设y=kx+b,过A(0,根号10),则b=根号10过原点且垂直于y=kx+b的直线方程为y=-x/k垂足为B满足-x/k=kx+b 即-(k+1/k)x=b,x=-bk/(k^2+1),y=b/(k^2+1),垂足到原点距离为根号5,则[-bk/(k^2+1)]^2+[b/(k^2+1)]^2=5,化简得k等于...