Question 93786


If you want to find the equation of line with a given a slope of {{{-1/2}}} which goes through the point ({{{8}}},{{{-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=(-1/2)(x-8)}}} Plug in {{{m=-1/2}}}, {{{x[1]=8}}}, and {{{y[1]=-5}}} (these values are given)



{{{y+5=(-1/2)(x-8)}}} Rewrite {{{y--5}}} as {{{y+5}}}



{{{y+5=(-1/2)x+(-1/2)(-8)}}} Distribute {{{-1/2}}}


{{{y+5=(-1/2)x+4}}} Multiply {{{-1/2}}} and {{{-8}}} to get {{{4}}}


{{{y=(-1/2)x+4-5}}} Subtract 5 from  both sides to isolate y


{{{y=(-1/2)x-1}}} Combine like terms {{{4}}} and {{{-5}}} to get {{{-1}}} 

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Answer:



So the equation of the line with a slope of {{{-1/2}}} which goes through the point ({{{8}}},{{{-5}}}) is:


{{{y=(-1/2)x-1}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-1/2}}} and the y-intercept is {{{b=-1}}}


Notice if we graph the equation {{{y=(-1/2)x-1}}} and plot the point ({{{8}}},{{{-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, -1, 17, -14, 4,
graph(500, 500, -1, 17, -14, 4,(-1/2)x+-1),
circle(8,-5,0.12),
circle(8,-5,0.12+0.03)
) }}} Graph of {{{y=(-1/2)x-1}}} through the point ({{{8}}},{{{-5}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-1/2}}} and goes through the point ({{{8}}},{{{-5}}}), this verifies our answer.