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

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

12x + 72 = 5y – 10

12x – 5y + 82 = 0 … (1)

Now, compare equation (1) with general equation of line Ax + By + C = 0, where 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

The given point is \( (x_1,y_1) \)=(-1,1)

Thus, the distance of point (-1,1) from the given line is,

\( d= \dfrac{|12\times (-1)+(-5)\times 1+82|}{\sqrt{12^2+(-5)^2}} \)

= \( d= \dfrac{|12-5+82|}{\sqrt{144+25}} \)

= \( d= \dfrac{|65|}{\sqrt{169}} \)

= \( d= \dfrac{65}{13} \) = 5 units

 The distance is 5units.