Question 537672

First let's find the slope of the line through the points *[Tex \LARGE \left(1,7\right)] and *[Tex \LARGE \left(3,3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(1,7\right)]. So this means that {{{x[1]=1}}} and {{{y[1]=7}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,3\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=3}}}.



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



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



{{{m=(-4)/(3-1)}}} Subtract {{{7}}} from {{{3}}} to get {{{-4}}}



{{{m=(-4)/(2)}}} Subtract {{{1}}} from {{{3}}} to get {{{2}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(1,7\right)] and *[Tex \LARGE \left(3,3\right)] is {{{m=-2}}}



Now let's use the point slope formula:



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



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



{{{y-7=-2x+-2(-1)}}} Distribute



{{{y-7=-2x+2}}} Multiply



{{{y=-2x+2+7}}} Add 7 to both sides. 



{{{y=-2x+9}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(1,7\right)] and *[Tex \LARGE \left(3,3\right)] is {{{y=-2x+9}}}



 Notice how the graph of {{{y=-2x+9}}} goes through the points *[Tex \LARGE \left(1,7\right)] and *[Tex \LARGE \left(3,3\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x+9),
 circle(1,7,0.08),
 circle(1,7,0.10),
 circle(1,7,0.12),
 circle(3,3,0.08),
 circle(3,3,0.10),
 circle(3,3,0.12)
 )}}} Graph of {{{y=-2x+9}}} through the points *[Tex \LARGE \left(1,7\right)] and *[Tex \LARGE \left(3,3\right)]



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