Question

If the pair of straight lines $x^2-2pxy-y^2=0$and $x^2-2qxy-y^2=0$, be such that each pair bisects the angle between the other pair, then

• A. $pq+1=0$
• B. $pq–1=0$
• C. $p+q=0$
• D. $p–q=0$

Answer 1

The correct option is A

$pq+1=0$

Explanation for correct option:

Equation of angle bisector for $a{x}^2+2hxy+b{y}^2=0$ is $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

For$x^2-2pxy-y^2=0$the equation of the angle bisector will be

$$\begin{gathered} \Rightarrow \frac{x^2-y^2}{1-\left(-1\right)}=\frac{xy}{-p} \\ \Rightarrow x^2-y^2+\frac{2xy}{p}=0 \end{gathered}$$

Given equation of angle bisector is $x^2-2qxy-y^2=0$

Comparing both the equation s

$$\begin{gathered} \Rightarrow \frac{2}{p}=-2q \\ \Rightarrow pq=-1 \end{gathered}$$

Thus, $pq+1=0$

Hence, option (A) $pq+1=0$, is the correct answer.