Question 102884
First lets find the slope through the points ({{{-1}}},{{{1}}}) and ({{{2}}},{{{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}}},{{{1}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{2}}},{{{3}}}))


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


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

  

So the slope is

{{{m=2/3}}}


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



{{{y-1=(2/3)(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y-1=(2/3)x+(2/3)(1)}}} Distribute {{{2/3}}}


{{{y-1=(2/3)x+2/3}}} Multiply {{{2/3}}} and {{{1}}} to get {{{2/3}}}


{{{y=(2/3)x+2/3+1}}} Add {{{1}}} to  both sides to isolate y


{{{y=(2/3)x+5/3}}} Combine like terms {{{2/3}}} and {{{1}}} to get {{{5/3}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)



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Answer:



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


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=2/3}}} and the y-intercept is {{{b=5/3}}}


Notice if we graph the equation {{{y=(2/3)x+5/3}}} and plot the points ({{{-1}}},{{{1}}}) and ({{{2}}},{{{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, -8.5, 9.5, -7, 11,
graph(500, 500, -8.5, 9.5, -7, 11,(2/3)x+5/3),
circle(-1,1,0.12),
circle(-1,1,0.12+0.03),
circle(2,3,0.12),
circle(2,3,0.12+0.03)
) }}} Graph of {{{y=(2/3)x+5/3}}} through the points ({{{-1}}},{{{1}}}) and ({{{2}}},{{{3}}})


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