Question 146260
Are you trying to find the equation of a line with a slope of {{{1/3}}} that goes through (1,6)?





If you want to find the equation of line with a given a slope of {{{1/3}}} which goes through the point ({{{1}}},{{{6}}}), 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-6=(1/3)(x-1)}}} Plug in {{{m=1/3}}}, {{{x[1]=1}}}, and {{{y[1]=6}}} (these values are given)



{{{y-6=(1/3)x+(1/3)(-1)}}} Distribute {{{1/3}}}


{{{y-6=(1/3)x-1/3}}} Multiply {{{1/3}}} and {{{-1}}} to get {{{-1/3}}}


{{{y=(1/3)x-1/3+6}}} Add 6 to  both sides to isolate y


{{{y=(1/3)x+17/3}}} Combine like terms {{{-1/3}}} and {{{6}}} to get {{{17/3}}} (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 with a slope of {{{1/3}}} which goes through the point ({{{1}}},{{{6}}}) is:


{{{y=(1/3)x+17/3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=1/3}}} and the y-intercept is {{{b=17/3}}}


Notice if we graph the equation {{{y=(1/3)x+17/3}}} and plot the point ({{{1}}},{{{6}}}),  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, -8, 10, -3, 15,
graph(500, 500, -8, 10, -3, 15,(1/3)x+17/3),
circle(1,6,0.12),
circle(1,6,0.12+0.03)
) }}} Graph of {{{y=(1/3)x+17/3}}} through the point ({{{1}}},{{{6}}})

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 ({{{1}}},{{{6}}}), this verifies our answer.