关于平面直角坐标系

题目

关于平面直角坐标系
在平面直角坐标系中,A(x1,y1) B(x2,y2) C(x3,y3)
求△ABC的面积
注意绝对值

答案

底边=BC=√[(x2-x3)^2+(y2-y3)^2]
BC所在直线是(y-y3)/(y2-y3)=(x-x3)/(x2-x3)
x/(x2-x3)-y/(y2-y3)+y3/(y2-y3)-x3/(x2-x3)=0
高=A到BC距离=|x1/(x2-x3)-y1/(y2-y3)+y3/(y2-y3)-x3/(x2-x3)|/√[1/(x2-x3)^2+1/(y2-y3)^2]
=√[(x2-x3)^2*(y2-y3)^2]|(x1-x3)/(x2-x3)+(y3-y1)/(y2-y3)|/√[(x2-x3)^2+(y2-y3)^2]
所以 S=√[(x2-x3)^2+(y2-y3)^2]*{√[(x2-x3)^2*(y2-y3)^2]|(x1-x3)/(x2-x3)+(y3-y1)/(y2-y3)|/√[(x2-x3)^2+(y2-y3)^2]}÷2
=|(x2-x3)*(y2-y3)|*|(x1-x3)/(x2-x3)+(y3-y1)/(y2-y3)|/2
=|(x1-x3)(y2-y3)+(y3-y1)(x2-x3)|/2