Question 538942


First let's find the slope of the line through the points *[Tex \LARGE \left(-2,-8\right)] and *[Tex \LARGE \left(-5,-9\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-2,-8\right)]. So this means that {{{x[1]=-2}}} and {{{y[1]=-8}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-5,-9\right)].  So this means that {{{x[2]=-5}}} and {{{y[2]=-9}}}.



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



{{{m=(-9--8)/(-5--2)}}} Plug in {{{y[2]=-9}}}, {{{y[1]=-8}}}, {{{x[2]=-5}}}, and {{{x[1]=-2}}}



{{{m=(-1)/(-5--2)}}} Subtract {{{-8}}} from {{{-9}}} to get {{{-1}}}



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



{{{m=1/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-2,-8\right)] and *[Tex \LARGE \left(-5,-9\right)] is {{{m=1/3}}}



Now let's use the point slope formula:



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



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



{{{y--8=(1/3)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y+8=(1/3)(x+2)}}} Rewrite {{{y--8}}} as {{{y+8}}}



{{{y+8=(1/3)x+(1/3)(2)}}} Distribute



{{{y+8=(1/3)x+2/3}}} Multiply



{{{y=(1/3)x+2/3-8}}} Subtract 8 from both sides. 



{{{y=(1/3)x-22/3}}} Combine like terms. note: If you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>.



So the equation that goes through the points *[Tex \LARGE \left(-2,-8\right)] and *[Tex \LARGE \left(-5,-9\right)] is {{{y=(1/3)x-22/3}}}



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