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.
Solution:
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
$P Q=Q R \Rightarrow P Q^{2}=Q R^{2}$
$\Rightarrow(0+4)^{2}+(y-0)^{2}=8^{2}$
$\Rightarrow y^{2}=64-16=48$
$\Rightarrow y=\pm 4 \sqrt{3}$
Hence, the required coordinates are $R(4,0)$ and $P(0,4 \sqrt{3})$ or $P(0,-4 \sqrt{3})$.