bearing in mind that parallel lines have exactly the same slope, hmmmm what's the slope of the line on that table anyway?
well, let's just pick two points off the table to get that slope, say (5,6) and (11,18)
[tex]\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{11}~,~\stackrel{y_2}{18}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{18}-\stackrel{y1}{6}}}{\underset{run} {\underset{x_2}{11}-\underset{x_1}{5}}}\implies \cfrac{12}{6}\implies 2[/tex]
so we're really looking for the equation of a line whose slope is 2 and runs through (0,10),
[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{10})~\hspace{10em} \stackrel{slope}{m}\implies 2 \\\\\\ \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-\stackrel{y_1}{10}=\stackrel{m}{2}(x-\stackrel{x_1}{0}) \\\\\\ y-10=2x\implies y=2x+10[/tex]
