Question

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3).
The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find
the coordinates of another point D such that ABCD is a rhombus.

Answer 1

Let (0, y) be the coordinates of B. Then
$0=\frac{-3+y}{2}\Rightarrow y=3$
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
$$\begin{gathered} AB=BC\Rightarrow A{B}^2=B{C}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(0-3\right)}^2=6^2 \\ \Rightarrow x^2=36-9=27 \\ \Rightarrow x=\pm 3\sqrt{3} \end{gathered}$$
If the coordinates of point A are $\left(3\sqrt{3},0\right)$, then the coordinates of D are $\left(-3\sqrt{3},0\right)$.
If the coordinates of point A are $\left(-3\sqrt{3},0\right)$, then the coordinates of D are $\left(3\sqrt{3},0\right)$.
Hence, the required coordinates are $A\left(3\sqrt{3},0\right),B\left(0,3\right)\mathrm{and}D\left(-3\sqrt{3},0\right)$ or $A\left(-3\sqrt{3},0\right),B\left(0,3\right)\mathrm{and}D\left(3\sqrt{3},0\right)$.