Question 660551


First let's find the slope of the line through the points *[Tex \LARGE \left(-2,-1\right)] and *[Tex \LARGE \left(-4,-3\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(-4,-3\right)].  So this means that {{{x[2]=-4}}} and {{{y[2]=-3}}}.



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



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



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



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



{{{m=1}}} Reduce



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



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



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


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