Locus problem for vertex of equilateral triangle

Question

Question

Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$

My attempt

1. Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taking $\tan \angle Q$ and $\tan\angle R.$

2. Similarly for $\angle P$ to equate all 3 angles. Got the equation.

3. Also use the distance method and then finally get the required locus.

My question

Is there any other method which is easier?

Answer 1

I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:

$$\pmatrix{x'\\y'} = \pmatrix{x\\y}-\pmatrix{1\\ 3}$$

Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $\pm\frac {2\pi}6$ about the origin $P$.

For $i \in \{0, 1\}$:

$$\begin{align*} R_i=\pmatrix{x'=h'_i\\y'=k'_i} &= \pmatrix{\cos\frac {2\pi}6 & -\sin\left[(-1)^i\frac {2\pi}6\right]\\ \sin\left[(-1)^i\frac {2\pi}6\right] & \cos\frac {2\pi}6} \pmatrix{2\\p'}\\ &= \pmatrix{\frac12 & -(-1)^i\frac{\sqrt3}2\\ (-1)^i\frac{\sqrt3}2 & \frac12} \pmatrix{2\\p'}\\ &= \pmatrix{1-(-1)^i\sqrt3\frac{p'}2\\ (-1)^i\sqrt3+\frac{p'}2} \end{align*}$$

One way to eliminate the $p'$ is to note that

$$\begin{align*} \frac{p'}2 &= k'_i-(-1)^i\sqrt3\\ h'_i &= 1-(-1)^i\sqrt3\frac{p'}2\\ &= 1-(-1)^i\sqrt3\left[k'_i-(-1)^i\sqrt3\right]\\ &= 4- (-1)^i\sqrt3 k'_i \end{align*}$$

i.e. the loci are $x'=4-(-1)^i\sqrt3 y'$.

The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.

$$\begin{align*} (x-1) &= 4-(-1)^i\sqrt3(y-3)\\ (x-1) + (-1)^i\sqrt3(y-3) - 4 &= 0 \end{align*}$$

Lastly, if you prefer having one equation representing two straight lines:

$$\begin{align*} \left[(x-1) + \sqrt3(y-3) - 4\right]\left[(x-1) - \sqrt3(y-3) - 4\right] &= 0\\ (x-5)^2-3(y-3)^2 &= 0 \end{align*}$$

Answer 2

Moving the origin so that $P = (0,0)$ instead of $P = (1,3)$ and taking as parameters (to eliminate) the side, say $s$, of the triangle and the angle $\alpha$ determined by $R = (x, y)$, we have the equations $$\begin{cases}(x-2)^2+(y-s\sin(60^{\circ}+\alpha))^2=s^2\\x=s\cos (\alpha)\space\space y=s\sin(\alpha)\end{cases}$$

from which $$(x-2)^2+\left(\frac{y-\sqrt3x}{2}\right)^2=x^2+y^2$$ or $$3x^2-3y^2-2\sqrt3xy-16x+16=0$$ or $$(3x+\sqrt3y-4)(x-\sqrt3y-4)=0$$ Then the locus is given by the two straight lines $$3x+\sqrt3y-4=0\\x-\sqrt3y-4=0$$ The task of returning to the original coordinate system is immediate.