Question 133009
First lets find the slope through the points (-4,3) and (5,6)


{{{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 ({{{-4}}},{{{3}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{5}}},{{{6}}}))


{{{m=(6-3)/(5--4)}}} Plug in {{{y[2]=6}}},{{{y[1]=3}}},{{{x[2]=5}}},{{{x[1]=-4}}}  (these are the coordinates of given points)


{{{m= 3/9}}} Subtract the terms in the numerator {{{6-3}}} to get {{{3}}}.  Subtract the terms in the denominator {{{5--4}}} to get {{{9}}}

  


{{{m=1/3}}} Reduce

  

So the slope is

{{{m=1/3}}}


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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-3=(1/3)(x--4)}}} Plug in {{{m=1/3}}}, {{{x[1]=-4}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=(1/3)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y-3=(1/3)x+(1/3)(4)}}} Distribute {{{1/3}}}


{{{y-3=(1/3)x+4/3}}} Multiply {{{1/3}}} and {{{4}}} to get {{{4/3}}}


{{{y=(1/3)x+4/3+3}}} Add {{{3}}} to  both sides to isolate y


{{{y=(1/3)x+13/3}}} Combine like terms {{{4/3}}} and {{{3}}} to get {{{13/3}}} (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>)






So the equation of the line which goes through the points ({{{-4}}},{{{3}}}) and ({{{5}}},{{{6}}})  is:



{{{y=(1/3)x+13/3}}} (note: this equation is in slope intercept form)



So lets convert it into standard form:



{{{3y=x+13}}} Multiply both sides by 3 to clear the fractions



{{{3y-x=13}}} Subtract x from both sides



{{{-x+3y=13}}} Rearrange the equation




So the equation in standard form that goes through the points (-4,3) and (5,6) is 



{{{-x+3y=13}}}