Let point P move at the speed of V on the PF section. Then the moving speed of FE segment is √2v/2.

Let the coordinate of point F be (1, y), then we can know A( 1, 3) e (2,0).

From the formula of the distance between two points:

pe=√[( 1- 1)^2+(3-y)^2]=3-y

fe=√[(2- 1)^2+(y-0)^2]=√(y^2+ 1)

So the time from p to e

t=PF/v+FE/(√2v/2)

=(3-y)/v+√(y^2+ 1)/(√2v/2)

=√(2y^2+2)-y+3

Now we just need to find the value of y when the function t(y) takes the minimum value.

Using the knowledge of function derivatives.

Derive a function. t'=√2y/(√y^2+ 1)- 1

Let t'=0.

The solution is y= 1. Or y=- 1 (truncation).

When y is at (-∞, 1) t'

Added (1, +∞) function.

So y= 1 is the minimum point of the function.

Because there is only one extreme point, y= 1 is the minimum point of the function.

So the coordinates of point F are (1, 1).