一次函数y=kx+b与反比例函数y=2/x交于点A(1,2) 与x轴交于点M,与y轴交于点N

设点M的坐标为（m,o）,点A(1,2)分别代入y=kx+b,
得0=km+b和2=k+b,以这二式为方程组,解得k=2/(1-m),b=-2m/(1-m)
一次函数为y=2(x-m)/(1-m),用x=0代入得y=-2m/(1-m),所以点N(0,-2m/(1-m))
OA^2=1^2+2^2=5,AM=√[(1-m)^2+2^2],
AN=√[1^2+(2+ 2m/(1-m))^2]=√[(1-m)^2+(2-2m+2m)^2]/(1-m)^2= √[(1-m)^2+4]/(1-m)^2
令（m-1）^2=t,AM^2=t+4,AN^2=(t+4)/t
25=OA^4=AM^2*AN^2,25t=(t+4)^2,
t^2-17t+16=0,t=1或t=16
(m-1)=±1,m=2或0,（m-1）=±4,m=5或-3
综上m=0,2,5,-3,点M坐标为（0,0）或（2,0）或（5,0）或（-3,0）