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(A) The general equation of a pair of lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$. Here, $a = 1$, $h = -p$, and $b = -1$.The equation

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$pq + 1 = 0$

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(A) The general equation of a pair of lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$. Here, $a = 1$, $h = -p$, and $b = -1$.The equation of the angle bisectors is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.Substituting the values: $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-p}$.$\frac{x^2 - y^2}{2} = \frac{xy}{-p}$.$-p(x^2 - y^2) = 2xy$.$-px^2 + py^2 = 2xy$.$px^2 + 2xy - py^2 = 0$.Comparing this with the given equation $x^2 - 2qxy - y^2 = 0$, we divide our derived equation by $p$:$x^2 + \frac{2}{p}xy - y^2 = 0$.Comparing coefficients: $-2q = \frac{2}{p}$.$-q = \frac{1}{p}$, which implies $pq = -1$ or $pq + 1 = 0$.

If the combined equation of the angle bisectors of the pair of lines $x^2 - 2pxy - y^2 = 0$ is $x^2 - 2qxy - y^2 = 0$, then which of the following is true?

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