两条线段是否相交，计算交点公式。

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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));

    发现x和y的求解公式，结构相同，只要把x换成y下标不变，就是求解y值的公式，

原理部分来自 http://zhidao.baidu.com/question/191530048.html?push=ql

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
        }
    }
}