若直线kx
解:kx-y+1-k=0 (1) x-ky+2k=0 (2) (2)*k得:kx-k^2y+2k=0 (3) (3)-(1)得:kx-k^2y+2k-kx+y-1+k=0 k^2y-y+1-3k=0 解得:y=(3k-1)/(k^2-1) 把y=(3k-1)/(k^2-1)代进(1)得:x=(k^2+k)/(k^2-1) ∵交点位于第二象限 ∴x<0,y>0 即(k^2+k)/(k^2-1)<0,(3k-1)/(k^2-1)>0 解得:0