If the lines $(p - q) x^{2} + 2 (p + q) x y + (q - p) y^{2} = 0$ are mutually perpendicular, then

p = q

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q = 0

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p = 0

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p and q may have any value

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Solution

The correct option is D
p and q may have any value

Pair of straight lines represented by a second degree equation with coefficient of $x^{2}$ as a and coefficient of $y^{2}$ as b are perpendicular if a+b = 0. Herere a + b = 0 for every p and q.

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