[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-8}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{-2}-\underset{x_1}{1}}}\implies \cfrac{-8+6}{-3}\implies \cfrac{-2}{-3}\cfrac{2}{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}{(-6)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{1})\implies y+6=\cfrac{2}{3}x-\cfrac{2}{3}[/tex]
hmmm now let's put that 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{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y+6)~~ = ~~3\left( \cfrac{2}{3}x-\cfrac{2}{3} \right)}\implies 3y+18~~ = ~~2x-2 \\\\\\ 3y=2x-20\implies -2x+3y=-20\implies 2x-3y=20[/tex]
