If P0= (x0, y0) is the point on the line 2x+y =5?

Question

closest to the origin, what is the value of x0?

Answer 1

(x0, y0) = (x0, 5-2x0)
d^2 = x0^2+y0^2 = x0^2+(5-2x))^2
2dd' = 2x0 + 2(5-2x0)(-2)
Solve d' = 0 for x0,
x0 = 2

Answer 2

The square of the distance from (x, y) to the origin is x^2 + y^2, and on the line 2x + y = 5 we have y = 5 - 2x, so we get
D = x^2 + 25 - 20x + 4x^2 (D is the square of the distance)
= 5(x^2 - 4x + 5)
= 5((x-2)^2 + 1)
which has a minimum at x = 2.
So x0 = 2 (and y0 = 1 and the distance is √5).