一次函数：y=k2x+b,与y轴交点为A(0,b),这里b=3
与反比例函数交点满足：k2x+b=k1/x,即k2x^2+bx-k1=0
B(x1,y1),C(x2,y2),其中x1+x2=-b/k2,x1x2=-k1/k2,y1=k2x1+b,y2=k2x2+b
AB^2=x1^2+(y1-b)^2=x1^2+(k2x1)^2=(1+k2^2)x1^2
BC^2=(x1-x2)^2+(y1-y2)^2 =(x1-x2)^2(1+k2^2)
AB=BC,则有：x1^2=(x1-x2)^2
即x1=x1-x2,或x1=x2-x1
即x2=0(舍去,因为反比例函数定义域不能为0）,或x2=2x1
故当x2=2x1时,k2=-b/(3x1),k1=-k2x1x2=-2bx1/3.
k1k2=2b^2/9=2 为定值