If the pairs of straight lines $x^2-2 q x y-y^2=0$ and $x^2-2 p x y-y^2=0$ bisect the angles between each other, then which of the following is correct?

Question

If the pairs of straight lines $x^2-2 q x y-y^2=0$ and $x^2-2 p x y-y^2=0$ bisect the angles between each other, then which of the following is correct?, $1-p q=0$, $p q-1=0$, $p q+1=0$, $p q=0$

MARKS App · Accepted Answer

It is given that the pair of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ bisect, the angles between each other, then equations $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$ and $x^2-2 q x y-y^2=0$ represents same line, so $x^2+\frac{2}{p} x y-y^2=0$ represents $x^2-2 q x y-y^2=0$ on comparing, we get $\frac{2}{p}=-2 q$ $\Rightarrow \quad p q+1=0$ Hence, option (c) is correct.

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