3
7
··············································································································· (10分) Q 1 . ·
3
③若PC GC,则(3 t) 2 2,
解得t 3, P(3,2),此时PC GC 2,△PCG是等腰直角三角形. 过点Q作QH⊥x轴于点H, 则QH GH,设QH h,
2
2
2
Q(h 1,h).
513
(h 1)2 (h 1) 1 h.
66
7
解得h1 ,h2 2(舍去).
5
127 Q . ··············································· (12分)
55
综上所述,存在三个满足条件的点Q,
x
即Q(2,2)或Q 1 或Q
7
3 127
.
55
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