[tex](\stackrel{x_1}{4}~,~\stackrel{y_1}{-1})\qquad \qquad m = -3 \\\\\\ \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}{(-1)}=\stackrel{m}{-3}(x-\stackrel{x_1}{4})\implies y+1=-3(x-4) \\\\\\ y+1=-3x+12\implies \stackrel{\textit{slope-intercept form}}{y=-3x+11}[/tex]
does it pass through the origin? well, we could draw it, or we can just check that, the origin is at 0,0, or namely when x = 0, y = 0, let's see if that's true, let's make x = 0, let's see what we get for "y".
y = -3(0) + 11 => y = 11
woops, no dice, y ≠ 0, so nope, doesn't pass through the origin.
