PHP的正则表达式来检查一个数字是否由5位数组成(php regular expression to check whether a number consists of 5 digits)

 两点之间的距离（毕达哥拉）： 
 
dx = x1 - x2;
dy = y1 - y2;

dist = sqrt (dx*dx + dy*dy);
 
 两条水平线之间的距离： 
 
d = y1 - y2;
 
 如果你的线被定义为y1 = k1x + n1和y2 = k2x + n2 ，那么（它们是水平的， k1和k2是0）它们之间的距离是n2 - n1 。 
 
 编辑：好的，在你编辑了你的问题之后，它现在变得更有意义了。 但仍然是：既然你（或用户）正在添加行，你的代码总是知道它们在哪里。 他们的结束坐标是： 
 
 line1：{（0，y1）:( picture.width，y1）} line2：{（0，y2）:( picture.width，y2）} distance：| y2-y1 | 
 
 因为它们都是水平的，所以它们永远不会交叉。 
 
 您应该在适当的空间中保留对y1和y2（来自放线代码）的引用。 由于您的问题是针对Android和iOS的，因此答案是：在与MVC中的model相对应的那部分代码中。 

Distance between two points (Pythagora):
 
dx = x1 - x2;
dy = y1 - y2;

dist = sqrt (dx*dx + dy*dy);
 
Distance between two horizontal lines:
 
d = y1 - y2;
 
If your lines are defined as y1 = k1x + n1 and y2 = k2x + n2, then (they're horizontal, k1 and k2 are 0) the distance between them is n2 - n1.
 
EDIT: ok, after you edited your question it makes a bit more sense now. But still: since you (or user) is adding the lines, your code always knows where they lie. Their end coordinates would be:
 
line1: {(0,y1):(picture.width,y1)} line2: {(0,y2):(picture.width,y2)} distance: |y2-y1|
 
Since they're both horizontal they ofcourse never cross.
 
You should just keep a reference to y1 and y2 (from the line-placing code) in an appropriate space. Since your question is for Android and iOS the answer is: in that part of code that would correspond to model in MVC.