Question 118262
First lets find the slope through the points ({{{-1}}},{{{-7}}}) and ({{{1}}},{{{3}}})


{{{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}}},{{{-7}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{1}}},{{{3}}}))


{{{m=(3--7)/(1--1)}}} Plug in {{{y[2]=3}}},{{{y[1]=-7}}},{{{x[2]=1}}},{{{x[1]=-1}}}  (these are the coordinates of given points)


{{{m= 10/2}}} Subtract the terms in the numerator {{{3--7}}} to get {{{10}}}.  Subtract the terms in the denominator {{{1--1}}} to get {{{2}}}

  


{{{m=5}}} Reduce

  

So the slope is

{{{m=5}}}


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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--7=(5)(x--1)}}} Plug in {{{m=5}}}, {{{x[1]=-1}}}, and {{{y[1]=-7}}} (these values are given)



{{{y+7=(5)(x--1)}}} Rewrite {{{y--7}}} as {{{y+7}}}



{{{y+7=(5)(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y+7=5x+(5)(1)}}} Distribute {{{5}}}


{{{y+7=5x+5}}} Multiply {{{5}}} and {{{1}}} to get {{{5}}}


{{{y=5x+5-7}}} Subtract {{{7}}} from  both sides to isolate y


{{{y=5x-2}}} Combine like terms {{{5}}} and {{{-7}}} to get {{{-2}}} 

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Answer:



So the equation of the line which goes through the points ({{{-1}}},{{{-7}}}) and ({{{1}}},{{{3}}})  is:{{{y=5x-2}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=5}}} and the y-intercept is {{{b=-2}}}


Notice if we graph the equation {{{y=5x-2}}} and plot the points ({{{-1}}},{{{-7}}}) and ({{{1}}},{{{3}}}),  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, -11, 7,
graph(500, 500, -9, 9, -11, 7,(5)x+-2),
circle(-1,-7,0.12),
circle(-1,-7,0.12+0.03),
circle(1,3,0.12),
circle(1,3,0.12+0.03)
) }}} Graph of {{{y=5x-2}}} through the points ({{{-1}}},{{{-7}}}) and ({{{1}}},{{{3}}})


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