设m(x,y)
直线ob过原点,则直线ob解析式为正比例函数
则设直线ob:y=kx
又b=(0,0)+(5,1)=(5,1)
则将b(5,1)代入y=kx
得k=1/5,则y=1/5*x
又m在直线上ob上
则m(x,y)也满足y=1/5*x
又ma=(1,7)-(x,y)=(1-x,7-y)
mb=(5,1)-(x,y)=(5-x,1-y)
则ma*mb=(1-x,7-y)*(5-x,1-y)
=(1-x)*(5-x)+(7-y)*(1-y)
又y=1/5*x
则ma*mb=(1-x)*(5-x)+(7-y)*(1-y)
=x^2-6x+5+(7- x/5)*(1- x/5)
=x^2-6x+5+ x^2/25- 8x/5 +7
=26x^2/25 -38x/5 +12
则当x=-(-38/5)/[2*(26/25)]=85/26
ma*mb最小=1026/13
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