若三条直线2x+3y+8=0,x-y-1=0和x+ky+k+12=0相交于一点,则k=( ) A.-2 B.-12 C.2 D.12
问题描述:
若三条直线2x+3y+8=0,x-y-1=0和x+ky+k+
=0相交于一点,则k=( )1 2
A. -2
B. -
1 2
C. 2
D.
1 2
答
由
得交点为(-1,-2),代入x+ky+k+
2x+3y+8=0 x-y-1=0
=0,得k=-1 2
.1 2
故选B