Given Point A and Point B in 2D space, how can I find the angle Point B is from Point A? 0° can be any direction; it doesn't matter. For example, Point A is at (0, 10) and Point B is at (10, 20). The angle is 45° in this example (assuming 0° is up).


I believe that the accepted answer does not correctly solve the problem for many people visiting this question. I am doing 2D simulations and the answer above does not solve my problem. Imagine the following circle:


Assume that the middle of the circle is point A. Point B is at some angle from A according to the angles of the circle (so 0°) is right.

$\hskip2in$The atan2 function is what you need!

$$ atan2(y, x) $$ $\hskip3.2in$ Where

$$ y = y_B - y_A $$ $$ x = x_B - x_A $$

Read more about it here.

  • 2
    $\begingroup$ You made my day. I spent hours trying to figure this out and got nothing but convoluted solutions that never return a proper angle. Thank you! $\endgroup$ Aug 28 '18 at 17:19
  • $\begingroup$ Is x1,y1 point A, and x2,y2 point B? $\endgroup$ May 16 '20 at 15:21
  • $\begingroup$ @SteveSmith yes. $\endgroup$ Jul 2 '20 at 12:21
  • $\begingroup$ Maybe this is useful to someone: justingolden.me/angleto $\endgroup$
    – Justin
    Mar 27 at 2:41

From what I understood about your question, you want to find the angle between two points given their coordinates. In that case, first find the slope of the line using the slope formula: $$m=\dfrac{y_2-y_1}{x_2-x_1}$$ where $(x_1,y_1)$ and $(x_2,y_2)$ are coordinates on the line. Next, use this formula: $$\tan(\theta)=m$$ where $\theta$ is the angle. Therefore, the angle $\theta$ equals: $$\theta=\tan^{-1}(m)$$

Let's use the points $(0,10)$ and $(10,20)$ as an example (you mentioned it in your question). The slope is: $$m=\dfrac{10-20}{0-10}$$ $$m=\dfrac{-10}{-10}$$ $$m=1$$ Now we will find $\theta$. $$\tan(\theta)=1$$ $$\theta=\tan^{-1}(1)$$ $$\theta=45^\circ$$
Note: The $\tan(\theta)=m$ formula only gives the angle facing the positive $x$-axis (i.e. facing the "right"). So for a negative slope, you should get an angle that is greater than $90^\circ$.

  • $\begingroup$ In your example using my two points you used the Xs over the Ys. Is that intentional? Why does it differ from the equation above? $\endgroup$
    – Keavon
    Mar 14 '14 at 2:08
  • $\begingroup$ @Keavon Nice observation, it is supposed to be the Ys over the Xs, sorry $\endgroup$ Mar 14 '14 at 3:19
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    $\begingroup$ What about the scenario where A = (10, 0) and B = (10, 10), and so x2-y2 = 0 and you get a divide by 0 error? This isn't a paticularly helpful answer imo. $\endgroup$ Jul 15 '18 at 2:42
  • $\begingroup$ Does this answer work if point B is to the left of point A? $\endgroup$ May 16 '20 at 15:24

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