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