[tex]\bf C(\stackrel{x_1}{3}~,~\stackrel{y_1}{-5})\qquad
D(\stackrel{x_2}{6}~,~\stackrel{y_2}{0})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-(-5)}{6-3}\implies \cfrac{0+5}{6-3}\implies \cfrac{5}{3}
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-5)=\cfrac{5}{3}(x-3)\implies y+5=\cfrac{5}{3}(x-3)[/tex]
[tex]\bf \stackrel{\textit{multiplying both sides by LCD of 3}}{3(y+5)=3\left[ \cfrac{5}{3}(x-3) \right]}\implies 3y+15=5(x-3)
\\\\\\
3y+15=5x-15\implies -5x+3y=-30\implies \stackrel{\textit{multiplying by -1}}{5x-3y=30}[/tex]
bearing in mind the standard form uses all integers, and the x-variable cannot have a negative coefficient.
