Question 212930
First let's find the slope of the line through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(3,-2\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,2\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=2}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,-2\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=-2}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



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



{{{m=(-4)/(3--3)}}} Subtract {{{2}}} from {{{-2}}} to get {{{-4}}}



{{{m=(-4)/(6)}}} Subtract {{{-3}}} from {{{3}}} to get {{{6}}}



{{{m=-2/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(3,-2\right)] is {{{m=-2/3}}}



Now let's use the point slope formula:



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



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



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



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



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



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



{{{y=(-2/3)x+0}}} Combine like terms. 



{{{y=(-2/3)x}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(3,-2\right)] is {{{y=(-2/3)x}}}



 Notice how the graph of {{{y=(-2/3)x}}} goes through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(3,-2\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(-2/3)x),
 circle(-3,2,0.08),
 circle(-3,2,0.10),
 circle(-3,2,0.12),
 circle(3,-2,0.08),
 circle(3,-2,0.10),
 circle(3,-2,0.12)
 )}}} Graph of {{{y=(-2/3)x}}} through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(3,-2\right)]