Question

直线(2m^2+m-3)x + (m^2+m)y=4m-1与直线2x-3y=5平行,求m的值

Answer 1

直线(2m²+m-3)x + (m²+m)y=4m-1与直线2x-3y=5平行,求m的值

y=[-(2m²+m-3)/(m²+m)]x+(4m-1)/(m²+m).(1)

y=(2/3)x-5/3.(2)

两线平行,斜率相等,故有-(2m²+m-3)/(m²+m)=2/3

即有-3(2m²+m-3)=2(m²+m)

故有8m²+5m-9=0

∴m=(-5±√313)/16.

【你的m=-9/8或m=1是怎么算出来的?】