Question 605330

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



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

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



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



{{{m=(-3)/(-1)}}} Subtract {{{3}}} from {{{2}}} to get {{{-1}}}



{{{m=3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(3,8\right)] and *[Tex \LARGE \left(2,5\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-3)}}} Plug in {{{m=3}}}, {{{x[1]=3}}}, and {{{y[1]=8}}}



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



{{{y-8=3x-9}}} Multiply



{{{y=3x-9+8}}} Add 8 to both sides. 



{{{y=3x-1}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(3,8\right)] and *[Tex \LARGE \left(2,5\right)] is {{{y=3x-1}}}



 Notice how the graph of {{{y=3x-1}}} goes through the points *[Tex \LARGE \left(3,8\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, 10,3x-1),
 circle(3,8,0.08),
 circle(3,8,0.10),
 circle(3,8,0.12),
 circle(2,5,0.08),
 circle(2,5,0.10),
 circle(2,5,0.12)
 )}}} Graph of {{{y=3x-1}}} through the points *[Tex \LARGE \left(3,8\right)] and *[Tex \LARGE \left(2,5\right)]




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