Question 530138


First let's find the slope of the line through the points *[Tex \LARGE \left(-8,-1\right)] and *[Tex \LARGE \left(-2,23\right)]



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-2,23\right)].  So this means that {{{x[2]=-2}}} and {{{y[2]=23}}}.



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



{{{m=(23--1)/(-2--8)}}} Plug in {{{y[2]=23}}}, {{{y[1]=-1}}}, {{{x[2]=-2}}}, and {{{x[1]=-8}}}



{{{m=(24)/(-2--8)}}} Subtract {{{-1}}} from {{{23}}} to get {{{24}}}



{{{m=(24)/(6)}}} Subtract {{{-8}}} from {{{-2}}} to get {{{6}}}



{{{m=4}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-8,-1\right)] and *[Tex \LARGE \left(-2,23\right)] is {{{m=4}}}



Now let's use the point slope formula:



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



{{{y--1=4(x--8)}}} Plug in {{{m=4}}}, {{{x[1]=-8}}}, and {{{y[1]=-1}}}



{{{y--1=4(x+8)}}} Rewrite {{{x--8}}} as {{{x+8}}}



{{{y+1=4(x+8)}}} Rewrite {{{y--1}}} as {{{y+1}}}



{{{y+1=4x+4(8)}}} Distribute



{{{y+1=4x+32}}} Multiply



{{{y=4x+32-1}}} Subtract 1 from both sides. 



{{{y=4x+31}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-8,-1\right)] and *[Tex \LARGE \left(-2,23\right)] is {{{y=4x+31}}}



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