# Find the bearing angle between two points in a 2D space

I continue developing a 2D Collision Detection System in a programming language (Javascript) and one of the last things I need to sharpen it is to know a formula to find this angle:

NOTE: X and Y increase their value FROM LEFT TO RIGHT AND TOP TO BOTTOM

As you can see the angle is relative to the 0° degree or north pole of the 2D space.

Knowing the coordinates of the two points, how can I know that angle? I might have an idea of finding the bearing to rectangle vertices and stuff like that (I just used them for the system) but I want to know if there is already a simple formula for this.

Thank you beforehand!

• You probably want to look at the $atan2$ function. Commented Jan 1, 2016 at 22:48
• Commented Mar 8, 2022 at 1:24

Define the bearing angle $\theta$ from a point $A(a_1,a_2)$ to a point $B(b_1,b_2)$ as the angle measured in the clockwise direction from the north line with $A$ as the origin to the line segment $AB$.

Then,

$$(b_1,b_2) = (a_1 + r\sin\theta, a_2 + r\cos\theta),$$

where $r$ is the length of the line segment $AB$. It follows that $\theta$ satisfies the equation

$$\tan\theta = \frac{b_1 - a_1}{b_2 - a_2}$$

As suggested by @rogerl we can use the $\mathrm{atan2}$ function to compute $\theta$. Let

$$\hat{\theta} = \mathrm{atan2}(b_1 - a_1, b_2 - a_2) \in (-\pi,\pi]$$

Then the bearing angle $\theta\in[0,2\pi)$ is given by

$$\theta = \left\{ \begin{array}{ll} \hat{\theta}, & \hat{\theta} \geq 0\\ 2\pi + \hat{\theta}, & \hat{\theta} < 0 \end{array}\right.$$

Note that the equations are given in terms of Cartesian coordinates, so it is necessary to transform to screen coordinates. I believe the formula for $\hat{\theta}$ in terms of screen coordinates $(a_1,a_2)$ and $(b_1,b_2)$ is $\hat{\theta} = \mathrm{atan2}(b_1 - a_1,a_2 - b_2)$.

You could code this function in C++ as follows.

#include <cmath>

// Computes the bearing in degrees from the point A(a1,a2) to
// the point B(b1,b2). Note that A and B are given in terms of
// screen coordinates.
double bearing(double a1, double a2, double b1, double b2) {
static const double TWOPI = 6.2831853071795865;
static const double RAD2DEG = 57.2957795130823209;
// if (a1 = b1 and a2 = b2) throw an error
double theta = atan2(b1 - a1, a2 - b2);
if (theta < 0.0)
theta += TWOPI;
}

• Excuse me good sir, I'm not really used to Math terms. Is there any way you can show me the formula in a way I can understand (as a programmer). Thank you for your quick reply! Commented Jan 1, 2016 at 22:45
• Thank you! I'm gonna try to code this formula and let you know how it went! thanks! Commented Jan 1, 2016 at 22:54
• I just realised that if you have the point to the RIGHT or tothe LEFT, for example, it wont show 90° or 270° but it will always show between 0° and 180°. Any idea why? Commented Jan 2, 2016 at 4:26
• It depends on how you define the angle. Should the angle be in the range $[0,2\pi]$? Relative to what point is it measured from? It is measured from the positive $y$-axis, but do you want the clockwise or counterclockwise direction to be positive? Commented Jan 2, 2016 at 13:22
• I understand now, what we had before was a plain with negative and positive axes. Well, in my case, this software doesn't have that, we go from the top-left corner of the screen 0,0 to whatever the screen size is (: I'm gonna check and try the new formulas and let you know! Thank you again sir! Commented Jan 2, 2016 at 14:45

I was trying to do the same thing and K. Miller's answer helped me. Since I did it in R, I thought I'd post my code here. I'm not 100% sure that I did the right thing by returning 360 when it could also be 0. This is the first I've touched trig since high school, make sure it's correct before you use it.

bearing = function(x1=10, y1 = 10, x2=3, y2=3){
require(NISTunits)
if((x1 == x2) & (y1 > y2)){
return(360)
}else if((y1 == y2) & (x1 < x2)){
return(90)
} else if((y1==y2 & x1 > x2)){
return(270)
}else if(y1 == y2 & x1 < x2){
return(180)
}else if((x1 == x2) & (y1==y2)){
return(NaN)
}
else
theta = atan2(x2 - x1, y1 - y2)
if(theta < 0){
theta = theta + 2*pi
}