Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1).
Given points are A (5, 2), B (2, 3) and C (3, -1)
Firstly, we find the slope of the line joining the points $(2,3)$ and $(3,-1)$ Slope of the line joining two points $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$\therefore \mathrm{m}_{\mathrm{BC}}=\frac{-1-3}{3-2}=-\frac{4}{1}=-4$
It is given that line passing through the point $(5,2)$ is perpendicular to $B C$
$\because m_{1} m_{2}=-1$
$\Rightarrow-4 \times m_{2}=-1$
$\Rightarrow m_{2}=1 / 4$
Therefore slope of the required line $=1 / 4$
Now, we have to find the equation of line passing through point $(5,2)$ Equation of line: $y-y_{1}=m\left(x-x_{1}\right)$
$\Rightarrow y-2=\frac{1}{4}(x-5)$
$\Rightarrow 4 y-8=x-5$
$\Rightarrow x-5-4 y+8=0$
$\Rightarrow x-4 y+3=0$
Hence, the equation of line passing through the point $(5,2)$ is $x-4 y+3=0$