JEE Main · 2014 · 06 Apr 2014 · Offline · Straight Lines

Let

a, b, c

and d be non-zero numbers. If the point of intersection of the lines

4 a x + 2 a y + c = 0

and

5 b x + 2 b y + d = 0

lies in the fourth quadrant and is equidistant from the two axes then

(A) $2 b c - 3 a d = 0$
(B) $2 b c + 3 a d = 0$
(C) $3 b c - 2 a d = 0$ ✓ Correct
(D) $3 b c + 2 a d = 0$

Answer: (C)

Solution

Since the intersection lies in the fourth quadrant and is equidistant from the axes, it has the form

(h, - h)

with

h > 0

. Putting this in the first line,

4 ah - 2 ah + c = 0

, so

2 ah + c = 0

and

h = - \frac{c}{2 a}

. In the second line,

5 bh - 2 bh + d = 0

, so

3 bh + d = 0

and

h = - \frac{d}{3 b}

. Equating the two values of

h

gives

\frac{c}{2 a} = \frac{d}{3 b}

, that is

3 b c - 2 a d = 0

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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