Question 750887


First let's find the slope of the line through the points *[Tex \LARGE \left(8,-39\right)] and *[Tex \LARGE \left(1,3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(8,-39\right)]. So this means that {{{x[1]=8}}} and {{{y[1]=-39}}}.

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--39)/(1-8)}}} Plug in {{{y[2]=3}}}, {{{y[1]=-39}}}, {{{x[2]=1}}}, and {{{x[1]=8}}}



{{{m=(42)/(1-8)}}} Subtract {{{-39}}} from {{{3}}} to get {{{42}}}



{{{m=(42)/(-7)}}} Subtract {{{8}}} from {{{1}}} to get {{{-7}}}



{{{m=-6}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(8,-39\right)] and *[Tex \LARGE \left(1,3\right)] is {{{m=-6}}}



Now let's use the point slope formula:



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



{{{y--39=-6(x-8)}}} Plug in {{{m=-6}}}, {{{x[1]=8}}}, and {{{y[1]=-39}}}



{{{y+39=-6(x-8)}}} Rewrite {{{y--39}}} as {{{y+39}}}



{{{y+39=-6x+-6(-8)}}} Distribute



{{{y+39=-6x+48}}} Multiply



{{{y=-6x+48-39}}} Subtract 39 from both sides. 



{{{y=-6x+9}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(8,-39\right)] and *[Tex \LARGE \left(1,3\right)] is {{{y=-6x+9}}}