解(1)S △ABC=3×3-(
1
2 ×3×1+
1
2 ×2×1+
1
2 ×2×3)=
7
2 ;
(2)AC=
2 2 +1 2 =
5 ;
(3)设点B到AC边的距离为h,则S △ABC=
1
2 ×AC×h=
7
2 ,
解得:h=
7
5
5 .
解(1)S △ABC=3×3-(
1
2 ×3×1+
1
2 ×2×1+
1
2 ×2×3)=
7
2 ;
(2)AC=
2 2 +1 2 =
5 ;
(3)设点B到AC边的距离为h,则S △ABC=
1
2 ×AC×h=
7
2 ,
解得:h=
7
5
5 .