当m为何值时,直线(2m^2+m-3)x+(m^2-m)y=4m-1(1)倾斜角为π/4(2)在x轴上的截距为1

Question

当m为何值时,直线(2m^2+m-3)x+(m^2-m)y=4m-1(1)倾斜角为π/4(2)在x轴上的截距为1

Answer 1

①(2m^2+m-3)x+(m^2-m)y=4m-1
斜率k=-(2m^2+m-3)/(m^2-m)
=-(m-1)(2m+3)/m(m-1)
=-(2m+3)/m
∵倾斜角为π/4 ∴tanπ/4=-(2m+3)/m 得 1=-(2m+3)/m m=-1
②在x轴上的截距为1即y=0时x=1
∴(2m^2+m-3)×1+(m^2-m)×0=4m-1 即2m^2+m-3=4m-1 解得m=2或者m=-1/2