Question

已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量O

Answer 1

设坐标分别为：P(x,y);A(x1,y1);B(x2,y2);C(x3,y3)

则有：

x=1/3[(1-λ)x1+(1-λ)x2+(1+2λ)x3=(x1+x2+x3)/3-(x1+x2-2x3)λ/3

y=1/3[(1-λ)y1+(1-λ)y2+(1+2λ)y3=(y1+y2+y3)/3-(y1+y2-2y3)λ/3

将λ消去可得：

y-(y1+y2+y3)/3=(y1+y2-2y3)/(x1+x2-2x3)*[x-(x1+x2+x3)/3]

所以,当x=(x1+x2+x3)/3时,y=(y1+y2+y3)/3.

因此,过重心.