在△AOB中,已知点O(0,0),A(0,5),B(4,3),OC=1/4OA,OD=1/2OB,AD与BC交于点M,求点M的坐标.

问题描述:

在△AOB中,已知点O(0,0),A(0,5),B(4,3),

OC
=
1
4
OA
OD
=
1
2
OB
,AD与BC交于点M,求点M的坐标.

∵O(0,0),A(0,5),OC=14OA,∴设点C(x1,y1),可得(x1,y1)=14(0,5),解之得x1=0,y1=54,即C(0,54)同理,可得点D坐标为(2,32)由此可得直线AD方程为y−532−5=x−02−0,化简得y=-74x+5…①同理...