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(A) Let $m_1$ be the slope of the line passing through $(3,2)$ and $m_2$ be the slope of $x-2y=3$. $m_2 = \frac{-1}{-2} = \frac{1}{2}$. The angle betw

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$3x-y=7$ and $x+3y=9$

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(A) Let $m_1$ be the slope of the line passing through $(3,2)$ and $m_2$ be the slope of $x-2y=3$. $m_2 = \frac{-1}{-2} = \frac{1}{2}$. The angle between the lines is $45^{\circ}$,so $	an 45^{\circ} = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. $1 = \left| \frac{m_1 - 1/2}{1 + m_1/2} ight| = \left| \frac{2m_1 - 1}{2 + m_1} ight|$. Case $1$: $\frac{2m_1 - 1}{2 + m_1} = 1$ $\Rightarrow 2m_1 - 1 = 2 + m_1$ $\Rightarrow m_1 = 3$. The equation is $y - 2 = 3(x - 3)$ $\Rightarrow y - 2 = 3x - 9$ $\Rightarrow 3x - y = 7$. Case $2$: $\frac{2m_1 - 1}{2 + m_1} = -1$ $\Rightarrow 2m_1 - 1 = -2 - m_1$ $\Rightarrow 3m_1 = -1$ $\Rightarrow m_1 = -1/3$. The equation is $y - 2 = -\frac{1}{3}(x - 3)$ $\Rightarrow 3y - 6 = -x + 3$ $\Rightarrow x + 3y = 9$. Thus,the equations are $3x - y = 7$ and $x + 3y = 9$.

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

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