to get the equation of any straight line we only need two points off of it, hmmm let's use P and Q here and then let's set the equation in standard form, that is
standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
[tex](\stackrel{x_1}{6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{6}}}\implies \cfrac{2}{-3}\implies -\cfrac{2}{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}{0}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{6})[/tex]
[tex]\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-0)~~ = ~~3\left( -\cfrac{2}{3}(x-6) \right)}\implies 3y=-2(x-6) \\\\\\ 3y=-2x+12 \implies \stackrel{a}{2} x+\stackrel{b}{3} y=12[/tex]
