Question

If the point of intersection of the lines $2px+3qy+r=0$ and $px–2qy–2r=0$ lies strictly in the fourth quadrant and is equidistant from the two axes, then

• A. $5p–4q=0$
• B. $4p+5q=0$
• C. $4p–5q=0$
• D. $5p+4q=0$

Answer 1

The correct option is A

$5p–4q=0$

Let the point of intersection be $(h,-h);h>0$ then it will satisfy both the lines.

$\Rightarrow 2ph-3qh+r=0$ ... (1) and $ph+2qh-2r=0$ ... (2)

$\Rightarrow h(2p-3q)=-r\Rightarrow h=\frac{-r}{2p-3q}$ [From (1)]

and $h(p+2q)=2r\Rightarrow h=\frac{2r}{p+2q}$ [From (2)]

$\Rightarrow \frac{-r}{2p-3q}=\frac{2r}{p+2q}$
$\Rightarrow 5p-4q=0$