Question 389363
First let's find the slope of the line through the points *[Tex \LARGE \left(-1,0\right)] and *[Tex \LARGE \left(3,10\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-1,0\right)]. So this means that {{{x[1]=-1}}} and {{{y[1]=0}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,10\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=10}}}.



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



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



{{{m=(10)/(3--1)}}} Subtract {{{0}}} from {{{10}}} to get {{{10}}}



{{{m=(10)/(4)}}} Subtract {{{-1}}} from {{{3}}} to get {{{4}}}



{{{m=5/2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-1,0\right)] and *[Tex \LARGE \left(3,10\right)] is {{{m=5/2}}}



Now let's use the point slope formula:



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



{{{y-0=(5/2)(x--1)}}} Plug in {{{m=5/2}}}, {{{x[1]=-1}}}, and {{{y[1]=0}}}



{{{y-0=(5/2)(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y=(5/2)(x+1)}}} Simplify



{{{2y=5(x+1)}}} Multiply both sides by 2.



{{{2y=5x+5}}} Distribute.



{{{-5x+2y=5}}} Subtract 5x from both sides.



{{{5x-2y=-5}}} Multiply EVERY term by -1 to make the x coefficient positive.



So the equation in standard form is {{{5x-2y=-5}}}



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