Question 385761

First let's find the slope of the line through the points *[Tex \LARGE \left(2,-7\right)] and *[Tex \LARGE \left(4,-13\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(2,-7\right)]. So this means that {{{x[1]=2}}} and {{{y[1]=-7}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(4,-13\right)].  So this means that {{{x[2]=4}}} and {{{y[2]=-13}}}.



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



{{{m=(-13--7)/(4-2)}}} Plug in {{{y[2]=-13}}}, {{{y[1]=-7}}}, {{{x[2]=4}}}, and {{{x[1]=2}}}



{{{m=(-6)/(4-2)}}} Subtract {{{-7}}} from {{{-13}}} to get {{{-6}}}



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



{{{m=-3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(2,-7\right)] and *[Tex \LARGE \left(4,-13\right)] is {{{m=-3}}}



Now let's use the point slope formula:



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



{{{y--7=-3(x-2)}}} Plug in {{{m=-3}}}, {{{x[1]=2}}}, and {{{y[1]=-7}}}



{{{y+7=-3(x-2)}}} Rewrite {{{y--7}}} as {{{y+7}}}



{{{y+7=-3x+-3(-2)}}} Distribute



{{{y+7=-3x+6}}} Multiply



{{{y=-3x+6-7}}} Subtract 7 from both sides. 



{{{y=-3x-1}}} Combine like terms. 




So the equation that goes through the points *[Tex \LARGE \left(2,-7\right)] and *[Tex \LARGE \left(4,-13\right)] is {{{y=-3x-1}}}



 Notice how the graph of {{{y=-3x-1}}} goes through the points *[Tex \LARGE \left(2,-7\right)] and *[Tex \LARGE \left(4,-13\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -15, 5,
 graph( 500, 500, -10, 10, -15, 5,-3x-1),
 circle(2,-7,0.08),
 circle(2,-7,0.10),
 circle(2,-7,0.12),
 circle(4,-13,0.08),
 circle(4,-13,0.10),
 circle(4,-13,0.12)
 )}}} Graph of {{{y=-3x-1}}} through the points *[Tex \LARGE \left(2,-7\right)] and *[Tex \LARGE \left(4,-13\right)]

 

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