Area s =1/2 (OA+BC) * oc =1/2 (6+4) oc = 20, OC = 4.

Then b (4 4,4)

AB:Y= - 2X+ 12

(2)

P(t，0)

PQ:Y= - 2X+2t

Q: (0,2t)

QC = Y = 4-2t(0 & lt； = t & lt=2)

(3)

D(0，2)

2 & ltDP & lt2√2

2√5 & lt； DG & lt2√ 10

8/5√5 & lt； PG< six

DP is the smallest, and the smallest values of DG and PG are also greater than DP, so an isosceles right triangle cannot be formed.