Question

If the three points $(0,1)$, $(0,-1)$, and $(x,0)$ are vertices of an equilateral triangle, then the values of $x$ are

• A. $\sqrt{3},\sqrt{2}$
• B. $\sqrt{3},-\sqrt{3}$
• C. $-\sqrt{5},\sqrt{3}$
• D. $\sqrt{2},-\sqrt{2}$
• E. $\sqrt{5},-\sqrt{5}$

Answer 1

The correct option is B

$\sqrt{3},-\sqrt{3}$

Explanation for the correct option.

Step 1. Understanding the problem.

Let the points be denoted as $A(0,1),B(0,-1),$ and $C(x,0)$. These points form an equilateral triangle $ABC$.

As the triangle $ABC$ is equilateral so the lengths of the three sides of the triangle would all be the same so, $AB=BC=CA$.

Step 2. Find the distance $AB$ and $BC$

The distance between the points $A(0,1)$ and $B(0,-1)$ is given as:

$$\begin{gathered} AB=\sqrt{{(0-0)}^2+(1-{(-1))}^2} \\ \Rightarrow AB=\sqrt{0+2^2} \\ \Rightarrow AB=\sqrt{2^2} \\ \Rightarrow AB=2 \end{gathered}$$

The distance between the points $B(0,-1)$ and $C(x,0)$ is given as:

$$\begin{gathered} BC=\sqrt{{(0-x)}^2+(0-{(-1))}^2} \\ \Rightarrow BC=\sqrt{{(-x)}^2+{\left(1\right)}^2} \\ \Rightarrow BC=\sqrt{x^2+1} \end{gathered}$$

Step 3. Form and solve the equation

The distance $AB$ and $BC$ would be the same, so:

$\begin{array}{rcl}AB& =& BC\\ 2& =& \sqrt{x^2+1}\end{array}$

Now square both sides of the equation and solve for $x$.

$\begin{array}{rcl}{2}^2& =& {\left(\sqrt{x^2+1}\right)}^2\Rightarrow 4=x^2+1\\ & \Rightarrow & 4-1=x^2+1-1\\ & \Rightarrow & 3=x^2\\ & \Rightarrow & \pm \sqrt{3}=x\end{array}$

So, the values of $x$ satisfying the condition are $\sqrt{3},-\sqrt{3}$.

Hence, the correct option is B.