The base QR of an equilateral triangle PQR lies on x-axis.

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-axisLet 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 $P(0,-4 \sqrt{3})$.

 

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