已知两条直线L1:x+(1+m)y=2-m,L2:2mx+4y=-16 m为何值时L1与L2 1、相交 2、平行

2025-03-14 19:44:46
推荐回答(3个)
回答1:

L1:A1x+B1y+C1=0
L2:A2x+B2y+C2=0
若L1//L2,则A1B2-A2B1=0
若L1与L2相交,则A1B2-A2B1≠0(但要补充不重合的条件)
若L1与L2重合,则A1/A2=B1/B2=C1/C2,(还要结合各系数是否为0具体分析一下)

L1:x+(1+m)y=2-m
L2:2mx+4y=-16
(1)若L1与L2相交,则有
1×4-2m×(1+m)≠0
2m²+2m-4≠0
2(m+2)(m-1)≠0
m≠-2 且 m≠1
即当m≠-2且m≠1时,L1与L2相交
(2)若L1//L2,则有
1×4-2m(1+m)=0
2m²+2m-4=0
2(m+2)(m-1)=0
m=-2 或 m=1
当m=-2时
L1:x-y=4
L2:-4x+4y=-16,化简为x-y=4
L1与L2表示同一直线,即L1与L2重合了
当m=1时
L1:x+2y=1
L2:2x+4y=-16
两直线不重合
所以只有当m=1时,L1//L2

回答2:

看斜率,相交斜率不等,平行斜率相等,不过注意斜率不存在的时候

回答3:

首先要把Y的表达式做出来
L1y=-x/1*m+(2-m)/(1+m)
L2y=-mx/2-4
如果是平行的话,那么只要X的系数相等,则1/(1+m)=m/2,
求解m=-2,m=1这时还不能下结论,还要下一步
相同的话就是后面(2-m)/(1+m)=-4
求解m=-2
综m=1时平行
m=-2时相等

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