Question 627630
First let's find the slope of the line through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(3,4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,-4\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=-4}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,4\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=4}}}.



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



{{{m=(4--4)/(3--3)}}} Plug in {{{y[2]=4}}}, {{{y[1]=-4}}}, {{{x[2]=3}}}, and {{{x[1]=-3}}}



{{{m=(8)/(3--3)}}} Subtract {{{-4}}} from {{{4}}} to get {{{8}}}



{{{m=(8)/(6)}}} Subtract {{{-3}}} from {{{3}}} to get {{{6}}}



{{{m=4/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(3,4\right)] is {{{m=4/3}}}



Now let's use the point slope formula:



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



{{{y--4=(4/3)(x--3)}}} Plug in {{{m=4/3}}}, {{{x[1]=-3}}}, and {{{y[1]=-4}}}



{{{y--4=(4/3)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y+4=(4/3)(x+3)}}} Rewrite {{{y--4}}} as {{{y+4}}}


So the answer is {{{y+4=(4/3)(x+3)}}}


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