Question 609962


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



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,9\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=9}}}.

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



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



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



{{{m=(-8)/(1--3)}}} Subtract {{{9}}} from {{{1}}} to get {{{-8}}}



{{{m=(-8)/(4)}}} Subtract {{{-3}}} from {{{1}}} to get {{{4}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,9\right)] and *[Tex \LARGE \left(1,1\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-9=-2(x--3)}}} Plug in {{{m=-2}}}, {{{x[1]=-3}}}, and {{{y[1]=9}}}



{{{y-9=-2(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-9=-2x+-2(3)}}} Distribute



{{{y-9=-2x-6}}} Multiply



{{{y=-2x-6+9}}} Add 9 to both sides. 



{{{y=-2x+3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,9\right)] and *[Tex \LARGE \left(1,1\right)] is {{{y=-2x+3}}}



 Notice how the graph of {{{y=-2x+3}}} goes through the points *[Tex \LARGE \left(-3,9\right)] and *[Tex \LARGE \left(1,1\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x+3),
 circle(-3,9,0.08),
 circle(-3,9,0.10),
 circle(-3,9,0.12),
 circle(1,1,0.08),
 circle(1,1,0.10),
 circle(1,1,0.12)
 )}}} Graph of {{{y=-2x+3}}} through the points *[Tex \LARGE \left(-3,9\right)] and *[Tex \LARGE \left(1,1\right)]

 

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