如何檢查Point是否在對角線上？

Question

我有一個帶lines的canvas 。 click我想檢查單擊是否在我的行上以突出顯示它。

我也有一些rectangles ，只需使用正方形的start和end point 。 但是對於一條diagonal line我不能使用相同的技術，因為一條線不會填充矩形。

但是我還能怎么實現呢？ 此外，我還希望有一些“偏移”，這樣如果一條點擊接近該線，它也會被標記，否則可能很難點擊細線。

可能我缺少正確的關鍵字，因為我確定不是第一個想要這樣做的關鍵字。 希望你能幫忙。

Answer 1

寫出該行的等式：

a*x + b*y + c = 0

然后將您點擊的坐標放入以下等式中：

distance = a*x1 + b*y1 + c 

其中(x1, y1)是您單擊的點。 如果distance < threshold ，則單擊該行。

Answer 2

Gabor是正確的，很容易計算兩個點之間的距離並使用它。 根據Roger建議的鏈接，這里有一些來自AWT源代碼的提取代碼，用於測量兩點之間的距離。 http://developer.classpath.org/doc/java/awt/geom/Line2D-source.html

所以，你的代碼就像是

if (ptLineDist(lineX1,lineY1,lineX2,lineY2,clickX,clickY) < someLimit) 
   clicked=true; 
else clicked=false;

這是AWT代碼（請查看上面的鏈接以獲得許可證）

 521:   /**
 522:    * Measures the square of the shortest distance from the reference point
 523:    * to a point on the infinite line extended from the segment. If the point
 524:    * is on the segment, the result will be 0. If the segment is length 0,
 525:    * the distance is to the common endpoint.
 526:    *
 527:    * @param x1 the first x coordinate of the segment
 528:    * @param y1 the first y coordinate of the segment
 529:    * @param x2 the second x coordinate of the segment
 530:    * @param y2 the second y coordinate of the segment
 531:    * @param px the x coordinate of the point
 532:    * @param py the y coordinate of the point
 533:    * @return the square of the distance from the point to the extended line
 534:    * @see #ptLineDist(double, double, double, double, double, double)
 535:    * @see #ptSegDistSq(double, double, double, double, double, double)
 536:    */
 537:   public static double ptLineDistSq(double x1, double y1, double x2, double y2,
 538:                                     double px, double py)
 539:   {
 540:     double pd2 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
 541: 
 542:     double x, y;
 543:     if (pd2 == 0)
 544:       {
 545:         // Points are coincident.
 546:         x = x1;
 547:         y = y2;
 548:       }
 549:     else
 550:       {
 551:         double u = ((px - x1) * (x2 - x1) + (py - y1) * (y2 - y1)) / pd2;
 552:         x = x1 + u * (x2 - x1);
 553:         y = y1 + u * (y2 - y1);
 554:       }
 555: 
 556:     return (x - px) * (x - px) + (y - py) * (y - py);
 557:   }
 558: 
 559:   /**
 560:    * Measures the shortest distance from the reference point to a point on
 561:    * the infinite line extended from the segment. If the point is on the
 562:    * segment, the result will be 0. If the segment is length 0, the distance
 563:    * is to the common endpoint.
 564:    *
 565:    * @param x1 the first x coordinate of the segment
 566:    * @param y1 the first y coordinate of the segment
 567:    * @param x2 the second x coordinate of the segment
 568:    * @param y2 the second y coordinate of the segment
 569:    * @param px the x coordinate of the point
 570:    * @param py the y coordinate of the point
 571:    * @return the distance from the point to the extended line
 572:    * @see #ptLineDistSq(double, double, double, double, double, double)
 573:    * @see #ptSegDist(double, double, double, double, double, double)
 574:    */
 575:   public static double ptLineDist(double x1, double y1,
 576:                                    double x2, double y2,
 577:                                    double px, double py)
 578:   {
 579:     return Math.sqrt(ptLineDistSq(x1, y1, x2, y2, px, py));
 580:   }
 581: