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

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$p$ and $q$ may have any value

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(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.For these lines to be mutually perpendicular,the condition is $a + b = 0$.Comparing the given equation $(p - q)x^2 + 2(p + q)xy + (q - p)y^2 = 0$ with the general equation,we have $a = (p - q)$ and $b = (q - p)$.Substituting these into the condition $a + b = 0$,we get $(p - q) + (q - p) = 0$,which simplifies to $0 = 0$.This identity holds true for all values of $p$ and $q$. Therefore,$p$ and $q$ may have any value.

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

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