Question

Prove that the points A(2, 4), B(2, 6) and $C\left(2+\sqrt{3},5\right)$ are the vertices of an equilateral triangle.

Answer 1

The given points are A(2, 4), B(2, 6) and $C\left(2+\sqrt{3},5\right)$. Now
$$\begin{gathered} AB=\sqrt{{\left(2-2\right)}^2+{\left(4-6\right)}^2}=\sqrt{{\left(0\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{0+4}=2 \\ BC=\sqrt{{\left(2-2-\sqrt{3}\right)}^2+{\left(6-5\right)}^2}=\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(1\right)}^2} \\ =\sqrt{3+1}=2 \end{gathered}$$
$$\begin{gathered} AC=\sqrt{{\left(2-2-\sqrt{3}\right)}^2+{\left(4-5\right)}^2}=\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{3+1}=2 \end{gathered}$$
Hence, the points A(2, 4), B(2, 6) and $C\left(2+\sqrt{3},5\right)$ are the vertices of an equilateral triangle.