题目内容
如图,△AB
C中,AD=DB,AE=EC,CD与BE交于F,设
=a,
=b,
=x a+y b,则(x,y)为( )
![]()
(A)(
,
) (B)(
,
)
(C)(
,
) (D)(
,
)
C.
=a,
=b,得
=
b-a,
=b-
a.因为B,F,E三点共线,令
=t
,则
=
+t
=(1-t)a+
t b.因为D,F,C三点共线,令
=s
,则
=
+s
=
(1-s)a+s b.根据平面向量基本定理得
,解得t=
,s=
,得x=
,y=
,即(x,y)为(
,
),故选C.
练习册系列答案
相关题目