Question

If the points $(0,0),(2,2\sqrt{3})$, and $(p,q)$ are the vertices of an equilateral triangle, then $(p,q)$ is

• A. $(0,-4)$
• B. $(4,4)$
• C. $(4,0)$
• D. $(5,0)$

Answer 1

The correct option is A $(0,-4)$

Let the three given vertices be $A(0,0),B(2,2\sqrt{3})\text{and}C(p,q)$

We know that, the distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Also, we know that for an equilateral triangle, all the sides are equal.

$\therefore AB=BC=AC$

$\Rightarrow \sqrt{(0-2{)}^2+(0-2\sqrt{3}{)}^2}=\sqrt{(p-2{)}^2+(q-2\sqrt{3}{)}^2}=\sqrt{(0-p{)}^2+(0-q{)}^2}$

$\therefore p^2+q^2=2^2+(2\sqrt{3}{)}^2$

$\Rightarrow p^2+q^2=16.....\left(i\right)$

Also,

$(p-2{)}^2+(q-2\sqrt{3}{)}^2=16$

$\Rightarrow p^2+4-4p+q^2+12-4\sqrt{3}q=16$

Now since $p^2+q^2=16$,

$16-4p-4\sqrt{3}q+16=16$

Hence,

$p+\sqrt{3}q=4$

$\Rightarrow p=4-\sqrt{3}q$

Substituting the value of $p$ in $\left(i\right)$, we get:

$(4-\sqrt{3}q{)}^2+q^2=16$

$\Rightarrow 16-8\sqrt{3}q+3q^2+q^2=16$

$\Rightarrow q(4q-8\sqrt{3})=0$

$\Rightarrow q=0,q=2\sqrt{3}$

For $q=0,$

$p=4-\sqrt{3}\left(0\right)=4$

and for $q=2\sqrt{3},$

$p=4-\sqrt{3}\times 2\sqrt{3}=4-6=-2$

So, $p=4,-2$

Therefore, the required point is either $(4,0)$ or $(-2,2\sqrt{3})$.

Hence, option A is correct.