) & (iii) the given lines
equi - distant distant from from both the lines, lines, the
form isosceles right angled triangle.
… (ii)
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Q-11) ∆ OAB is
7
formed
by
the
lines
x – 4xy + y = 0 and the line AB is given 2
2
by 2x
3y – 1 = 0. Find the equation of the
+
median passing through O.
16 3y = 1 – 37 3y =
y= (x1,y 1)
Ans.
21 37
7 37
∴ D ≡
8 7 37 , 37
slope of OD, A
D
m =
7/37 8/37
m =
7 8
(x2,y 2) B O
equation of line OD is, y = mx Let A
≡
( x1 , y 1 ) , B
≡
( x 2 , y 2 )
y =
Since,
7x 8
7x – 8y = 0
D is the mid-point of AB By mid-point formula
Q-12)Show x 2 + 4xy + y 2 = 0 and x – y = = 4 contain
x + x 2 y1 + y2 D≡ 1 , ≡ ( x , y ) 2 2
the sides of an equilateral triangle. Find the area of the triangle.
The combine d equation of the li nes passing passin g 2
2
through origin is x – 4xy + y = 0
Ans. Let slope of OA be m and slope OB be 1
The e quation o f line AB is, i s,
tan θ =
2x + 3y = 1 y =
1 – 2x 3
...(iii)
∴
∵ points A and B are intersecting point of
lines passing through origin and line AB substituting (iii) in (i) x 2 – 4x
(1 – 2x ) 3
+
(1 – 2x ) 9
∴
3(m 3(m 2 + 2m + m + 1) = m 2 – 2m + + 1
∴
3m 2 + 6m + m + 3 = m 2 – 2m + + 1
∴
2m 2 + 8m + + 2 = 0 m 2 + 4m + + 1 = 0 2
∴
37x 2 – 16x + 1 = 0
16 x1 + x 2 = 37 x1 + x 2 2 D
≡
=
m – 1 1 + m
3 (1 + m )2 = m 2 – 2m + + 1
2
Let, x 1 and x 2 be the roots of the equation
3 =
∴
9x 2 – 12x (1 – 2x ) + (1 – 2x ) = 0 9x 2 – 12x + 24x + 1 – 4x + 4x 2 = 0
1 + m1m 2
squaring
2
=0
m1 – m 2
∴
y y + 4 +1 = 0 x x y 2 + 4xy + + x 2 = 0
is the joint equation of
lines 0
through (0,0) and making 60 with x – – y – – 4 = 0
8 37 A
8 y1 + y 2 8 37 , 2 ≡ 37 , y
Point int D lies on the line 2x + 3y =1
... [By (i)]
M O
C
passing
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OM =
GROUP (B) – HOME WORK PROBLEMS
–4
=2 2
2
Q -1) -1) Determine Determine the nature of lines represente represented d 2
∴
P
A (∆ OAB ) =
3
8
=
3
sq uni units
by following equations : i) x 2 + 2xy – y 2 = 0
Q-13)Show Q-13) Show that the eqation eqation x 2 – 16xy – 11y 2 = 0
ii) 4x 2 + 4 xy + y 2 = 0
represents a pair of lines through the
iii) x 2 + y 2 = 0
origin, each making an angle of 30 0 with
iv) x 2 + 7xy + 2y 2 = 0
the line x + 2y – 1 = 0
v) px 2 + qy 2 = 0 , where p & q are real numbers. Ans. i) Compa Compari ring ng the the equa equatio tion n x 2 + 2xy – y 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, 0
A
0 30 30 x + 2 y – 1= 0
B
a = = 1, 2h 2 h = = 2, i.e., h = = 1 and b = = –1 ∴h
Ans. Let us find the joint equation OA and and OB each each makes an angle of 30 0 with x + + 2y 2y –1 –1 = 0 Let slope of OA /OB be OB be m 1 (say) (m (m ) 1 ∴ slope of AB = = m 2 = – 2
tan θ =
∴
∴
x 2 + 2xy – y 2 = 0 are real and distinct. ii ) Compari Comparing ng the the equatio equation n
4x 2 + 4xy + y2 = 0 with,
a = = 4, 2h 2 h = = 4, i.e. h = = 2 and b = = 1
1 2 = 3 1 + – 1 m 2 =
2m + 1 2 – m
2
2
( 2m +1) y = 2 x ( 2 – m )
(2 – m )2 = 3 (2m (2m + + 1)2
4 – 4m + + m 2 = 3(4m 3(4m 2 + 4m + + 1) 2 2 ∴ 12m 12m – m + 12m 12m + 4m + 3 –1 = 0 2 ∴ 11m 11m + 16m 16m –1 –1 = 0 Lines pass through origin ∴
∴
lines represented by
1 + m1m 2
Squaring both sides
∴
∴ the
ax 2 + 2hxy + by 2 = 0 we get,
m +
3
– ab = 12 – 1 ( –1) = 1 +1 = 2 > 0
m1 – m 2
1
1
2
y = mx is the equation is
∴h
2
2
– ab = ( 2) – 4 (1) = 4 – 4 = 0
∴ the
lines represented by
4x 2 + 4xy + y2 = 0 are real and coincident. iii) Comparing the equation x 2 – y 2 = 0 with = 1, 2h 2h = = 0, ax 2 + 2hxy + by 2 = 0 , we get, a = i.e. h = = 0 and b = = –1 ∴h
2
2
– ab = ( 0 ) – 1( –1) = 0 +1 = 1 > 0
∴ the
lines represented by x 2 – y 2 = 0 are
real and distinct. iv) Comparing Comparing the equation
x 2 + 7xy + 2y 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, a = = 1, 2h 2h = = 7, b = = 2, i.e. h = =
2
∴
∴
y y 11 + 16 – 1 = 0 x x 11y 2 2
x ∴ ∴
y −1 = 0 x
2
49 7 –2 ∴h – ab = – (1 )( ) ( 2) = 4 2 2
+ 16
11y 11y 2 + 16xy 16xy – – x x2 = 0 x 2 16xy 16xy –11 –11y y 2 = 0
7 2
= ∴ The
49 – 8 4
=
41 4
lines represented by
>
0
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9
v) Compa Comparin ring g the equatio equation n px 2 – qy 2 = 0 with ax 2 + 2hxy + by 2 = 0 , we get, a = p = p ,
2
2
– ab = ( 0 ) – p ( –q ) = pq
– m 2 + 2m + 2 = 0 ∴
2h = = 0, i.e. h = = 0 and b = = – q q ∴h
The auxilary aux ilary equations is
m 2 – 2m – 2 = 0 2
∴
If p If p and and q have have the same sign, then
m =
h 2 – ab = pq > 0 ∴ the
=
lines represented by px 2 – qy 2 = 0
are real and distinct.
∴
If p If p and and q have have different signs, then
2 ± ( –2) – 4 (1) ( –2) 2 ×1
=
2± 4+8 2
2±2 3 =1 ± 3 2
m 1 = 1 + 3 and m 2 = 1 – 3 are the
slopes of the lines.
2
h – ab = pq < 0 ∴ the
∴ 2
2
equation px – qy = 0 does not
represent the lines.
their separate equations are
y = m 1x and y = m 2x
(
)
(
)
i.e., y = 1 + 3 x and y = 1 – 3 x
Q-2) Find separate s eparate equations of lines represented
v) x 2 + 2xy cos ec α + y 2 = 0
by following equations
y x
i) 6x 2 – 5xy – 6y 2 = 0
2 +
y +1 = 0 x
2 cos ec α
ii) x 2 – 4 y 2 = 0 iii) 3x 2 – y 2 = 0
Put
iv) 2x 2 + 2xy – y 2 = 0
(m )
v) x 2 + 2xy ( cos α ) + y 2 = 0 Ans. i)
6x 2 – 5xy – 6y 2 = 0
y = m x 2
m =
∴
6x 2 – 9xy + + 4xy – – 6y 2 = 0
∴
3x (2x (2x – – 3y ) + 2y (2x (2x – – 3y ) = 0
∴
(2x (2x – – 3y )(3x )(3 x + + 2y ) = 0
∴
the separate equations of the lines are
2x – – 3y = 0 and 3x 3x + + 2y = 0.
m =
m
x 2 – (2y (2y )2 = 0
∴
(x – – 2y ) (x (x + + 2y ) = 0
∴
the separate equations of the lines are
m =
m =
x – – 2y = 0 and x + + 2y = 0. iii) 3x 3x 2 – y 2 = 0 2
∴
(
3x
)
∴
(
3x
−y
∴
the separate equations of the lines are
3x
− y =
2
− y =
)(
0 and
+
y
−2 cos ec α ±
3x
+ y =
0
2 cot α
2
=
−1
±
sin α − (1 ±
cos α sin α
cos α )
sin α − (1 ± cos α )
x
sin α
y sin α
= − (1 ±
& sin α y
) = 0
4 cos ec 2α − 4 × 1 ×1 2
−2 cos ec α ±
cos α ) x
∴ sin α y = − (1 +
0
3x
y
2 cos ec α (m ) + 1 = 0
= − cos ec α ± cot α
ii ) x 2 – 4y 4y 2 = 0 ∴
+
cos α ) x
= − (1 − cos α ) x
are the separate separate equations.
Mahesh Tutorials Science
10
Q-3) If lines ines x 2 + 2hxy 2hxy + + 4y 4y 2 = 0 are coincident find the values of h Ans. Comparing Comparing the equation 2
m1. m 2 =
a = –1 b
The required lines are perpendicular perpendi cular to
2
2
2
x + 2hxy + hxy + 4y 4y = 0 with ax + 2hxy + hxy + by = 0,
these lines
we get, a = = 1, h = = h and b = = 4. ∴ The
Since the lines represented by x 2 + 2hxy + + 4y 2 = 0 are coincident, h 2 – ab = 0 ∴
h 2 – 1
∴
h 2 = 4
∴
h = = + 2.
×
4=0
–1 m 1
and
–1 m 2
Th es e li ne s are ar e pas si ng th roug ro ugh h th e origin, their separate equations are y=
Q-4) Find the combined equations of lines through the origin which are perpendicular to the lines represented by
–1 –1 x an d y = x m1 m 2
i.e., m 1y = – x and m 2y = – x x + m 1y = 0 and x + m 2y = 0 The combine d equation is
i) x 2 + 4xy + 5y 2 = 0 xy + 5y 2
slopes are
(x + m 1y ) (x (x + m 2y ) = 0
2
ii) x ii) x + xy – – y = 0
x 2 + (m 1+ m 2) xy + m 1m 2 y 2 = 0
Comparin ring g the equat equation ion Ans. i) Compa
x 2 + xy – xy – y 2 = 0
x 2 + 4xy + xy + 5y 2 = 0 with ax 2 + 2hxy + hxy + by 2 = 0, Q-5) Find k if if sum of the slope of lines represented
we get, a = = 1, 2h 2h = = 4, b 4, b = = 5 Let m 1 and m 2 be the slopes of the lines
Ans. Comparing the equation
represented by x 2 + 4xy 4xy + + 5y 5y 2 = 0
3x 2 + kxy + y 2 = 0 with ax 2 + 2hxy + by 2 = 0,
–2h –4 = b 5 a 1 and m1m 2 = = b 5
∴ m1
by the equation 3x 3 x 2 + kxy + + y 2 = 0 is 0.
+ m 2 =
we get, a = a = 3, 2h = = k , b = = 1 Let m 1 and m 2 be the slopes of the lines
... (i)
represented by 3x 3x 2 + kxy + y 2 = 0.
Now required lines are perpendicular to
∴ m1
+ m2 =
these lines ∴
their slopes are – 1/m 1/m 1 and – 1/m 1/m 2
Since these lines are passing through the origin, their separate equations are –1 –1 y= x and y = x m1 m 2
∴
... [By (i)]
Q-6) F in in d k if sum of the slope of lines
we get, a = a = 2, 2h = = k , b = = 3 Let m 1 and m 2 be the slopes of the lines represented by 2x 2 x 2 + kxy + 3y 3y 2 = 0.
ii ) Compari Comparing ng the the equatio equation n 2
2
x + xy – – y = 0 with ax + 2hxy + + by = 0,
∴ m1
+ m 2 =
we get, a = = 1, h = =
1 2
and m1m 2 = , b = = – 1
Let m 1 and m 2 be the slopes of the lines represented by x 2 + xy – – y 2 = 0 h
k = 0.
Ans. Comparing Comparing the equation
5x 2 – 4xy + + y 2 = 0. 2
∴
2x 2 + kxy + 3y 2 = 0, with ax 2 + 2hxy + by 2 = 0,
4 1 xy + y 2 = 0 5 5
2
k=0
product.
their combined equations is
x 2 –
∴ –
... (Given)
2x 2 + kxy + + 3y 2 = 0 is 0 is equal to their
(x + + m 1y ) (x + x + m 2y ) = 0 ∴
Now, m 1 + m 2 = 0
represented by the equation
i.e., m 1y = – x and and m 2y = – 0 ∴
–2h – k = = –k b 1
–2h – k = b 3 a 2 = b 3
Now, m 1 + m 2 = m 1m 2 ∴ –
k 2 = 3 3
... (Given)
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Q-7) Q-7) Show that that the differenc difference e between between slope of of lines repersented by the equation 3x 2 + 4xy + + y 2 = 0 is 2. Ans. The given give n equation is is 2
2
3x + 4xy + y = 0,
... (i) 2
∴
45 3k – +3=0 25 5
∴
45 – 15k 15k + + 75 = 0
∴
15k 15k = = 120
∴
k = = 8
2
Comparing it with ax + 2hxy + + by = 0, we get, a = a = 3, 2h = = 4, b = = 1
Q-10)The slope of one of the lines given by the
Let m 1 and m 2 be the slopes of lines given by(i) Then m1 + m 2 = –
2h –4 = = –4 b 1
equation 3x 2 + 4xy + + λy 2 = 0 is 3 times the slope of the other line. Find λ. Ans. Comparing the equation 3x 2 + 4xy + + λy 2 = 0 with ax 2 + 2hxy + + by 2 = 0,
a 3 and m1m 2 = = = 3 b 1
we get,
∴
a = = 3, 2h 2h = = 4, b = = λ
(m 1 – m 2 )2 = (m 1 + m 2 )2 – 4m 1m 2 = (– 4)2 – 4(3) = 16 – 12 = 4
∴ ∴
Let m 1 and m 2 be the slopes of the lines represented by 3x 3 x 2 + 4xy 4xy + + λy 2 = 0. 0.
m1 – m 2 = 2 the slopes of lines represented by
∴ m1
+ m 2 = –
3x 2 + 4xy + y 2 = 0, differ by 2. Q-8) Find the value value of k , if slope of one of the lines given by 3x 2 – 4xy + + ky 2 = 0 is 1.
and m1m 2 =
∴
2
3x – 4xy + ky = 0 is
a 3 = . b λ
We are given that m 2 = 3m 3m 1
Ans. The auxiliary equation of the line s given by 2
2h –4 = b λ
m1 + 3m 1 =
–4 λ
2
km – 4m + + 3 = 0. Given, slope of one of the lines i s 1. ∴
∴ 4m 1
=
–4 λ
∴ m 1
1 is the root of the auxiliary equation
Let m 1 and m 2 be the slopes of lines given by
Also, m 1 (3m (3m 1) =
km 2 – 4m + + 3 = 0. ∴
k (1) (1)2 – 4(1) + 3 = 0
∴
k – k – 4 + 3 = 0
∴
k = k = 1.
2
∴ m 1
=
1 = – λ
... (i)
3 λ
1 λ
2
Q-9) Q-9) Find Find the the va value lue of of k , if one of the lines given by 3x 2 – kxy + + 5y 2 = 0 is perpendicular, to the line 5x + + 3y = = 0 Ans. Th e au x i l i ar y e qu a t i o n o f th e l i n e s
1 1 ∴ – = λ λ ∴
1 λ
∴λ
2
=
... [By (i)]
1 λ
= 1, as λ ≠ 0.
represented by Q-11) Q-11) If slopes of lines given by kx 2 + 5xy + y 2 = 0
3x 2 – kxy + 5y 5y 2 = 0 is
differ by 1 then find the value of k .
5m 2 – km + + 3 = 0. Now, one line is perpendicular to the line 5x + + 3y = 0, whose slope is
–3 5
3 ∴ slope of that line = m = 5 ∴
3 is the root of the auxiliary auxil iary equation 5
5m 2 – km + + 3 = 0.
Ans. Comparing the equation kx 2 + 5xy + + y 2 = 0 with ax 2 + 2hxy + + by 2 = 0, we get, a = = k , 2h = = 5, b = = 1. Let m 1 and m 2 be the slopes of the lines represented by kx 2 + 5xy 5xy + + y 2 = 0 ∴ m1
+ m 2 = –
2h –5 = = –5 b 1
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12
(m 1 – m 2)2 = (m 1 + m 2)2 – 4m 4m 1m 2
∴
= (– 5) 2 – 4(k 4(k ) = 25 – 4k 4 k But m1 − m 2
=
... (i)
∴
Th e s e p ar at e e q u a ti o n o f t h e l i n e s
represented by 2x 2 x 2 + 3xy + + y 2 = 0 are
1
2x + y = 0 and x + y = 0
(m 1 – m 2)2 = 1
∴
(2x (2x + + y )(x )(x + + y ) = 0
From (i) and (ii), 25 – 4k 4 k = = 1 4k = = 24
∴
k = = 6.
∴
Y
Angle isector
Q-12) Q-12) Find the value of k , if one of the lines given
P(x,y )
by 6x 2 + kxy + + y 2 = 0 is 2 x + + y = 0 Ans. Th e au x i l i ar y e qu a t i o n o f th e l i n e s represented by 6x 2 + kxy + y 2 = 0 is
X Angle isector
m 2 + km + + 6 = 0. Since one of the line is 2 x + y = 0 whose slope is m = = – 2 – 2 is the root of the auxiliary equation
∴
m 2 + km + + 6 = 0. ∴
(– 2)2 + k (– 2) + 6 = 0
∴
4 – 2k + + 6 = 0
∴
2k = 10
Let P (x (x , y ) be any point on one angle bisector. Since the points on the angle bisectors are equidistant from from both the lines, the distance of P (x ( x , y ) from the line 2x 2 x + + y = y = 0 ∴ the distance of P (x ( x , y ) from the line x + + y =0 =0
k = 5.
∴
∴
2x + y 4 +1
Q-13)If Q-13)If the sum of the slopes of the lines given by 2x 2 + kxy – 9y 2 = 0 is equal to five times
∴
( 2x + y ) 5
their product, find the value of k . Ans. Comparing the equation
x +y
=
1 +1
2
( x + y)
=
2
2
2
2
∴
2 ( 2x + y ) = 5 ( x + y )
we get,
∴
2 4x 2 + 4xy + y 2 = 5 x 2 + 2xy + y 2
k a = = 2, h = = , b = – = – 9 2
8x + 8xy + 2y = 5x + 10xy + 5y
2x 2 + kxy – – 9y 2 = 0, with ax 2 + 2hxy + + by 2 = 0, 0,
2
Let m 1 and m 2 be the slopes of the lines represented by 2x 2 x 2 + kxy – – 9y 2 = 0
∴
m1 + m 2 = –
m1m 2 =
2h b
k 2 = k
–2 =
–9
9
2
3x 2 – 2xy – 3y 2 = 0 This is the required requ ired join t equation equa tion of the lines which bisect the angle between the lines represented by 2x 2 + 3xy + y2 = 0
Q-15 Q-15)) Find the joint equation equation of the pair p air of lines triangle with the line x = = 3
Now, m 1 + m 2 = 5m 5 m 1m 2
... (given)
and OB be be the lines through the origin Ans. Let OA and making an angle of 60 0 with the line x = = 3.
k = = –10 Y
Q-14)Find Q-14) Find the joint equation of lines which bisect
150
angles betweeen lines represented by 2
+ y = 0 Ans. 2x + 3xy + 2x 2 + 2xy + + xy + y 2 = 0
0
A
60
2x 2 + 3xy + + y 2 = 0 2
2
through origin and making an equilateral
a –2 = b 9
k –2 = 5 ; 9 9
2
30 X'
O
30
0
0
X
0
60
0
Mahesh Tutorials Science
∴
13
OA and OB make an angle of 30 0 and 1500
Now, (m (m 1 – m 2)2 = (m (m 1 + m 2)2 – 4m1m2 = k 2 – 4(–3)
with the positive direction of X-axis ∴
∴
∴
slope of OA = tan 30 0 = 1
= k 2 + 12
3 1
equation of the line OA is y =
3y = x
∴
3
But |m |m 1 – m 2| = 4
x
∴ ∴
3
equation of the line OB is y =
3y = – x
∴
–1 3
x
) (x +
From (i) and (ii), we get
∴
k = = 4
∴
k = = ± 2
4xy – – y 2 = 0 exceeds the slops of the kx 2 + 4xy other by 8, find k Ans. Given equation is kx 2 + 4xy – – y 2 = 0
x + 3y = 0
is 3y
∴
Q-18) Q-18) If the slope of one of the lin es given by
required combined equation of the lines
( x –
(m 1 – m 2)2 = 16
...(given)
k 2 + 12 = 16
Slope of OB = tan1500 = tan (1800 – 300)
∴
∴
x – 3y = 0
= – tan30 0 = –1
...(i)
Comparing it with ax 2 + 2hxy + + by 2 = 0 we get, a = = k , , 2h 2h = = 4 , b = = –1 Let m 1 and m 2 be the slopes of the lines
)
3y = 0
represented by kx 2 + 4xy – – y 2 = 0 ∴
i.e., x 2 – 3y 3y 2 = 0.
m 1 + m 2 = 4 and m 1m 2 = – k k
According to the given condition Q-16 Q-16)) Find k , if the sum of the slopes of the lines represented by the equation x 2 + kxy – kxy – 3y 2 = 0 is twice their product – 3y 3y 2 = 0 Ans. Given equation is x 2 + kxy – Comparing it with ax with ax 2 + 2hxy + + by 2 = 0 we get, a = = 1 , 2h 2 h = = k , , b = = –3 Let m 1 and m 2 be the slopes of the lines represented by x 2 + kxy – – 3y 2 = 0 ∴
m 1 + m 2 =
–2h k = b 3
∴
=
m 1 + (m 1 + 8) = 4
∴
2m 1 = – 4
∴
m 1 = – 2
Now, m 1 (m (m 1 + 8 ) = – k k (– 2) (–2 + 8 ) = – k k k ∴ (–2) (6) = – k k ∴ –12 = – k = 12 ∴ k =
may be perpendicular to one of the lines
a 1 = – b 3 According to the given condition. m 1m 2 = 2 (m 1m 2) k 3
∴
Q-19) Q-19) Find the condition that the line 3 3x +y = = 0 x +y represented by ax 2 + 2 hxy + by 2 = 0
and m 1m 2 =
∴
m 2 = m 1 + 8
Ans. Th e au x i l i ar y e qu a t i o n o f th e l i n e s represented by ax 2 + 2hxy + + by 2 = 0 is bm 2 + 2hm + + a = = 0 Since one line is perpendicular to the line
1 2
2 −
3x + + y = = 0 whose slope is –
k = = – 3
Q-17)Find k , if the slopes of the lines
∴
slope of that line = m =
∴
m =
represented by the equation 3x 2 + kxy – – y 2 = 0 differ by 4 Ans. Given equation 3x 3 x 2 + kxy + kxy – – y 2 = 0 Comparing it with ax with ax 2 + 2hxy + + by 2 = 0
bm 2 + 2hm + + a = = 0 ∴
1 1 b + 2h + a = 0 3 3
∴
b
Let m 1 and m 2 be the slopes of the lines represented by 3x 3 x 2 + kxy + kxy – – y 2 = 0 –2h –2h
k
a
+
∴ 9a +
1 3
1 is the root of the auxiliary equation 3
2
we get, a = = 3 , 2h 2 h = = k , , b = = –1
3 = –3 1
6h + 9a = 0 6h
+ b =
0
Mahesh Tutorials Science
14
Q-20)Show Q-20)Show that 3x + 4y + + 5 = 0 and the lines (3x +4y ) )2 –3 (4x +3y )2 = 0 form the side an equilateral triangle.
Q-21)Show Q-21)Show that
4xy + + y 2 = 0 and the lines
+ y – 6 = 0 form an equilateral triangle
x
Find its area and perimeter
Ans. The slope of the line 3x + 4y + + 5 = 0 is m 1 =
–3 4
Ans. Let
O A
Y
and OB be be
B 0
Let m be be the slope of one of the line making
the lines
an angle of 60 0 with the line 3 x + + 4y + + 5 = 0
through
Since the angle between the lines having
origin
slope m and m 1 is 600
Let M be
tan 6 60 00 =
x 2 –
m – m 1
the slope
1 + m.m 1
of OA or
6 0
M x + y =
60
0
A O
X
OB which which
–3 m – 4 3 = –3 1 + m 4
from equilateral triangle with line x + y = 6 whose slope is Its equation is y = = mx
∴
tan 6 60 00 =
4m + 3 3 = 4 – 3m
3=
Squaring both sides, we get 2
( 4m + 3) 3= 2 ; ( 4m – 3)
6
2
3(4 – 3m ) = (4m + 3) 3)
3 (16 – 24m + 9m 2) = 16m 2 + 24 24m + 9
∴
48 – 72m + 27m 2 = 16 16m 2 + 24m + 9
∴
11m 2 – 96m + + 39 = 0
1 – m
;
3=
m +1
1 – m
m 2 + 2m + 1 m 2 – 2m + 1
∴
3m 2 – 6m + + 3 = m 2 + 2m + + 1
∴
2m 2 – 8m – – 2 = 0
∴
m 2 – 4m – – 1 = 0
2
∴
m +1
...(i)
Joint equation of the pair of lines is given by y 2 2
– 4
y
+ 3 = 0
...[From (i)]
Since, required lines pass through the origin,
x
their equations is of the form
x 2 – 4xy + + y 2 = 0 which is given in question
y y = = mx i.e. m = x
Hence, joint equation from equilateral triangle with x + y = 6
2
y y 11 – 96 + 3 9 = 0 x x 11
y
2
x 2
⊥
distance of AB from origin is
OM =
2
–
x
96y + 39 = 0 x
– 6 1 +1
= 3
2
11y – 96xy + + 39x = 0
∴
39x 2 – 96xy + + 11y 2 = 0
Area of
∆
OAB = =
is the required joint equation which can be writ writte ten n as as 39x 2 – 96xy + + 11y 2 = 0
sin 60 600 =
∴
i.e. (9x 2 + 24 24xy + 16y 2) –3(16x 2 – 24xy + 9y 2) = 0 i.e. (3x + + 4y )2 –3(4x – – 3y ) = 0 Hence, the line 3 x + + 4y + + 5 = 0 and the lines
=
( 3)
2
3
In right – angled triangle OAM
i.e. (9x 2 + 24 24xy + 16 16y 2) – (48x 2 – 72xy + 27y 2) = 0
3
2
= 3 sq units.
i.e. (9x 2 – 48x 2) + (24xy + 72xy ) + (16y 2 – 27y 2) = 0
( OM)
∴ ∴
OM OA
3 3 = = 2 OA = ∴ 2 OA Length of the each side of the equilateral triangle OAB = = 2 units perimeter of ∆ OAB = 3 × length of each side
Mahesh Tutorials Science
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Q-22)Find Q-22) Find the joint equation of lines lines through the
GROUP (C) – CLASS WORK PROBLEMS
origin, each of which makes angle of 30 0 with the line 3x + + 2y + + 66 = 0
Q -1) -1) Find the acute angle between between the lines given by x 2 + y 2 = 2xy cosec cosec α
e3 x + + 2y 2y + + 66 = 0 is Ans. The slope of the lin e3x
Ans. Comparing the equation
3 m 1 = – 2
x 2 +y 2 = 2xy cossec cossec α with a with ax x 2 + by 2 + 2hxy = hxy = 0
Let m be the slope of one of the lines making
we get a = = 1 , b = = 1 , h = = – cosec α
an angle of 300 with the 3x 3x + + 2y 2y + + 66 = 0. The
Let θ be the acute anle between the lines
angle between the lines having slopes m and m and m 1 is 30
30 0 ∴ tan 3
∴
tan
0
0
=
m – m 1
, where tan 3 30 00 =
1 + m.m 1
3 m – – 1 2 = 3 1 + m – 3 2
∴
1 3
=
θ
=
1 3
2m + 3 2 – 3m
tan θ =
tan θ =
On squaring both the sides, we get,
2 h 2 – ab a + b 2 cosec 2 α – 1 2 cot2 α
tan
θ
= cot
tan
θ
= tan – α 2
α
π
2
1 ( 2m + 3 ) = 3 ( 2 – 3m )2
θ
∴
(2 – 3m 3m )2 = 3(2m 3(2m + + 3)2
∴
4 – 12m 12m + 9m 9m 2 = 3(4m 3(4m 2 + 12m 12m + 9)
∴
4 – 12m 12m + 9m 9m 2 = 12m 12m 2 + 36m 36m + 27
∴
3m 2 + 48m 48m + 23 = 0
2
– α
linesgiven linesgiven by (a 2 – 3b 2) x 2 + 8abxy + + (b 2 – 3a 2)y 2 = 0
lines and their joint e quation is obtained by
∴
π
Q -2) Find the acute angle between the
This is the auxi liary liar y equatio equ ation n of the two
putting m = m =
=
y x
Ans. Combined equation of lines is,
(a 2 – 3b 2 ) x 2 + 8abxy + (b 2
– 3a 2 ) y 2 = 0
comparing with, Ax 2 + 2Hxy + By 2 = 0 So,
the joint equation of the two lines is
A = a 2 – 3b 2 ; H = 4a ; B = b 2 – 3a 2
2
y y 3 + 4 8 + 23 = 0 x x ∴
3y
2
2
x
+
ta n θ =
48y + 23 = 0 x
2
2
2
48xy + + 23x 23x = 0; ∴ 23x 23x ∴ 3y + 48xy
2
+ 48xy 48xy + + 3y = 0.
tan θ =
2 h 2 – ab
a + b 2 16a 2b 2 – (a 2 – 3b 2 ) (b 2 – 3a 2 ) a 2 – 3b 2 + b 2 – 3a 2
Q-23) Q-23) Find the condition that the line 4x +5y = 0 coincides with one of the lines given by 2
2
+ by = 0 ax + 2hxy +
tan θ =
2 16a 2b 2 – 10a 2b 2 + 3a 4 + 3b 4 –2 (a 2 + b 2 )
hxy + by 2 = 0. Ans. ax 2 + 2hxy + 2
y y ax + + 2h + b = 0 x x
tan θ =
4x +5y +5y = = 0 ; 5y 5y = = –4x –4x 2
–4 y –4 –4 = ; a + + 2h 2h + b = 0 5 x 5 5
tan θ =
2 3a 4 + 6a 2b 2 + 3b 4 –2 (a + b 2
2
)
– 3 (a 2 + b 2 )
(a 2 + b 2 )
; t an θ = 3
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Q -3) -3) If the angle angle between between the lines lines + by 2 = 0 is equal to the angle ax 2 + 2hxy + between the lines 2x 2 – 5xy + 3y 2 = 0 prove
2 tan45 =
that 100(h 2 – ab ) = (a + + b )2
h 2 –6 4 5
Ans. 1st combined equation is, ax 2 + 2hxy + by 2 = 0
... (i)
h 2 – 24
5=2
So,
4
5 = h 2 – 24
2 h 2 – ab tan θ = a + b
h 2 – 24 = 25
2nd combined equation is,
h 2 = 2 5 + 2 4
2x 2 – 5xy + 3y 2 = 0
h 2 = 49 h = ±7
–5 , B = 3 A = 2, H = 2
Q -5) -5) Show that one of the straight straight lines given given 2 tan θ =
25 –6 4 5
by ax 2 + 2hxy + by 2 = 0 bisects an angle between the co-ordinate axes if (a + + b )2 = 4h 2 e quation of a straight straig ht line li ne is give n by, by , Ans. The equation
ax 2 + 2hxy + by 2 = 0
1 2 4 tan θ = 5
tan θ =
Divide by x 2 , 2
y y a + 2h + b = 0 x x put y = mx
1 5
∴ m =
As per given, The angle be tween twee n these two lines li nes is e qual ∴
2 h 2 – ab 1 = a + b 5
2
2
Q -4) -4) If acute acute angle angle between between 3x + hxy + + 2y = 0 is of measure 45 0, find h . Ans. The combined equation is,
comparing with, Ax 2 + 2Hxy + By 2 = 0 h A = 3, H = , B = 2 2 2 h 2 – ab tan θ = a + b
+ 2hm + a = 0
ordinate axes. 2
3x + hxy + 2y = 0
2
As the lines bisects an angle between co-
100 (h 2 – ab ) = (a + b ) ... Hence proved
2
a + 2hm + bm bm 2 = 0 ∴ bm
10 h 2 – ab = a + b
2
y x
So, θ = 45
0
∴ m =
tan θ
∴ m =
tan450
∴ m = 1
∴b
2
(1) + 2h (1) + a = 0
b + 2h + a = 0
(a + b ) = –2h squaring on both sides, 2
(a + b ) = 4h 2
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GROUP (C) – HOME WORK PROBLEMS ∴ ta n θ
Q -1) Find the measure of the acute angle between the lines represented by : i) 3x 2 – 4 3xy
+
3y 2 = 0
∴θ =
=
3 5
3 tan –1 . 5
iii) Comparing the equation
ii) 4 x 2 + 5xy + y 2 = 0
2x 2 + 7xy + 3y 2 = 0 with
iii) 2x 2 + 7xy + 3y 2 = 0
ax 2 + 2hxy + by 2 = 0 , we get,
Ans. i) Comp Compar arin ing g the the equat equation ion 2
2
3x – 4 3xy + 3y = 0 with ax 2 + 2hxy + by 2 = 0 , we get, a = = 3, 2h = –4 3 , i.e., h = –2 = –2 3 and b = 3
Let θ be the acute angle between the lines. ∴ ta n θ
∴ ta n θ
=
(
–2 3
)
2
2 h 2 – ab a + b
2+3
– 3 (3)
3+3
2 12 12 – 9 2 3 = 6 6 1
∴ tan θ = ∴θ θ =
49 –6 4 5
3
=
= tan 300 =
300
2 49 – 24 24 5×2
ii ) Compa Comparin ring g the equation equation 2
ax 2 + 2hxy + by 2 = 0 , we get, a = = 4, 2h = = 5, i.e., h = =
5 2
= and b = = 1.
Let θ be the acute angle between the lines. ∴ ta n θ
2 h 2 – ab a + b
=
2
5 2 – 4 (1) 2 4 +1
25 5
=
2
4x + 5xy + y = 0 with
=
=
7 2 – ( 2) (3) 2
2
=
, b = = 3
2
a + b
=
2
2
Let θ be the acute angle between the lines.
2 h 2 – ab
=
7
a = = 2, h = =
5 5
∴ tan θ
θ=
=1
450
Q -2) Find the value of b, if the angle between between the lines given by 6x 2 + xy + + by 2 = 0 is 45 0 Ans. Comparing the equation 6x 2 + xy + + by 2 = 0 with Ax 2 + 2Hxy + By 2 = 0 , we get, A = = 6, 2 H = = 1, i.e., H = =
25 3 2 –4 2× 4 2 = = 5 5
1 2
and B = = b
Since the angle between the lines is 45 0, 0
tan 4 45 5 =
2 H 2 – AB A + B
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Q -4) -4) If the lines lines given given by ax 2 + 2hxy + + by 2 = 0 2
1 – 6 (b ) 2
form an equilateral triangle with the line
2 ∴1 =
= 1, show that (3 a + b ) (a +3b ) lx + my =
6 + b
–4h 2 = 0 + by 2 = 0 form an Ans. Since the lines ax 2 + 2hxy + equilateral triangle with the line
2 ∴1 =
= 1, the angle between the line s lx + my =
1 – 6b 4 6 + b
+ by 2 = 0 is 600 ax 2 + 2hxy + ∴
∴
1 – 6b 4
(6 + b )2 = 4
∴
36 + 12b + b 2 = 1 – 24b ∴ b 2 + 36b + 35 = 0 35 = 0 ∴ b 2 + 35b + b + 35 1(b + 35) = 0 ∴ b (b + 35) + 1( 1(b + 35) = 0 ∴ (b + 1) + 1( 0 or b + 35 = 0 0 ∴ b + 1 = 0 – 1 or b = – – 35. ∴ b = – ∴
Q -3) -3) Show that the angle angle betwee between n (
(4b 2 – 3a 2)y 2 = 0 a 2 – 12b 2)x 2 + 16abxy + (4
is 600 Ans. Given combined equation is,
(a
tan 60 600 =
2
– 12b
2
)x
+ 16abxy + ( 4b – 3a
2
2
2
)y
2
= 0
3=
2 h 2 – ab a + b
2 h 2 – ab a + b
∴
3(a + +b )2 = 4 (h 2 – ab )
∴
3(a 2 + 2ab + b 2) = 4h 2 – 4ab
∴
3a 2 + 6ab + 3b 2 + 4ab = = 4h 2
∴
3a 2 + 10ab + 3b 2 = 4h 2
∴
3a 2 + 9ab + ab + 3b 2 = 4h 2
∴
3a (a + + 3b ) + b (a + + 3b )= )= 4h 2
(3a + + b ) (a + + 3b ) – 4h 2 =0 This is the required re quired condition ∴
Q-5) Find the value value of k , if the lines x 2 + kxy + + + y = = 1 contain the sides of an y 2 = 0 and x +
comparing with,
equilateral triangle. Also, find the area of
Ax 2 + 2Hxy + By 2 = 0
the triangle. Ans. Since the lines x 2 + kxy + y 2 = 0 form an
A = a 2 – 12b 2
equilateral triangle with the line
H = 8ab
+ y = = 1, the angle between the lines x +
B = 4b 2 – 3a 2 =
+ y 2 = 0 is 600 x 2 + kxy +
2 64a 2b 2 – 40a 2b 2 + 3a 2 + 48b 2 a 2 – 12b 2 + 4b 2 – 3a 2
Comparing the equation x 2 + kxy + y 2 = 0 with
ax 2 + 2hxy + by 2 = 0 = 1, 2h = = k , b = = 1 a =
=
=
2 24a 2b 2 + 3a 2 + b 2 + 48b 4 2
2
– 8b – 2a
tan 60 =
2 3 ( 4b 2 + a 2 ) –2 ( 4b + a 2
2
)
2 h 2 – ab a + b 2
tan 60 =
2 ( k 2 ) – (1) (1) 1 +1
= – 3
3=
tan θ = 3 –1
2
k 2
4
–1
2
( 3)
θ
= t an
θ
= 60 ... Hence proved
3=
k 2 – 4 2
On squaring both sides
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∴
k = ±4
2x 1 + y 1 = 0 and 2x 1 – y 1 + 1 = 0
Let the length of the perpendicular from from the
Solving these e quations simultaneously, simultaneously, we
given to the line x + + y = = 1 be p ‘ ’ ’
get, x 1 = – 1/4 and y 1 = 1/2
Area of the equilateral triangle =
=
p 2 3
sq units =
1 2 3
1 1 3 2
sq units
Q -6) -6) Find the value value of b, if if θ is the measure of angle between the line 3x 2 + 4xy + + by 2 = 0 and tan θ =
1 2
∴
the point of intersection is P (– (– 1/4, 1/2)
Q -2) -2) Find combined equation equation of a pair of lines lines through the point (2, -3) and perpendicular to the lines given by x 2 + xy – – 2y 2= 0. Ans. x 2 + xy – 2y 2 = 0 x 2 + 2xy – xy – 2y 2 = 0 x ( x + 2y ) – y ( x + 2y ) = 0
( x – y ) ( x + 2y ) = 0
Comparing the equation 3x 3x 2 + 4xy 4xy + + by 2 = 0 Ans. Comparing
x – y = 0 or x + 2y = 0
with Ax 2 + 2Hxy 2Hxy + + By 2 = 0
Slope of x – y = 0 is 1.
we get A = = 3, 2H 2H = = 4 i.e. H = = 2 B = = b
Slope of line
Now, tanθ =
tan θ =
∴
1 2
[gn]
2 H 2 – AB A + B
1 = 2
2 22 – 3b 3 + b
1 = 2
2 4 – 3b 3 + b
On squaring both sides
1 = 4
4 ( 4 – 3b )
( 3 + b )
2
9 + 6b 6b + + b 2 = 64 – 48b 48b
r
⊥
to to x – y = 0 is –1
equation of line
r
⊥
to x – y = 0
and passing through ( 2, –3) is,
y + 3 = –1( x – 2) y + 3 = – x + 2 x + y + 1 = 0
... (i)
–1 Slope of x + 2y = 0 is 2 r
∴
Slope of line
∴
equation of line
⊥
to x + 2y = 0 is 2 r
⊥
to x + 2y = 0
and passing through ( 2, –3) is,
y + 3 = 2 ( x – 2) y + 3 = 2x – 4
b 2 + 54b 54b – – 55 = 0
2x – y – 7 = 0
b 2 + 55b 55b – – b – – 55 = 0
combined equations of lines is,
(b + + 55) (b (b – – 1) = 0
( x + y + 1) ( 2x – y – 7 ) = 0
∴
from (i) and (ii)
b = = –55 & b = = 1
GROUP (D) – CLASS WORK PROBLEMS
... (ii)
2x 2 – xy – 7x + 2xy – y 2 – 7y + 2x – y – 7 = 0 2x 2 + xy – y 2 – 5x – 8y – 7 = 0
Q -1) Find the separate separate equation equation of the the lines represented by 4x 2 – y 2 + 2 x – – y = 0. Also y = find the point of intersection inters ection of these lines. li nes is Ans. The combine d equation of the lines 2
2
4x – y + 2x – – y y = = 0 ∴
(2x (2x + + y )(2x )(2 x – – y ) + (2x (2x + + y ) = 0
∴
(2x (2x + + y )(2x )(2 x – – y + + 1) = 0
∴
their separate equations are
2x + + y = = 0 and 2x 2 x y + + 1 = 0
Q -3 -3) Find Find k if kx 2 – xy xy + 2y 2 + 19x – 6y – 20 = 0 represents pairs of lines. Further find whether these lines are parallel or intersecting Ans. Given equation is kx 2 – xy xy + 2y 2 + 19x – 6y – 20 = 0
Mahesh Tutorials Science
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we get a = k,h = ∴
–1 2
, b = 2, g =
19 2
, f = –3, c = –2 0
Equation represents a pair of lines
abc + 2 fgh – af –40k + ( –6 ) ×
2
2
2
– bg – ch = 0
19
2
px 2 – 8xy + + 3y 2 + 14 14x +2 +2y + + q = 0 ax 2 + 2hxy + by + by 2 +2gx +2gx + + 2 fy + + c = = 0, we get
1
a = p = p , h = = –4 , b = = 3, g = = 7 f = = 1 and c = = q Since the given equation represents a pair of line perpendicular to each other, ∴
57 3 61 –49k + – +5 = 0 2 2 –49k –
Ans. Given equation is Comparing it with
– k × 9 – 2 × 2 2 36 1 1 + 20 × = 0 × 4 4
304
Q -5) Find Find p and and q , if the equation 2 px – 8xy + 3y 2 + 14x +2y + q = 0 represents a pair of perpendicula p erpendicularr lines
a + b = = 0
∴ p
+3=0
∴ p
= –3
Also the equation represents a pair of straight
+5 =0
lines
–49k – 152 + 5 = 0 –49k = 147
∴
k = –3 2
–1 Consider h 2 – ab = – k ×2 2 =
h
g
h
b
f = 0
g
f
c
–3 –4 7 ∴
1 25 +6= ≠0 4 4
∴
a
lines are intersecting
–4
3
1 = 0
7
1
q
∴ –
3 (3q (3q – – 1) + 4 (–4q (–4q – – 7) + 7 (–4 – 21) = 0
∴ –9q –9q + +
Q -4 -4) Find Find λ λ so that equation λ = 0. (x –1)2 – (y –1)2 – (x –1) – 5 (y –1) + λ
represents two straight lines. Ans. Given equation is
3 – 16q 16 q – – 28 – 175 = 0
∴ –25q –25q – –
200 = 0
∴ –
25(q 25(q + + 8) = 0 ∴ q = = –8 ∴ p = = –3 and q = = –8
λ = 0. (x –1)2 – (y –1)2 – (x (x –1) – 5 (y (y –1) + λ
...(i)
( x 2 – 2x + 1) – ( y 2
by ax 2 + 2hxy + by 2 = 0 bisects an angle
– 2y + 1)
between the co-ordinate axes if
– x + 1 –5y –5y + 5 +λ= 0 x –2x –2x + 1 – y y +2y +2y – 1 – x + 1 –5y –5y + 5 +λ = 0 2
2
x 2 –3x –3x – y y 2 –3y –3y + 6 +λ = 0 x 2 – y y 2 –3x –3x –3y –3y + 6 +λ = 0
...(ii)
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
∴
(a + + b )2 = 4h 2 Ans. The equation of a straight line is given give n by, ax 2 + 2hxy + by 2 = 0 2 Divide by x ,
Comparing with
a = 1, h 1, h = 0, b 0, b = –1, g =
Q -5) -5) Show that one of the straight straight lines given given
–3 2
, f =
–3 2
2
, c = 6 +λ
y y a + 2h + b = 0 x x put y = mx
Equation (ii) represents a pair of lines 2
2
abc + 2 fgh – fgh – af af – bg bg 2 – ch ch = 0 1 × –1 × (6 + λ ) + 0 –1 × 9 9 λ – –6 + λ + =0 4 4 λ = 0 –6 – λ
9 9 + 1 × – 0 = 0 4 4
∴ m =
y x
a + 2hm + bm 2 = 0 ∴ bm
2
+ 2hm + a = 0
As the lines bisects an angle between coordinate axes. So,
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tanθ = 1
∴ m =
tan θ
∴ m =
tan45 0
θ=
∴ m = 1
∴ b (1)
2
+ 2h (1) + a = 0
π
4
GROUP (D) – HOME WORK PROBLEMS
b + 2h + a = 0 Q -1) Find the combined equation equation if the pair of
(a + b ) = –2h
lines through (2,3) one of wh ich is parallel
squaring on both sides, 2
to 2x + 3y = = 5 and other perpendicular to
2
(a + b ) = 4h
– x –
Q -6) -6) Show Show that that 2x 2 + 7xy +3 +3y 2 – 5x – – 5y + + 2 = 0 represents a pair of lines and find f ind the acute angle between them. Ans. Comparing
parallel Ans. Let L 1 be the line throught (2,3) and parallel to the line 2 x + + 3y = = 5 2 The slope of the line 2x + + 3y = = 5 is – 3
2x 2 + 7xy +3 +3y 2 – 5x – – 5y + + 2 = 0 with 2
2
ax + 2hxy + by + by +2 +2gx + gx + 2 fy + + c = c = 0 we get a = = 2 ,h = =
7 –5 –5 , b = = 3, g = = , f = = and c = = 2 2 2 2 2
∆ =
2
0 2
LHS = abc = abc + 2 fgh – fgh – af – bg – ch
–5 7 = 2 × 3 × 2 + (–5) 2 2 ×
∴
2 the slope of the line L 1is – and it passes 3
through (2,3)
condition for pair of lines is
– 2
4y =7. =7.
25 –3 4
×
25 – 2 4
×
∴
2 equation of the line L 1is y – – 3 = – (x – 2) 3
∴
3y – – 9 = – 2x + 4
∴
2x + + 3y – 13 = 0
Let L 2 be the line through (2,3) and perpendicular to the line x – – 4y = = 7.
49 4
The slope of the l ine x – – 4y = = 7 is
25 × 7 50 75 98 = 12 + – – – 4 4 4 4
∴
the slope of the line L 2 is – 4 and it passes
through (2,3) ∴
175 – 125 98 = 12 + = 4 4
–1 1 = . –4 4
equation of the line L 2 is – 3 = – 4(x – – 2) y –
50 98 = 12 + – 4 4
∴
– 3 = – 4x + + 8 y –
∴
4x + + y – – 11= 0 0
Hence, the equations of required lines are
48 = 12 – 4
2x + + 3y – – 13 = 0 0 and 4x + + y – – 11= 0. 0. ∴
= 12 – 12
their combined equation is
(2x + + 3y – 13) (4x + + y – – 11) = 0. 0.
=0 ∴ The
equation represents a pair of straight lines If θ is the acute angle between the lines then tan θ =
2
=
2+3
=
8x 2 + 2xy – – 22x + 12xy + 3y 2 – 33y – 52x – 13y + 143 = 0
∴
8x 2 + 14xy + + 3y 2 – 74x – 46y + 143 = 0.
Q -2) -2) Find the joint equation equation of the pair of lines through (– 2, 2), one of wh ich is parallel to the line x + 2 y – 3 = 0 and other is
2 h 2 – ab a + b
7 2 – 2 (3) 2
∴
49 2 –6 4 5
perpendicular perpendicular to the line y = = 3 Ans. Let L 1 be the line passes throught (– 2,2) and parallel to the line x + + 2y – – 3 = 0 whose slope –1
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∴
∴
Slope of line L 1 is
Q -4 -4) Find Find k if if x 2 – 3xy + 2y 2 + x – y + k = 0
–1 2
represents a pair of lines
equation of the line L 1 is
( y – 2) =
Ans. x 2 + 3xy + + 2y 2 + x – – y + + k = 0
–1 ( x + 2) 2
a = = 1, b = = 2, h = =
2y – – 4 = – x – – 2
Condition for pair of lines.
x + + 2 + 2y 2 y – – 4 = 0
abc + + 2 fgh – fgh – af 2 – bg 2 – ch 2 = 0
x + + 2y – – 2 = 0 Let L 2 be the line passes throught (– 2, 2) and
1 1 9 3 – 1 – 2 – k = 0 4 4 4 8
2k + + 2
perpendicular to the line y = = 3 ∴
equation of the line L 2 is of the form x = = a
∴
2k – –
Since L 2 passes throug (– 2, 2), 2), – 2 = a ∴
equation of the line L 2 is x = = – 2
i.e., x + + 2 = 0 Hence the equation of the required line are x + + 2y – 2 = 0 and x + + 2 = 0
3 1 1 9k – – – =0 4 4 2 4
8k – – 3 – 1 – 2 – 9k 9 k = = 0 – k k = 6 ∴ k = – 6 Q -5) -5) Find the separat separate e equations equations for for each of the the
The joint equation is
∴
following lines
(x + + 2y – 2) 2) (x (x + + 2) = 0
10(x + + 1)2 + (x + + 1) (y – – 2) – 3 (y – – 2)2 = 0
x 2 + 2x + 2xy 2xy + 4y – – 2x – – 4 = 0
Ans. The given gi ven equation is
x 2 + 2xy + 4y 4y – – 4 = 0
10(x 10(x +1) + (x (x +1) +1) (y (y – 2) – 3 (y – – 2) = 0 ...(i) Let x + + 1 = X = X and and y – – 2 = Y
Q -3) -3) Show that that the equat equation ion 2
3 1 –1 , g = = , f = = , c = = k 2 2 2
∴
2
The given equation becomes
9x – 6xy + y + 18x – 6y + 8 = 0
10X 10X 2 + XY – – 3Y 2
= 0
represents two lines parallel to each other. oth er.
10X 10X 2 + 6XY – XY – 5XY 5XY – – 3Y 3Y 2
= 0
Ans. Given equation is 2
2X (5X (5X + + 3Y ) – Y (5X (5X + + 3Y 3Y ) = 0 (2X (2X – – Y ) (5X (5X + + 3Y ) = 0
2
9x – 6xy + y + 18x – 6y + 8 = 0
∴
Comparing with ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
Either 2X 2X – – Y = 0 ⇒ 2(x 2(x + 1) – (y ( y – – 2)= 0
⇒ 2x – – y + + 4
= 0
OR, 5X 5X + + 3Y
= 0 ⇒ 5(x 5(x + + 1)+ 3(y 3( y – – 2) 2) = 0
a = 9, h = –3 –3, b = 1, g = 9, f = –3 –3, c = 8.
⇒ 5x + + 3y – – 1 = 0
Consider
∴
abc + 2 fgh – af 2 – bg 2 – ch 2
straight lines
= 9 × 1 × 8 + ( –6) × 9 × –3 – 9 × 9 – 1 × 81 – 8 × 9 = 72 + 16 162 – 81 – 8 1 – 72
2x – – y + + 4 = 0 and 5x 5 x + + 3y – – 1 = 0 Q -6) -6) Find p and q if the the equati equation on
=0
12x 2 + 7xy – – py 2 – 18x + + qy + + 6 = 0
The given equation e quation represents a pair of
∴
lines
Th e g i v e n e q u at i o n re pr e s e n t s t wo
… (i)
represent a pair of perpendicular perpendicular lines. Ans. 12x 12x 2 + 7xy – xy – py 2 – 18x 18x + + qy + + 6 = 0
2
Now, h 2 – ab = ( –3 ) – 9 × 1 = 9 – 9 = 0 As h 2 – ab = 0 line lines s ar are pa parallel llel.. ∴
a = = 12, h = =
…(ii) (ii)
From (i) & (ii)
9x 2 – 6xy + y 2 + 18x – 6y + 8 = 0 represents a pair of parallel straight lines.
7 q , b = = – p , c = c = 6, g = = –9, f –9, f = = 2 2
Conditions of perpendicular lines, a + + b
= 0
12 – p – p
= 0
p = 12 Condition for pair of lines, abc + + 2 fgh – – af 2 – bg 2 – ch 2 = 0
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q (–9) 2
12 × (– )(6) + 2 p )(6)
×
– (– p ) 81 – 6 72 × – p – p –
108 –
×
2q 2 – 4q + 2 = 0
49 4
=
0
147 63q – 3q 2 + 81p – = 0 2 2
72 × – 12 – 12 –
– 864 – 864 –
–240 + 2 – 2q 2 + 240 + 4q = 0
2 7 q – 12 2 2
147 63q – 3q 2 + 81 × 12 – = 0 2 2
147 63q – 3q 2 + 972 – = 0 2 2
+ 1 = 0 q 2 – 2q + (q – – 1) (q – – 1) = 0 i.e. q = = 1
q 2 = 1 ;
8, q = 1 q = 1 ; p = 8,
Q -8) -8) Find the the valu value e of k , if each of the following equations represents a pair of lines : i) 3x 2 + 10xy +3y 2 +16y + +k = 0. ii) kxy + + 10x +6y +4 = 0 iii) x 2 + 3xy +2y 2 +x – y +k = = 0 y +
147 63q – 3q 2 – = 0 2 2
Given n equa equati tion on is is Ans. i) Give
6q 2 + 63q – – 69 = 0
3x 2 + 10xy +3y 2 +16y +k = 0.
2q 2 + 21q – – 23 = 0
Comparing it with
2q 2 + 23q – – 2q – – 23 = 0
+ by 2 + 2gx + + 2 fy + + c = = 0 ax 2 + 2hxy +
(2q + 23) –1 (2 q + + 23) = 0 q (2
we get, a = = 3, h = = 5 b = = 3 , g = = 0, f = = 8 c = = k
–23 q= or q = = 1 2
Since the equation represents a pair of lines
= 12 p =
∴
+ 2 fgh – – af 2 – bg abc + bg 2 – ch 2 = 0
∴
(3)(3)(k ) + 2 (8) (0) (5) –3(8) 2 –3 (0) 2 – k (5)2 =0
Q -7 -7) Find Find p and q if the following equations 9k + + 0 – 192 – 0 – 25k
represent a pair of parallel lines
∴
2x 2 + 8xy + + py 2 + qx + + 2y – – 15 = 0
∴
–16k – 192 = 0
∴
–16k – 192
Ans. The equation is,
Hence k = –12
2x 2 + 8xy + py 2 + qx + 2y – 15 = 0
ii) Given Given equat equation ion is is kxy + + 10x + 6y + + 4 = 0
comparing with 2
Comparing it with
2
ax + 2hxy + by + 2gx + 2 fy + c = 0 a = 2, h = 4, b = p, g =
+ by 2 + 2gx + + 2 fy + + c = = 0 ax 2 + 2hxy +
q , f = 1, c = –1 5 2
we get, a = = 0, h = =
The lines line s are parallel
= 4 c = Since the equation represents a pair of
h 2 – ab = 0
lines
2
( 4 ) – 2 ( p ) = 0
∴
+ 2 fgh – – af 2 – bg abc + bg 2 – ch 2 = 0
∴
(0)(0)(4) + 2 (3) (5)
2
( 4 ) = 2 p p =
k = 0 , g = = 5, f = = 3 b = 2
16 2
k –0 (3)2 – 0 (5)2 2
2
k – 4 = 0 2
p = 8 The abo ve equation equ ation re presents prese nts pair of line lin e
abc + 2 fgh – af
2
– bg 2 – ch 2 = 0
2 q q 2 ( 8 ) ( –15 ) + 2 (1) ( 4) – 2 (1) – 8 2 2 2
– ( –15 ) ( 4) = 0
2
∴
0 + 15k – – 0 – 0 – k 2 = 0
∴
15k – – k k 2 = 0
∴ ∴
– k (k – – 15) = 0 k ( = 0 or k = = 15 k =
If k = = 0 then the given equation becomes 10x + 6y + + 4 = 0 which does not represent a
Mahesh Tutorials Science
24
Hence k = = 15 iii) iii)
Given equ equation is
∴
+k = = 0 x 2 + 3xy +2y 2 +x – y y + Comparing it with
∴
+ by 2 + 2gx + + 2 fy + + c = 0 ax 2 + 2hxy + 3 1 1 we get, a = = 1, h = = = 2 , g = = , f = = – b = 2 2 2
c2 c bc =0 – 0 + 0 – 4 2 2
a 0 –
∴
0–
ac 2 4
– 0+0 –
4
=0
ac 2 bc 2 – =0 4 4
–
ac 2 + bc 2 = 0
= k c = Since the e quation represents represents a pair of
c 2 (a (a + + b ) = 0
lines
a + + b = = 0 or c or c = = 0
∴
bc 2
This is the th e required requi red condition co ndition
a
h
g
h
b
f = 0
g
f
c
Q-10)Find p and and q if if the equation = 0 represent px 2 – 8xy + 3y 2 + 14x + 2y + q = two straight lines which are perpendicular. perpendicular.
3 2
1
3 2 2 1 1 – 2 2
∴
1 2 1 – =0 2 k
Also find the co-ordinate of their point of intersection. Ans. Given equation is px 2 – 8xy + + 3y 2 + 14x 14x +2y +2y + + q = 0 Comparing it with ax 2 + 2hxy + by + by 2 +2gx +2gx + + 2 fy + + c = = 0, we get
2 1 3 2 1
∴
∴
∴ ∴
3
1
4
–1 = 0
–1 2k
2
3
1
3
4
–1 = 0
1
–1 2k
2 (8k (8k –1) –1) –3 (6k (6k +1 +1 ) + 1 (–3 –4) = 0 16k 16k – – 2 – 18k 18k – – 3 – 7 = 0
∴
–2k –2k –12 –12 = 0
∴
–2k –2k = 12
∴
k = –6
a = p = p , h = = –4 , b = = 3, g = = 7 f 7 f = = 1 and c = = q Since the given equation represents a pair of line perpendicular to each other, ∴
a + b = = 0
∴
p + 3 = 0
∴
p = –3
Also the equation represents rep resents a pair of straight lines
∴
a
h
g
h
b
f = 0
g
f
c
Q -9) -9) Find the conditi condition on that equati equation on +cx +cy = = 0 may represent a l pair ax 2 +by + of lines
∴
Ans. Given equation is
−
3
−
4
7
4 7
−
3 1
1=0 q
2
ax +by +by +cx +cx +cy = = 0 with Ax 2 + 2Hxy 2Hxy + + By 2 + 2Gx 2Gx + + 2Fy + Fy + C = = 0, we get
∴
– 3 (3q (3q – – 1) + 4 (–4q (–4q – – 7) + 7 (–4 – 21) = 0
c c A = = a , H = = 0, B = = b , g = = , f = = , C = = 0 2 2
∴
–9q –9q + + 3 – 16q 16q – – 28 – 175 = 0
∴
–25q –25q – – 200 = 0
∴
– 25(q 25(q + + 8) = 0
The given equation represents represe nts a pair of lines
A
H
H
B
c a 0 2 G c F = 0 i.e, if 0 b =0 2
∴
∴
q = = –8 p = = –3 and q = = –8
Mahesh Tutorials Science
25
BASIC ASSIGNMENTS (BA) :
Q-2) Find the separat separate e equati equations ons of the the lines lines
BA – 1
represented by the following equations : = 0 x 2 – 4xy =
Q-1) Q-1) Find the joint equation equation of the following following pair pair
4xy = = 0 Ans. Given equation is x 2 – 4xy
of lines
∴
i) 2x + + y = = 0 and 3 x – – 5y = 0
∴
x (x – – 4y ) = 0 The separate equations of lines are x = = 0 and x – – 4y = = 0.
ii) Passing through through (2, (2, 3) and and perpendicular perpendicular to the lines 3 x + 2 y – 1 = 0 and
Q-3) Q-3) Find the the joint joint equation equation of of pair pair of lines through through
– 3y + + 2 = 0 x –
the origin, which are perpendicular to the
Given n equa equati tion ons s: Ans. i) Give
lines represented by 5x 2 + 2xy – 3y 2 = 0
2x + + y = = 0 and 3x 3x – – 5y = = 0 Joint equation of the two lines u = = 0 and
Comparing it with ax 2 + 2hxy + + by 2 = 0, we
v = = 0 is uv = = 0 ∴
∴ ∴
5 x 2 + 2xy – – 3y 3y 2 = 0. Ans. Given equation is 5x
Joint equation equation of 2x + x + y = = 0 and 3x 3x – – 5y 5y = = 0
get a = = 5, h = = 1, and b = = – 3
is
Let m 1 and m 2 be the slopes of the lines
(2x (2x + + y )(3x )(3x – – 5y 5y ) = 0
represented by 5x 5 x 2 + 2xy 2xy – – 3y 2 = 0
6x 2 – 10xy 10xy + + 3xy 3xy – – 5y 5y 2 = 0 2
2
6x – 7xy 7xy – – 5y 5y = 0 is the joint equation.
∴
m 1.m 2 =
ii ) Given equa equation tion of line line is 3x 3x + + 2y 2y – – 1 = 0 ∴
∴
y = = –
3 1 x + 2 2
Slope is
m 1 + m 2 =
– 2h 2 = and b 3
a – 5 = b 3
Since, the required lines are perpendicular to these lines
– 3 2
Slopes of the required lines are
–1 –1 and m 1 m 2
And equation of line is x – – 3y 3y + + 2 = 0
Required lines also pass through the origin,
∴
3y = = x + + 2
therefore their equations are of the form
∴
y = =
∴
x 2 + 3 3
Slope is
y = =
1 3
∴ ∴
Let u and and v be be the lines passing through (2, 3) and perpendicular to the above lines 2 and v is is –3 3
∴
Slope of u is is
∴
Equation of u is is y – – 3 =
∴
3y – – 9 = 2x 2x – – 4
∴
2x – – 3y + + 5 = 0
2 (x – – 2) 3
y – – 3 = –3x –3x + + 6
∴
3x + + y – – 9 = 0
∴
∴
m 1
x and and y = =
–1 x m 2
x + + m 1y = = 0 and x + + m 2y = 0 The joint equation of the line s is given by (x + x + m 1y )(x )(x + + m 2y ) = 0
∴
x 2 + (m 1 + m 2) xy + xy + m 1m 2y 2 = 0
∴
x 2 +
∴
3x 2 + 2xy – – 5y 2 = 0
2 5 xy – – y 2 = 0 3 3
Q-4) Find Find k if, if, slope of one of the lines given by ....(i)
– y 2 = 0 exceeds ex ceeds the slope of the kx 2 + 4xy – other by 8.
Equation of u is is y – – 3 = –3(x –3(x – – 2) ∴
–1
4xy – – y 2 = 0 Ans. Given equation is kx 2 + 4xy ....(ii)
Joint equation o f (i) and (ii) is
Comparing it with ax 2 + 2hxy + + by 2 = 0, we get a = = k , 2h 2h = = 4, and b = = – 1
(2x (2x – – 3y + + 5)(3x 5)(3 x + + y – – 9) = 0
Let m 1 and m 2 be the slopes of the lines
6x 2 + 2xy 2xy – – 18x 18x – – 9xy 9xy – – 3y 3y 2 + 27y 27y + + 15x 15x + +
represented by kx 2 + 4xy – – y 2 = 0. ∴
5y – – 45 = 0 6 2 – 7
– 3
2
– 3 + 32 – 45 = 0 is the
m 1 + m 2 = 4 and m 1m 2 = – k According to the given condition,
Mahesh Tutorials Science
26
∴ ∴
∴
2m 1 = – 4
∴
b 2h + + a = = 0 9 3
Now, m 1(m 1 + 8) = – k
∴
+ 6h + + 9a = = 0 b +
(–2)(–2 + 8) = – k
∴
9b + + b + + 6h = = 0
m 1 = – 2
∴ –2(6) ∴ –12 ∴
= – k
This is the required condition.
= – k
Q-2) If the slope slope of of one of the lines lines given given by ax 2
= 12 k =
+ 2 hxy + + by 2 = 0 is four times the other,
Q-5) Find the joint equa equation tion of of the lines lines passing passing through the origin having inclinations 60 0 and 1200 with X –axis
show that 16h 2 = 25ab Ans. Let m 1 and m 2 be the slopes of the lines represented by ax 2 + 2hxy 2hxy + + by 2 = 0.
Ans. Slope of the line having inclination 60 0 is ∴
3
m 1 = tan 600 =
And Slope of the line having inclinat inclination ion 120 0
According to the given condition, m 2 = 4m 1
is m 2 = tan 1200 = tan (180 – 60 0) = – tan 60 0 = – 3
∴
2h m 1 + 4m 1 = – b
∴
5m 1 =
∴
2h m 1 = – 5b
Required lines pass through the origin, therefore their equations are of the form y = y = mx ∴
Equations Equations of the required lines are y = y = m 1x and y = = m 2x
i.e. y = = i.e.
( ∴
+ y = = 0 3 x +
Also, m 1(4m (4m 1) =
joint joint equation equation of these these lines lines is given by
3x – y
)(
)
m 12 =
∴
a –2h 5b = 4b
∴
a 4h 2 2 = 4b 25b
∴
4h 2 a = , as b ≠ 0 25b 4
a 4b 2
may be perpendicular to one of the lines given by ax 2 + 2hxy + + by 2 = 0 + by 2 = 0 is Ans. Auxiliary equation of ax 2 + 2hxy + 2
bm + 2hm + + a = = 0. Slope of the lines 3x 3 x + + y = = 0 is – 3
∴
Now, one of the lines represented by ax 2 + 2hxy 2hxy + + by 2 = 0 is perpendicular to the line 3x 3x + + y = = 0 1 Slope of that line is m = = 3 1 m = = is a root of the auxiliary equation 3 bm 2 + 2hm + + a = = 0. 2
∴
a b
∴
Q-1) Find the the conditi condition on that that the line 3x + y = 0
∴
a b
4m 12 =
3x + y = 0
3x 2 – y 2 = 0
... (i)
∴
BA – 2
∴
– 2h b
3 x and and y = = – 3 x
– y = = 0 and 3 x –
The ∴ The
2h a m 1 + m 2 = – and m 1m 2 = b b
1 1 + 2h 2h + a = = 0 3 3
... [From (ii)]
16h 2 = 25ab
Q-3) Find the meas measure ure of acute acute angle angle betwee between n the lines represented by 2x 2 + 7xy + 3y 2 = 0 2 x 2 + 7xy 7xy + + 3y 2 = 0. Ans. Given equation is 2x Comparing with ax 2 + 2hxy + + by 2 = 0, we get a = = 2, h = =
7 and b = = 3 2
Let θ be the acute angle between the lines. ∴
tan
θ
b
2
2
49
–6
2×
5
Mahesh Tutorials Science
∴
tan
∴ θ
θ
27
=1
2
1 2 – 2 × ( – 3) 2
= 450
Q-4) Show Show tha that 2x 2 – xy – 3y 2 – 6x + 19y – – 20 = 0
2 + ( – 3)
=
represents a pair of lines, also find the acute angle between them. Ans. Given equation of straight line is 2x 2 – xy – – 3y 2 – 6x + + 19y – – 20 = 0
2
Comparing with ax 2 + 2hxy + by 2 + 2gx + 2 fy +
=
= 0, we get c = = 2, h = = – a =
1 , b = = – 3, g = = – 3, 2
= – 2 ×
19 , c = = – 20 2
f =
1 +6 4 –1
∴
tan
= 5;
θ
25
= – 2
4
5 2
∴ θ =
tan –1(5)
If the equation represents a pair of lines then
Q-5) Find the the value value of k , if the following
a
h
g
h
b
f = 0
g
f
c
∴
D =
=
equations represents a pair of lines 3x 2 + 10xy + + 3y 2 + 16y + + k = 0 Ans. Given equation is 3 x 2 + 10xy + 3y 2 + 16y + k = 0
a
h
g
h
b
f
g
f
c
2 60 –
Comparing it with ax 2 + 2hxy + by 2 + 2gx + 2 fy
1 – 2
–3
1 = – 2
–3
19 2
– 3
19 2
– 20
2
36 1 – 1 57 – 10 + 4 2 2
–19 + ( – 3) – 9 4
=
=
2×
240 – 36 3 61 4
+
1
×
+ c = = 0, we get = 3, h = = 5, g = = 0, f = 8 and c = = k a = Since the euation represents a pair of lines ∴
+ 2 fgh – – af 2 – bg 2 – ch 2 = 0 abc +
∴
3(3)(k ) + 2(8)(0)(5) – 3(8) 2 – 3(0)2 – k (5) (5) 2 = 0
∴
9k + + 0 – 192 – 0 – 25k = = 0
∴ –16k – –
192 = 0
– ∴ –16k –
192
Hence, k = = –12
BA – 3
20 + 57
2
2 –19 – 36 – 3 × 4
–121 77 165 + + 2 4 4
– 242 + 242 = = 0 4 Since, determinant of the given equation is
Q-1) If one one of the the lines lines give given n by 3x 2 – kxy + + ky 2 = 0 is perpendicular to the line 5 x + + 3y = 0 + 5y 2 = 0. Ans. Given equations is 3 x 2 – kxy + Its auxiliary equation is 5 m 2 – km + + 3 = 0 Slope of the line 5 x + + 3y = = 0 is ∴
zero, it represents a pair of lines. Let
θ be
lines. ∴ ∴
tan
θ
=
2 h 2 – ab a + b
3
Slope of the perpendicular to 5x + + 3y = = 0 is
the acute angle between the pair
– 5
3 . 5
3 is a root of auxiliary equation 5 5m 2 – km + + 3 = 0 2
3 5
3 +3=0
k
Mahesh Tutorials Science
28
∴
45 25
–
3k 5
Q-4) If the lines lines repre represent sented ed by by ax 2 + 2hxy + by 2
+3=0
∴
45 – 15k 15k + + 75 = 0
∴
15k 15k = = 120
∴
k = = 8
= 0 make angles of equal measure with the t he coordinate axes, then show that a = ± b. Ans. Let two lines make equal angle
α with
two
coordinate axes. π
Q-2) Find the joint equation equation of pair pair of lines
Thus, Thus, these these lines lines will will form form angle angle
through the origin and perpendicular to the lines given by 2 x 2 – 3xy – – 9y 2 = 0 2
2
Ans. Given equation is 2 2x x – 3xy – – 9y = 0 i.e. 2x 2x 2 – 6xy + + 3xy – – 9y 2 = 0 i.e. (x (x – – 3y )(2x )(2x + + 3y ) = 0 Seperate equations of 2 2x x 2 – 3xy 3xy – – 9y 9y 2 = 0 are x – – 3y = = 0 and 2x 2x + + 3y = = 0 ∴
Slopes of these lines are
1 – 2 and 3 3
Now, required lines are perpendicular to
2
– 2
α
+ α with the positive direction of
X -axis. -axis.
π slope of the other line is m 2 = tan – α or 2 π tan + α 2 ∴
m 2 = cot
∴
m 1m 2 = tan
these lines, therefore their slopes are – 3 and
π
and
Slope of the first line is m 1 = tan α and
i.e. 2x 2x (x – – 3y ) + 3y (x – – 3y ) = 0 ∴
or α and
α
3 . 2
α×
∴ ∴
cot α = 1
α
(–cot α) = –1
m 1m 2 = ± 1
But m 1m 2 =
origin, their equations are of the form 3 y = = – 3x 3x and and y = = x 2
or m 2 = –cot α
or m 1m 2 = tan ∴
Since these lines also pass through the
α
a b
a = ± 1 b
a = =
±
b
i.e. 3x 3x + + y = = 0 and 3x 3 x – – 2y = = 0 ∴ The
∴
combin ed equation equ ation of the line s is
Q-5) Find the joint equati equation on of pair of of lines lines
given by
through (2, –3) and parallel to
(3x (3x + + y )(3x )(3x – – 2y 2y ) = 0
– y 2 = 0 x 2 + xy –
9x 2 – 3xy – 2y 2y 2 = 0
Q-3) Determine Determine the the nature nature of lines lines represent represented ed 2
2
by x + 7xy + + 2y = 0
Ans. Ans. Solution Solution is is attac attached hed at at end ADVANCED ASSIGNMENTS ASSIGNM ENTS (AA) : AA – 1
Ans. Given equation is x 2 + 7xy + + 2y 2 = 0 Comparing with ax 2 + 2hxy + + by 2 = 0,
Q-1) Find the joint equation equation of pair of of lines lines
we get a = = 1, 2h 2h = = 7
through the origin and making an
7 i.e. h = = and b = = 2 2
equilateral triangle with the line y = = 3
49 49–8 41 Now, h 2 – ab = = – 2 = = > 0 4 4 4 ∴ The
lines represented by x 2 + 7xy + xy + 2y 2 = 0
Ans. Let OA and and OB be be the lines through the origin making an angle of 60 0 with the line y = = 3 ∴
OA and and OB make make an angle of 60 0 and and 1200 with the positive direction of X-
are real and distinct.
axis. ∴
Slope of OA = = tan 60 0 =
∴
Equation of line OA is is y = =
∴
3 x
y = = 0
3 3 x
Mahesh Tutorials Science
29
∴
Y
∴
(x – – 1)(x – – 6) = 0 The separate equations equatio ns of x 2 – 7x + + 6 = 0 are ar e
y = =
B
60
0
60
3
A
0
and
x – – 1 = 0
... (i)
x – – 6 = 0
... (ii)
Let the equation of side AB be be x – – 1 = 0 and equation of side CD be be x – – 6 = 0. 120 60 ′
Consider, y 2 – 14y + + 40 = 0.
0
60 X
0
∴ 0
∴
X
O
(y – – 4)(y – – 10) = 0 The separate equations of y 2 – 14y + + 40 = 0 are ar e
and Y
′
– 10 = 0 y –
... (iv)
equation of side AD be be y – – 10 = 0.
= –tan 600 = – 3
Y A(1,10)
D(6,10)
B( 1,4)
C(6, 4)
Equation of line OB is is = – 3 x i.e. y =
∴
... (iii)
Let the equation of side BC be be y – – 4 = 0 and
Slope of OB = = tan 1200 = tan (1800 – 600)
∴
y – – 4 = 0
y = =
10
y = =
4
3 x + y = = 0
The joint equation of the lines is given by
(
3x – y
)(
)
3x + y = 0
i.e. 3x 2 – y 2 = 0
Q-2) Find the sepaeat sepaeate e equations equations of the following lines
O
2
2
(x – – 2) – 3(x – – 2)(y + 1) + 2(y + + 1) = 0
X x = =
1
x = =
6
Ans. Given equation of line is (x – – 2)2 – 3(x – – 2)(y + + 1) + 2( y + + 1)2 = 0 ∴
(x – – 2)2 – 2(x – – 2)( y + + 1) – (x – – 2)( y + + 1) + 2(y + + 1)2 = 0
∴
(x – – 2) [( x – – 2) – 2(y + + 1)] – (y + + 1)[(x – – 2)
Solving (i), (ii), (iii) and (iv), co-ordinates co-ordinates of vertices of the parallelogram are A (1,10), (1,10), (1,4), C (6,4) (6,4) and D (6,10). (6,10). B (1,4), ∴
– 2(y + + 1)] = 0
Equation of the diagonal AC is y – 10
=
10 – 4 6 = 1– 6 –5
∴
[( x – – 2) – ( y + + 1)] [x – – 2) – 2(y + + 1)] = 0
∴
y – 1)(x – (x – – 2 – y – 2 – 2 – 2y – 2) = 0
∴
–5y + + 50 = 6 x – – 6
∴
(x – – y – 3)(x – – 2y – – 4) = 0
∴
6x + + 5y – – 56 = 0
Thus, separate equation of the lines are
x – 1
Equation of diagonal BD is
– y – 3 = 0 and x – – 2y – 4 = 0 x –
y – 4 x – 1
Q-3) Equation Equation of pairs of opposite opposite side of a parallelogram are y 2
x 2
– 7x + 6 = 0 and
– 14y + + 40 = 0. Find the joint equation e quation of
its diagonals.
4 – 10 –6 6 = = 1– 6 –5 5
∴
5y – – 20 = 6x – – 6
∴
6x – – 5y + + 14 = 0
∴
Equations of the diagonals of the parallelogram are 6x + 5y – 56 = 0 and
be the parallelogram such that the Ans. Let ABCD be
6x – – 5y + + 14 = 0 respectively.
combined equation of sides AB and CD is + 6 = 0 x 2 – 7x +
=
∴
Joint equation e quation is (6x + + 5y – – 56)(6x – – 5y + + 14) = 0
And the combined equation of sides BC and and ∴
6x (6 (6x 5y + + 14) + 5 y (6 (6x 5y + + 14)
Mahesh Tutorials Science
30
∴
36x 2 – 30xy + 84x + 30xy – – 25y 2 + 70y ∴
– 3 m – 4 3 = – 3 1 + m – 4
∴
4m + 3 3 = 4 – 3m
– 336x + + 280y – – 784 = 0 ∴
36x 2 – 25y 2 – 252x + + 350y – – 784 = 0
Q-4) Show that that the joint joint equat equation ion of a pair pair of lines through the origin and each making an angle of
with x + + y = = 0 is
α
+ y 2 = 0 x 2 + 2sec 2axy + and OB be be the required lines. Ans. Let OA and
2
( 4m + 3) 3= 2 ( 4 – 3m )
Let m be be the slope of OA or OB . ∴
Its equation is y = = mx
it makes an angle
... (i)
with the line x + + y = = 0
α
whose slope is –1. ∴
tan
m +1 = 1 + m ( –1)
α
(m +1) 2 (1 – m )
=
3(4 – 3m 3m )2 = (4m (4m + + 3)2
∴
3(16 – 24m 24m + 9m 9m 2) = 16m 16m 2 + 24m 24m + +9
∴
48 – 72m 72m + 27m 27m 2 = 16m 16m 2 + 24m 24m + +9
∴
11m 11m 2 – 96m 96m + + 39 = 0
... (i)
their equations are of the form y = mx i.e.
2
α
∴
Since, required lines pass through the origin,
Squaring both sides, we get tan2
Squaring both sides, we get
m = =
∴
tan2 α (1 – 2m 2m + + m 2) = m 2 + 2m 2m + + 1
∴
tan2 α – 2m tan2 α + m 2 tan2 α = m 2 + 2m 2m + +
y x 2
∴
y y 11 – 96 + 39 = 0 ... [From [From (i)] x x
1 ∴
(tan 2
– 1)m 1)m 2 – 2(1 + tan 2 α) m +
α
(tan2 ∴
1+ tan2 α 2 m – 2 + 1 = 0 m + 2 tan α – 1
∴
1+ tan α m + 2 + 1 = 0 m + 2 1 – t an α
∴
m 2 + 2(sec 2α) m + + 1 = 0
∴
– 1) = 0
α
∴
∴ ∴
2
x
+ 2(sec 2 α)
2
y + 1 = x
96y + 39 = 0 x
11y 11y 2 – 96xy 96xy + + 39x 39x 2 = 0
∴
39x 39x 2 – 96xy 96xy + + 11y 11y 2 = 0 is the required joint. equation which can be written as – 39x 39x 2 + 96xy 96xy – – 11y 11y 2 = 0
2
y
x 2
–
∴
2
2
11y 2
i.e. (9x (9x 2 + 48x 48x 2) + (24xy (24xy + + 72xy 72xy ) + (16y (16y 2 – 27y 27y 2) = 0 i.e. (9x (9x 2 + 24xy 24xy + 16y 16y 2) – (48x (48x 2 – 72xy 72xy + 27y 27y 2) = 0
... [From (i)]
i.e. (9x (9x 2 + 24xy 24xy + 16y 16y 2)
2
y + 2xy 2xy sec 2α + x = 0 2
– 3(16x 3(16x 2 – 24xy 24xy + 9y 2) = 0
2
x + 2xy sec 2α + y = 0 is the required required joint equation.
Hence, the line 3x 3 x + + 4y + + 5 = 0 and the lines (3x (3x + 4y )2 – 3(4x 3(4x – 3y )2 form an equilateral
Q-5) Show Show that that 3x + 4y + + 5 = 0 and
triangle.
(3x +4 +4y )2 – 3(4x – – 3y )2 = 0
AA – 2
form an equilateral triangle. 3 x + + 4y 4y = = –5 is m 1 = Ans. Slope of the line 3x
– 3 4
Let m be be the slope of one of the line making an angle of 60 0 with the line 3x 3x + + 4y 4y = = –5. Since, the angle between the lines having slope m and and m 1 is 600,
i.e. (3x (3x + + 4y )2 – 3(4x 3(4x – – 3y )2 = 0
Q-1) If the slope slope of of one of the lines lines given given by ax 2 + ab 2 + 8h 3 = 6abh is is square of the slope of the other line, then show that a 2b + + ab 2 + 8h 3 = 6abh be the slope of one of the line s given by Ans. Let m be + by 3 = 0 ax 2 + 2hxy +
Mahesh Tutorials Science
∴
m + + m 2 =
31
Comparing the coefficients of (i) and (ii), we
–2h –2h b
... (i) get
a a and m .m = i.e. m 3 = ... (ii) b b
a 1
=
b
=
2
2b 2c = and 2a 2a + + c = = 0 4 k
2
∴
(m + + m 2) = m 3 + (m 2)3 + 3(m 3(m )(m )(m 2)(m )(m + + m 2)
∴
a a 2 a –2h –2h b = b + b 2 + b b
3
∴
– 8h b
3
2
=
3
a a 6ah + 2 – 2 b b b
Multiplying both sides by b 3, we get 3
2
∴
2
a=
∴
a=
∴
1=
∴
k=–4
2 ( – 2a ) k – 4 k
Q-4) Find the joint equation equation of of bisector bisectors s of
2
–8h –8h = ab + a b – – 6abh 6abh 2
2c and c = = –2a –2a k
∴
angles between the lines represented by
3
5x 2 + 6xy – y 2 = 0
a b + + ab + 8h 8h = 6abh
5 x 2 + 6xy 6xy – – y 2 = 0 Ans. Given equation is 5x 2
2
Q-2) If the the line lines s give given n by ax + 2hxy + + by = 0
Comparing it with ax 2 + 2hxy + + by 2 = 0, we
form an equilateral triangle with the line
get a = = 5, 2h 2h = = 6 and b = = – 1
= 1 then show that lx + my =
Let m 1 and m 2 be the slopes of the lines
2
(3a + + b )( )(a + + b )– )– 4h = 0 2
represented by 5x 5 x 2 + 6xy 6xy – – y 2 = 0.
2
hxy + by = 0 form an equilateral Ans. Lines ax + 2hxy + triangle with the line lx + + my = = 1 ∴
∴
2
Angle between the lines ax + 2hxy 2hxy + + by
y = = m 1x and and y = = m 2x , where m 1 ≠ m 2 ∴
m 1x – y y = 0 and m 2x – y y = 0
Let (x ( x ,y ) be any point on one of the bisector
2 h 2 – ab tan 600 = a + b
of the angles between the lines. ∴
Distance of P from the line m 1x – y y = 0 is equal to the distance of P from line
2 h 2 – ab 3 = a + b
m 2x – y y = 0.
3(a 3(a + + b )2 = 4(h 4(h 2 – ab ) 2
2
...(i)
The separate se parate e quations of the lines are
2
= 0 is 600
∴
m 1 + m 2 = 6 and m 1.m 2 = –5
Y 2
∴
3(a 3(a + 2ab 2ab + b ) = 4h 4h – 4ab
∴
3a 2 + 6ab 6ab + 3b 3b 2 + 4ab 4ab = = 4h 2
∴
3a 2 + 10ab 10ab + 3b 3b 2 = 4h 4h 2
∴
3a 2 + 9ab 9ab + ab + ab + 3b 3b 2 = 4h 2
∴
3a (a (a + 3b 3b ) + b (a + + 3b 3b ) = 4h 4h 2
∴
(3a (3a + b )(a )( a + + 3b 3b ) – 4h 4h 2 = 0
Angle isector P(x ,y )
× ×
Q-3) If the the line line x + + 2 = 0 coincides with with one of
X
the lines represented by the Equation
Angle bisector
2
= 0 then prove that k = – 4 x + 2xy + 4y + k = Ans. Given equation is x 2 + 2xy + + 4y + + k = = 0
... (i)
One of the lines represented by given equation is x + + 2 = 0. Let the other line be ax + + by + + c = = 0 ∴
The combine d equation of o f the lines lin es is given by (x (x + + 2) (ax ( ax + + by + + c ) = 0
∴
m1x – y 2
=
m1 + 1
m2 x – y m 22 + 1
Squaring both sides, we get
(m1 x – y ) m2 +1
2
=
(m2 x – y ) m 2 + 1
2
Mahesh Tutorials Science
32
∴
[(m [( m 22 + 1)(m 1)(m 12x 2 – 2 – 2m 1xy + y 2) = (m 12 + 1)(m 1)(m 22x 2 – 2 – 2m 2xy + y 2)
∴
m 12m 22x 2 – 2 – 2m 1m 22xy + m 22y 2 + m 12x 2 – 2m 1xy + y 2 – m 12m 22x 2 – 2m 12m 2xy + m 12y 2 + m 22x 2 – 2m 2xy + y 2
∴
(m 12 – m 22)x 2 + 2m 1m 2(m 1 – m 2)xy – 2(m 2(m 1 – m 2)xy – xy – (m (m 12 – m 22)y 2 = 0
Dividing throughout by ‘ m m 1 – m 2’, we get, (m 1 + m 2)x 2 + 2m 2m 1m 2xy – xy – 2xy – xy – (m (m 1 – m 2)y 2 = 0 ∴
6x 2 – 10xy 10xy – – 2xy – – 6y 2 = 0
∴
6x 2 – 12xy 12xy – – 6y 2 = 0
∴
x 2 – 2xy – y 2 = 0 is the required joint
...[By (i)]
equation.
Q-5) Find the measure measure of the acute acute angle angle between the lines represented by (a 2 – 3b 3) + (b 2 – 3a 2)y 2 = 0 x 2 + 8abxy + Ans. Given equation is (a 2 – 3b 3b 3) x 2 + 8abxy 8abxy + + (b 2 – 3a 2)y 2 = 0 Comparing with Ax 2 + 2Hxy + Hxy + By 2 = 0, we get A = = a 2 – 3b 2, H = = 4ab and and B = = b 2 – 3a 2 Now, H 2 – AB = 16a 16a 2b 2 – (a 2 – 3b 2)(b )(b 2 – 3a 2) = 16a 16a 2b 2 – (a 2 – 3b 2)(3a )(3 a 2 – b 2) = 16a 16a 2b 2 – 3a 3a 4 – 10a 10a 2b 2 + 3b 3b 4 = 3(a 3(a 2 + b 2)2 H 2 – AB =
∴
3 (a 2 + b 2)
Also, A + A + B = (a 2 – 3b 2)(b )( b 2 – 3a 2) = –2(a 2 + b 2) Let θ be the acute angle between the lines. ∴
=
tan
θ
2 3 (a 2 + b 2 ) 2 H2 – AB = = 3 A+AB –2 (a 2 + b 2 )
∴ θ
= 600
