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JavaScript: Find three points in a line, that are equally distant from a reference point

Hello I have three JavaScript points and I want to know if there is a way to calculate what points are (approximately) equally distant from a specific point, so I can put them in a array.
I found this answer and thought that it’d do the trick, but it doesn’t produce the desired result in all cases. For example, given a line of the form:
((0, -1), (-1, 0), (-1, 1))

It should return (0, -1) because it is the closest. However, in the case of:
((0, 0), (0, 0), (1, 0))

It should return (0, 0) or (1, 0) as the closest as per that answer, but it returns (0, 0) twice, whereas I would expect it to return only once.
Can anyone suggest the best way to do this?

A:

Assuming that you want only the “closest” 3 points in the line, you could do this by walking along the line (previously found) until you have reached 3 points.

const point1 = {x:0, y:-1};
const point2 = {x:-1, y:0};
const point3 = {x:-1, y:1};
const line = [point1, point2, point3];

const closest

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