# Homework Help: Find the third coordinate of a vertex of an equilateral triangle

Tags:
1. Apr 3, 2017

### parshyaa

1. The problem statement, all variables and given/known data

Q. Prove that If (x1,y1) and (x2,y2) are the coordinates of the two vertices of an Equilateral Triangle then the coordinates of the 3rd vertex (X,Y) are
$$X=\frac{x1+x2\pm\ √3(y1-y2)}{2},$$
$$Y=\frac{y1+y2\pm\ √3(x1-x2)}{2},$$

3. The attempt at a solution
I used distance formula,cos rule,sine rule, equated determinant formula of area of triangle to (√3/4)((side)^2)
Please give me a really helpful hint or not so complex proof.

Last edited: Apr 3, 2017
2. Apr 3, 2017

### Mastermind01

I don't think you can do it using just one formula. Take the unknown coordinate as something and then use any of the two formulas to get two equations and solve them together.

3. Apr 3, 2017

### parshyaa

I am not asking how to find the coordinates of the third vertex if coordinates of two vertices are given, i am asking how do i prove that (X,Y) coordinates of third vertex are as given in question.

4. Apr 3, 2017

### Mastermind01

Aren't they the same? I mean if you find the third vertices and they came out to be the same as given then that's the proof.

Anyway you can assume that the ones given are the third vertices and prove that they're equidistant from each other.

5. Apr 3, 2017

### parshyaa

Thats the question how do i find the third vertex, tell me which equations i have to use, do you know the proof.

6. Apr 3, 2017

### Mastermind01

While the calculation will be complicated, take the third vertex to be $(X,Y)$ and use the distance formula twice to get two equations. Doesn't that work?

A little less calculation might be to find the equation of the perpendicular to the line joining $(x1, y1)$ and $(x2,y2)$ and passing through it's midpoint. Then find points on the line such that they are equidistant from the two vertices and the distance is equal to the distance between them.

A third method might be to find the equation as in the previous method. Given two vertices you can surely find the length of height of the triangle. Find points on the line at a distance of the height lying on your line.

Last edited: Apr 3, 2017
7. Apr 3, 2017

### parshyaa

No its not working, try it yourself first

8. Apr 3, 2017

### Buffu

WLOG assume $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (a, 0)$. where $a$ is the length of a side.

Then the question becomes,

$$X=\frac{a}{2}$$ and $$Y=\frac{\pm\ \sqrt{3}(a)}{2}$$

Which is easy enough to prove.

9. Apr 3, 2017

### Buffu

For the WLOG, the reason is that you can always transform origin to $(x_1, x_2)$ and then perform a rotation for an angle equal to arctangent of slope of line passing through two given point.

10. Apr 4, 2017

### LCKurtz

Here's another approach. To save typing I will call the first point $(a,b)$ and the second $(c,d)$. Express them as complex numbers $w = a +bi$, and $v=c+di$. The vector side of the triangle can be represented as $v-w$. Rotate it by $\pm \frac \pi 3$ and add it to $w$. So the point(s) you seek for the third vertex are represented by the complex number(s)$$w + e^{\pm \frac {i\pi} 3}(v-w)$$Express the exponential using Euler's formula and calculate it directly.

Last edited: Apr 4, 2017
11. Jun 20, 2018

### Ray Vickson

I have tried it, and it works, but by equating the squares of the distances, rather than the distances themselves. It really does work, but may take several pages of algebraic manipulation; I just used a computer algebra package instead, but in principle it can all be done manually.

12. Jun 20, 2018

All you really need to do is show $(X-x_1)^2+(Y-y_1)^2=(X-x_2)^2+(Y-y_2)^2=(x_1-x_2)^2+(y_1-y_2)^2$. They give you the answer, so you really don't need to solve for $(X,Y)$. The $\pm$ is because there are two ways (=two separate points) to complete the equilateral triangle.