Question 127008
If you want to find the equation of line with a given a slope of {{{-2/7}}} which goes through the point ({{{-4}}},{{{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=(-2/7)(x--4)}}} Plug in {{{m=-2/7}}}, {{{x[1]=-4}}}, and {{{y[1]=5}}} (these values are given)

{{{y-5=(-2/7)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}

{{{y-5=(-2/7)x+(-2/7)(4)}}} Distribute {{{-2/7}}}

{{{y-5=(-2/7)x-8/7}}} Multiply {{{-2/7}}} and {{{4}}} to get {{{-8/7}}}

{{{y=(-2/7)x-8/7+5}}} Add 5 to  both sides to isolate y

{{{y=(-2/7)x+27/7}}} Combine like terms {{{-8/7}}} and {{{5}}} to get {{{27/7}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)

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So the equation of the line with a slope of {{{-2/7}}} which goes through the point ({{{-4}}},{{{5}}}) is:

{{{y=(-2/7)x+27/7}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-2/7}}} and the y-intercept is {{{b=27/7}}}

Notice if we graph the equation {{{y=(-2/7)x+27/7}}} and plot the point ({{{-4}}},{{{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, -13, 5, -4, 14,
graph(500, 500, -13, 5, -4, 14,(-2/7)x+27/7),
circle(-4,5,0.12),
circle(-4,5,0.12+0.03)
) }}} Graph of {{{y=(-2/7)x+27/7}}} through the point ({{{-4}}},{{{5}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-2/7}}} and goes through the point ({{{-4}}},{{{5}}}), this verifies our answer.