=(a-2,-2),AB
=(-2,b-2),AC
依题意知
∥AB
,AC
有(a-2)?(b-2)-4=0
即ab-2a-2b=0
所以
+1 a
=1 b
1 2
故答案为
1 2
解答:解:∵三点A(a,0),B(0,b),C(2,2)共线,
∴kAC=kBC,即
2-0
2-a
=
2-b
2-0
,
可得(a-2)(b-2)=4,化简得ab-2a-2b=0
∵ab≠0,∴等式的两边都除以ab,可得1-
2
a
-
2
b
=0
整理可得
1
a
+
1
b
=
1
2
故答案为:
1
2