Question 718936


First let's find the slope of the line through the points *[Tex \LARGE \left(-1,3\right)] and *[Tex \LARGE \left(2,-5\right)]



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

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



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



{{{m=(-5-3)/(2--1)}}} Plug in {{{y[2]=-5}}}, {{{y[1]=3}}}, {{{x[2]=2}}}, and {{{x[1]=-1}}}



{{{m=(-8)/(2--1)}}} Subtract {{{3}}} from {{{-5}}} to get {{{-8}}}



{{{m=(-8)/(3)}}} Subtract {{{-1}}} from {{{2}}} to get {{{3}}}



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



Now let's use the point slope formula:



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



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



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



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



{{{y-3=(-8/3)x-8/3}}} Multiply



{{{y=(-8/3)x-8/3+3}}} Add 3 to both sides. 



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