Question 640632

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



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-5,-9\right)]. So this means that {{{x[1]=-5}}} and {{{y[1]=-9}}}.

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



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



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



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



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



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



{{{y--9=(1/3)(x+5)}}} Rewrite {{{x--5}}} as {{{x+5}}}



{{{y+9=(1/3)(x+5)}}} Rewrite {{{y--9}}} as {{{y+9}}}



{{{y+9=(1/3)x+(1/3)(5)}}} Distribute



{{{y+9=(1/3)x+5/3}}} Multiply



{{{y=(1/3)x+5/3-9}}} Subtract 9 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(-5,-9\right)] and *[Tex \LARGE \left(-2,-8\right)] is {{{y=(1/3)x-22/3}}}