Question 668989
First let's find the slope of the line through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(14,8\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(6,-2\right)]. So this means that {{{x[1]=6}}} and {{{y[1]=-2}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(14,8\right)].  So this means that {{{x[2]=14}}} and {{{y[2]=8}}}.



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



{{{m=(8--2)/(14-6)}}} Plug in {{{y[2]=8}}}, {{{y[1]=-2}}}, {{{x[2]=14}}}, and {{{x[1]=6}}}



{{{m=(10)/(14-6)}}} Subtract {{{-2}}} from {{{8}}} to get {{{10}}}



{{{m=(10)/(8)}}} Subtract {{{6}}} from {{{14}}} to get {{{8}}}



{{{m=5/4}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(14,8\right)] is {{{m=5/4}}}



Now let's use the point slope formula:



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



{{{y--2=(5/4)(x-6)}}} Plug in {{{m=5/4}}}, {{{x[1]=6}}}, and {{{y[1]=-2}}}



{{{y+2=(5/4)(x-6)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{4(y+2)=5(x-6)}}} Multiply both sides by 4



{{{4y+8=5x-30}}} Distribute



{{{4y=5x-30-8}}} Subtract 8 from both sides.



{{{4y-5x=-30-8}}} Subtract 5x from both sides.



{{{-5x+4y=-38}}} Combine like terms.



{{{5x-4y=38}}} Multiply everything by -1 to make the x coefficient positive



So the answer is {{{5x-4y=38}}}