What is the equation of the line that is parallel to the line 2x+3y=-8 and passes through the point (2,-2)?

Equation of line passing through (2, -2) and parallel to 2x+3y = -8 is [tex]y=\frac{-2 x}{3}+\frac{-2}{3}[/tex]Solution:Need to write equation of line parallel to 2x+3y=-8 and passes through the point (2, -2)Generic slope intercept form of a line is given by y = mx + cwhere "m" = slope of the line and "c" is the y - interceptLet’s first find slope intercept form of 2x+3y=-8 to get slope of line [tex]\begin{array}{l}{2 x+3 y=-8} \\\\ {=>y=\frac{-2 x-8}{3}} \\\\ {\Rightarrow y=-\frac{2}{3} x-\frac{8}{3}}\end{array}[/tex]On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c,[tex]\text {for line } 2 x+3 y=-8, \text { slope } m=-\frac{2}{3}[/tex]We know that slopes of parallel lines are always equalSo the slope of line passing through (2, -2) is also [tex]m=-\frac{2}{3}[/tex]Equation of line passing through [tex](x_1 , y_1)[/tex] and having slope of m is given by[tex]\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)[/tex][tex]\text { In our case } x_{1}=2 \text { and } y_{1}=-2[/tex]Substituting the values in equation of line we get[tex](y-(-2))=-\frac{2}{3}(x-2)[/tex][tex]\begin{array}{l}{\Rightarrow y+2=\frac{-2 x+4}{3}} \\\\ {=>3(y+2)=-2 x+4} \\\\ {=>3 y+6=-2 x+4} \\\\ {3 y=-2 x-2}\end{array}[/tex][tex]y=\frac{-2 x}{3}+\frac{-2}{3}[/tex]Hence equation of line passing through (2 , -2) and parallel to 2x + 3y = -8 is given as [tex]y=\frac{-2 x}{3}+\frac{-2}{3}[/tex]