设A(x1,y1),B(x2,y2),向量OA(x1,y1),OB( x2,y2),
∵OA⊥OB,
∴x1x2+y1y2=0,
设AB方程为:y=kx-2,(AB经过P点,在Y轴截距为-2)
x=(y+2)/k,
y^2=-2(y+2)/k,
ky^2+2y+4=0,
根据韦达定理,
y1+y2=-2/k,
y1y2=4/k,
x1=(y1+2)/k,
x2=(y2+2)/k,
(y1+2)(y2+2)/k^2+y1y2=0,
y1y2+2(y1+y2)+4+y1y2k^2=0,
4/k+2(-2)/k+4+4k^2/k=0,
k=-1,
∴AB直线方程为:y=-x-2.