1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the third coordinate of a vertex of an equilateral triangle

  1. Apr 3, 2017 #1
    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. jcsd
  3. Apr 3, 2017 #2
    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.
     
  4. Apr 3, 2017 #3
    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.
     
  5. Apr 3, 2017 #4
    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.
     
  6. Apr 3, 2017 #5
    Thats the question how do i find the third vertex, tell me which equations i have to use, do you know the proof.
     
  7. Apr 3, 2017 #6
    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
  8. Apr 3, 2017 #7
    No its not working, try it yourself first
     
  9. Apr 3, 2017 #8
    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.
     
  10. Apr 3, 2017 #9
    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.
     
  11. Apr 4, 2017 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find the third coordinate of a vertex of an equilateral triangle
  1. Equilateral Triangles? (Replies: 2)

Loading...