Question 315602


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



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

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



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



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



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



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



{{{m=1}}} Reduce



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



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



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



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



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



{{{y=1x+4}}} Combine like terms. 



{{{y=x+4}}} Simplify



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



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

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