- (c) : Since equation of
diagonal 11x + 7y = 9 does
not pass through origin, so
it cannot be the equation
of the diagonal OB. Thus
on solving the equation AC
with the equations OA and
OC, we get
AC^5
3
4
3
2
3
7
3
⎛⎜⎝ ,,− ⎞⎠⎟ and ⎛⎝⎜− ⎞⎠⎟
Therefore, the middle point M is^1
2
1
2
⎛⎝⎜ , ⎞⎠⎟
Hence the equation of OB is y = x i.e., x – y = 0.
- (b) : Let p be the length of the perpendicular
from the origin on the required line. Then its
equation in normal form is xcos30° + ysin30° = p or
32 xy p+=
This meets the coordinate axes at A^2 p
3
⎛⎝⎜ , 0 ⎞⎠⎟ and
B(0, 2p)
∴ Area of ΔOAB =^1 ⎝⎜⎛ ⎞⎠⎟ =
2
2
3
2 2
3
p^2
p p
#ZIZQPUIFTJT
^2
3
50
3
5
p^2
= ⇒ p=±
∴ The required equation of line is 310 xy+=±
- (a) : The median passes through A, i.e., the
intersection of the given lines. Its equation is given by
(px + qy – 1) + λ(qx + py – 1) = 0, where λ is some
real number. Also, since the median passes through the
point (p, q), we have (p^2 + q^2 – 1) + λ(qp + pq – 1) = 0.
⇒ =− + −
−
λ pq
pq
(^221)
21
and the equation of median
through A is ()px qy pq ()
pq
- −− + − qx py
−
1 1 + − =
21
10
22
⇒−()()()()21pq px qy+ − =+p^22 q − qx+py− 1 1 1
- (a) : Take two perpendicular lines as the coordinate
axes. If a, b be the intercepts made by the moving line
on the coordinate axes, then the equation of the line is
x
a
y
b
+= 1 ....(i)
According to the question, 11 1
ab k
+= (say)
i.e., k
a
k
b
+= 1 ....(ii)
The result (ii) shows that the straight line (i) passes
through a fixed point (k, k).
- (a) : u ≡ a 1 x + b 1 y + c 1 = 0, v ≡ a 2 x + b 2 y + c 2 = 0
and a
a
b
b
c
c
(^1) c
2
1
2
1
2
=== (say)
⇒ a ===a
c
b b
c
c c
(^2) c
(^1) ,, 2 1 2 1
Given that u + kv = 0
⇒ a 1 x + b 1 y + c 1 + k(a 2 x + b 2 y + c 2 ) = 0
⇒ ax by c+++ + + =ka
c
xkb
c
ykc
(^111) c
(^1110)
⇒ ax⎝⎜⎛ +k⎟⎠⎞++⎛⎝⎜ ⎟⎠⎞++⎝⎜⎛ ⎞⎠⎟=
c
by k
c
c k
(^111) c
1110
⇒ a 1 x + b 1 y + c 1 = 0
⇒ u = 0
- (a) : #ZUIFIFMQPGHJWFODPOEJUJPOPGa + b + c = 0,
the three lines reduce to xyp
a
− = or p
b
p
c
or (p≠ 0 ).
All these lines are parallel. Hence they do not intersect
in finite plane.
- (a) : Here a + b = –1.
Required line is x
a
y
a
−
+
=
1
1 ....(i)
Since line (i) passes through (4, 3)
∴−
+
(^43) = ⇒ + − =+
1
1443 2
aa
aaaa
⇒ a^2 = 4 ⇒ a = ± 2
∴ Required lines are xy x y
23
1
21
− = 1
−
and +=
- (c) : Let the equation of line parallel to x-axis be
y = λ ....(i)
Solving (i) with the cuve yx= ....(ii)
We get P(λ^2 , λ) which is the point of intersection at P
∴ Slope of (ii) is m dy
dx P
=⎛⎝⎜ ⎞⎠⎟ =
at
1
2 λ
∵ (i) and (ii) intersect at 45°
∴ −
+ ⋅
⎛
⎝⎜
⎞
⎠⎟=± °
tan−^10
10
m 45
m
⇒ m=⎛⎝⎜^1 ⎟⎞⎠=± ⇒ =±
2
1 1
λ 2
λ
∴ The equation of line is yy==
1 −
2
1
2
or
- (c) : The lines passing through the intersection of
the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0 is
ax+++^23 by b λ()bx−−^230 ay a=
⇒ ()( )abx b a y b++λλλ 22 − +33 0− a= ...(i)