Question 189711

First let's find the slope of the line through the points *[Tex \LARGE \left(4,2\right)] and *[Tex \LARGE \left(12,3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(4,2\right)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(12,3\right)].



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



{{{m=(3-2)/(12-4)}}} Plug in {{{y[2]=3}}}, {{{y[1]=2}}}, {{{x[2]=12}}}, and {{{x[1]=4}}}



{{{m=(1)/(12-4)}}} Subtract {{{2}}} from {{{3}}} to get {{{1}}}



{{{m=(1)/(8)}}} Subtract {{{4}}} from {{{12}}} to get {{{8}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(4,2\right)] and *[Tex \LARGE \left(12,3\right)] is {{{m=1/8}}}



Now let's use the point slope formula:



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



{{{y-2=(1/8)(x-4)}}} Plug in {{{m=1/8}}}, {{{x[1]=4}}}, and {{{y[1]=2}}}



{{{y-2=(1/8)x+(1/8)(-4)}}} Distribute



{{{y-2=(1/8)x-1/2}}} Multiply



{{{y=(1/8)x-1/2+2}}} Add 2 to both sides. 



{{{y=(1/8)x+3/2}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(4,2\right)] and *[Tex \LARGE \left(12,3\right)] is {{{y=(1/8)x+3/2}}}