12)
[tex](\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{4}}}\implies \cfrac{-2}{-8}\implies \cfrac{1}{4}[/tex]
now, I used y₂ - y₁ and x₂ - x₁, is pretty much the same slope the other way.
[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}{5}=\stackrel{m}{\cfrac{1}{4}}(x-\stackrel{x_1}{4}) \\\\\\ y-5=\cfrac{1}{4}x-1\implies y=\cfrac{1}{4}x+4[/tex]
13)
[tex](\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-2)}}}\implies \cfrac{-2}{-4+2}\implies \cfrac{-2}{-2}\implies 1[/tex]
[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{1}(x-\stackrel{x_1}{(-4)}) \\\\\\ y=1(x+4)\implies y=x+4[/tex]
notice, for the point-slope form, we can use either x₁ , y₁ or x₂ , y₂ , doesn't quite matter, so long we use the corresponding coordinates.
