Question 512031


First let's find the slope of the line through the points *[Tex \LARGE \left(8,3\right)] and *[Tex \LARGE \left(5,-7\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(8,3\right)]. So this means that {{{x[1]=8}}} and {{{y[1]=3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(5,-7\right)].  So this means that {{{x[2]=5}}} and {{{y[2]=-7}}}.



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



{{{m=(-7-3)/(5-8)}}} Plug in {{{y[2]=-7}}}, {{{y[1]=3}}}, {{{x[2]=5}}}, and {{{x[1]=8}}}



{{{m=(-10)/(5-8)}}} Subtract {{{3}}} from {{{-7}}} to get {{{-10}}}



{{{m=(-10)/(-3)}}} Subtract {{{8}}} from {{{5}}} to get {{{-3}}}



{{{m=10/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(8,3\right)] and *[Tex \LARGE \left(5,-7\right)] is {{{m=10/3}}}



Now let's use the point slope formula:



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



{{{y-3=(10/3)(x-8)}}} Plug in {{{m=10/3}}}, {{{x[1]=8}}}, and {{{y[1]=3}}}



{{{y-3=(10/3)x+(10/3)(-8)}}} Distribute



{{{y-3=(10/3)x-80/3}}} Multiply



{{{y=(10/3)x-80/3+3}}} Add 3 to both sides. 



{{{y=(10/3)x-71/3}}} Combine like terms. 



So the answer is {{{y=(10/3)x-71/3}}}