Question

Find the equation of the line passing through the point $(-4,-3)$ and perpendicular to the straight line joining $(1,3)$ and $(2,7)$.

Answer 1

Slope between two given points $(1,3)$ and $(2,7)$ is

$m=\frac{7-3}{2-1}=4$

Then perpendicular slope is $m_1=\frac{-1}{4}$

The line equation with this slope is given by

$y=\frac{-1}{4}x+c$.......$\left(1\right)$

Now, the above line passes through the point $(-4,-3)$

$\Rightarrow -3=\frac{-1}{4}\times (-4)+c$

$\Rightarrow c=-4$

Therefore required line is $4y+x+16=0$ (Substitute ${}^{\prime }{c}^{\prime }$ in $\left(1\right)$ and simplify)