Question 578759
If you want to find the equation of the line that goes through those points, then...




First let's find the slope of the line through the points *[Tex \LARGE \left(-4,1\right)] and *[Tex \LARGE \left(2,-3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-4,1\right)]. So this means that {{{x[1]=-4}}} and {{{y[1]=1}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(2,-3\right)].  So this means that {{{x[2]=2}}} and {{{y[2]=-3}}}.



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



{{{m=(-3-1)/(2--4)}}} Plug in {{{y[2]=-3}}}, {{{y[1]=1}}}, {{{x[2]=2}}}, and {{{x[1]=-4}}}



{{{m=(-4)/(2--4)}}} Subtract {{{1}}} from {{{-3}}} to get {{{-4}}}



{{{m=(-4)/(6)}}} Subtract {{{-4}}} from {{{2}}} to get {{{6}}}



{{{m=-2/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-4,1\right)] and *[Tex \LARGE \left(2,-3\right)] is {{{m=-2/3}}}



Now let's use the point slope formula:



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



{{{y-1=(-2/3)(x--4)}}} Plug in {{{m=-2/3}}}, {{{x[1]=-4}}}, and {{{y[1]=1}}}



{{{y-1=(-2/3)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y-1=(-2/3)x+(-2/3)(4)}}} Distribute



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



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



{{{y=(-2/3)x-5/3}}} 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(-4,1\right)] and *[Tex \LARGE \left(2,-3\right)] is {{{y=(-2/3)x-5/3}}}