找到线和高度之间的交点

Question

我试图找到一条线与该线的高度之间的交点。因此，鉴于信息，创建一条线的两个点和一个绘制高度的点，我想找到最接近该点的线上的点，（所以线上的一个点与给定点。）

（希望没有说明是有道理的）

给定一条线和一个点，我如何找到这个新点？

public static Point2D findPointOnTwoLines(Line2D line, Point2D point) {
    double slope = line.getSlope();
    double oppRec = line.getOppRecip(); // returns the opposite reciprocal of the slope
    double x = ((slope * line.getPoint1().getX()) - (line.getPoint1().getY())) 
            / ((slope) - (oppRec));
    double y = (slope * ((oppRec * point.getX()) - point.getY())) - (oppRec * ((slope * point.getX()) - point.getY()))
            / ((slope) - (oppRec));

    return new Point2D(x, y);
}

这是我尝试使用确定式求解方程，当我通过时它未能给出正确的坐标：

Line2D line = new Line2D(2, 3, 1, 1); // line from (2, 3) to (1, 1)
Point2D point = new Point2D(0, 2);

如果您知道如何使用此方法（或任何其他方法）找到正确的点，将不胜感激！

我的意思是 ASKII 艺术，如果你可以制作一个真实的图像并将其发送给我，我会很乐意使用它来代替。

1
 \       3
  \     /    
   \   /      the number points are thrown in as arguments, and the "X" point is returned
    \ /
     X    <---- this is a 90 degree angle
      \
       \        the "X" is suppose to represent the point on the line closest to the point "3"
        \
         2

Answer 1

double x = (point.getY() - oppRec * point.getX() - line.getPoint1().getY() + slope * line.getPoint1().getX()) / (slope - oppRec);
double y = oppRec * x + point.getY() - oppRec * point.getX();