Question 745226


First let's find the slope of the line through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(3,4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-6,-2\right)]. So this means that {{{x[1]=-6}}} and {{{y[1]=-2}}}.

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--2)/(3--6)}}} Plug in {{{y[2]=4}}}, {{{y[1]=-2}}}, {{{x[2]=3}}}, and {{{x[1]=-6}}}



{{{m=(6)/(3--6)}}} Subtract {{{-2}}} from {{{4}}} to get {{{6}}}



{{{m=(6)/(9)}}} Subtract {{{-6}}} from {{{3}}} to get {{{9}}}



{{{m=2/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(3,4\right)] is {{{m=2/3}}}



Now let's use the point slope formula:



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



{{{y--2=(2/3)(x--6)}}} Plug in {{{m=2/3}}}, {{{x[1]=-6}}}, and {{{y[1]=-2}}}



{{{y--2=(2/3)(x+6)}}} Rewrite {{{x--6}}} as {{{x+6}}}



{{{y+2=(2/3)(x+6)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(2/3)x+(2/3)(6)}}} Distribute



{{{y+2=(2/3)x+4}}} Multiply



{{{y=(2/3)x+4-2}}} Subtract 2 from both sides. 



{{{y=(2/3)x+2}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(3,4\right)] is {{{y=(2/3)x+2}}}