Question 235287

First let's find the slope of the line through the points *[Tex \LARGE \left(4,3\right)] and *[Tex \LARGE \left(2,-5\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(4,3\right)]. So this means that {{{x[1]=4}}} and {{{y[1]=3}}}.

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



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



{{{m=(-5-3)/(2-4)}}} Plug in {{{y[2]=-5}}}, {{{y[1]=3}}}, {{{x[2]=2}}}, and {{{x[1]=4}}}



{{{m=(-8)/(2-4)}}} Subtract {{{3}}} from {{{-5}}} to get {{{-8}}}



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



{{{m=4}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(4,3\right)] and *[Tex \LARGE \left(2,-5\right)] is {{{m=4}}}



Now let's use the point slope formula:



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



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



{{{y-3=4x+4(-4)}}} Distribute



{{{y-3=4x-16}}} Multiply



{{{y=4x-16+3}}} Add 3 to both sides. 



{{{y=4x-13}}} Combine like terms. 




So the equation that goes through the points *[Tex \LARGE \left(4,3\right)] and *[Tex \LARGE \left(2,-5\right)] is {{{y=4x-13}}}



 Notice how the graph of {{{y=4x-13}}} goes through the points *[Tex \LARGE \left(4,3\right)] and *[Tex \LARGE \left(2,-5\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,4x-13),
 circle(4,3,0.08),
 circle(4,3,0.10),
 circle(4,3,0.12),
 circle(2,-5,0.08),
 circle(2,-5,0.10),
 circle(2,-5,0.12)
 )}}} Graph of {{{y=4x-13}}} through the points *[Tex \LARGE \left(4,3\right)] and *[Tex \LARGE \left(2,-5\right)]