Question 499813


First let's find the slope of the line through the points *[Tex \LARGE \left(-1,4\right)] and *[Tex \LARGE \left(5,-8\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-1,4\right)]. So this means that {{{x[1]=-1}}} and {{{y[1]=4}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(5,-8\right)].  So this means that {{{x[2]=5}}} and {{{y[2]=-8}}}.



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



{{{m=(-8-4)/(5--1)}}} Plug in {{{y[2]=-8}}}, {{{y[1]=4}}}, {{{x[2]=5}}}, and {{{x[1]=-1}}}



{{{m=(-12)/(5--1)}}} Subtract {{{4}}} from {{{-8}}} to get {{{-12}}}



{{{m=(-12)/(6)}}} Subtract {{{-1}}} from {{{5}}} to get {{{6}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-1,4\right)] and *[Tex \LARGE \left(5,-8\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-4=-2(x--1)}}} Plug in {{{m=-2}}}, {{{x[1]=-1}}}, and {{{y[1]=4}}}



{{{y-4=-2(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y-4=-2x+-2(1)}}} Distribute



{{{y-4=-2x-2}}} Multiply



{{{y=-2x-2+4}}} Add 4 to both sides. 



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



So the equation that goes through the points *[Tex \LARGE \left(-1,4\right)] and *[Tex \LARGE \left(5,-8\right)] is {{{y=-2x+2}}}



 Notice how the graph of {{{y=-2x+2}}} goes through the points *[Tex \LARGE \left(-1,4\right)] and *[Tex \LARGE \left(5,-8\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x+2),
 circle(-1,4,0.08),
 circle(-1,4,0.10),
 circle(-1,4,0.12),
 circle(5,-8,0.08),
 circle(5,-8,0.10),
 circle(5,-8,0.12)
 )}}} Graph of {{{y=-2x+2}}} through the points *[Tex \LARGE \left(-1,4\right)] and *[Tex \LARGE \left(5,-8\right)]

 

If you need more help, feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>


Thanks,


Jim