两条线段是否相交计算交点公式.doc

1、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

2、)/(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 -

3、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

4、- 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)

5、*(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Java 代码import java.awt.Point;public classAlgorithmUtil public static void main(String args)AlgorithmUtil.GetIntersection(new Point(1, 2),new Point(1, 2),ne

6、w Point(1, 2), newPoint(1, 2);AlgorithmUtil.GetIntersection(newPoint(1, 2), new Point(1, 2),new Point(1,4), new Point(1, 4);AlgorithmUtil.GetIntersection(new Point(100, 1), newPoint(100, 100),new Point(100, 101), newPoint(100, 400);AlgorithmUtil.GetIntersection(new Point(5, 5), new Point(100,100),ne

7、w 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.

8、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！ ); elseSystem.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

9、(A 、B 是一个点， 且在 CD 线段上！ ); elseSystem.out.println(A 、B 是一个点， 且不在 CD 线段上！ );return0;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！ ); elseSystem.out.println(C 、D 是一个点， 且不在 AB 线段上！ );return0;if (b.y

10、- 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.

11、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&& (intersecti

12、on.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 elseSystem.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

13、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 Java 代码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

14、); Param pm2 = CalParam(p3, p4);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*/publicstatic Param CalParam(Point2D p1, Point2D p2)doublea,b,c;double x1 = p1.getX(), y1 = p1.getY(), x2 =p2.getX(), y2 = p2.getY();a = y2 - y1;b = x1 - x2;

15、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*/publicstatic Point2D getIntersectPoint(Param pm1, Parampm2)return getIntersectPoint(pm1.a, pm1.b, pm1.c,pm2.a, pm2.b, pm2.c);public static Point2DgetIntersectPoint(double a1, double b1, double c1, double a2,double b2, double c2)Point2D p = null;double m= a1 * b2 - a2 * b1;if (m = 0) returnnull;double x = (c2 * b1 - c1 * b2) / m;double