Question 140258
First lets find the slope through the points ({{{2}}},{{{-3}}}) and ({{{-5}}},{{{4}}})


{{{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 ({{{-5}}},{{{4}}}))


{{{m=(4--3)/(-5-2)}}} Plug in {{{y[2]=4}}},{{{y[1]=-3}}},{{{x[2]=-5}}},{{{x[1]=2}}}  (these are the coordinates of given points)


{{{m= 7/-7}}} Subtract the terms in the numerator {{{4--3}}} to get {{{7}}}.  Subtract the terms in the denominator {{{-5-2}}} to get {{{-7}}}

  


{{{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 {{{y--3}}} as {{{y+3}}}



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


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


{{{y=-x+2-3}}} Subtract {{{3}}} from  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 ({{{-5}}},{{{4}}})  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 ({{{-5}}},{{{4}}}),  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, -10.5, 7.5, -8.5, 9.5,
graph(500, 500, -10.5, 7.5, -8.5, 9.5,(-1)x+-1),
circle(2,-3,0.12),
circle(2,-3,0.12+0.03),
circle(-5,4,0.12),
circle(-5,4,0.12+0.03)
) }}} Graph of {{{y=-x-1}}} through the points ({{{2}}},{{{-3}}}) and ({{{-5}}},{{{4}}})


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