precalc!!!!!
posted by L.Bianchessi .
find the point on the line y=2x1 which is closest to the point (2,1).
I used distAnce formula and got
D= (square root 5x^2 4x +4)
I did....
D=square root (2x)^2 + (1(2x1))^2
What do I do now?

The point would be the intersection of y = 2x1 with the perpendicular to that line passing through (2,1)
slope of y = 2x1 is 2
so the slope of the perpendicuar is 1/2
and the perpendicular is
y = (1/2)x + b, with (2,1) on it
so 1 = (1/2)(2) + b
b =0
so intersect y = 2x1 with y = (1/2)x
2x  1 = (1/2)x
4x  2 = x
5x = 2
x = 2/5
then y = (1/2)(2/5) = 1/5
the point is (2/5 , 1/5)
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