Given: Two coordinates of an equilateral triangle P(-3, 2) and Q(5, 2).
To find: Coordinate of third vertex R.
Solution: Let us take coordinate of R is (x,y).
Because we are given PQR is an equilateral triangle, all side of the triangle PQR are equal.
We can say PQ = QR = RS.
Let us find equations by using distance formula.
Distance between two coordinates is given by formula



.
We know, PQ= QR and PQ= RS.
\sqrt{x^2+y^2-10x-4y+29}[/tex] = 8
Squaring both sides, we get
-------- equation (1)
and \sqrt{x^2+y^2+6x-4y+13}[/tex] = 8
Squaring both sides, we get
---------- equation (2).
Subtracting equation 2 from equation 1.
[/tex]x^2+y^2-10x-4y+29[/tex] = 64
[/tex]x^2+y^2+6x-4y+13[/tex] = 64
----------------------------------------------------
-16x +16 = 0
Subtracting 16 on both sides ,
-16x +16-16 = 0-16
-16x = -16
Dividing -16 on both sides.
-16x/-16 = -16/-16
x=1.
Plugging x=1 in first equation

(Simplifying)

Subtracting 64 from both sides.


By using quadratic formula..

![y=\frac{-(-4)-\sqrt{(-4)^2-4(1)(-44)} }{2(1)} [tex]y=2\left(1+2\sqrt{3}\right),\:y=2\left(1-2\sqrt{3}\right)](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B-%28-4%29-%5Csqrt%7B%28-4%29%5E2-4%281%29%28-44%29%7D%20%7D%7B2%281%29%7D%20%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%5Btex%5Dy%3D2%5Cleft%281%2B2%5Csqrt%7B3%7D%5Cright%29%2C%5C%3Ay%3D2%5Cleft%281-2%5Csqrt%7B3%7D%5Cright%29)
In decimal form
y=8.928, -4.928.
Therefore, coordinates of vertex R could be (1, 8.928) or (1, -4.928).