blocks|key|1136956|text|计算角度xyz和yxz。如果其中任何一个大于90，则它是外部的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1136957|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Compute the angle xyz and yxz. If either is > 90 then it is outside.

blocks|key|4922114|text|既然你用java标记了你的问题，那就开始吧：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4922115|import+javafx.geometry.Point2D;
....
//+is+z+between+parallel+lines+
boolean+betweenLines(Point2D+x,+Point2D+y,+Point2D+z)+{+
++++return++x.angle(y,z)+<+90+&&+y.angle(x,z)+<+90;
}|code-block|syntax|javascript|4922116|entityMap^0|4|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Since you tagged your question with <code>java</code>, here you go:

<pre><code>import javafx.geometry.Point2D;
....
// is z between parallel lines 
boolean betweenLines(Point2D x, Point2D y, Point2D z) { 
 return x.angle(y,z) &lt; 90 &amp;&amp; y.angle(x,z) &lt; 90;
}
</code></pre>

blocks|key|3453462|text|要确定Z点是否落在所需的条纹中，可以确定Z到X-Y线的投影是否落在这些点之间。定义向量v+=Y-+X和w+=Z-+X。如果参数b落在范围0..1内，则投影位于XY段。非常简单的公式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3453463|B=+DotProduct(w，v)+/+DotProduct(v，v)|3453464|​|3453465|📷|atomic|3453466|3453467|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/dbc4adbbf11e97149b1ba4780c5e0cb0.gif|imageAlt^0|17|7|1F|7|0|E|3|W|3|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|17|8|@]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|M|3|I|5|6|7|1B|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U|V|-4]]]]

To find if Z point falls in needed stripe, you can determine if projection of Z to X-Y line falls between these points.
Define vectors v = Y - X and w = Z - X. Projection lies in XY segment if
parameter b falls in range 0..1. Very simple formula:
b = DotProduct(w, v) / DotProduct(v, v)
<img src="https://i.stack.imgur.com/7rclc.gif" alt="enter image description here" />

blocks|key|4922100|text|JavaScript中的示例代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4922101|//+JavaScript+function+to+determine+if+infinite+strip+generated
//+by+x+and+y+contains+the+point+z
//+point+structure+is:
//++{
//++++++double+x;
//++++++double+y;
//++}
//+returns+true+or+false
function+stripContainsPoint(x,+y,+z)+{
++++var+distXZ+=+(x.x+-+z.x)+*+(x.x+-+z.x)+%2B+(x.y+-+z.y)+*+(x.y+-+z.y),
++++++++distXY+=+(x.x+-+y.x)+*+(x.x+-+y.x)+%2B+(x.y+-+y.y)+*+(x.y+-+y.y),
++++++++distYZ+=+(y.x+-+z.x)+*+(y.x+-+z.x)+%2B+(y.y+-+z.y)+*+(y.y+-+z.y);

++++//+if+triangle+is+right+or+acute,+or+obtuse+with+hypotenuse+XY,+returns+true
++++return+(distXZ+%2B+distXY+>=+distYZ)+&&+(distYZ+%2B+distXY+>=+distXZ);
}|code-block|syntax|javascript|4922102|变量dist??用词不当，因为它们实际上是每个变量的距离的平方。|offset|length|style|CODE|4922103|entityMap^0|0|0|2|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Example code in JavaScript:

<pre><code>// JavaScript function to determine if infinite strip generated
// by x and y contains the point z
// point structure is:
// {
// double x;
// double y;
// }
// returns true or false
function stripContainsPoint(x, y, z) {
 var distXZ = (x.x - z.x) * (x.x - z.x) + (x.y - z.y) * (x.y - z.y),
 distXY = (x.x - y.x) * (x.x - y.x) + (x.y - y.y) * (x.y - y.y),
 distYZ = (y.x - z.x) * (y.x - z.x) + (y.y - z.y) * (y.y - z.y);

 // if triangle is right or acute, or obtuse with hypotenuse XY, returns true
 return (distXZ + distXY &gt;= distYZ) &amp;&amp; (distYZ + distXY &gt;= distXZ);
}
</code></pre>

The variables <code>dist??</code> are misnomers, as they are actually the distance squared of each.

blocks|key|4922004|text|所以一行可以用y+=+mx+%2B+b来描述。假设您的第一行由y+=+2x-1表示。然后对于任何点(u，v)，你可以插入x=u，这样y+=+2u-1。这使您可以确定行上x位置的y。因此，如果为v>y，则您的点位于线的上方。否则，你的观点就会低于这条线。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4922005|通过使用两条平行线执行此操作，您将得到从您的图片中显而易见的三种情况：|4922006|4922007|y大于两条线上的对应点|ordered-list-item|4922008|y大于一条线上的对应点|4922009|y小于两条线上的对应点|4922010|4922011|编辑：|4922012|阅读你的帖子上的一些评论，听起来可能有比这更好的方法:)|4922013|entityMap^0|7|A|T|8|1M|3|1S|8|2E|1|2M|3|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|1B|8|@]|D|@]|E|$]]|$1|L|3|M|5|K|7|1C|8|@]|D|@]|E|$]]|$1|N|3|O|5|K|7|1D|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1F|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1G|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|V|$]]

So a line can be described by <code>y = mx + b</code>. Lets say your first line is represented by <code>y = 2x-1</code>. Then for any point (u,v), you can plug <code>x=u</code> so that <code>y = 2u-1</code>. That allows you to determine <code>y</code> for the x position on your line. Hence, if <code>v&gt;y</code>, then your point is above the line. Otherwise, your point is below the line.

By doing this with two parallel lines, you'll end up with the three cases obvious from your picture:

<ol>
<li>y is greater than the corresponding points on both lines</li>
<li>y is greater than the corresponding point on one of your lines</li>
<li>y is less than the corresponding point on both lines</li>
</ol>

EDIT:

Reading some of the comments on your post, it sounds like there may be better ways than this :)

blocks|key|1164349|text|基本上，这是一个数学问题，而不是编程问题。(或者换句话说，一旦你理解了数学，编程就变得微不足道了。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1164350|我可以想到两种方法来做到这一点：|1164351|1164352|算出通过x和y的两条垂直线。如果z在x垂直线之上，y垂直线之下。|ordered-list-item|1164353|计算y-x-z之间的夹角和x-y-z之间的夹角。如果两个角度都小于90度，则z在两条线之间。|1164354|1164355|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|O|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|P|8|@]|9|@]|A|$]]|$1|H|3|I|5|G|7|Q|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|L|$]]

Basically, this is a maths problem not a programming problem. (Or to put it another way, once you understand the maths, the programming is trivial.)

I can think of two ways to do this:

<ol>
<li>Work out the 2 perpendicular lines passing through x and y. If z is above the x perpendicular line and below the y perpendicular line.</li>
<li>Work out the angle between y-x-z and the angle between x-y-z. If both angles are less than 90 degrees, then z is between the lines.</li>
</ol>

After Googling and looking on Stack for a long time, I've only managed to find methods for determining whether or not a point falls on the line that connects two points. This, unfortunately, isn't what I need.
Please see the image at the end of this question. I apologize in advance for the terrible picture, but it gets the point (get it?) across.
I need to create two perpendicular lines to the one that joins the points x and y. They need to intersect with the perpendicular line at points x and y. I then need to then tell if point z appears between those two lines or not.
Any help is appreciated. Thanks for your time!
<img src="https://i.stack.imgur.com/J74nd.png" alt="Point graph." />

Is a point between two points?

33 天实现自己的 AI 进化论？发文瓜分机械键盘、耳机等万元奖池！点击参加活动：<a href="https://cloud.tencent.com/developer/article/2503022" target="_blank">https://cloud.tencent.com/developer/article/2503022</a> 
 <img src="https://qcloudimg.tencent-cloud.cn/raw/01543e0407db10898a7ee9f420794432.jpg"/>

技术创作特训营有奖征文

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋 

腾讯云AI代码助手

CODING DevOps

Cloud Studio

SDK中心

API中心

命令行工具

文章&问答评论现已支持表情

全新交互，全新视觉，新增快捷键、悬浮工具栏、高亮块等功能并同时优化现有功能，全面提升创作效率和体验

社区富文本编辑器全新改版！诚邀体验～ 

在谷歌搜索和查看Stack很长一段时间后，我只找到了一些方法来确定一个点是否落在连接两个点的那条线上。不幸的是，这不是我需要的。请参阅此问题末尾的图像。我为这张糟糕的照片提前道歉，但它抓住了要点(明白了吗？)穿过。我需要创建两条垂直线来连接x点和y点，它们需要与x点和y点的垂直线相交，然后我需要判断z点是否出现在这两条线之间。任何帮助都是非常感谢的。耽误您时间，实在对不起!​​

问是两个点之间的点吗？
EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问是两个点之间的点吗？EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问是两个点之间的点吗？
EN