Question 318569


First let's find the slope of the line through the points *[Tex \LARGE \left(-3,1\right)] and *[Tex \LARGE \left(5,4\right)]



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(5,4\right)].  So this means that {{{x[2]=5}}} and {{{y[2]=4}}}.



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



{{{m=(4-1)/(5--3)}}} Plug in {{{y[2]=4}}}, {{{y[1]=1}}}, {{{x[2]=5}}}, and {{{x[1]=-3}}}



{{{m=(3)/(5--3)}}} Subtract {{{1}}} from {{{4}}} to get {{{3}}}



{{{m=(3)/(8)}}} Subtract {{{-3}}} from {{{5}}} to get {{{8}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,1\right)] and *[Tex \LARGE \left(5,4\right)] is {{{m=3/8}}}



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



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



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



{{{y-1=(3/8)x+9/8}}} Multiply



{{{y=(3/8)x+9/8+1}}} Add 1 to both sides. 



{{{y=(3/8)x+17/8}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,1\right)] and *[Tex \LARGE \left(5,4\right)] is {{{y=(3/8)x+17/8}}}