定点C(5,0)到线段AB的距离 h CM S ABC
2
(5 2)2 (0 y0)2 9 y0.
11222AB h (9 y0)(12 y0) 9 y0 23
11222
(9 y0)(24 2y0)(9 y0) 32
222 24 2y0 9 y0119 y0
()3
323
14
. 3
22
当且仅当9
y0,即y0
,A 24 2y0
B
或A B时等号成立. 所以, ABC面积的最大值为
14
7. 3
解法二:同解法一,线段AB的垂直平分线与x轴的交点C为定点,且点C坐标为(5,0).
51222
设x1 t1,x2 t2,t1 t2,t12 t2 4,则S ABC t12
22t2
S ABC ((56t1 6t1t2 6t1t2 56t2))
2
06t1的绝对值, 6t212
222
3
(t1 t2)2(t1t2 5)2 23
(4 2t1t2)(t1t2 5)(t1t2 5)
2
3143
(),
23
所以S ABC
142
, 当且仅当(t1 t2)2 t1t2 5且t12 t2 4,即t1 3
,A(
7 56
,
t
2
7 6 6B(或
33
A B时等号成立.