题目内容
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.
(i)相交;
(ii)平行;
(iii)重合.
(1)若m=0时,l1:x=-6,l2:2x-3y=0,此时l1与l2相交;
(2)若m≠0,由
=
有m=-1或m=3,
由
=
有m=±3;
故(i)当m≠-1且m≠3时,
≠
,l1与l2相交;
(ii)当m=-1时,
=
≠
,l1与l2平行;
(iii)当m=3时
=
=
,l1与l2重合.
(2)若m≠0,由
| m-2 |
| 1 |
| 3 |
| m |
由
| 3 |
| m |
| 2m |
| 6 |
故(i)当m≠-1且m≠3时,
| m-2 |
| 1 |
| 3 |
| m |
(ii)当m=-1时,
| m-2 |
| 1 |
| 3 |
| m |
| 2m |
| 6 |
(iii)当m=3时
| m-2 |
| 1 |
| 3 |
| m |
| 2m |
| 6 |
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