if two vertices of equilateral triangle is (0,0) (3,3^1/2)find the third vertex
A(0,0), B(3,sqrt(3)), need to find C and C'.
Using the general method:
By Pythagoras theorem,
Length of one side:
L = sqrt(3^2+sqrt(3)^2)=sqrt(12)=2sqrt(3)
Draw a circle with (0,0) as centre, and L as radius:
C1(x,y)=x²+y²=L²=12
Find the intersection with another circle centred at B:
C2(x,y)=(x-3)²+(y-sqrt(3))^2=12
Solve for x,y in
C1(x,y)=0 and
C2(x,y)=0
by elimination to get
(0,2sqrt(3)), or (3,-sqrt(3))
In particular, we note that the line AB makes with the x-axis is 30° (because y/x=sqrt(3)/3), therefore
the other two points are at θ=-30° and at 90° which correspond to
(3,-sqrt(3)) and (0,2sqrt(3)) respectively.