设Q(a,4a),PQ：y-4a=4（a-1）/(a-6)·（x-a）与x轴的交点为(5a/(a-1),0),a>1 S=1/2·5a/(a-1)·4a =10a²/(a-1)=[10·（a-1）²+20（a-1）+10]/(a-1)=10(a-1)+10/(a-1)+20≥40,a=2,Q(2,8), Smin=40 P(6,4),Q(2,8)k=-1L:y-4=-(x-6),即y=-x+10
