Question 293607


First let's find the slope of the line through the points *[Tex \LARGE \left(5,-3\right)] and *[Tex \LARGE \left(8,4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(5,-3\right)]. So this means that {{{x[1]=5}}} and {{{y[1]=-3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(8,4\right)].  So this means that {{{x[2]=8}}} and {{{y[2]=4}}}.



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



{{{m=(4--3)/(8-5)}}} Plug in {{{y[2]=4}}}, {{{y[1]=-3}}}, {{{x[2]=8}}}, and {{{x[1]=5}}}



{{{m=(7)/(8-5)}}} Subtract {{{-3}}} from {{{4}}} to get {{{7}}}



{{{m=(7)/(3)}}} Subtract {{{5}}} from {{{8}}} to get {{{3}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(5,-3\right)] and *[Tex \LARGE \left(8,4\right)] is {{{m=7/3}}}



Now let's use the point slope formula:



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



{{{y--3=(7/3)(x-5)}}} Plug in {{{m=7/3}}}, {{{x[1]=5}}}, and {{{y[1]=-3}}}



{{{y+3=(7/3)(x-5)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=(7/3)x+(7/3)(-5)}}} Distribute



{{{y+3=(7/3)x-35/3}}} Multiply



{{{y=(7/3)x-35/3-3}}} Subtract 3 from both sides. 



{{{y=(7/3)x-44/3}}} Combine like terms. note: If you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>.



So the equation that goes through the points *[Tex \LARGE \left(5,-3\right)] and *[Tex \LARGE \left(8,4\right)] is {{{y=(7/3)x-44/3}}}