to get the equation of any straight line we only need two points off of it.
1st one
well, we have four points on the table, let's use a couple from it hmmm say (10 , -25) and (13 , -34)
[tex](\stackrel{x_1}{10}~,~\stackrel{y_1}{-25})\qquad (\stackrel{x_2}{13}~,~\stackrel{y_2}{-34}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-34}-\stackrel{y1}{(-25)}}}{\underset{run} {\underset{x_2}{13}-\underset{x_1}{10}}}\implies \cfrac{-34+25}{3}\implies \cfrac{-9}{3}\implies -3[/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}{(-25)}=\stackrel{m}{-3}(x-\stackrel{x_1}{10}) \\\\\\ y+25=-3x+30\implies y=-3x+5[/tex]
4th one
well, from the picture itself we already have a couple of points we can use hmmm (0 , -4) and (5 , 0)
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{0}}}\implies \cfrac{0+4}{5}\implies \cfrac{4}{5}[/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}{(-4)}=\stackrel{m}{\cfrac{4}{5}}(x-\stackrel{x_1}{0}) \\\\\\ y+4=\cfrac{4}{5}x\implies y=\cfrac{4}{5}x-4[/tex]
