Locus problem for vertex of equilateral triangle 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


*

*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 $\tan\angle R.$

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

*Also use the distance method and then finally get the required locus. 
My question
Is there any other method which is easier? 
 A: 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:
$$\begin{align*}
\left[(x-1) + \sqrt3(y-3) - 4\right]\left[(x-1) - \sqrt3(y-3) - 4\right] &= 0\\
(x-5)^2-3(y-3)^2 &= 0
\end{align*}$$
A: 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.
