假设存在直线L:y=x+λ
x-y+λ=0
联立
｛x²+y²-2x+4y-4=0
｛y=x+λ
x²+(x+λ)²-2x+4(x+λ)-4=0
x²+(x²+2λx+λ²)-2x+4(x+λ)-4=0
2x²+2(λ+1)x+(λ²+4λ-4)=0
x1+x2=-(λ+1)
x1x2=(λ²+4λ-4)/2
根据题意：
OA⊥OB
x1x2+y1y2=0
x1x2+(x1+λ)(x2+λ)=0
2x1x2+λ(x1+x2)+λ²=0
(λ²+4λ-4)-λ(λ+1)+λ²=0
λ²+3λ-4=0
λ1=1
λ2=-4
L1:y=x+1
L2:y=x-4
如果发现答案不对,那就是计算错方法是对的可以套改