Question 92561


If you want to find the equation of line with a given a slope of {{{-5}}} which goes through the point ({{{2}}},{{{4}}}), 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-4=(-5)(x-2)}}} Plug in {{{m=-5}}}, {{{x[1]=2}}}, and {{{y[1]=4}}} (these values are given)



{{{y-4=-5x+(-5)(-2)}}} Distribute {{{-5}}}


{{{y-4=-5x+10}}} Multiply {{{-5}}} and {{{-2}}} to get {{{10}}}


{{{y=-5x+10+4}}} Add 4 to  both sides to isolate y


{{{y=-5x+14}}} Combine like terms {{{10}}} and {{{4}}} to get {{{14}}} 

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Answer:



So the equation of the line with a slope of {{{-5}}} which goes through the point ({{{2}}},{{{4}}}) is:


{{{y=-5x+14}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-5}}} and the y-intercept is {{{b=14}}}


Notice if we graph the equation {{{y=-5x+14}}} and plot the point ({{{2}}},{{{4}}}),  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, -7, 11, -5, 13,
graph(500, 500, -7, 11, -5, 13,(-5)x+14),
circle(2,4,0.12),
circle(2,4,0.12+0.03)
) }}} Graph of {{{y=-5x+14}}} through the point ({{{2}}},{{{4}}})

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 ({{{2}}},{{{4}}}), this verifies our answer.