已知A,B,C是平面上不共线上三点,O为△ABC外心,动点P满足: OP = 1 3 [(1-λ

2025-05-08 16:03:17
推荐回答(1个)
回答1:

取AB的中点D,则 2
OD
=
OA
+
OB

OP
=
1
3
[(1-λ)
OA
+(1-λ)
OB
+(1+2λ)
OC
]

OP
=
1
3
[(1-λ)(2
OD
)+(1+2λ)
OC
]

=
2(1-λ)
3
OD
+
1+2λ
3
OC

2(1-λ)
3
+
1+2λ
3
=1

∴P、C、D三点共线,
∵λ≠0
∴点P的轨迹一定不经过△ABC的重心.
故选D.