Check the picture below.
let's notice something, the equation of line B is in slope-intercept form, thus
[tex]y = 2x + \underset{\stackrel{\uparrow }{b}}{2}\qquad \impliedby \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}[/tex]
so using that grid scale, we can just get two points from the line A to get its equation, let's use those in the picture below
[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-6)}}}\implies \cfrac{0+6}{6+6}\implies \cfrac{1}{2}[/tex]
[tex]\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}{(-6)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y+6=\cfrac{1}{2}(x+6)\implies y+6=\cfrac{1}{2}x+3\implies y=\cfrac{1}{2}x-3[/tex]
