Question 275934

First let's find the slope of the line through the points *[Tex \LARGE \left(4,-2\right)] and *[Tex \LARGE \left(3,8\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(3,8\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=8}}}.



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



{{{m=(8--2)/(3-4)}}} Plug in {{{y[2]=8}}}, {{{y[1]=-2}}}, {{{x[2]=3}}}, and {{{x[1]=4}}}



{{{m=(10)/(3-4)}}} Subtract {{{-2}}} from {{{8}}} to get {{{10}}}



{{{m=(10)/(-1)}}} Subtract {{{4}}} from {{{3}}} to get {{{-1}}}



{{{m=-10}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(4,-2\right)] and *[Tex \LARGE \left(3,8\right)] is {{{m=-10}}}



Now let's use the point slope formula:



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



{{{y--2=-10(x-4)}}} Plug in {{{m=-10}}}, {{{x[1]=4}}}, and {{{y[1]=-2}}}



{{{y+2=-10(x-4)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=-10x+-10(-4)}}} Distribute



{{{y+2=-10x+40}}} Multiply



{{{y=-10x+40-2}}} Subtract 2 from both sides. 



{{{y=-10x+38}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(4,-2\right)] and *[Tex \LARGE \left(3,8\right)] is {{{y=-10x+38}}}