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

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Solution

Let the slope of the required line be m.

Then, its equation is $\frac{y - 2}{x - 3} = m$

Given line is $x - 2 y = 3 \Rightarrow y = \frac{1}{2} x - \frac{3}{2}$

Clearly, the slope of this line is $\frac{1}{2}$

It is given that the angle between (i) and (ii) is 45 $^{\circ}$ .

$∣ ∣ ∣ \begin{matrix} \frac{m - \frac{1}{2}}{1 + \frac{1}{2} m} \end{matrix} ∣ ∣ ∣$ = $tan 45^{\circ}$

$[using tan θ = ∣ ∣ \begin{matrix} \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} \end{matrix} ∣ ∣]$

$\Leftrightarrow \frac{2 m - 1}{2 + m} = 1 or \frac{2 m - 1}{2 + m} = - 1 \Leftrightarrow m = 3 or m = \frac{- 1}{3}$

$∴$ the required equation is $\frac{y - 2}{x - 3} = 3 or \frac{y - 2}{x - 3} = \frac{- 1}{3}$

i.e., $3 x - y - 7 = 0 o r x + 3 y - 9 = 0$

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