Question 696675


First let's find the slope of the line through the points *[Tex \LARGE \left(-3,105\right)] and *[Tex \LARGE \left(6,-3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,105\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=105}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(6,-3\right)].  So this means that {{{x[2]=6}}} and {{{y[2]=-3}}}.



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



{{{m=(-3-105)/(6--3)}}} Plug in {{{y[2]=-3}}}, {{{y[1]=105}}}, {{{x[2]=6}}}, and {{{x[1]=-3}}}



{{{m=(-108)/(6--3)}}} Subtract {{{105}}} from {{{-3}}} to get {{{-108}}}



{{{m=(-108)/(9)}}} Subtract {{{-3}}} from {{{6}}} to get {{{9}}}



{{{m=-12}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,105\right)] and *[Tex \LARGE \left(6,-3\right)] is {{{m=-12}}}



Now let's use the point slope formula:



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



{{{y-105=-12(x--3)}}} Plug in {{{m=-12}}}, {{{x[1]=-3}}}, and {{{y[1]=105}}}



{{{y-105=-12(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-105=-12x+-12(3)}}} Distribute



{{{y-105=-12x-36}}} Multiply



{{{y=-12x-36+105}}} Add 105 to both sides. 



{{{y=-12x+69}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,105\right)] and *[Tex \LARGE \left(6,-3\right)] is {{{y=-12x+69}}}