Question 628257


First let's find the slope of the line through the points *[Tex \LARGE \left(-1,-3\right)] and *[Tex \LARGE \left(-2,5\right)]



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

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



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



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



{{{m=(8)/(-2--1)}}} Subtract {{{-3}}} from {{{5}}} to get {{{8}}}



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



{{{m=-8}}} Reduce



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



Now let's use the point slope formula:



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



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



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



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



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



{{{y+3=-8x-8}}} Multiply



{{{y=-8x-8-3}}} Subtract 3 from both sides. 



{{{y=-8x-11}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-1,-3\right)] and *[Tex \LARGE \left(-2,5\right)] is {{{y=-8x-11}}}


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