设A(x1,y1),B(x2,y2),弦AB中点M(x0,y0)
把A,B坐标代入抛物线x^2=y,得(x1)^2=y1,(x2)^2=y2,两式两边相减得,(x1)^2-x2)^2=y1-y2,即
[(y1)-(y2)]/[(x1)-(x2)]=(x1)+(x2),即kAB=2(x0),AB中垂线斜率为-1,则kAB=1,x0=1/2
,AB中点M(x0,y0)在AB中垂线上,把x0=1/2代入y0=-（x0）+3,得y0=5/2,M(1/2,5/2)
由点斜式得AB所在的直线方程为y-5/2=x-1/2,即x-y+2=0