用行列式求.
|1..3..1|
|3..1..1|
|-1.0..1|=
|2..3..0|
|4..1..0|
|-1.0..1|=
2-12=-10,
∴△ABC的面积=5.
解2：AB:x+y-4=0,
点C到AB的距离h=5/√2,
|AB|=4√2,
∴△ABC的面积=5.