Question 197569


First let's find the slope of the line through the points *[Tex \LARGE \left(5,8\right)] and *[Tex \LARGE \left(3,2\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(5,8\right)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,2\right)].



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



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



{{{m=(-6)/(3-5)}}} Subtract {{{8}}} from {{{2}}} to get {{{-6}}}



{{{m=(-6)/(-2)}}} Subtract {{{5}}} from {{{3}}} to get {{{-2}}}



{{{m=3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(5,8\right)] and *[Tex \LARGE \left(3,2\right)] is {{{m=3}}}



Now let's use the point slope formula:



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



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



{{{y-8=3x+3(-5)}}} Distribute



{{{y-8=3x-15}}} Multiply



{{{y=3x-15+8}}} Add 8 to both sides. 



{{{y=3x-7}}} Combine like terms. 



{{{y=3x-7}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(5,8\right)] and *[Tex \LARGE \left(3,2\right)] is {{{y=3x-7}}}



 Notice how the graph of {{{y=3x-7}}} goes through the points *[Tex \LARGE \left(5,8\right)] and *[Tex \LARGE \left(3,2\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,3x-7),
 circle(5,8,0.08),
 circle(5,8,0.10),
 circle(5,8,0.12),
 circle(3,2,0.08),
 circle(3,2,0.10),
 circle(3,2,0.12)
 )}}} Graph of {{{y=3x-7}}} through the points *[Tex \LARGE \left(5,8\right)] and *[Tex \LARGE \left(3,2\right)]