已知直线l1:x+m^2y+6=0,l2:(m-2)x+3my+2m=0
当m为何值时,l1与l2:(1)相交(2)平行(3)重合?
人气:468 ℃ 时间:2020-05-21 03:24:49
解答
L1:x+m^2y+6=0,L2:(m-2)x+3my+2m=0
m=0,L1:x=-6,L2:x=0,L1//L2
m≠0,k(L1)=-1/m^2,k(L2)=(2-m)/(3m)
L1与L2:
(1)相交
k(L1)≠k(L2)
-1/m^2≠(2-m)/(3m)
m≠0,-1,3
(2)平行
k(L1)=k(L2)
-1/m^2=(2-m)/(3m)
m=0,-1
(3)重合
1/(m-2)=m^2/(3m)=6/(2m)
m=3
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