已知平行四邊形ABCD的頂點(diǎn)A(3,-1)、C(2,-3),點(diǎn)D在直線3x-y+1=0上移動(dòng),則點(diǎn)B的軌跡方程為( ?。?br/>A. 3x-y-20=0(x≠3)
B. 3x-y-10=0(x≠3)
C. 3x-y-9=0(x≠2)
D. 3x-y-12=0(x≠5)
設(shè)B點(diǎn)的坐標(biāo)為(x,y),取直線上D點(diǎn)的坐標(biāo)為(x
1,y
1).
∵
=
,
∴
,即
.
代入3x-y+1=0得:3x
1-y
1+1=0,即3(5-x)-(-4-y)+1=0.
整理得:3x-y-20=0(x≠3).
當(dāng)x=3時(shí)A,B,C,D共線.
即B點(diǎn)的軌跡方程為3x-y-20=0(x≠3).
故選:A.