Question 117204


If you want to find the equation of line with a given a slope of {{{-5}}} which goes through the point ({{{-5}}},{{{18}}}), 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-18=(-5)(x--5)}}} Plug in {{{m=-5}}}, {{{x[1]=-5}}}, and {{{y[1]=18}}} (these values are given)



{{{y-18=(-5)(x+5)}}} Rewrite {{{x--5}}} as {{{x+5}}}



{{{y-18=-5x+(-5)(5)}}} Distribute {{{-5}}}


{{{y-18=-5x-25}}} Multiply {{{-5}}} and {{{5}}} to get {{{-25}}}


{{{y=-5x-25+18}}} Add 18 to  both sides to isolate y


{{{y=-5x-7}}} Combine like terms {{{-25}}} and {{{18}}} to get {{{-7}}} 

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Answer:



So the equation of the line with a slope of {{{-5}}} which goes through the point ({{{-5}}},{{{18}}}) is:


{{{y=-5x-7}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-5}}} and the y-intercept is {{{b=-7}}}


Notice if we graph the equation {{{y=-5x-7}}} and plot the point ({{{-5}}},{{{18}}}),  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, -14, 4, 9, 27,
graph(500, 500, -14, 4, 9, 27,(-5)x+-7),
circle(-5,18,0.12),
circle(-5,18,0.12+0.03)
) }}} Graph of {{{y=-5x-7}}} through the point ({{{-5}}},{{{18}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-5}}} and goes through the point ({{{-5}}},{{{18}}}), this verifies our answer.