Q.Find the distance of P(−1,1) from the line 12(x+6) = 5(y−2).

The perpendicular distance from a point P(x1,y1) to the line ax + by + c = 0 is p = |ax1 + by1 + c| / √(a² + b²); setting (x1,y1) = (0,0) gives the special case of the distance from the origin, p = |c| / √(a² + b²). Both are proved by computing the area of a triangle formed by the line and two reference points in two different ways — once using the perpendicular distance as a height, and once using a direct coordinate formula for area — and equating the two expressions. A closely related result is the distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 (same a, b), which is p = |c1 − c2| / √(a² + b²): pick any …