The equation of the lines through the point $(3,2)$ which makes an angle of $45^{\circ}$ with the line $x-2…
The equation of the lines through the point $(3,2)$ which makes an angle of $45^{\circ}$ with the line $x-2 y=3$ are
$3 x-y=7$ and $x+3 y=9$
$x-3 y=7$ and $3 x+y=9$
$x-y=3$ and $x+y=2$
$2 x+y=7$ and $x-2 y=9$
Solution
Let $m_1$ be the slope of the line, which passes through the point $(3,2)$ and $m_2$ be the slope of $x-2 y=3$
$\therefore \quad m_2=\frac{- \text { Coefficient of } x}{\text { Coefficient of } y}$
$m_2=\frac{-1}{-2}$
$\Rightarrow \quad m_2=\frac{1}{2}$
Given that, the angle between lines is $45^{\circ}$
$\therefore \quad \tan 45^{\circ}=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$
$1= \pm\left|\frac{m_1-\frac{1}{2}}{1+m_1\left(\frac{1}{2}\right)}\right|$
When, $\quad 1=+\left(\frac{2 m_1-1}{2+m_1}\right)$
$\begin{aligned} \Rightarrow & & 2+m_1 & =2 m_1-1 \\ \Rightarrow & & m_1 & =3\end{aligned}$
and when $\quad 1=-\left(\frac{2 m_1-1}{2+m_1}\right)$
$\Rightarrow \quad m_1=\frac{-1}{3}$
Hence, equation a line passing through $(3,2)$ and having slope 3 is
