Question 117626
a)


First lets find the slope through the points ({{{-2}}},{{{3}}}) 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 ({{{-2}}},{{{3}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{1}}},{{{3}}}))


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


{{{m= 0/3}}} Subtract the terms in the numerator {{{3-3}}} to get {{{0}}}.  Subtract the terms in the denominator {{{1--2}}} to get {{{3}}}

  


{{{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-3=(0)(x--2)}}} Plug in {{{m=0}}}, {{{x[1]=-2}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=(0)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-3=0x+(0)(2)}}} Distribute {{{0}}}


{{{y-3=0x+0}}} Multiply {{{0}}} and {{{2}}} to get {{{0/0}}}. Now reduce {{{0/0}}} to get {{{0}}}


{{{y=0x+0+3}}} Add {{{3}}} to  both sides to isolate y


{{{y=0x+3}}} Combine like terms {{{0}}} and {{{3}}} to get {{{3}}} 



{{{y=3}}} Remove the zero term

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Answer:



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


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=3}}}


Notice if we graph the equation {{{y=3}}} and plot the points ({{{-2}}},{{{3}}}) 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, -9.5, 8.5, -6, 12,
graph(500, 500, -9.5, 8.5, -6, 12,(0)x+3),
circle(-2,3,0.12),
circle(-2,3,0.12+0.03),
circle(1,3,0.12),
circle(1,3,0.12+0.03)
) }}} Graph of {{{y=3}}} through the points ({{{-2}}},{{{3}}}) and ({{{1}}},{{{3}}})


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



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b)



*[invoke equation_parallel_or_perpendicular "parallel", "1/3", -2, 3, 1]




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c)



*[invoke equation_parallel_or_perpendicular "perpendicular", "1/3", -2, 3, 1]