// 在两点所在直线上，以从startPoint到endPoint为方向，离startPoint的距离disToStartPoint的点
    public static Point2D extentPoint(Point2D startPoint, Point2D endPoint, double disToStartPoint) {
        double disX = endPoint.getX() - startPoint.getX();
        double disY = endPoint.getY() - startPoint.getY();
        double dis = Math.sqrt(disX * disX + disY * disY);
        double sin = (endPoint.getY() - startPoint.getY()) / dis;
        double cos = (endPoint.getX() - startPoint.getX()) / dis;
        double deltaX = disToStartPoint * cos;
        double deltaY = disToStartPoint * sin;

        return new Point2D.Double(startPoint.getX() + deltaX, startPoint.getY() + deltaY);
    }

------解决方案--------------------

探讨


一楼的大侠啊 我尝试过判断角度 但是求反三角函数 得到的值有重复的 而且貌似还没有规律 后来就放弃了

但是你后来提出的那个算法是什么意思呢???没看懂你的中文说明 嘿嘿

什么叫   计算在两线所在方向上离第一个点指定距离的点的函数


请大侠指教 谢谢


------解决方案--------------------

利用反正切可以很容易求出点在坐标系中与x轴正向的夹角
Java code
    // 求点与x轴正向的夹角
    public static double degreeOfPoint(Point2D p) {
        return degreeOfPoint(p.getX(), p.getY());
    }

    public static double degreeOfPoint(double x, double y) {
        // 先求在一或者四象限的角度值，再判断是否在二三象限
        double radian = Math.atan(y / x);
        if ((x < 0 && y < 0) || (x < 0 && y > 0)) {
            radian += Math.PI;
        }

        return Math.toDegrees(radian);
    }

    public static void main(String[] args) {
        System.out.println(degreeOfPoint(20, 20));
        System.out.println(degreeOfPoint(-20, 20));
        System.out.println(degreeOfPoint(-20, -20));
        System.out.println(degreeOfPoint(20, -20));

        System.out.println(degreeOfPoint(40, 20));
        System.out.println(degreeOfPoint(-20, 30));
        System.out.println(degreeOfPoint(-50, -230));
        System.out.println(degreeOfPoint(120, -60));
        System.out.println(degreeOfPoint(120, -0));
    }