The distance of the point (-1,1) from the line 12(x+6)=5(y-2) is

The given line is $12(x+6)=5(y-2)$ (\Rightarrow 12 x+72=5 y-10) or (12 x-5 y+72+10=0 \Rightarrow 12 x-5 y+82=0) The perpendicular distance from ($\left(x_1, y_1\right))$ to the line $a x+b y+c=0$ is (\frac{\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}\right)}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}$. The point (\left(x_1, y_1\right)) is$(-1,1)$, therefore, perpendicular distance from$$(-1,1)$ to the line $12 x-5 y+82=0$ is (=\frac{|-12-5+82|}{\sqrt{12^2+(-5)^2}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=$5)