Question 116469


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



{{{y+4=(-1)(x-1)}}} Rewrite {{{y--4}}} as {{{y+4}}}



{{{y+4=-1x+(-1)(-1)}}} Distribute {{{-1}}}


{{{y+4=-1x+1}}} Multiply {{{-1}}} and {{{-1}}} to get {{{1}}}


{{{y=-1x+1-4}}} Subtract 4 from  both sides to isolate y


{{{y=-1x-3}}} Combine like terms {{{1}}} and {{{-4}}} to get {{{-3}}} 

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Answer:



So the equation of the line with a slope of {{{-1}}} which goes through the point ({{{1}}},{{{-4}}}) is:


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


Notice if we graph the equation {{{y=-1x-3}}} and plot the point ({{{1}}},{{{-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, -8, 10, -13, 5,
graph(500, 500, -8, 10, -13, 5,(-1)x+-3),
circle(1,-4,0.12),
circle(1,-4,0.12+0.03)
) }}} Graph of {{{y=-1x-3}}} through the point ({{{1}}},{{{-4}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-1}}} and goes through the point ({{{1}}},{{{-4}}}), this verifies our answer.