Question

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equdistant from the two axes, then

• A. $2bc-3d=0$
• B. $2bc+3ad=0$
• C. $2ad-3bc=0$
• D. $3bc+2ad=0$

Answer 1

The correct option is C $2ad-3bc=0$

Let coordinate of the intersection point in fourth quadrant be $(\alpha ,-\alpha )$.
Since $(\alpha ,-\alpha )$ lies on both lines $4ax+2ay+c=0$ and $5bx+2by+d=0$
$\therefore 4a\alpha -2a\alpha +c=0\Rightarrow \alpha =\frac{-c}{2a}$ …..(i)
And $5b\alpha -2b\alpha +d=0\Rightarrow \alpha =\frac{-d}{3b}$ ….(ii)
From Eqs. (i) and (ii), we get
$\frac{-c}{2a}=\frac{-d}{3b}\Rightarrow 3bc=2ad$
$\Rightarrow 2ad-3bc=0$