If the lines represented by (p - q) x^2 + 2 ( | vedclass

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(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.The condition for the lines to be

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$p$ and $q$ can take any value.

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(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.The condition for the lines to be perpendicular is that the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a + b = 0$.In the given equation $(p - q) x^2 + 2 (p + q) xy + (q - p) y^2 = 0$,we have $a = (p - q)$ and $b = (q - p)$.Applying the condition $a + b = 0$:$(p - q) + (q - p) = 0$$0 = 0$.Since the condition $0 = 0$ is always true regardless of the values of $p$ and $q$,the lines are perpendicular for any values of $p$ and $q$ (provided $p 
eq q$ so that the equation represents a pair of lines).

If the lines represented by $(p - q) x^2 + 2 (p + q) xy + (q - p) y^2 = 0$ are perpendicular to each other,then:

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