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已知直线L1:x+m^2*y+6=0,L2:(m-2)x+3my+2m=0,当m为何值时,两线相交:平行:重合
人气:250 ℃ 时间:2020-05-27 20:46:56
解答
L1:x+m^2*y+6=0.y=-x/m^2-6/m^2
L2:(m-2)x+3my+2m=0.y=-(m-2)x/(3m)-2/3
1:两线重合时,-1/m^2=-(m-2)/(3m),-6/m^2=-2/3.m=3
2:两线平行时,-1/m^2=-(m-2)/(3m),-6/m^2≠-2/3.m=-1
3:两线相交时,m≠3,m≠-1
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