[tex]y = \stackrel{\stackrel{m}{\downarrow }}{5}x-3\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]
well, what's the equation of a line whose slope is 5 and passes through (-2 , -13) ?
[tex](\stackrel{x_1}{-2}~,~\stackrel{y_1}{-13})\qquad \qquad \stackrel{slope}{m}\implies 5 \\\\\\ \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}{(-13)}=\stackrel{m}{5}(x-\stackrel{x_1}{(-2)})\implies y+13=5(x+2)[/tex]
