find coordinates of an equilateral triangle(two points given)

alextm

New member

Coordinates of A and B are known and I pick a point C such that AB = AC = BC. I thought it'd be very simple to calculate the coordinates of C, but when I tried, I couldn't do it.:?

MarkFL

Super Moderator
Staff member
What did you try?

alextm

New member

I calculated the length of AB and then tried to find the sides of the red rectangle where CB is a diagonal. The only thing I know is the length of the diagonal, but I assumed it should still be possible to calculate since there are only 2 possible points where C can be.

ksdhart2

Full Member
For now, I'll work with your two example points. Using the distance formula and knowing that the two points A and B must be equidistant from C, we get:

$$\displaystyle \sqrt{(2 - X_C)^2 + (1 - Y_C)^2} = \sqrt{(3 - X_C)^2 + (2 - Y_C)^2}$$

If we expand everything out, we get:

$$\displaystyle \sqrt{(X_C)^2 - 4X_C + (Y_C)^2 - 2Y_C + 5} = \sqrt{(X_C)^2 - 6X_C + (Y_C)^2 - 4Y_C + 13}$$

Now, for any positive x and y, we have the property that $$\displaystyle \sqrt{x} = \sqrt{y} \implies x = y$$. Since distance is always positive, we can just "delete" the square roots:

$$\displaystyle (X_C)^2 - 4X_C + (Y_C)^2 - 2Y_C + 5 = (X_C)^2 - 6X_C + (Y_C)^2 - 4Y_C + 13$$

Collecting all the variable terms on one side and the constants on the other, we see that the squared terms cancel out and we're left with:

$$\displaystyle 2X_C + 2Y_C = 8 \implies X_C + Y_C = 4$$

It should be easy enough for you to do the last step and figure out where this line intersects your circles. In fact, because you know the circles also intersect with each other at these points, it would be sufficient to find where the line intersects with one of them. And finally, what would happen if we left the coordinates of A and B as unknown? Well, we'd have:

$$\displaystyle (X_A - X_C)^2 + (Y_A - Y_C)^2 = (X_B - X_C)^2 + (Y_B - Y_C)^2$$

Try continuing from here and see what you can come up with.

MarkFL

Super Moderator
Staff member
I would let the coordinates of $$\displaystyle C$$ be $$\displaystyle (x,y)$$. Now, if we equate the square of $$\displaystyle \overline{AC}$$ to the square of $$\displaystyle \overline{BC}$$, there results:

$$\displaystyle (x-2)^2+(y-1)^2=(x-3)^2+(y-2)^2$$

You should be able to show this reduces to:

$$\displaystyle y=-x+4$$

We should not be surprised to find this it the line perpendicular to $$\displaystyle \overline{AB}$$ which passes through the midpoint of this same segment. The slope of $$\displaystyle \overline{AB}$$ is:

$$\displaystyle m=\frac{2-1}{3-2}=1$$

And the midpoint is:

$$\displaystyle \left(\frac{2+3}{2},\frac{1+2}{2}\right)= \left(\frac{5}{2},\frac{3}{2}\right)$$

And so the line we want is:

$$\displaystyle y-\frac{3}{2}=-\left(x-\frac{5}{2}\right)$$

$$\displaystyle y=-x+4$$

So, now we know that $$\displaystyle C$$ is at $$\displaystyle (x,-x+4)$$. Knowing that $$\displaystyle \overline{AC}=\overline{AB}$$ we may write:

$$\displaystyle (x-2)^2+((-x+4)-1)^2= (3-2)^2+(2-1)^2$$

$$\displaystyle (x-2)^2+(x-3)^2=2$$

What do you get when you solve this quadratic in $$\displaystyle x$$?

pka

Elite Member
View attachment 11060
Coordinates of A and B are known and I pick a point C such that AB = AC = BC. I thought it'd be very simple to calculate the coordinates of C, but when I tried, I couldn't do it.
I assume that you must know the coordinates of $$\displaystyle A~\&~B$$. The midpoint of $$\displaystyle \overline{AB}$$ is $$\displaystyle \left(\dfrac{x_a+x_b}{2},\dfrac{y_a+ y_b}{2}\right)$$
Now the point $$\displaystyle C$$ is on the perpendicular bisector of $$\displaystyle \overline{AB}$$ It at the same distance as the length $$\displaystyle \ell(\overline{AB})$$.

Dr.Peterson

Elite Member
View attachment 11060

Coordinates of A and B are known and I pick a point C such that AB = AC = BC. I thought it'd be very simple to calculate the coordinates of C, but when I tried, I couldn't do it.:?
If you are familiar with vectors, that would be another way to go. Vector AC is just vector AB rotated by 60 degrees.