Question 149473

If you want to find the equation of line with a given a slope of {{{-1/3}}} which goes through the point ({{{12}}},{{{-9}}}), 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--9=(-1/3)(x-12)}}} Plug in {{{m=-1/3}}}, {{{x[1]=12}}}, and {{{y[1]=-9}}} (these values are given)



{{{y+9=(-1/3)(x-12)}}} Rewrite {{{y--9}}} as {{{y+9}}}



{{{y+9=(-1/3)x+(-1/3)(-12)}}} Distribute {{{-1/3}}}


{{{y+9=(-1/3)x+4}}} Multiply {{{-1/3}}} and {{{-12}}} to get {{{4}}}


{{{y=(-1/3)x+4-9}}} Subtract 9 from  both sides to isolate y


{{{y=(-1/3)x-5}}} Combine like terms {{{4}}} and {{{-9}}} to get {{{-5}}} 

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Answer:



So the equation of the line with a slope of {{{-1/3}}} which goes through the point ({{{12}}},{{{-9}}}) is:


{{{y=(-1/3)x-5}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-1/3}}} and the y-intercept is {{{b=-5}}}


Notice if we graph the equation {{{y=(-1/3)x-5}}} and plot the point ({{{12}}},{{{-9}}}),  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, -5, 21, -18, 5,
graph(500, 500, -5, 21, -18, 5,(-1/3)x+-5),
circle(12,-9,0.12),
circle(12,-9,0.12+0.03)
) }}} Graph of {{{y=(-1/3)x-5}}} through the point ({{{12}}},{{{-9}}})

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 ({{{12}}},{{{-9}}}), this verifies our answer.