parallel lines have the same exact slopes, thefore "q" must also have the same slope as "p", wait a second, what's p's slope anyway? let's see the equation, low and behold, the equation is already in point-slope form, therefore,
[tex]\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}~\hspace{10em}y+2=\stackrel{\stackrel{slope}{\downarrow }}{\cfrac{-9}{7}}(x-3)[/tex]
so we're really looking for the equation of a line whose slope is -9/7 and runs through (2, -4).
[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})~\hspace{10em}
slope = m\implies \cfrac{-9}{7}
\\\\\\
\begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-(-4)=-\cfrac{9}{7}(x-2)
\implies y+4=-\cfrac{9}{7}x+\cfrac{18}{7}
\\\\\\
y=-\cfrac{9}{7}x+\cfrac{18}{7}-4\implies y=-\cfrac{9}{7}x-\cfrac{10}{7}\impliedby \textit{slope-intercept form}[/tex]
