Question

if (3,2) and (-3,2) r 2 vertices of an equilateral triangle which contains the origin , find its 3rd vertex.

Answer 1

HI,

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).

HOPE IT HELPS