Question 122562
First lets find the slope through the points ({{{-2}}},{{{3}}}) and ({{{2}}},{{{-1}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{-2}}},{{{3}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{2}}},{{{-1}}}))


{{{m=(-1-3)/(2--2)}}} Plug in {{{y[2]=-1}}},{{{y[1]=3}}},{{{x[2]=2}}},{{{x[1]=-2}}}  (these are the coordinates of given points)


{{{m= -4/4}}} Subtract the terms in the numerator {{{-1-3}}} to get {{{-4}}}.  Subtract the terms in the denominator {{{2--2}}} to get {{{4}}}

  


{{{m=-1}}} Reduce

  

So the slope is

{{{m=-1}}}


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Now let's use the point-slope formula to find the equation of the line:




------Point-Slope Formula------
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(\textrm{x_{1},y_{1}}\right)] is one of the given points


So lets use the Point-Slope Formula to find the equation of the line


{{{y-3=(-1)(x--2)}}} Plug in {{{m=-1}}}, {{{x[1]=-2}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=(-1)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-3=-x+(-1)(2)}}} Distribute {{{-1}}}


{{{y-3=-x-2}}} Multiply {{{-1}}} and {{{2}}} to get {{{-2}}}


{{{y=-x-2+3}}} Add {{{3}}} to  both sides to isolate y


{{{y=-x+1}}} Combine like terms {{{-2}}} and {{{3}}} to get {{{1}}} 

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Answer:



So the equation of the line which goes through the points ({{{-2}}},{{{3}}}) and ({{{2}}},{{{-1}}})  is:{{{y=-x+1}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=-1}}} and the y-intercept is {{{b=1}}}


Notice if we graph the equation {{{y=-x+1}}} and plot the points ({{{-2}}},{{{3}}}) and ({{{2}}},{{{-1}}}),  we get this: (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -9, 9, -8, 10,
graph(500, 500, -9, 9, -8, 10,(-1)x+1),
circle(-2,3,0.12),
circle(-2,3,0.12+0.03),
circle(2,-1,0.12),
circle(2,-1,0.12+0.03)
) }}} Graph of {{{y=-x+1}}} through the points ({{{-2}}},{{{3}}}) and ({{{2}}},{{{-1}}})


Notice how the two points lie on the line. This graphically verifies our answer.