If I have equilateral $\Delta ABC$ with A being $(-x,0)$ and B being $(x,0)$, how can I solve for the coordinates of C in terms of $x$?
I tried the following:
$2x^2 = x^2 + b^2 $ -- pythagorean thm, since we know that one side of the triangle is two times the length of half of the base.
$3x^2 = b^2$ -- simplification
$ x\sqrt3 = b $ -- so this says that the height of point C must be x\sqrt3 units.
This gives me an end result with the coordinates $(0, 2)$. However, frankly, this doesn't seem logical to me...what I'm basically saying with that result is that all equilateral triangles are two units tall...what?! This sounds completely incorrect.
Any help would be great; thanks.
edited: I actually had the right idea, I just made a stupid mistake with my simplification. ._.