l1:(2m^2+m-3)x+(m^2-m)y=2m直线L2:x-y=1当m取何值时两直线垂直
问题描述：
l1:(2m^2+m-3)x+(m^2-m)y=2m直线L2:x-y=1当m取何值时两直线垂直
答
(2m^2+m-3)x+(m^2-m)y=2m
y=-(2m^2+m-3)x/(m^2-m)+2m/(m^2-m)
L1⊥L2则1*[-(2m^2+m-3)/(m^2-m)]=-1
解得m1=-3 ;m2=1（舍）
答
L2y=x-1斜率=1垂直则L1斜率=-1L1y=-(2m²+m-3)/(m²-m) x+2m/(m²-m)-(2m²+m-3)/(m²-m)=-12m²+m-3=m²-mm²+2m-3=0m=1,m=-3m=1时分母m²-m=0所以m=-3