问题
解答题
已知两条直线l1:x+my+6=0,l2:(m-2)x+3y+2m=0,问:当m为何值时,l1与l2
(i)相交;
(ii)平行;
(iii)重合.
答案
(1)若m=0时,l1:x=-6,l2:2x-3y=0,此时l1与l2相交;
(2)若m≠0,由
=m-2 1
有m=-1或m=3,3 m
由
=3 m
有m=±3;2m 6
故(i)当m≠-1且m≠3时,
≠m-2 1
,l1与l2相交;3 m
(ii)当m=-1时,
=m-2 1
≠3 m
,l1与l2平行;2m 6
(iii)当m=3时
=m-2 1
=3 m
,l1与l2重合.2m 6