Question 190944

{{{-3x + 2y = 9}}} Start with the given equation.



{{{2y=9+3x}}} Add {{{3x}}} to both sides.



{{{2y=3x+9}}} Rearrange the terms.



{{{y=(3x+9)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=(3/2)x+9/2}}} Break up the fraction.



We can see that the equation {{{y=(3/2)x+9/2}}} has a slope {{{m=3/2}}} and a y-intercept {{{b=9/2}}}.



Since parallel lines have equal slopes, this means that we know that the slope of the unknown parallel line is {{{m=3/2}}}.

Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=3/2}}}  and the coordinates of the given point *[Tex \LARGE \left\(-2,1\right\)].



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y-1=(3/2)(x--2)}}} Plug in {{{m=3/2}}}, {{{x[1]=-2}}}, and {{{y[1]=1}}}



{{{y-1=(3/2)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-1=(3/2)x+(3/2)(2)}}} Distribute



{{{y-1=(3/2)x+3}}} Multiply



{{{y=(3/2)x+3+1}}} Add 1 to both sides. 



{{{y=(3/2)x+4}}} Combine like terms. 



So the equation of the line parallel to {{{-3x + 2y = 9}}} that goes through the point *[Tex \LARGE \left\(-2,1\right\)] is {{{y=(3/2)x+4}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(3/2)x+9/2,(3/2)x+4),
circle(-2,1,0.08),
circle(-2,1,0.10),
circle(-2,1,0.12))}}}Graph of the original equation {{{-3x + 2y = 9}}} (red) and the parallel line {{{y=(3/2)x+4}}} (green) through the point *[Tex \LARGE \left\(-2,1\right\)].