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. ._.
Best Answer
Since the triangle is equilateral, the $x$-coordinate of $C$ must be $x=0$, so that it falls in the middle of the $x$-coordinates of $A$ and $B$.
The base has length $|AB|=2x$ and so all of the sides have length $2x$. The point $C$ will lie on the circle, centre $A$ and radius $2x$. The whole circle is parametrized by $(-x,0) + 2x(\cos\theta,\sin\theta)$, where $\theta = 0$ corresponds to the base. Since we have an equilateral triangle, we have $\theta = 60^{\circ}$:
\begin{array}{ccc} (-x,0)+2x(\cos 60, \sin 60) &=& (-x,0) + 2x(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}) \\ \\ &=& (0,x\sqrt{3}) \end{array}
If you want an "inverted" triangle then let $C=(0,-x\sqrt{3})$.