Question 117836
First lets find the slope through the points ({{{-1}}},{{{3}}}) and ({{{4}}},{{{-2}}})


{{{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 ({{{-1}}},{{{3}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{4}}},{{{-2}}}))


{{{m=(-2-3)/(4--1)}}} Plug in {{{y[2]=-2}}},{{{y[1]=3}}},{{{x[2]=4}}},{{{x[1]=-1}}}  (these are the coordinates of given points)


{{{m= -5/5}}} Subtract the terms in the numerator {{{-2-3}}} to get {{{-5}}}.  Subtract the terms in the denominator {{{4--1}}} to get {{{5}}}

  


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



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



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


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


{{{y=-x-1+3}}} Add {{{3}}} to  both sides to isolate y


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

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Answer:



So the equation of the line which goes through the points ({{{-1}}},{{{3}}}) and ({{{4}}},{{{-2}}})  is:{{{y=-x+2}}}


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=2}}}


Notice if we graph the equation {{{y=-x+2}}} and plot the points ({{{-1}}},{{{3}}}) and ({{{4}}},{{{-2}}}),  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, -7.5, 10.5, -8.5, 9.5,
graph(500, 500, -7.5, 10.5, -8.5, 9.5,(-1)x+2),
circle(-1,3,0.12),
circle(-1,3,0.12+0.03),
circle(4,-2,0.12),
circle(4,-2,0.12+0.03)
) }}} Graph of {{{y=-x+2}}} through the points ({{{-1}}},{{{3}}}) and ({{{4}}},{{{-2}}})


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