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(C) The given line is $3x + y = 0$,which can be written as $y = -3x$. The slope of this line is $m_1 = -3$. Let the slope of the required lines be $m$

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$2x^2 + 3xy - 2y^2 = 0$

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(C) The given line is $3x + y = 0$,which can be written as $y = -3x$. The slope of this line is $m_1 = -3$. Let the slope of the required lines be $m$. Since the angle between the lines is $45^{\circ}$,we use the formula $	an 	heta = |\frac{m - m_1}{1 + m m_1}|$. $	an 45^{\circ} = |\frac{m - (-3)}{1 + m(-3)}| = |\frac{m + 3}{1 - 3m}|$. $1 = |\frac{m + 3}{1 - 3m}| \Rightarrow 1 - 3m = \pm(m + 3)$. Case $1$: $1 - 3m = m + 3$ $\Rightarrow 4m = -2$ $\Rightarrow m = -1/2$. Case $2$: $1 - 3m = -(m + 3)$ $\Rightarrow 1 - 3m = -m - 3$ $\Rightarrow 2m = 4$ $\Rightarrow m = 2$. The lines pass through the origin,so their equations are $y = -\frac{1}{2}x$ and $y = 2x$. These can be written as $x + 2y = 0$ and $2x - y = 0$. The combined equation is $(x + 2y)(2x - y) = 0$. $2x^2 - xy + 4xy - 2y^2 = 0 \Rightarrow 2x^2 + 3xy - 2y^2 = 0$.

The combined equation of two lines passing through the origin and making an angle of $45^{\circ}$ with the line $3x + y = 0$ is:

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