Question 114962


If you want to find the equation of line with a given a slope of {{{1/3}}} which goes through the point ({{{3}}},{{{3}}}), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


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



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


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


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


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

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Answer:



So the equation of the line with a slope of {{{1/3}}} which goes through the point ({{{3}}},{{{3}}}) is:


{{{y=(1/3)x+2}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=1/3}}} and the y-intercept is {{{b=2}}}


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

and we can see that the point lies on the line. Since we know the equation has a slope of {{{1/3}}} and goes through the point ({{{3}}},{{{3}}}), this verifies our answer.