在三角形ABC中,向量AB=(X1,Y1),AC=(X2,Y2),求证:三角形ABC的面积S=1/2X1Y2-X2Y1的绝对值



向量AB=(X1,Y1),AC=(X2,Y2)
向量AB*AC=(X1,Y1)*(X2,Y2)
=x2x2+y1y2
设向量AB与向量AC的夹角为a
cosa=(AB*AC)/(|AB||AC|)
sina=√[1-(AB*AC)^2/(|AB||AC|)^2]
所以三角形ABC的面积S=1/2*|AB||AC|sina
=1/2*√[(|AB||AC|)^2]-(AB*AC)^2]
又[(|AB||AC|)^2=(x1x2)^2+(x1y2)^2+(x2y1)^2+(y1y2)^2
所以S=1/2*√[(x1x2)^2+(x1y2)^2+(x2y1)^2+(y1y2)^2-(x1x2)^2-(y1y2)^2-2x1x2y1y2]
=1/2*√(x1y2-x2y1)^2
=1/2|x1y2-x2y1|


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