两条线段是不是相交，计算交点公式

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：

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?


65.71428571428571) ?

?

1 楼 Terence.Wang 2011-08-26   有木有求椭圆和直线相交交点坐标的code？