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
问题描述:
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
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