怎样判断一个坐标是否在已知三角形内？

根据直线Ax+By+C=0和点P的关系知识可知，把点P代入式子Ax+By+C的值的正负，来判断点P在直线的位置。

又知三角形的内心必在三角形内部，所以我们可以把测试点和内心比较，即可得出该点是否在三角形内部。

源代码：（在BCB6.0测试通过）

void __fastcall TForm1::Button1Click(TObject *Sender)
{
float fPointX1,fPointY1;
float fPointX2,fPointY2;
float fPointX3,fPointY3;
float k1,b1,k2,b2,k3,b3;

char cFlag1 = 0;
char cFlag2 = 0;

float fTestPX =232,fTestPY = 192;
float fInX,fInY;

fPointX1 = 72;
fPointY1 = 56;

fPointX2 = 104;
fPointY2 = 208;

fPointX3 = 280;
fPointY3 = 136;

float temp1 = sqrt( Power(fPointX3-fPointX2,2)+Power(fPointY3-fPointY2,2) );
float temp2 = sqrt( Power(fPointX3-fPointX1,2)+Power(fPointY3-fPointY1,2) );
float temp3 = sqrt( Power(fPointX2-fPointX1,2)+Power(fPointY2-fPointY1,2) );

//得到该三角形的内心，内心必在三角形内部
fInX = ((temp1*fPointX1)+(temp2*fPointX2)+(temp3*fPointX3))/(temp1+temp2+temp3);
fInY = ((temp1*fPointY1)+(temp2*fPointY2)+(temp3*fPointY3))/(temp1+temp2+temp3);

//得到三角形 三个边的直线方程系数
k1 = (fPointY2-fPointY1)/(fPointX2-fPointX1);
b1 = fPointY1-k1*fPointX1;

k2 = (fPointY3-fPointY2)/(fPointX3-fPointX2);
b2 = fPointY2-k2*fPointX2;

k3 = (fPointY3-fPointY1)/(fPointX3-fPointX1);
b3 = fPointY3-k3*fPointX3;

//得到内心的 对应数的值
if (k1*fInX-fInY+b1 > 1e-6)
cFlag1 = cFlag1 | 0200;
else
cFlag1 = cFlag1 | 0100;

if (k2*fInX-fInY+b2 > 1e-6)
cFlag1 = cFlag1 | 0020;
else
cFlag1 = cFlag1 | 0010;

if (k3*fInX-fInY+b3 > 1e-6)
cFlag1 = cFlag1 | 0002;
else
cFlag1 = cFlag1 | 0001;

//检验对应点

if (k1*fTestPX-fTestPY+b1 > 1e-6)
cFlag2 = cFlag2 | 0200;
else
cFlag2 = cFlag2 | 0100;

if (k2*fTestPX-fTestPY+b2 > 1e-6)
cFlag2 = cFlag2 | 0020;
else
cFlag2 = cFlag2 | 0010;

if (k3*fTestPX-fTestPY+b3 > 1e-6)
cFlag2 = cFlag2 | 0002;
else
cFlag2 = cFlag2 | 0001;

if (cFlag1 == cFlag2)
ShowMessage("inside");
else
ShowMessage("outside");

}