blocks|key|1268126|text|让我们将你试图寻找的点称为P，坐标为px，py，而起点A的坐标为ax和ay。斜率m是Y的变化与X的变化的比率，所以如果你的点P到A的距离s，那么它的坐标是px+=+ax+%2B+s，py+=+ay+%2Bm*s。现在使用毕达哥拉斯，从A到P的距离d将是d=+sqrt(s+*s%2B+(m+*+s)+*+(m+*+s))。要使P是远离A的特定D单位，请找到s+as+s=+D/sqrt(1+%2Bm*+m)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1268127|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|$]]

Let's call the point you are trying to find P, with coordinates px, py, and your starting point A's coordinates ax and ay. Slope m is just the ratio of the change in Y over the change in X, so if your point P is distance s from A, then its coordinates are px = ax + s, and py = ay + m * s. Now using Pythagoras, the distance d from A to P will be d = sqrt(s * s + (m * s) * (m * s)). To make P be a specific D units away from A, find s as s = D/sqrt(1 + m * m).

blocks|key|1482758|text|我认为这是一个很棒且易于理解的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1482759|http://www.physicsforums.com/showpost.php?s=f04d131386fbd83b7b5df27f8da84fa1&p=2822353&postcount=4|offset|length|1482760|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|2Q|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

I thought this was an awesome and easy to understand solution:

<a href="http://www.physicsforums.com/showpost.php?s=f04d131386fbd83b7b5df27f8da84fa1&amp;p=2822353&amp;postcount=4" rel="nofollow">http://www.physicsforums.com/showpost.php?s=f04d131386fbd83b7b5df27f8da84fa1&amp;p=2822353&amp;postcount=4</a>

blocks|key|1482808|text|Let+me+explain+the+answer+in+a+simple+way。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1482809|起点-+(x0,+y0)|style|CODE|1482810|端点-+(x1,+y1)|1482811|我们需要在从起点到终点的距离dt处找到一个点(xt,+yt)。|1482812|​|1482813|📷|atomic|1482814|1482815|起点和终点之间的距离由d+=+sqrt((x1+-+x0)%5E2+%2B+(y1+-+y0)%5E2)给出|1482816|让距离之比，t+=+dt+/+d|1482817|然后是(xt,+yt)+=+(((1+-+t)+*+x0+%2B+t+*+x1),+((1+-+t)+*+y0+%2B+t+*+y1))点|1482818|如果为0+<+t+<+1，则点在线上。|1482819|如果为t+<+0，则该点位于(x0,+y0)附近的线外。|1482820|如果为t+>+1，则该点位于(x1,+y1)附近的线外。|1482821|entityMap|0|LINK|mutability|MUTABLE|url|https://math.stackexchange.com/questions/175896/finding-a-point-along-a-line-a-certain-distance-away-from-another-point/1630886#1630886|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/8cb59c5b18777a2369573237787d61f5.png|imageAlt^0|0|15|0|0|4|8|0|4|8|0|M|8|0|0|0|1|1|0|0|B|Z|0|6|A|0|3|1P|0|3|9|0|3|5|E|8|0|3|5|E|8|0^^$0|@$1|2|3|4|5|6|7|1H|8|@]|9|@$A|1I|B|1J|1|1K]]|C|$]]|$1|D|3|E|5|6|7|1L|8|@$A|1M|B|1N|F|G]]|9|@]|C|$]]|$1|H|3|I|5|6|7|1O|8|@$A|1P|B|1Q|F|G]]|9|@]|C|$]]|$1|J|3|K|5|6|7|1R|8|@$A|1S|B|1T|F|G]]|9|@]|C|$]]|$1|L|3|M|5|6|7|1U|8|@]|9|@]|C|$]]|$1|N|3|O|5|P|7|1V|8|@]|9|@$A|1W|B|1X|1|1Y]]|C|$]]|$1|Q|3|M|5|6|7|1Z|8|@]|9|@]|C|$]]|$1|R|3|S|5|6|7|20|8|@$A|21|B|22|F|G]]|9|@]|C|$]]|$1|T|3|U|5|6|7|23|8|@$A|24|B|25|F|G]]|9|@]|C|$]]|$1|V|3|W|5|6|7|26|8|@$A|27|B|28|F|G]]|9|@]|C|$]]|$1|X|3|Y|5|6|7|29|8|@$A|2A|B|2B|F|G]]|9|@]|C|$]]|$1|Z|3|10|5|6|7|2C|8|@$A|2D|B|2E|F|G]|$A|2F|B|2G|F|G]]|9|@]|C|$]]|$1|11|3|12|5|6|7|2H|8|@$A|2I|B|2J|F|G]|$A|2K|B|2L|F|G]]|9|@]|C|$]]|$1|13|3|-4|5|6|7|2M|8|@]|9|@]|C|$]]]|14|$15|$5|16|17|18|C|$19|1A]]|1B|$5|1C|17|1D|C|$1E|1F|1G|-4]]]]

<a href="https://math.stackexchange.com/questions/175896/finding-a-point-along-a-line-a-certain-distance-away-from-another-point/1630886#1630886">Let me explain the answer in a simple way</a>.

Start point - <code>(x0, y0)</code>

End point - <code>(x1, y1)</code>

We need to find a point <code>(xt, yt)</code> at a distance dt from start point towards end point.

<a href="https://i.stack.imgur.com/3yDtI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3yDtI.png" alt="Point on a line at a distance"></a>

The distance between Start and End point is given by <code>d = sqrt((x1 - x0)^2 + (y1 - y0)^2)</code>

Let the ratio of distances, <code>t = dt / d</code>

Then the point <code>(xt, yt) = (((1 - t) * x0 + t * x1), ((1 - t) * y0 + t * y1))</code>

When <code>0 &lt; t &lt; 1</code>, the point is on the line.

When <code>t &lt; 0</code>, the point is outside the line near to <code>(x0, y0)</code>.

When <code>t &gt; 1</code>, the point is outside the line near to <code>(x1, y1)</code>.

I have a question i know a line i just know its slope(m) and a point on it A(x,y) How can i calculate the points(actually two of them) on this line with a distance(d) from point A ???
I m asking this for finding intensity of pixels on a line that pass through A(x,y) with a distance .Distance in this case will be number of pixels.

Finding points on a line with a given distance

我有一个问题，我知道一条线，我只知道它的斜率(M)和它上面的点A(x，y)，我如何计算这条线上的点(实际上是两个点)，离点A的距离(D)？我问这个是为了找出通过A(x，y)的直线上的像素强度，在这种情况下，距离.Distance将是像素数。

问在给定距离的直线上查找点
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问在给定距离的直线上查找点
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