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(A) The given equation is $2x^2 - xy - y^2 = 0$.Divide by $x^2$ to get the slopes $m = \frac{y}{x}$:$2 - m - m^2 = 0 \implies m^2 + m - 2 = 0$.Let the

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$x^2 - xy - 2y^2 = 0$

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(A) The given equation is $2x^2 - xy - y^2 = 0$.Divide by $x^2$ to get the slopes $m = \frac{y}{x}$:$2 - m - m^2 = 0 \implies m^2 + m - 2 = 0$.Let the slopes of the given lines be $m_1$ and $m_2$. The slopes of the lines perpendicular to these will be $-\frac{1}{m_1}$ and $-\frac{1}{m_2}$.Replacing $m$ with $-\frac{1}{m}$ in the equation $2 - m - m^2 = 0$:$2 - (-\frac{1}{m}) - (-\frac{1}{m})^2 = 0$$2 + \frac{1}{m} - \frac{1}{m^2} = 0$Multiply by $m^2$:$2m^2 + m - 1 = 0$.Substitute $m = \frac{y}{x}$:$2(\frac{y}{x})^2 + \frac{y}{x} - 1 = 0$$2y^2 + xy - x^2 = 0$Multiplying by $-1$ gives $x^2 - xy - 2y^2 = 0$.

The combined equation of the pair of lines passing through the origin and perpendicular to the pair of lines $2x^2 - xy - y^2 = 0$ is:

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