兩條線段是否相交，計算交點公式

    A本身無限長，假設B也無限長，直接求得AB的交點坐標，然后再判斷該坐標是否在定長線段B的內部就可以了啊 
    AB本身就是兩條直線，知道兩端點就可以知道其直線方程，B也是一樣，兩個方程聯立，
    得到一個坐標，再看該坐標是否在B的定義域內就可以啊 
    
    A的兩點為(x1,y1),(x2,y2)
    則A的直線方程為l1:y-y1=(y2-y1)(x-x1)/(x2-x1)
    B的兩點為(x3,y3),(x4,y4)
    則B的直線方程為l2:y-y3=(y4-y3)(x-x3)/(x4-x3)
    
    聯立解出交點坐標為的橫坐標為：
    x=(k2x3-y3-k1x1+y1)/(k2-k1)
    其中k1=(y2-y1)/(x2-x1)
          k2=(y4-y3)/(x4-x3)    
    可以推導出來
    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));

    同理也可以推導出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));

總結：

//第一條直線  
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("求出該直線方程為: y=" + a + "x + " + b);  
  
//第二條  
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("求出該直線方程為: 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("他們的交點為: (" + x + "," + y + ")"); 

下面附上java的實現，

前提是：a 線段1起點坐標

            b 線段1終點坐標

            c 線段2起點坐標

            d 線段2終點坐標

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));  
    }  
  
    /** 
     * 判斷兩條線是否相交 a 線段1起點坐標 b 線段1終點坐標 c 線段2起點坐標 d 線段2終點坐標 intersection 相交點坐標 
     * reutrn 是否相交: 0 : 兩線平行 -1 : 不平行且未相交 1 : 兩線相交 
     */  
  
    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是同一個點！");  
            } else {  
                System.out.println("AB是一個點，CD是一個點，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是一個點，且在CD線段上！");  
            } else {  
                System.out.println("A、B是一個點，且不在CD線段上！");  
            }  
            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是一個點，且在AB線段上！");  
            } else {  
                System.out.println("C、D是一個點，且不在AB線段上！");  
            }  
            return 0;  
        }  
  
        if ((b.y - a.y) * (c.x - d.x) - (b.x - a.x) * (c.y - d.y) == 0) {  
            System.out.println("線段平行，無交點！");  
            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("線段相交於點(" + intersection.x + "," + intersection.y + ")！");  
            return 1; // '相交  
        } else {  
            System.out.println("線段相交於虛交點(" + intersection.x + "," + intersection.y + ")！");  
            return -1; // '相交但不在線段上  
        }  
    }  
}  

第二種方法: 利用斜率公式, 直線方程為ax+bx+c=0, 先求出a,b,c, 然后再求出交點 

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("他們的交點為: (" + rp.getX() + "," + rp.getY() + ")");  
}  
  
/** 
 * 計算兩點的直線方程的參數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);  
}  
  
/** 
 * 計算兩條直線的交點 
 * @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;  
}  

輸出的結果為: 

求出該直線方程為: y=2.0x + -0.0
求出該直線方程為: y=-2.6666666666666665x + 153.33333333333334
他們的交點為: (32.857142857142854,65.71428571428571)
他們的交點為: (32.857142857142854,65.71428571428571)

轉：  http://263229365.iteye.com/blog/1155745