Question 274720

First let's find the slope of the line through the points *[Tex \LARGE \left(-3,-1\right)] and *[Tex \LARGE \left(-9,-6\right)]



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-9,-6\right)].  So this means that {{{x[2]=-9}}} and {{{y[2]=-6}}}.



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



{{{m=(-6--1)/(-9--3)}}} Plug in {{{y[2]=-6}}}, {{{y[1]=-1}}}, {{{x[2]=-9}}}, and {{{x[1]=-3}}}



{{{m=(-5)/(-9--3)}}} Subtract {{{-1}}} from {{{-6}}} to get {{{-5}}}



{{{m=(-5)/(-6)}}} Subtract {{{-3}}} from {{{-9}}} to get {{{-6}}}



{{{m=5/6}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,-1\right)] and *[Tex \LARGE \left(-9,-6\right)] is {{{m=5/6}}}



Now let's use the point slope formula:



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



{{{y--1=(5/6)(x--3)}}} Plug in {{{m=5/6}}}, {{{x[1]=-3}}}, and {{{y[1]=-1}}}



{{{y--1=(5/6)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y+1=(5/6)(x+3)}}} Rewrite {{{y--1}}} as {{{y+1}}}



{{{y+1=(5/6)x+(5/6)(3)}}} Distribute



{{{y+1=(5/6)x+5/2}}} Multiply



{{{y=(5/6)x+5/2-1}}} Subtract 1 from both sides. 



{{{y=(5/6)x+3/2}}} 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(-3,-1\right)] and *[Tex \LARGE \left(-9,-6\right)] is {{{y=(5/6)x+3/2}}}