Question 530233


First let's find the slope of the line through the points *[Tex \LARGE \left(-2,2\right)] and *[Tex \LARGE \left(2,4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-2,2\right)]. So this means that {{{x[1]=-2}}} and {{{y[1]=2}}}.

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



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



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



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



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



{{{m=1/2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-2,2\right)] and *[Tex \LARGE \left(2,4\right)] is {{{m=1/2}}}



Now let's use the point slope formula:



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



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



{{{y-2=(1/2)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-2=(1/2)x+(1/2)(2)}}} Distribute



{{{y-2=(1/2)x+1}}} Multiply



{{{y=(1/2)x+1+2}}} Add 2 to both sides. 



{{{y=(1/2)x+3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-2,2\right)] and *[Tex \LARGE \left(2,4\right)] is {{{y=(1/2)x+3}}}



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