Question

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

Answer 1

The given equation of the line is $12(x + 6) = 5(y -2).$
$⇒ 12x + 72 = 5y -10$
$⇒12x \- 5y + 82 = 0 … (1)$
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 $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
The given point is $(x_1, y_1) = (-1, 1)$.
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.