问题 填空题
OA
=(1,-2),
OB
=(a,-1),
OC
=(-b,0)
且a≥0,b≥0,O为坐标原点,若A、B、C三点共线,则4a+21+b的最小值为______.
答案

AB
=
OB
-
OA
=(a-1,1),
AC
=
OC
-
OA
=(-b-1,2).

又∵A、B、C三点共线,∴

AB
AC
,从而(a-1 )×2-1×(-b-1)=0,

∴2a+b=1.

4a+21+b=22a+21+b≥2

22a+1+b
=2
4
=4

故4a+21+b的最小值是4,

故答案为:4.

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