Question 110378
{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{13}}},{{{16}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{9}}},y))


{{{2=(y-16)/(9-13)}}} Plug in {{{y[2]=y}}},{{{y[1]=16}}},{{{x[2]=9}}},{{{x[1]=13}}}  (these are the coordinates of given points). Now plug in {{{m=2}}} (this is the given slope)


{{{2= (y-16)/-4}}} Subtract the terms in the denominator {{{9-13}}} to get {{{-4}}}

  


{{{-8=y-16}}} Multiply both sides by -4



{{{8=y}}} Add 16 to both sides




So the value of y is {{{y=8}}}




Notice if we find the slope of ({{{13}}},{{{16}}}) and ({{{9}}},{{{8}}}), we get



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{13}}},{{{16}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{9}}},{{{8}}}))


{{{m=(8-16)/(9-13)}}} Plug in {{{y[2]=8}}},{{{y[1]=16}}},{{{x[2]=9}}},{{{x[1]=13}}}  (these are the coordinates of given points)


{{{m= -8/-4}}} Subtract the terms in the numerator {{{8-16}}} to get {{{-8}}}.  Subtract the terms in the denominator {{{9-13}}} to get {{{-4}}}

  


{{{m=2}}} Reduce

  

So the slope is

{{{m=2}}}



So our answer is verified