Question 913051

First let's find the slope of the line through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(2,-8\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,2\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=2}}}.

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=(-8-2)/(2--3)}}} Plug in {{{y[2]=-8}}}, {{{y[1]=2}}}, {{{x[2]=2}}}, and {{{x[1]=-3}}}



{{{m=(-10)/(2--3)}}} Subtract {{{2}}} from {{{-8}}} to get {{{-10}}}



{{{m=(-10)/(5)}}} Subtract {{{-3}}} from {{{2}}} to get {{{5}}}



{{{m=-2}}} Reduce



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



Now let's use the point slope formula:



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



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



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



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



{{{y-2=-2x-6}}} Multiply



{{{y=-2x-6+2}}} Add 2 to both sides. 



{{{y=-2x-4}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(2,-8\right)] is {{{y=-2x-4}}}



 Notice how the graph of {{{y=-2x-4}}} goes through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(2,-8\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,0,-2x-4),
 circle(-3,2,0.08),
 circle(-3,2,0.10),
 circle(-3,2,0.12),
 circle(2,-8,0.08),
 circle(2,-8,0.10),
 circle(2,-8,0.12)
 )}}} Graph of {{{y=-2x-4}}} through the points *[Tex \LARGE \left(-3,2\right)] and *[Tex \LARGE \left(2,-8\right)]

 
Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,


Jim