Question 583793
First let's find the slope of the line through the points *[Tex \LARGE \left(-3,-8\right)] and *[Tex \LARGE \left(-9,-5\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,-8\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=-8}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-9,-5\right)].  So this means that {{{x[2]=-9}}} and {{{y[2]=-5}}}.



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



{{{m=(-5--8)/(-9--3)}}} Plug in {{{y[2]=-5}}}, {{{y[1]=-8}}}, {{{x[2]=-9}}}, and {{{x[1]=-3}}}



{{{m=(3)/(-9--3)}}} Subtract {{{-8}}} from {{{-5}}} to get {{{3}}}



{{{m=(3)/(-6)}}} Subtract {{{-3}}} from {{{-9}}} to get {{{-6}}}



{{{m=-1/2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,-8\right)] and *[Tex \LARGE \left(-9,-5\right)] is {{{m=-1/2}}}



Now let's use the point slope formula:



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



{{{y--8=(-1/2)(x--3)}}} Plug in {{{m=-1/2}}}, {{{x[1]=-3}}}, and {{{y[1]=-8}}}



{{{y--8=(-1/2)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y+8=(-1/2)(x+3)}}} Rewrite {{{y--8}}} as {{{y+8}}}



{{{y+8=(-1/2)x+(-1/2)(3)}}} Distribute



{{{y+8=(-1/2)x-3/2}}} Multiply



{{{y=(-1/2)x-3/2-8}}} Subtract 8 from both sides. 



{{{y=(-1/2)x-19/2}}} Combine like terms. note: If you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>.



So the equation that goes through the points *[Tex \LARGE \left(-3,-8\right)] and *[Tex \LARGE \left(-9,-5\right)] is {{{y=(-1/2)x-19/2}}}


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