设点O(x,y)
OF=√[(x-1)^2+(y-1)^2]
O到直线2X-Y-1=0的距离d
d=│2x-y-1│/√(2*2+1*1)
OF=d
[(x-1)^2+(y-1)^2]=(2x-y-1)^2/5整理得
x^2+4y^2+4xy-6x-12y+9=0
(x+2y-3)^2=0
x+2y-3=0轨迹是一条直线