Question 395379


First let's find the slope of the line through the points *[Tex \LARGE \left(-3,-2\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,-2\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=-2}}}.

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



{{{m=(6)/(5--3)}}} Subtract {{{-2}}} from {{{4}}} to get {{{6}}}



{{{m=(6)/(8)}}} Subtract {{{-3}}} from {{{5}}} to get {{{8}}}



{{{m=3/4}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,-2\right)] and *[Tex \LARGE \left(5,4\right)] is {{{m=3/4}}}



Now let's use the point slope formula:



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



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



{{{y--2=(3/4)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y+2=(3/4)(x+3)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(3/4)x+(3/4)(3)}}} Distribute



{{{y+2=(3/4)x+9/4}}} Multiply



{{{y=(3/4)x+9/4-2}}} Subtract 2 from both sides. 



{{{y=(3/4)x+1/4}}} 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(-3,-2\right)] and *[Tex \LARGE \left(5,4\right)] is {{{y=(3/4)x+1/4}}}



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