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 equidistant from the two axes then:

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

Answer 1

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

Let the point of intersection is $(h,-h)$, so
$4ah-2ah+c=\mathrm{0...}\left(1\right)$
$5bh-2bh+d=\mathrm{0...}\left(2\right)$
from equation (1), $h=-\frac{c}{2a}$
put the value of $h$ in equation (2)
$\Rightarrow 3b(-\frac{c}{2a})+d=0$
$\Rightarrow 3bc-2ad=0$