check the picture below.
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad
% (c,d)
&({{ -1}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-5}{-1-(-2)}\implies \cfrac{1-5}{-1+2}
\\\\\\
\cfrac{-4}{1}\impliedby\stackrel{slope~of}{PQ}[/tex]
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 1}})\quad
% (c,d)
&({{ 7}}\quad ,&{{ 3}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{3-1}{7-(-1)}\implies \cfrac{3-1}{7+1}
\\\\\\
\cfrac{2}{8}\implies \cfrac{1}{4}\impliedby \stackrel{slope~of}{QR}[/tex]
now, two lines who meet at a perpendicular angle, have a negative reciprocal slope of each other, to make it short, if you multiply their slopes, you should get -1 as the product, if they're indeed perpendicular.
[tex]\bf \stackrel{PQ}{-\cfrac{4}{1}}\cdot \stackrel{QR}{\cfrac{1}{4}}\implies -1[/tex]
low and behold, they're indeed perpendicular, thus that angle is indeed a right-angle.
