Finding the middle coordintes of each side of an equilateral triangle

Question

I have a problem with finding the center coordinates of each side in an equilateral triangle. I've linked an image below that shows exactly which coordinates I'm after.

I understand that I can work out the height of then triangle by using pythagoras theorem. I can split the triangle down the middle and calculate the opposite of one of the triangles.

However, I can't figure out how to work out the coordinates marked with a "?"

Is there any formula I can use to work this out? The center part of the triangle would be it's origin at (0,0) and the sides would be 3 units each.

Thankyou

Answer 1

The height of the equilateral triangle is $$h={\sqrt 3 \over 2}a$$

(you have $a=3$), and coordinates of its vertices are

$$A = \left(-\frac a 2, -\frac h 3\right),\quad B = \left(\frac a 2, -\frac h 3\right),\quad C = \left(0, {\frac 2 3}h\right).$$

The center of AC is ${A+C \over 2}$, the center of BC is ${B+C \over 2}$.

Answer 2

Assuming one of the vertices is on axis $y$, it's coordinates will be A$(0,\sqrt{3})$ (use Pythagoras and $30-60-90$ triangle). The bottom side will lie on the line $y=-0.5\sqrt{3}$. Thus, the coordinates of the other two vertices are: $B(-1.5, -0.5\sqrt{3})$ and $C(1.5, -0.5\sqrt{3})$. To find coordinates of a midpoint of a segment, you simply take arithmetic mean of the corresponding coordinates of the end points.

Answer 3

Hint: Draw a line parallel to the base through the centre. Then drop a perpendicular from one of the $?$-points to meet this line. You have a right triangle whose legs are the coordinates of that point.

To calculate these legs, use parallelisms to determine that this little triangle is a $30$-$60$-$90$ one, so that it is similar to one-half of the original. Then Bob's your uncle!