find the distance of the point (2,3) from the line 2x-3y+9=0 measured along a line making an angle of 45degreewith the x axis
eqn of line through (2,3) making 45 degrees from x-axis (i.e.with slope m = than 45 =1) is
y-3 = 1. (x-2) OR y = x+1 --(i)
solving above eqn and eqn 2x-3y +9 = 0, we get x = 6, y = 7 (i.e. intersection point is (6,7)
now distance = distance betweeeen (2,3) and (6,7) which is equal to sqrt(6-2)2+ (7-3)2 = sqrt (32 = 4v2