# Find the distance of the point (–1, 1) from the line 12(x + 6)

Question:

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

Solution:

The given equation of the line is 12(+ 6) = 5(– 2).

⇒ 12x + 72 = 5y – 10

⇒12x – 5y + 82 = 0 … (1)

On comparing equation (1) with general equation of line Ax By C = 0, we obtain A = 12, B = –5, and C = 82.

It is known that the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by $d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$.

The given point is $\left(x_{1}, y_{1}\right)=(-1,1)$.

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

$=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^{2}+(-5)^{2}}}$ units $=\frac{|-12-5+82|}{\sqrt{169}}$ units $=\frac{|65|}{13}$ units $=5$ units