Consider m_{1} as the slope of the required line

It can be written as

\( y=\dfrac{1}{2}x-\dfrac{3}{2}\) which is of the form y = mx + c

So the slope of the given line m_{2} = \( \dfrac{1}{2}\)

We know that the angle between the required line and line x – 2y = 3 is 45°

If θ is the acute angle between lines l_{1} and l_{2} with slopes m_{1} and m_{2}

Substitute the values,

= \( 1=|\dfrac{\dfrac{1}{2}-m_1}{1+\dfrac{m_1}{2}}|\)

Take LCM,

\( 1=|(\dfrac{\dfrac{1-2m_1}{2}}{\dfrac{2+m_1}{2}})|\)

\( 1=|\dfrac{1-2m_1}{2+m_1}|\)

\( 1=\pm|\dfrac{1-2m_1}{2+m_1}|\)

\( 1=|\dfrac{1-2m_1}{2+m_1}|\) or

\( 1=-(\dfrac{1-2m_1}{2+m_1})\)

= 2+m_{1}=1-2m_{1} or 2+m_{1} = -1+2m_{1}

\( m_1=-\dfrac{1}{3}\) or m_{1}= 3

Case 1: m_{1} = 3

The equation of the line passing through (3,2) and having a slope of 3 is,

y - 2 = 3(x -3)

Expand bracket,

y - 2 = 3x -9

3x - y = 7

Case 2: m_{1} = \(- \dfrac{1}{3}\)

The equation of the line passing through (3,2) and having a slope of \( - \dfrac{1}{3}\) is

\(y-2= - \dfrac{1}{3}(x-3)\)

Cross multiply and expand bracket,

3y - 6 = -x +3

x +3y = 9

Therefore, the equations of the line are 3x - y = 7 and x + 3y = 9 .

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