Question

If (3,2) and (-3,2) are two vertices of an equilateral triangle which contains the origin then find the third vertex.

Answer 1

Let an equilateral triangle Δ ABC : AB = BC = AC.
 vertices are A(3,2) and B(-3,2) and C(x,y).
The mid point of the side AB is M (0,2).
 (AB)² =  (3+3)² + (2-2)² = 6² + 0² = 36 
AB = 6 cm
AB = BC = AC = 6 unit.
AM = 3 unit.
As two vertices are A(-3,2) and B(3,0) so that Third vertex will be at
 Y-axis(x=0).
so that
Third vertex C(0,y) and it is located below the origin.
Hence it contains the origin.
Now,

    y² = (AC)² - (AM)² 
⇒ y² = 6² - 3² = 36 - 9 =27
    y = √27 = 3√3 unit
Hence Third vertices of the triangle are A(-3,2) , B(3,2) and C(0,3√3).

b1dcaf468bd2ff6f0a88411bb012ec42.docx

Answer 2

i think ths is the answer
c(0,3root3