How to chose a random point at a certain distance to a reference point

Question

I have point A (pointA = (x1, y1)) and I need to choose a random point B (pointB = (x2, y2)) such that the distance between the A and B is equal to K.

[This answer][1] has what you're looking for. [1]: http://stackoverflow.com/a/929781/4805137 — David Rans, Apr 18 '15 at 17:41
possible duplicate of [Calculating the distance between two points](http://stackoverflow.com/questions/929773/calculating-the-distance-between-two-points) — gioele, Apr 18 '15 at 17:43
I'm voting to close this question as off-topic because it is a geometry problem, not a coding/software problem. — Brett Okken, Apr 18 '15 at 18:53

Anubian Noob · Accepted Answer · 2015-04-18T17:51:12.800

7

Let's solve in polar form.

We'll need these doubles distance, x1, and y1.

First, we want the angle in radians:

double angle = Math.random()*2*Math.PI;

Then we want to get the x and y offsets from our point:

double xOff = Math.cos(angle)*distance;
double yOff = Math.sin(angle)*distance;

Then we add these to our first point:

double x2 = x1 + xOff;
double y2 = y1 + yOff;

This will get you a point a certain distance away from your first point.

edited Apr 18 '15 at 17:51

answered Apr 18 '15 at 17:45

Anubian Noob

13,166
6
49
73

Why angle = Math.random() * 2 * PI ? – Sep 25 '18 at 05:29
oh i see. max random is 1. 1 * 2 * 3.14 = 6.28 radians = 360°. – Sep 25 '18 at 05:32

score -1 · Answer 2 · answered Apr 18 '15 at 17:43

-1

This is a simple math question.

x2 = x1 - k + 2k*new Random().nextDouble()

Now you can calculate y2.

answered Apr 18 '15 at 17:43

Nick Allen

1,383
12
19

How to chose a random point at a certain distance to a reference point

2 Answers2