To place four points equidistant from each other, you would need to arrange them in the shape of a perfect square. This means that each point would be the same distance away from the other three points, forming equal sides of the square. The distance between each point can be calculated using the Pythagorean theorem if the coordinates of the points are known.
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you cannot do this on the plane. try proving this yourself. but a regular tetrahedron in space for example is an example where there four points equidistant from each other.
To find a point equidistant from three other points, construct perpendicular bisectors for two of the segments formed from three points. Note: this will be the center of the circle that has all three points on it's circumference. Three points, not in a straight line, form three pairs of points with each pair defining a different line. Take any pair of points and draw the perpendicular bisector of the line joining them. Repeat for one of the other pairs. These two perpendicular bisectors will meet at the point which is equidistant from all three points - the circumcenter of the triangle formed by the three points.
Congruent. If the two points are an equal distance from a third point, then those two points are congruent to each other, in respect to the third point. This is a true statement, but it may not be what the question is looking for.
Parallel lines are equidistant from each other and never intersect.
Once the circumcenter is found, each segment connecting each point of the triangle to the cirumcenter are equivalent, so you can put something equidistant to 3 places. Like a hospital equidistant to 3 cities.
Each point on the perimeter is equidistant from the centre. and because it is. what type of question is that anyways?