Question 543648

First let's find the slope of the line through the points *[Tex \LARGE \left(4,-3\right)] and *[Tex \LARGE \left(-4,7\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(4,-3\right)]. So this means that {{{x[1]=4}}} and {{{y[1]=-3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-4,7\right)].  So this means that {{{x[2]=-4}}} and {{{y[2]=7}}}.



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



{{{m=(7--3)/(-4-4)}}} Plug in {{{y[2]=7}}}, {{{y[1]=-3}}}, {{{x[2]=-4}}}, and {{{x[1]=4}}}



{{{m=(10)/(-4-4)}}} Subtract {{{-3}}} from {{{7}}} to get {{{10}}}



{{{m=(10)/(-8)}}} Subtract {{{4}}} from {{{-4}}} to get {{{-8}}}



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



So the slope of the line that goes through the points *[Tex \LARGE \left(4,-3\right)] and *[Tex \LARGE \left(-4,7\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--3=(-5/4)(x-4)}}} Plug in {{{m=-5/4}}}, {{{x[1]=4}}}, and {{{y[1]=-3}}}



{{{y+3=(-5/4)(x-4)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=(-5/4)x+(-5/4)(-4)}}} Distribute



{{{y+3=(-5/4)x+5}}} Multiply



{{{y=(-5/4)x+5-3}}} Subtract 3 from both sides. 



{{{y=(-5/4)x+2}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(4,-3\right)] and *[Tex \LARGE \left(-4,7\right)] is {{{y=(-5/4)x+2}}}



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim