Question

The base of an equilateral triangle with side 2a lies along the y-axis, such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer 1

Let ABC be an equilateral triangle, where BC = 2a. Let A(x, 0) be the third vertex of $∆$ABC.

In equilateral triangle ABC,
AB = BC = AC
$\Rightarrow$${\mathrm{AB}}^2{=\; BC}^2{=\; AC}^2$

$$\begin{gathered} \Rightarrow a^2+x^2={\left(2a\right)}^2\left(\because BC=2a\right) \\ \Rightarrow x^2=3a^2 \\ \Rightarrow x=\pm \sqrt{3}a \end{gathered}$$

So, the vertices of the triangle are $\left(0,-a\right),\left(0,a\right)\mathrm{and}\left(\sqrt{3}a,0\right)\mathrm{or}\left(0,-a\right),\left(0,a\right)\mathrm{and}\left(-\sqrt{3}a,0\right)$.