Find the perpendicular distance of (1 , 1) from x − y +√2=0.
Perpendicular distance of p(x1 , y1) from ax + by + c = 0 is given by,
d=|ax1 + by1 + c√a2+b2|
⇒ d = |1 − 1 + √2√1+1| = √2√2 = 1
Find the perpendicular distance from the origin of the line x + y – 2 = 0 is:
Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y-2)