∴AD=AB-BD=4

Let co = x, then CD = x and AC = 8-X.

From Pythagorean Theorem: x 2+4 2 = (8-x) 2

Solution: x = 3

So the coordinate of point C is (3,0).

Let the resolution function of line BC be: y = kx+b, and substitute the coordinates of b and c to get:

3k+b=0

b=6

∴k=-2,b=6

The resolution function of line BC is: y =-2x+6.

(2) easy BC = 3 √ 5, let the abscissa of point P be k (k(k>0), and the ordinate be -2k+6.

(1) When the point P is on the ray BC and does not coincide with the point B, if the passing point P is the PE⊥y axis at the point E, then PE = K, OE = |-2k+6 |.

∴BE=OB-OE=2k

From Pythagorean Theorem: k 2+(2k) 2 = t 2

Solution: k = √ 5/5 t

Therefore, when t = 4 √ 5, k = 4.

At this time, point P is on the vertical line of OA, so PO = PA, that is, M = PA-PO = 0.

When point p is at point b, pa-po = ba-bo = 4.

If PD is linked, then △ POB △ PDB.

∴PD=PO

∴PA-PO=PA-PD 4 √ 5, pa < po, but there is still po-pa = PD-pa < ad = 4.

Namely: 0

∴-4