well, according to the picture, we'll have to assume it's parallel to that line drawn and it passes through (1 , 2).
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the line above
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{0}}}\implies -\cfrac{1}{3}[/tex]
so we're really looking for the equation of a line whose slope is -1/3 an passes through (1 , 2)
[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \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}{2}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1}) \\\\\\ y-2=-\cfrac{1}{3}x+\cfrac{1}{3}\implies y=-\cfrac{1}{3}x+\cfrac{1}{3}+2\implies y=-\cfrac{1}{3}x+\cfrac{7}{3}[/tex]
