Question 121054
First lets find the slope through the points ({{{-2}}},{{{-6}}}) and ({{{4}}},{{{-6}}})


{{{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}}},{{{-6}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{4}}},{{{-6}}}))


{{{m=(-6--6)/(4--2)}}} Plug in {{{y[2]=-6}}},{{{y[1]=-6}}},{{{x[2]=4}}},{{{x[1]=-2}}}  (these are the coordinates of given points)


{{{m= 0/6}}} Subtract the terms in the numerator {{{-6--6}}} to get {{{0}}}.  Subtract the terms in the denominator {{{4--2}}} to get {{{6}}}

  


{{{m=0}}} Reduce

  

So the slope is

{{{m=0}}}


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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--6=(0)(x--2)}}} Plug in {{{m=0}}}, {{{x[1]=-2}}}, and {{{y[1]=-6}}} (these values are given)



{{{y+6=(0)(x--2)}}} Rewrite {{{y--6}}} as {{{y+6}}}



{{{y+6=(0)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y+6=0x+(0)(2)}}} Distribute {{{0}}}


{{{y+6=0x+0}}} Multiply {{{0}}} and {{{2}}} to get {{{0}}}


{{{y=0x+0-6}}} Subtract {{{6}}} from  both sides to isolate y


{{{y=0x-6}}} Combine like terms {{{0}}} and {{{-6}}} to get {{{-6}}} 


{{{y=-6}}} Remove the zero term

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Answer:



So the equation of the line which goes through the points ({{{-2}}},{{{-6}}}) and ({{{4}}},{{{-6}}})  is:{{{y=-6}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=0}}} and the y-intercept is {{{b=-6}}}


Notice if we graph the equation {{{y=-6}}} and plot the points ({{{-2}}},{{{-6}}}) and ({{{4}}},{{{-6}}}),  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, -8, 10, -15, 3,
graph(500, 500, -8, 10, -15, 3,(0)x+-6),
circle(-2,-6,0.12),
circle(-2,-6,0.12+0.03),
circle(4,-6,0.12),
circle(4,-6,0.12+0.03)
) }}} Graph of {{{y=-6}}} through the points ({{{-2}}},{{{-6}}}) and ({{{4}}},{{{-6}}})


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