Question 243589


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



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



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



{{{y=-3x+15+5}}} Add 5 to  both sides to isolate y



{{{y=-3x+20}}} Combine like terms {{{15}}} and {{{5}}} to get {{{20}}} 

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Answer:



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


{{{y=-3x+20}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-3}}} and the y-intercept is {{{b=20}}}


Notice if we graph the equation {{{y=-3x+20}}} and plot the point (5,5),  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, -4, 14,
graph(500, 500, -4, 14, -4, 14,(-3)x+20),
circle(5,5,0.12),
circle(5,5,0.12+0.03)
) }}} 


Graph of {{{y=-3x+20}}} through the point (5,5)

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}}},{{{5}}}), this verifies our answer.