check the picture below.
so, clearly UR is not parallel to ST, but is likely that UT is parallel to RS.
keeping in mind that parallel lines, have the same exact slope, let's check the slope for UT as well as RS, if they are the same, then indeed both are parallel,
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&R&(~ 1 &,& -3~)
% (c,d)
&S&(~ 4 &,& -1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-(-3)}{4-1}\implies \cfrac{-1+3}{4-1}\implies \cfrac{2}{3}\\\\
-------------------------------[/tex]
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&U&(~ -4 &,& -2~)
% (c,d)
&T&(~ 2 &,& 2~)
\end{array}
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-2)}{2-(-4)}\implies \cfrac{2+2}{2+4}\implies \cfrac{4}{6}\implies \cfrac{2}{3}[/tex]
there you have it, notice the slopes of each.
from the picture, clearly there are no right-angles at vertices U and R, however at S and T, it looks like, but we dunno.
well, we know UT || RS, now if ST ⟂ RS, then ST will have a negative reciprocal slope to RS, namely whatever the slope of RS is, ST will be the negative reciprocal of that, so let's check what is the slope of ST then,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{3}\\\\
negative\implies -\cfrac{2}{ 3}\qquad reciprocal\implies \boxed{- \cfrac{ 3}{2}}\\\\
-------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&S&(~ 4 &,& -1~)
% (c,d)
&T&(~ 2 &,& 2~)
\end{array}
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-1)}{2-4}\implies \cfrac{2+1}{2-4}\implies \boxed{-\cfrac{3}{2}}[/tex]
and if ST ⟂ RS, then ST ⟂ UT.
