Question 318423


First let's find the slope of the line through the points *[Tex \LARGE \left(\frac{1}{6},-\frac{1}{3}\right)] and *[Tex \LARGE \left(\frac{5}{6},5\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(\frac{1}{6},-\frac{1}{3}\right)]. So this means that {{{x[1]=1/6}}} and {{{y[1]=-1/3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(\frac{5}{6},5\right)].  So this means that {{{x[2]=5/6}}} and {{{y[2]=5}}}.



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



{{{m=(5--1/3)/(5/6-1/6)}}} Plug in {{{y[2]=5}}}, {{{y[1]=-1/3}}}, {{{x[2]=5/6}}}, and {{{x[1]=1/6}}}



{{{m=(16/3)/(5/6-1/6)}}} Subtract {{{-1/3}}} from {{{5}}} to get {{{16/3}}}



{{{m=(16/3)/(2/3)}}} Subtract {{{1/6}}} from {{{5/6}}} to get {{{2/3}}}



{{{m=(16/3)*(3/2)}}} Multiply the first fraction by the reciprocal of the second fraction.



{{{m=8}}} Multiply and reduce.



So the slope of the line that goes through the points *[Tex \LARGE \left(\frac{1}{6},-\frac{1}{3}\right)] and *[Tex \LARGE \left(\frac{5}{6},5\right)] is {{{m=8}}}



Now let's use the point slope formula:



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



{{{y--1/3=8(x-1/6)}}} Plug in {{{m=8}}}, {{{x[1]=1/6}}}, and {{{y[1]=-1/3}}}



{{{y+1/3=8(x-1/6)}}} Rewrite {{{y--1/3}}} as {{{y+1/3}}}



{{{y+1/3=8x+8(-1/6)}}} Distribute



{{{y+1/3=8x-4/3}}} Multiply



{{{y=8x-4/3-1/3}}} Subtract {{{1/3}}} from both sides. 



{{{y=8x-5/3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(\frac{1}{6},-\frac{1}{3}\right)] and *[Tex \LARGE \left(\frac{5}{6},5\right)] is {{{y=8x-5/3}}}