Question 553268
First let's find the slope of the line through the points *[Tex \LARGE \left(-2,-5\right)] and *[Tex \LARGE \left(-6,3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-2,-5\right)]. So this means that {{{x[1]=-2}}} and {{{y[1]=-5}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-6,3\right)].  So this means that {{{x[2]=-6}}} and {{{y[2]=3}}}.



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



{{{m=(3--5)/(-6--2)}}} Plug in {{{y[2]=3}}}, {{{y[1]=-5}}}, {{{x[2]=-6}}}, and {{{x[1]=-2}}}



{{{m=(8)/(-6--2)}}} Subtract {{{-5}}} from {{{3}}} to get {{{8}}}



{{{m=(8)/(-4)}}} Subtract {{{-2}}} from {{{-6}}} to get {{{-4}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-2,-5\right)] and *[Tex \LARGE \left(-6,3\right)] is {{{m=-2}}}



Now let's use the point slope formula:



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



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



{{{y--5=-2(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y+5=-2(x+2)}}} Rewrite {{{y--5}}} as {{{y+5}}}



So the answer is {{{y+5=-2(x+2)}}}



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