In the given figure, △PQR is equilateral. If the coordinates of the points Q and R are (0, 2) and (0, -2) respectively, find the coordinates of the point P.

Given, PQR is an equilateral triangle in which Q(0, 2) and R(0, -2) and O = (0, 0).

Let (x, 0) be the coordinates of P. [As, P lies on x axis, so y-coordinate is zero.]

By distance formula,

$\Rightarrow QR = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] = \sqrt{(0 - 0)^2 + (-2 - 2)^2} \\[1em] = \sqrt{0 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] \Rightarrow OQ = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\[1em] = \sqrt{(0 - 0)^2 + (2 - 0)^2} \\[1em] = \sqrt{0 + (2)^2} \\[1em] = \sqrt{4} \\[1em] = 2 \text{ units}.

PQ = PR = QR = 4.

OQ = 2

In right angle triangle POQ,

⇒ PQ² = OP² + OQ² [By pythagoras theorem]

⇒ 4² = OP² + 2²

⇒ 16 = OP² + 4

⇒ OP² = 16 - 4

⇒ OP² = 12

⇒ OP = $\sqrt{12} = 2\sqrt{3}$.

Since, P lies on x-axis and OP = $2\sqrt{3}$.

Hence, the coordinates of P are ($2\sqrt{3}$, 0).