Question

Find the equation of the line passing through the point $(2,-3)$ and perpendicular to the straight line joining the points $(5,7)$ and $(-6,3)$.

Answer 1

Slope of line joining two given points $(5,7)$ and $(-6,3)$ is

$m=\frac{3-7}{-6-5}=\frac{4}{11}$

Then, the slope of perpendicular line is $m_1=\frac{-11}{4}$

Thus, the line equation with this slope is given by

$y=\frac{-11}{4}x+c$.......$\left(1\right)$

Now, the above line passes through the point $(2,-3)$

$\Rightarrow -3=\frac{-11}{4}\times 2+c$

$\Rightarrow c=\frac{5}{2}$

Therefore required line is $4y+11x=10$ (Substitute ${}^{\prime }{c}^{\prime }$ in $\left(1\right)$ and simplify)