# precalc!!!!!

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find the point on the line y=2x-1 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 (2-x)^2 + (-1-(2x-1))^2

What do I do now?

• precalc!!!!! -

The point would be the intersection of y = 2x-1 with the perpendicular to that line passing through (2,-1)
slope of y = 2x-1 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 = 2x-1 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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