Let $a, b, c$ and be non-zero real numbers, if the point of intersection of the lines $4 a x+2 a y+c=0$ and…
Let $a, b, c$ and be non-zero real 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 $4^{\text {th }}$ quadrant and is equidistant from the two axes, then
$3 b c+2 a d=0$
$2 b c-3 a d=0$
$2 b c+3 a d=0$
$2 a d-3 b c=0$
Solution
$\begin{aligned} & \text { form (i) } \times b-(\mathrm{ii}) \times a \\ & \begin{array}{c}4 a b x+2 a b y+c b=0 \\ 5 a b x+2 a b y+a d=0 \\ -\quad-\quad- \\ \hline-a b x+c b-a d=0 \\ \Rightarrow x=\frac{c b-a d}{a b}\end{array}\end{aligned}$
Putting value of $x$ in (i) we get $y=\frac{4 a d-5 c b}{2 a b}$
$\begin{aligned} & A / Q x=-y \\ & \Rightarrow \frac{c b-a d}{a b}=-\frac{4 a d-5 c b}{2 a b} \\ & \Rightarrow 2 a d-3 b c=0\end{aligned}$