Question 214767

First let's find the slope of the line through the points *[Tex \LARGE \left(1,3\right)] and *[Tex \LARGE \left(3,4\right)]

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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,4\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=4}}}.

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

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

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

So the slope of the line that goes through the points *[Tex \LARGE \left(1,3\right)] and *[Tex \LARGE \left(3,4\right)] is {{{m=1/2}}}

Now let's use the point slope formula:

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

{{{y-3=(1/2)x+(1/2)(-1)}}} Distribute

{{{y-3=(1/2)x-1/2}}} Multiply

{{{y=(1/2)x-1/2+3}}} Add 3 to both sides.

{{{y=(1/2)x+5/2}}} Combine like terms.

So the equation that goes through the points *[Tex \LARGE \left(1,3\right)] and *[Tex \LARGE \left(3,4\right)] is {{{y=(1/2)x+5/2}}}