Mathematics_Today_-_October_2016

21. (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.

22. (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

#Z
IZQPUIFTJT

^2
3

50

3

5

p^2

= ⇒ p=±

∴ The required equation of line is 310 xy+=±

23. (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

24. (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).

25. (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

26. (a) : #Z
    UIF
    IFMQ
    PG
    HJWFO
    DPOEJUJPO
    PG
    a + 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.

27. (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 +=

28. (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

29. (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)