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

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