If the pair of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 p x y-y^2=0$ be such that each pair bisects the…
- $p q=-1$
- $\mathrm{p}=\mathrm{q}$
- $p=-q$
- $\mathrm{pq}=1$
Solution

$q x^2+2 x y-q y^2=0$

From (1) and (2). $\frac{\mathrm{q}}{1}=\frac{2}{-2 \mathrm{p}}=\frac{-\mathrm{q}}{-1} \Rightarrow \mathrm{pq}=-1$
Asked in: JEE Main 2003