计算 点到线段（直线）的距离

 

 方法一：使用向量计算

不叙述原理了，直接看代码 

//  param  : point  : 测试点
//           pLine  : 线段描述类，m_LTPoint、m_RBPoint分别为线段的两点
//  return : 点在直线上，true；否则，false
bool HitTestLine(CPoint point, CLineItem *pLine)
{
    int dxap = point.x - pLine->m_LTPoint.x;        // Vector AP
    int dyap = point.y - pLine->m_LTPoint.y;
    int dxab = pLine->m_RBPoint.x - pLine->m_LTPoint.x;    // Vector AB
    int dyab = pLine->m_RBPoint.y - pLine->m_LTPoint.y;

    int ab2 = dxab*dxab + dyab*dyab; // Magnitude of AB，AB²

    double t; // This will hold the parameter for the Point of projection of P on AB

    if (ab2 <= 2) 
    {
        t=0;   // A and B coincide
    } 
    else 
    {
        t = (double)(dxap*dxab + dyap*dyab) / ab2;  //  P的投影点在向量AB上的比例位置，为计算P的投影点准备
    }
    // Above equation maps to (AP dot normalized(AB)) / magnitude(AP dot normalized(AB))
    if (t<0)  t = 0;   // Projection is beyond A so nearest point is A  //  这个判断很重要
    if (t>1)  t = 1;   // Projection is beyond B so nearest point is B

    //  求出 点point在直线上的映射点
    int xf = pLine->m_LTPoint.x + t * dxab;  // Projection point on Seg AB  //  double->int,提高接下来计算的效率
    int yf = pLine->m_LTPoint.y + t * dyab;  //
    
    //  点point到映射点的向量
    int dxfp = point.x - xf;
    int dyfp = point.y - yf;

    int nDist2;
    nDist2 = dxfp*dxfp + dyfp*dyfp;  //  距离的平方
    if (nDist2 < 16) 
    {
        return true;
    }

    return false;
} 

 

 

 

方法二：

参考：求点到直线的距离：http://www.cnblogs.com/lauer0246/archive/2009/06/24/1510363.html

      求点到直线的距离：http://blog.csdn.net/cay22/article/details/6291462

 过点（x1,y1）和点（x2,y2）的直线方程为：KX - Y + (x2y1 - x1y2)/(x2-x1) = 0

        设直线斜率为: K = (y2-y1)/(x2-x1),  C=(x2y1 - x1y2)/(x2-x1)

        点P(x0,y0)到直线 AX + BY +C =0 的距离为：d=|Ax0 + By0 + C|/sqrt(A*A + B*B)

        点（x3,y3）到经过点（x1,y1）和点（x2,y2）的直线的最短距离为：

            distance = |K*x3 - y3 + C|/sqrt(K*K + 1)