Question

Find the perpendicular distance of the point $(-1,1)$ from the line $12(x+6)=5(y-2)$.

Answer 1

The given equation of the line is $12(x+6)=5(y-2)$.

$12x+72=5y-10$

$12x-5y+82=0$

On comparing this equation with general equation of line $Ax+By+C=0$, we get,

$A=12,B=-5,C=82$

We know, perpendicular distance ($d$) of a line $Ax+By+C=0$ from a point $(x_1,y_1)$ is given by

$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

Therefore, the distance of the point $(-1,1)$ from given line is :

$=\frac{|12(-1)+(-5)\left(1\right)+82|}{\sqrt{(12{)}^2+(-5{)}^2}}=\frac{65}{13}=5$ units