I think that there are infinitely many different combinations of points that could satisfy your criteria, but let's assume for the sake of this problem that we have a side length $s$ for our equilateral triangle and that we have our usual $xy$ coordinate system. We will also assume that that the base of the triangle is parallel to the $x$-axis.
What we can do is try to find the height of our triangle, and then we will have enough information to solve this problem. If we drop a height from the top vertex to the base, we will split the base in half, so the bottom part of the triangle will have length $\frac{s}{2}$ and the hypotenuse will be $s$. Since this a right triangle, we can apply the Pythagorean Theorem, which yields:$$(\frac{s}{2})^2+h^2=s^2$$ Solving for $h$ yields the following: $$h=\frac{s\sqrt{3}}{2}$$
So, our points are $(x\pm\frac{s}{2},y-\frac{s\sqrt{3}}{2}).$
A visual solution for what is described above can be seen
here.
I have searched the internet, but very different and complex things appear to what I want to know
– carlos cerda diaz Jan 08 '18 at 03:32