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


Given equation of the line 12(x + 6) = 5(y – 2).


12x + 72 = 5y – 10


12x – 5y + 82 = 0 … (1)


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


Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by



Given point (x1, y1) = (-1, 1)


distance of point (–1, 1) from the given line



d = 5 units


4
1