Question 729021


First let's find the slope of the line through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(-4,-5\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(1,5\right)]. So this means that {{{x[1]=1}}} and {{{y[1]=5}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-4,-5\right)].  So this means that {{{x[2]=-4}}} and {{{y[2]=-5}}}.



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



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



{{{m=(-10)/(-4-1)}}} Subtract {{{5}}} from {{{-5}}} to get {{{-10}}}



{{{m=(-10)/(-5)}}} Subtract {{{1}}} from {{{-4}}} to get {{{-5}}}



{{{m=2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(-4,-5\right)] is {{{m=2}}}



Now let's use the point slope formula:



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



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



{{{y-5=2x+2(-1)}}} Distribute



{{{y-5=2x-2}}} Multiply



{{{y=2x-2+5}}} Add 5 to both sides. 



{{{y=2x+3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(-4,-5\right)] is {{{y=2x+3}}}



 Notice how the graph of {{{y=2x+3}}} goes through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(-4,-5\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,2x+3),
 circle(1,5,0.08),
 circle(1,5,0.10),
 circle(1,5,0.12),
 circle(-4,-5,0.08),
 circle(-4,-5,0.10),
 circle(-4,-5,0.12)
 )}}} Graph of {{{y=2x+3}}} through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(-4,-5\right)]