求取点到直线的距离

问题描述：

已知点P(px,py)，直线L（P1,P2)，求点P到L的距离。

首先，推导直线公式：

点$$P_1(x_1,y_1)$$， 点$$P_2(x_2,y_2)$$ 可知直线方程为：

$$x(y_2-y_1)-y(x_2-x_1)+y_1(x_2-x_1)-x_1(y_2-y_1)=0$$

点$$P_0(x_0,y_0)$$ 到$$P_1P_2$$的距离如下：

$$\begin{array}{rcl}dist & = & \frac{\left|x_0(y_2-y_1)-y_0(x_2-x_1)+y_1(x_2-x_1)-x_1(y_2-y_1)\right|}{\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}} \\& = & \frac{\left|(y_2-y_1)*(x_0-x_1)-(x_2-x_1)*(y_0-y_1)\right|}{\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}}\end{array}$$

代码如下所示：

double getDistFromP2L(double px, double py, double p1x, double p1y, double p2x,
              double p2y)
{
    double y2_y1 = p2y - p1y;
    double x2_x1 = p2x - p1x;
    if (fabs(y2_y1) < EOPS && fabs(y2_y1) < EOPS) {
        return 0.0;
    }
    return fabs(y2_y1 * (px - p1x) -
            x2_x1 * (py - p1y)) / sqrt(y2_y1 * y2_y1 + x2_x1 * x2_x1);
}