Question 538943


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



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-3,4\right)].  So this means that {{{x[2]=-3}}} and {{{y[2]=4}}}.



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



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



{{{m=(5)/(-3-2)}}} Subtract {{{-1}}} from {{{4}}} to get {{{5}}}



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



{{{m=-1}}} Reduce



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



Now let's use the point slope formula:



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



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



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



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



{{{y+1=-1x+2}}} Multiply



{{{y=-1x+2-1}}} Subtract 1 from both sides. 



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



{{{y=-x+1}}} Simplify



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



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

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



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