# Coordinate geometry: calculating the height of an equilateral triangle

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. ._.

• $(2x)^2 = x^2 + b^2$. – Blue May 28 '13 at 17:18
• You guys are so fast... – newbie May 28 '13 at 17:21
• Hmmm ... Maybe the "$(2x)^2$" thing was a typo on your part. It seems like you might have gone from $4 x^2 = x^2 + b^2$ (a correct statement) to $4 = b^2$ (an incorrect simplification). To isolate $b^2$, subtract $x^2$ from both sides: $4x^2 - x^2 = b^2$, and note that the left-hand side reduces to $3x^2$, not $4$. ($4 x^2 - x^2 = 4 x^2 - 1x^2 = (4 - 1 )x^2 = 3 x^2$ ... That is: four copies of $x^2$, take-away one copy of $x^2$, leaves three copies of $x^2$.) – Blue May 28 '13 at 17:25
• @Blue, you're entirely correct--it was purely a typo on my part. However, the answers below are incredibly informative :) – Josh Hubert May 28 '13 at 17:27

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 $C=(0,-x\sqrt{3})$.

• There are 2 possibilities for the point $C$ ... – Calvin Lin May 28 '13 at 17:50
• @CalvinLin True enough. I've added an inverted triangle. – Fly by Night May 28 '13 at 18:44

Draw a diagram and you will how the triangle works out.

In an equilateral triangle all sides are equal. Thus, the hypotenuse of your right angle triangle is $2x$. Let the third coordinate be represented by $(0,c)$. You should reason out as to why the $X$ coordinate of point $C$ is $0$.

Therefore, it follows that:

$c^2 + x^2 = (2x)^2$

Therefore, $c$ = ----

Hope that is helpful.

You have the base vertices of your equilateral triangle at ( -x, 0 ) and ( x, 0 ) , so the altitude of your triangle lies on the y-axis at what you are calling ( 0, b ) . That makes the base of either of the right triangles you are using $\ x \$ , but then the hypotenuse of your triangle (a side of the equilateral triangle) has length $\ 2x \$ . Therefore, the set-up using the Pythagorean Theorem must be

$$(2x)^2 = x^2 + b^2 .$$

You need to square that 2x ...

Since the coordinates of the equilateral triangle is in terms of $x$, the answer will be as well, because it depends on the length of each sides, right ?

It's actually not too hard to find it, but let me show you a trick and an important theorem that we use in order to solve this problem.

1), The altitude of any isosceles triangle is the median, the angle bisector of top angle and the perpendicular bisector.

2), An equilateral triangle is a special case of an isosceles.

This should tell you that the altitude, which lies on the $y$-axis in your case, will not only be perpendicular to the base, but it will also bisect the top angle. This forms a 30-60-90 triangle which you must have learned in geometry.

So, if you want to find the height, just use the $1$-$2$-$\sqrt{3}$ ratio.

Does this help ?