Question 96874
First lets find the slope through the points ({{{1}}},{{{-1}}}) and ({{{-2}}},{{{-7}}})


{{{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}}},{{{-1}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{-2}}},{{{-7}}}))


{{{m=(-7--1)/(-2-1)}}} Plug in {{{y[2]=-7}}},{{{y[1]=-1}}},{{{x[2]=-2}}},{{{x[1]=1}}}  (these are the coordinates of given points)


{{{m= -6/-3}}} Subtract the terms in the numerator {{{-7--1}}} to get {{{-6}}}.  Subtract the terms in the denominator {{{-2-1}}} to get {{{-3}}}

  


{{{m=2}}} Reduce

  

So the slope is

{{{m=2}}}


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



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



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


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


{{{y=2x-2-1}}} Subtract {{{1}}} from  both sides to isolate y


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

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Answer:



So the equation of the line which goes through the points ({{{1}}},{{{-1}}}) and ({{{-2}}},{{{-7}}})  is:{{{y=2x-3}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=2}}} and the y-intercept is {{{b=-3}}}


Notice if we graph the equation {{{y=2x-3}}} and plot the points ({{{1}}},{{{-1}}}) and ({{{-2}}},{{{-7}}}),  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, -9.5, 8.5, -13, 5,
graph(500, 500, -9.5, 8.5, -13, 5,(2)x+-3),
circle(1,-1,0.12),
circle(1,-1,0.12+0.03),
circle(-2,-7,0.12),
circle(-2,-7,0.12+0.03)
) }}} Graph of {{{y=2x-3}}} through the points ({{{1}}},{{{-1}}}) and ({{{-2}}},{{{-7}}})


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