[tex]\bf \begin{cases}
2y=8x+4\\
\qquad y=\cfrac{8x+4}{2}\\\\
\qquad y=\stackrel{slope}{4}x+2\\\\
y=\stackrel{slope}{-\cfrac{1}{4}}x
\end{cases}[/tex]
now, looking at the first set, notice their slopes... are they the same? no.. ok, that means they're not parallel then.
are they perpendicular? well, only if their slopes are negative reciprocal. let's check
[tex]\bf slope\implies 4\implies \cfrac{4}{1}\qquad negative\implies \cfrac{-4}{1}\qquad reciprocal\implies -\cfrac{1}{4}[/tex]
low and behold, the negative reciprocal of 4 is the slope of the second equation, so they ARE indeed perpendicular.
now, let's check the second set.
[tex]\bf \begin{cases}
2x-3y=12\\
\qquad 2x-12=3y\\\\
\qquad \cfrac{2x-12}{3}=y\\\\
\qquad \stackrel{slope}{\cfrac{2}{3}}x-4=y\\\\
y=\stackrel{slope}{2}x+5
\end{cases}[/tex]
notice their slopes, are they the same slopes? no, so they're not parallel then, are they negative reciprocal of each other? well, negative reciprocal of 2 is -1/2, so no, they're not, so they're not perpendicular.
well, then, but the slopes differ, that means, they do intersect somewheres, not at a right-angle, but they do, so they're only intersecting.
btw, a coincident set, will be when both are exactly the same equation, none of them on either set are.
