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

We have,

$PQR$ is an equilateral triangle such that $QR$ lies on $x-axis$.

$O$ is the midpoint of $QR$.

So,

$OQ=QR=4units.$

$\Rightarrow$ Coordinates of point $R$ are $(4,0)$.

Now,

Point $P$ lies on the $y-axis$.

Let the coordinate of point $P$ be $(0,y).$

So,

$PQ=QR$

$\Rightarrow \sqrt{{(0+4)}^2+{(y-0)}^2}=\sqrt{{(4+4)}^2+0^2}$

$\Rightarrow \sqrt{16+y^2}=\sqrt{64}$

$\text{On}\text{squaring}\text{both}\text{side}\text{and}\text{we}\text{get,}$

$\Rightarrow 16+y^2=64$

$\Rightarrow y^2=64-16$

$\Rightarrow y^2=48$

$\Rightarrow y=\pm 4\sqrt{3}$

Hence, the coordinate of points $P$ are $(0,4\sqrt{3})$ above $x-axis.$

And $(0,-4\sqrt{3})$ below $x-axis$.

Hence, this is the answer.