//            利用直线的两点式方程：
        //(y-y0)/(x-x0) = (y1-y0)/(x1-x0)
        //具体到这个问题，就是(y-b)/(x-a) = (d-b)/(c-a)，即：y=(d-b)/(c-a)*(x-a)+b
        //新的2点。。。(y-news.y)(i-news.x)=(change.y-news.y)(change.x-news.x);
        //y-=(change.y-news.y)*(change.x-news.x)/(i-news.x)+news.y
        public static ArrayList xxx2点直线所以点(Point news,Point change)
        {
            ArrayList a = new ArrayList();
            for (int i = change.X + 1; i < news.X; i++)
            {
                double k = ((double)(change.X - news.X) / (i - news.X));
                int y =(int)((change.Y - news.Y) * k + news.Y);
                a.Add(new Point(i,y));
            }
            return a;
        }
帮忙看看哪里写错了改正下

结贴了，你还是没改对，我不需要那功能了