Posted by **L.Bianchessi** on Friday, March 9, 2012 at 7:45pm.

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!!!!! -
**Reiny**, Friday, March 9, 2012 at 9:07pm
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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