Question 530095


First let's find the slope of the line through the points *[Tex \LARGE \left(-4,-3\right)] and *[Tex \LARGE \left(-6,7\right)]



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-6,7\right)].  So this means that {{{x[2]=-6}}} and {{{y[2]=7}}}.



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



{{{m=(7--3)/(-6--4)}}} Plug in {{{y[2]=7}}}, {{{y[1]=-3}}}, {{{x[2]=-6}}}, and {{{x[1]=-4}}}



{{{m=(10)/(-6--4)}}} Subtract {{{-3}}} from {{{7}}} to get {{{10}}}



{{{m=(10)/(-2)}}} Subtract {{{-4}}} from {{{-6}}} to get {{{-2}}}



{{{m=-5}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-4,-3\right)] and *[Tex \LARGE \left(-6,7\right)] is {{{m=-5}}}



Now let's use the point slope formula:



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



{{{y--3=-5(x--4)}}} Plug in {{{m=-5}}}, {{{x[1]=-4}}}, and {{{y[1]=-3}}}



{{{y--3=-5(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y+3=-5(x+4)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=-5x+-5(4)}}} Distribute



{{{y+3=-5x-20}}} Multiply



{{{y=-5x-20-3}}} Subtract 3 from both sides. 



{{{y=-5x-23}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-4,-3\right)] and *[Tex \LARGE \left(-6,7\right)] is {{{y=-5x-23}}}



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