Check the picture below.
all we need to get the equation of the line is two points on it, in this case those would be (-3,2) and (1,1),
[tex]\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-3)}}}\implies \cfrac{-1}{1+3}\implies -\cfrac{1}{4}[/tex]
[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}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-\cfrac{1}{4}}[x-\stackrel{x_1}{(-3)}]\implies y-2=-\cfrac{1}{4}(x+3) \\\\\\ y-2=-\cfrac{1}{4}x-\cfrac{3}{4}\implies y=-\cfrac{1}{4}x-\cfrac{3}{4}+2\implies y=-\cfrac{1}{4}x+\cfrac{5}{4}[/tex]
