Question 555808


First let's find the slope of the line through the points *[Tex \LARGE \left(-3,12\right)] and *[Tex \LARGE \left(0,10\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,12\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=12}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(0,10\right)].  So this means that {{{x[2]=0}}} and {{{y[2]=10}}}.



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



{{{m=(10-12)/(0--3)}}} Plug in {{{y[2]=10}}}, {{{y[1]=12}}}, {{{x[2]=0}}}, and {{{x[1]=-3}}}



{{{m=(-2)/(0--3)}}} Subtract {{{12}}} from {{{10}}} to get {{{-2}}}



{{{m=(-2)/(3)}}} Subtract {{{-3}}} from {{{0}}} to get {{{3}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,12\right)] and *[Tex \LARGE \left(0,10\right)] is {{{m=-2/3}}}



Now let's use the point slope formula:



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



{{{y-12=(-2/3)(x--3)}}} Plug in {{{m=-2/3}}}, {{{x[1]=-3}}}, and {{{y[1]=12}}}



{{{y-12=(-2/3)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-12=(-2/3)x+(-2/3)(3)}}} Distribute



{{{y-12=(-2/3)x-2}}} Multiply



{{{y=(-2/3)x-2+12}}} Add 12 to both sides. 



{{{y=(-2/3)x+10}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,12\right)] and *[Tex \LARGE \left(0,10\right)] is {{{y=(-2/3)x+10}}}



The equation is now in y = mx+b form where {{{m=-2/3}}} and {{{b=10}}}



So the slope is {{{-2/3}}} and the y-intercept is (0,10)


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