一次函数y=ax+b的图像分别与x轴、y轴交于点M,N,与反比例函数y=k/x的图像相交于点A,B,过点A分别做AC⊥x轴,AE⊥y轴,垂足分别为C,E；过点B分别做BF⊥x轴,BD⊥y轴,垂足分别为F,D,AC与BD交于点K,连接EF.证EF∥AB
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由y=ax+b,y=k/x,可得,ax^2+bx-k=0,x1*x2=-k/a.
由题意可知,E（0,y1）,F（x2,0）,可以求出过EF的直线解析式为
y=-y1x/x2+y1.
点A在双曲线上,所以,y1=k/x1
于是 -y1x/x2=-k/x1*x2=a.
所以,直线EF∥AB.
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