Question 189702

First let's find the slope of the line through the points *[Tex \LARGE \left(4,1\right)] and *[Tex \LARGE \left(12,3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(4,1\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=(3-1)/(12-4)}}} Plug in {{{y[2]=3}}}, {{{y[1]=1}}}, {{{x[2]=12}}}, and {{{x[1]=4}}}



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



{{{m=(2)/(8)}}} Subtract {{{4}}} from {{{12}}} to get {{{8}}}



{{{m=1/4}}} Reduce



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



Now let's use the point slope formula:



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



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



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



{{{y-1=(1/4)x-1}}} Multiply



{{{y=(1/4)x-1+1}}} Add 1 to both sides. 



{{{y=(1/4)x+0}}} Combine like terms. 



{{{y=(1/4)x}}} Remove the trailing zero




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



 Notice how the graph of {{{y=(1/4)x}}} goes through the points *[Tex \LARGE \left(4,1\right)] and *[Tex \LARGE \left(12,3\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -5, 15, -10, 10,
 graph( 500, 500, -5, 15, -10, 10,(1/4)x),
 circle(4,1,0.08),
 circle(4,1,0.10),
 circle(4,1,0.12),
 circle(12,3,0.08),
 circle(12,3,0.10),
 circle(12,3,0.12)
 )}}} Graph of {{{y=(1/4)x}}} through the points *[Tex \LARGE \left(4,1\right)] and *[Tex \LARGE \left(12,3\right)]