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.