Question

The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of the point Q are (−4, 0)
and origin is the midpoint of the base. Find the coordinates of the points P and R.

Answer 1

Let (x, 0) be the coordinates of R. Then
$0=\frac{-4+x}{2}\Rightarrow x=4$
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
$$\begin{gathered} PQ=QR\Rightarrow P{Q}^2=Q{R}^2 \\ \Rightarrow {\left(0+4\right)}^2+{\left(y-0\right)}^2=8^2 \\ \Rightarrow y^2=64-16=48 \\ \Rightarrow y=\pm 4\sqrt{3} \end{gathered}$$
Hence, the required coordinates are $R\left(4,0\right)\mathrm{and}P\left(0,4\sqrt{3}\right)\mathrm{or}P\left(0,-4\sqrt{3}\right)$.