Question 613005


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



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-4,-2\right)]. So this means that {{{x[1]=-4}}} and {{{y[1]=-2}}}.

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



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



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



{{{m=1}}} Reduce



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



{{{y--2=1(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y+2=1(x+4)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=1x+1(4)}}} Distribute



{{{y+2=1x+4}}} Multiply



{{{y=1x+4-2}}} Subtract 2 from both sides. 



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



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



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



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