[tex]\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{7em}\boxed{y=\stackrel{\stackrel{m}{\downarrow }}{-1}x-6} \\\\\\ 5x+5y=10\implies 5(x+y)=10\implies x+y=\cfrac{10}{5} \\\\\\ x+y=2 \implies \boxed{y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+2}[/tex]
perpendicular lines have negative reciprocal slopes, so pretty much if one has a slope of say a/b the other has -b/a, so if their product is always -1.
these two, have actually the same slope of -1, that means they're parallel, not perpendicular.
