Question 215098


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



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

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



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



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



{{{m=-1}}} Reduce



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



{{{y-3=-1(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y-3=-1x+-1(1)}}} Distribute



{{{y-3=-1x-1}}} Multiply



{{{y=-1x-1+3}}} Add 3 to both sides. 



{{{y=-1x+2}}} Combine like terms. 



{{{y=-x+2}}} Simplify



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



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