题目内容

在△ABC中,已知A(1,4),B(4,1),C(0,-4),若P为△ABC所在平面一动点,则
PA
PB
+
PB
PC
+
PC
PA
的最小值是(  )
A.-
86
3
B.-
74
3
C.-
62
3
D.-
50
3
设P(x,y),可得
PA
=(1-x,4-y),
PB
=(4-x,1-y),
PC
=(-x,-4-y)
PA
PB
=(1-x)(4-x)+(4-y)(1-y)=x2+y2-5x-5y+8
PB
PC
=(4-x)(-x)+(1-y)(-4-y)=x2+y2-4x+3y-4
PC
PA
=(1-x)(-x)+(4-y)(-4-y)=x2+y2-x-16
因此,
PA
PB
+
PB
PC
+
PC
PA
=3x2+3y2-10x-2y-12
∵3x2+3y2-10x-2y-12=3(x-
5
3
2+3(y-
1
3
2-
62
3

∴当x=
5
3
且y=
1
3
时,
PA
PB
+
PB
PC
+
PC
PA
的最小值为-
62
3

故选:C
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