Respuesta :
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
[tex]y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x-\cfrac{4}{5}\qquad \impliedby \qquad \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]
so we're really looking for the equation of a line whose slope is 2/3 and that it passes through (-3 , 2) in standard form
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}{-3}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \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}{2}=\stackrel{m}{ \cfrac{2}{3}}(x-\stackrel{x_1}{(-3)}) \implies y -2= \cfrac{2}{3} (x +3)[/tex]
[tex]\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-2)=3\left( \cfrac{2}{3} (x +3) \right)}\implies 3y-6=2(x+3)\implies 3y-6=2x+6 \\\\\\ 3y=2x+12\implies -2x+3y=12\implies {\Large \begin{array}{llll} 2x-3y=-12 \end{array}}[/tex]
