Find the third coordinate of a vertex of an equilateral triangle

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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