Question

If $(3,2)$ and $(-3,2)$ are two vertices of an equilateral triangle which contains within it the origin, what are the coordinates of the third vertex ?

• A. $(2,2\sqrt{2})$
• B. $(1,2\sqrt{2})$
• C. $(0,-3\sqrt{3})$
• D. $Noneofthese$

Answer 1

The correct option is C $(0,-3\sqrt{3})$

Let $\Delta ABC$ be the equilateral triangle, where $AB=BC=AC$.

Vertices of this triangle be,

$A(3,2),B(-3,2)$ and $C(x,y)$.

The mid-point of the side $AB$ is $M(0,2)$.

$AB=\sqrt{{(3+3)}^2+{(2-2)}^2}$

$AB=6units$

Therefore,

$\Rightarrow AB=BC=AC=6units$

Also,

$\Rightarrow AM=3units$

As two vertices are $A(3,2)$ and $B(-3,2)$, so third vertex will be at $y-$axis.

Therefore,

$\Rightarrow C(0,y)$

It will be located below the origin as the triangle contains the origin.

Now,

$y^2={\left(AC\right)}^2-{\left(AM\right)}^2$

$y^2=36-9=27$

$y=3\sqrt{3}units$

Therefore,

$\Rightarrow C(0,-3\sqrt{3})$

Hence, this is the required point.