K为何值时,直线y=kx+3k-2与直线x+4y-4=o的交点在第一象限?

L2：x+4y-4=0y=-x/4+1交点kx+3k-2=-x/4+1x=(3-3k)/(k+1/4)=12(1-k)/(4k+1)代入y=-x/4+1=(7k-2)/(4k+1)第一象限,所以x>0且y>012(1-k)/(4k+1)>0所以(1-k)(4k+1)>0(k-1)(4k+1)2/7或k