已知三点A(x1,y1) B(x2,y2) C(x3,y3) 求三角形面积为
S1x22x3x1y11y21 y31B(x2,y2) ABC证明:如图所示
AB=(x2-x1,y2-y1)
H AC=(x3-x1,y3-y1)
SABC11ABACsinABAC 22A(x1,y1) C(x3,y3) 令ABAC=AD
x2x1y2y1x2x1y2y1ABACADx2x1y2y100i0jkkx3x1y3y1x3x1y3y1x3x1y3y10ijk
因此有:
SABC1111ABACsinABACAD222200(x2x1y2y121x2x1y2y1 )x3x1y3y12x3x1y3y1可以看出ADS的大小与i,j无关,现在取AD=(x1,y1,1),即i=x1:j=y1:k=1,则有
ABC1x2x1y2y111x2x1y2y10x22x3x1y3y122x3x1y3y10x31x22x3x1y11y21 得证。y31
x1y11x1y11y21 y31即SABC
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