Find the point on the curve y=2x+5 closest to the point (0,9).

Question

I know to use the distance formula but I cannot get a correct answer

Answer 1

the distance z from (x,y) to (0,9) is
z = √(x^2 + (y-9)^2) = √(x^2 + (2x-4)^2)
= √(5x^2 - 16x + 16)
dz/dx = (5x-8)/√(5x^2 - 16x + 16)
so, the minimum distance is 4/√5 when x = 8/5
The point on the line is (8/5,41/5)

Or, using the normal, we know that the closest point on the curve is where the normal goes through (0,9)
The slope of the normal is -1/2, so the line through (0,9) is
y-9 = -1/2 x
y = -1/2 x + 9
That line intersects y=2x+5 at (8/5,41/5)