Question 127143


Let's denote the first point (-8,-3) as *[Tex \Large \left(x_{1},y_{1}\right)]. In other words, *[Tex \LARGE x_{1}=-8] and *[Tex \LARGE y_{1}=-3]


Now let's denote the second point (-2,2) as *[Tex \Large \left(x_{2},y_{2}\right)]. In other words, *[Tex \Large x_{2}=-2] and *[Tex \Large y_{2}=2]




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


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



{{{m=5/6}}} Subtract the terms in the numerator {{{2--3}}} to get {{{5}}}.  Subtract the terms in the denominator {{{-2--8}}} to get {{{6}}}

  

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Answer:


So the slope of the line through the points (-8,-3) and (-2,2) is {{{m=5/6}}}