Question 223490

If you want to find the equation of line with a given a slope of {{{-8}}} which goes through the point ({{{6}}},{{{9}}}), 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-9=(-8)(x-6)}}} Plug in {{{m=-8}}}, {{{x[1]=6}}}, and {{{y[1]=9}}} (these values are given)



{{{y-9=-8x+(-8)(-6)}}} Distribute {{{-8}}}


{{{y-9=-8x+48}}} Multiply {{{-8}}} and {{{-6}}} to get {{{48}}}


{{{y=-8x+48+9}}} Add 9 to  both sides to isolate y


{{{y=-8x+57}}} Combine like terms {{{48}}} and {{{9}}} to get {{{57}}} 

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Answer:



So the equation of the line with a slope of {{{-8}}} which goes through the point ({{{6}}},{{{9}}}) is:


{{{y=-8x+57}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-8}}} and the y-intercept is {{{b=57}}}