当m=-1/2时x^2+y^2+x-y-1=0 ===>(x+1/2)^2+(y-1/2)^2=3/2
设所求直线L方程为y=kx+1 ;M(x1,y1),N(x2,y2)
OM垂直于ON则(y1/x1)*(y2/x2)=-1即x1x2+y1y2=0
把y=kx+1代入x^2+y^2+x-y-1=0
化简得（1+k^2)x^2+(k+1)x-1=0
x1+x2=-(k+1)/(1+k^2) ;x1x2=-1/(1+k^2)
y1+y2=kx1+kx2+2=k(x1+x2)+2=(k^2-k+2)/(k^2+1)
y1y2=(kx1+1)(kx2+1)=1+k(x1+x2)+x1x2k^2=(-k^2-k+1)/(k^2+1)
x1x2+y1y2=0
即-1/(1+k^2)+(-k^2-k+1)/(k^2+1)=0
解得k1=0 ,k2=-1
所以直线方程为y=1或x+y-1=0