Question 112351
Do you want to find the equation through these two points? If you do, then to find the equation of the line, we need to find the slope through the two points



{{{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 ({{{6}}},{{{1}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{4}}},{{{1}}}))


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


{{{m= 0/-2}}} Subtract the terms in the numerator {{{1-1}}} to get {{{0}}}.  Subtract the terms in the denominator {{{4-6}}} to get {{{-2}}}

  


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



{{{y-1=0x+(0)(-6)}}} Distribute {{{0}}}


{{{y-1=0x+0}}} Multiply {{{0}}} and {{{-6}}} to get {{{0/0}}}. Now reduce {{{0/0}}} to get {{{0}}}


{{{y=0x+0+1}}} Add {{{1}}} to  both sides to isolate y


{{{y=0x+1}}} Combine like terms {{{0}}} and {{{1}}} to get {{{1}}} 

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Answer:



So the equation of the line which goes through the points ({{{6}}},{{{1}}}) and ({{{4}}},{{{1}}})  is:{{{y=0x+1}}}


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=1}}}


Notice if we graph the equation {{{y=0x+1}}} and plot the points ({{{6}}},{{{1}}}) and ({{{4}}},{{{1}}}),  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, -4, 14, -8, 10,
graph(500, 500, -4, 14, -8, 10,(0)x+1),
circle(6,1,0.12),
circle(6,1,0.12+0.03),
circle(4,1,0.12),
circle(4,1,0.12+0.03)
) }}} Graph of {{{y=0x+1}}} through the points ({{{6}}},{{{1}}}) and ({{{4}}},{{{1}}})


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