Find the equation of the line passing
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 $x-4 y+3=0$