P is the point (3, −1) and Q is the point (−5, 5).

1. Find the horizontal and vertical changes between $P$ and $Q$.

For $P(3,-1)$ and $Q(-5,5)$,

$\Delta x=-5-3=-8$

$\Delta y=5-(-1)=6$

2. Apply the distance formula to find $|PQ|$.

$|PQ|=\sqrt{(-8)^2+6^2}$

$=\sqrt{64+36}$

$=\sqrt{100}=10$

1. Find the gradient of line $PQ$.

For $P(3,-1)$ and $Q(-5,5)$,
$m=\frac{5-(-1)}{-5-3}=\frac{6}{-8}=-\frac{3}{4}.$

2. Use the equation of the line and set $y=0$ to find the $x$-intercept.

Using $P(3,-1)$ in $y=mx+c$,
$-1=-\frac{3}{4}(3)+c$
$c=\frac{5}{4}.$
So the line is
$y=-\frac{3}{4}x+\frac{5}{4}.$
At the $x$-axis, $y=0$:
$0=-\frac{3}{4}x+\frac{5}{4}$
$3x=5$
$x=\frac{5}{3}.$
Hence $R=\left(\dfrac{5}{3},0\right)$.

1. Add the translation vector to the coordinates of point P.

$P=(3,-1)$ and the translation vector is $\begin{pmatrix}-6\\1\end{pmatrix}$.

So
$S=(3+(-6),\,-1+1)=(-3,\,0).$