Question

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

Answer 1

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