Question 149948


If you want to find the equation of line with a given a slope of {{{-3}}} which goes through the point ({{{5}}},{{{0}}}), 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-0=(-3)(x-5)}}} Plug in {{{m=-3}}}, {{{x[1]=5}}}, and {{{y[1]=0}}} (these values are given)



{{{y-0=-3x+(-3)(-5)}}} Distribute {{{-3}}}


{{{y-0=-3x+15}}} Multiply {{{-3}}} and {{{-5}}} to get {{{15}}}


{{{y=-3x+15+0}}} Add 0 to  both sides to isolate y


{{{y=-3x+15}}} Combine like terms {{{15}}} and {{{0}}} to get {{{15}}} 

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Answer:



So the equation of the line with a slope of {{{-3}}} which goes through the point ({{{5}}},{{{0}}}) is:


{{{y=-3x+15}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-3}}} and the y-intercept is {{{b=15}}}


Notice if we graph the equation {{{y=-3x+15}}} and plot the point ({{{5}}},{{{0}}}),  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, -4, 14, -9, 9,
graph(500, 500, -4, 14, -9, 9,(-3)x+15),
circle(5,0,0.12),
circle(5,0,0.12+0.03)
) }}} Graph of {{{y=-3x+15}}} through the point ({{{5}}},{{{0}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-3}}} and goes through the point ({{{5}}},{{{0}}}), this verifies our answer.