Question 367342

First let's find the slope of the line through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(-3,-2\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(3,0\right)]. So this means that {{{x[1]=3}}} and {{{y[1]=0}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-3,-2\right)].  So this means that {{{x[2]=-3}}} and {{{y[2]=-2}}}.



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



{{{m=(-2-0)/(-3-3)}}} Plug in {{{y[2]=-2}}}, {{{y[1]=0}}}, {{{x[2]=-3}}}, and {{{x[1]=3}}}



{{{m=(-2)/(-3-3)}}} Subtract {{{0}}} from {{{-2}}} to get {{{-2}}}



{{{m=(-2)/(-6)}}} Subtract {{{3}}} from {{{-3}}} to get {{{-6}}}



{{{m=1/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(-3,-2\right)] is {{{m=1/3}}}



Now let's use the point slope formula:



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



{{{y-0=(1/3)(x-3)}}} Plug in {{{m=1/3}}}, {{{x[1]=3}}}, and {{{y[1]=0}}}



{{{y-0=(1/3)x+(1/3)(-3)}}} Distribute



{{{y-0=(1/3)x-1}}} Multiply



{{{y=(1/3)x-1}}} Simplify





So the equation that goes through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(-3,-2\right)] is {{{y=(1/3)x-1}}}



 Notice how the graph of {{{y=(1/3)x-1}}} goes through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(-3,-2\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(1/3)x-1),
 circle(3,0,0.08),
 circle(3,0,0.10),
 circle(3,0,0.12),
 circle(-3,-2,0.08),
 circle(-3,-2,0.10),
 circle(-3,-2,0.12)
 )}}} Graph of {{{y=(1/3)x-1}}} through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(-3,-2\right)]

 

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


Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>


Jim