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

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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).

I think ths is the answer

c(0,3root3

c(0,3root3