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(C) Let the slope of the required line be $m$. The slope of the given line $x-2y-3=0$ is $m_1 = \frac{1}{2}$.Since the angle between the lines is $45^

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$3x-y-7=0, x+3y-9=0$

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(C) Let the slope of the required line be $m$. The slope of the given line $x-2y-3=0$ is $m_1 = \frac{1}{2}$.Since the angle between the lines is $45^{\circ}$,we have $	an 45^{\circ} = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$.$1 = \left| \frac{m - 1/2}{1 + m/2} ight| = \left| \frac{2m - 1}{2 + m} ight|$.This gives two cases: $\frac{2m - 1}{2 + m} = 1$ or $\frac{2m - 1}{2 + m} = -1$.Case $1$: $2m - 1 = 2 + m \Rightarrow m = 3$.The equation of the line is $y - 2 = 3(x - 3) \Rightarrow 3x - y - 7 = 0$.Case $2$: $2m - 1 = -2 - m \Rightarrow 3m = -1 \Rightarrow m = -1/3$.The equation of the line is $y - 2 = -1/3(x - 3) \Rightarrow 3y - 6 = -x + 3 \Rightarrow x + 3y - 9 = 0$.Thus,the equations are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

The equations of the lines passing through the point $(3,2)$ and making an acute angle of $45^{\circ}$ with the line $x-2y-3=0$ are

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