Calculus
posted by Anonymous .
Find the point on the line 3 x + y  4 =0 which is closest to the point ( 5, 0 ).

find slope of original line
y = 3 x + 4
so slope = m = 3
so slope of line perpendicular to original = m' = 1/m = 1/3
so find line through (5,0) with slope m' of 1/3
y = x/3 + b
0 = 5/3 + b
b = 5/3
so y = x/3 5/3
3y= x  5 or x + 3y + 5 = 0
find intersection of lines
9x + 3y 12 = 0
+1x + 3y + 5 = 0

8 x 17 = 0
x = 17/8
find y
then find the difference in distance between the two points
d^2 = change in x^2 + change in y^2