两条线段是否相交，计算交点公式。

A本身无限长，假设B也无限长，直接求得AB的交点坐标，然后再判断该坐标是否在定长线段B的内部就可以了啊 

    AB本身就是两条直线，知道两端点就可以知道其直线方程，B也是一样，两个方程联立，
    得到一个坐标，再看该坐标是否在B的定义域内就可以啊 
    
    A的两点为(x1,y1),(x2,y2)
    则A的直线方程为l1:y-y1=(y2-y1)(x-x1)/(x2-x1)
    B的两点为(x3,y3),(x4,y4)
    则B的直线方程为l2:y-y3=(y4-y3)(x-x3)/(x4-x3)
    
    联立解出交点坐标为的横坐标为：
    x=(k2x3-y3-k1x1+y1)/(k2-k1)
    其中k1=(y2-y1)/(x2-x1)
          k2=(y4-y3)/(x4-x3)    
    可以推导出来
    x = ((x2 - x1) * (x3 - x4) * (y3 - y1) - 
            x3 * (x2 - x1) * (y3 - y4) + x1 * (y2 - y1) * (x3 - x4)) / 
            ((y2 - y1) * (x3 - x4) - (x2 - x1) * (y3 - y4));

    同理也可以推导出y的值：

    y = ((y2 - y1) * (y3 - y4) * (x3 - x1) - 
            y3 * (y2 - y1) * (x3 - x4) + y1 * (x2 - x1) * (y3 - y4)) / 
            ((y2 - y1) * (y3 - y4) - (y2 - y1) * (x3 - x4));

 

 

总结：

//第一条直线
double x1 = 10, y1 = 20, x2 = 100, y2 = 200;
double a = (y1 - y2) / (x1 - x2);
double b = (x1 * y2 - x2 * y1) / (x1 - x2);
System.out.println("求出该直线方程为: y=" + a + "x + " + b);

//第二条
double x3 = 50, y3 = 20, x4 = 20, y4 = 100;
double c = (y3 - y4) / (x3 - x4);
double d = (x3 * y4 - x4 * y3) / (x3 - x4);
System.out.println("求出该直线方程为: y=" + c + "x + " + d);

double x = ((x1 - x2) * (x3 * y4 - x4 * y3) - (x3 - x4) * (x1 * y2 - x2 * y1))
    / ((x3 - x4) * (y1 - y2) - (x1 - x2) * (y3 - y4));

double y = ((y1 - y2) * (x3 * y4 - x4 * y3) - (x1 * y2 - x2 * y1) * (y3 - y4))
    / ((y1 - y2) * (x3 - x4) - (x1 - x2) * (y3 - y4));

System.out.println("他们的交点为: (" + x + "," + y + ")");

********************************************************************

下面附上java的实现，

前提是：a 线段1起点坐标

            b 线段1终点坐标

            c 线段2起点坐标

            d 线段2终点坐标

 

import java.awt.Point;

public class AlgorithmUtil {

    public static void main(String[] args) {
        AlgorithmUtil.GetIntersection(new Point(1, 2), new Point(1, 2),
                new Point(1, 2), new Point(1, 2));
        AlgorithmUtil.GetIntersection(new Point(1, 2), new Point(1, 2),
                new Point(1, 4), new Point(1, 4));
        AlgorithmUtil.GetIntersection(new Point(100, 1), new Point(100, 100),
                new Point(100, 101), new Point(100, 400));
        AlgorithmUtil.GetIntersection(new Point(5, 5), new Point(100, 100),
                new Point(100, 5), new Point(5, 100));
    }

    /**
     * 判断两条线是否相交 a 线段1起点坐标 b 线段1终点坐标 c 线段2起点坐标 d 线段2终点坐标 intersection 相交点坐标
     * reutrn 是否相交: 0 : 两线平行 -1 : 不平行且未相交 1 : 两线相交
     */

    private static int GetIntersection(Point a, Point b, Point c, Point d) {
        Point intersection = new Point(0, 0);

        if (Math.abs(b.y - a.y) + Math.abs(b.x - a.x) + Math.abs(d.y - c.y)
                + Math.abs(d.x - c.x) == 0) {
            if ((c.x - a.x) + (c.y - a.y) == 0) {
                System.out.println("ABCD是同一个点！");
            } else {
                System.out.println("AB是一个点，CD是一个点，且AC不同！");
            }
            return 0;
        }

        if (Math.abs(b.y - a.y) + Math.abs(b.x - a.x) == 0) {
            if ((a.x - d.x) * (c.y - d.y) - (a.y - d.y) * (c.x - d.x) == 0) {
                System.out.println("A、B是一个点，且在CD线段上！");
            } else {
                System.out.println("A、B是一个点，且不在CD线段上！");
            }
            return 0;
        }
        if (Math.abs(d.y - c.y) + Math.abs(d.x - c.x) == 0) {
            if ((d.x - b.x) * (a.y - b.y) - (d.y - b.y) * (a.x - b.x) == 0) {
                System.out.println("C、D是一个点，且在AB线段上！");
            } else {
                System.out.println("C、D是一个点，且不在AB线段上！");
            }
            return 0;
        }

        if ((b.y - a.y) * (c.x - d.x) - (b.x - a.x) * (c.y - d.y) == 0) {
            System.out.println("线段平行，无交点！");
            return 0;
        }

        intersection.x = ((b.x - a.x) * (c.x - d.x) * (c.y - a.y) - 
                c.x * (b.x - a.x) * (c.y - d.y) + a.x * (b.y - a.y) * (c.x - d.x)) / 
                ((b.y - a.y) * (c.x - d.x) - (b.x - a.x) * (c.y - d.y));
        intersection.y = ((b.y - a.y) * (c.y - d.y) * (c.x - a.x) - c.y
                * (b.y - a.y) * (c.x - d.x) + a.y * (b.x - a.x) * (c.y - d.y))
                / ((b.x - a.x) * (c.y - d.y) - (b.y - a.y) * (c.x - d.x));

        if ((intersection.x - a.x) * (intersection.x - b.x) <= 0
                && (intersection.x - c.x) * (intersection.x - d.x) <= 0
                && (intersection.y - a.y) * (intersection.y - b.y) <= 0
                && (intersection.y - c.y) * (intersection.y - d.y) <= 0) {
            
            System.out.println("线段相交于点(" + intersection.x + "," + intersection.y + ")！");
            return 1; // '相交
        } else {
            System.out.println("线段相交于虚交点(" + intersection.x + "," + intersection.y + ")！");
            return -1; // '相交但不在线段上
        }
    }
}

 

 

 

========================下面是找到的另外的一种方法====================

 

 

第二种方法: 利用斜率公式, 直线方程为ax+bx+c=0, 先求出a,b,c, 然后再求出交点 

public static void main(String[] args) {
    Point2D p1 = new Point2D.Double(10, 20);
    Point2D p2 = new Point2D.Double(100, 200);

    Point2D p3 = new Point2D.Double(50, 20);
    Point2D p4 = new Point2D.Double(20, 100);

    Param pm1 = CalParam(p1, p2);
    Param pm2 = CalParam(p3, p4);
    Point2D rp = getIntersectPoint(pm1, pm2);
    System.out.println("他们的交点为: (" + rp.getX() + "," + rp.getY() + ")");
}

/**
 * 计算两点的直线方程的参数a,b,c
 * @param p1
 * @param p2
 * @return
 */
public static Param CalParam(Point2D p1, Point2D p2){
    double a,b,c;
    double x1 = p1.getX(), y1 = p1.getY(), x2 = p2.getX(), y2 = p2.getY();
    a = y2 - y1;
    b = x1 - x2;
    c = (x2 - x1) * y1 - (y2 - y1) * x1;
    if (b < 0) {
        a *= -1; b *= -1; c *= -1;
    }else if (b == 0 && a < 0) {
        a *= -1; c *= -1;
    }
    return new Param(a, b, c);
}

/**
 * 计算两条直线的交点
 * @param pm1
 * @param pm2
 * @return
 */
public static Point2D getIntersectPoint(Param pm1, Param pm2){
    return getIntersectPoint(pm1.a, pm1.b, pm1.c, pm2.a, pm2.b, pm2.c);
}

public static Point2D getIntersectPoint(double a1, double b1, double c1, double a2, double b2, double c2){
    Point2D p = null;
    double m = a1 * b2 - a2 * b1;
    if (m == 0) {
        return null;
    }
    double x = (c2 * b1 - c1 * b2) / m;
    double y = (c1 * a2 - c2 * a1) / m;
    p = new Point2D.Double(x, y);
    return p;
}

输出的结果为: 

求出该直线方程为: y=2.0x + -0.0
求出该直线方程为: y=-2.6666666666666665x + 153.33333333333334
他们的交点为: (32.857142857142854,65.71428571428571)
他们的交点为: (32.857142857142854,65.71428571428571)

 

评论

2 楼 davidwkj 2017-02-10  

Y坐标的公式不对。算出来的值是错的！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！

1 楼 Terence.Wang 2011-08-26  

有木有求椭圆和直线相交交点坐标的code？