Question 105510


If you want to find the equation of line with a given a slope of {{{-2/3}}} which goes through the point ({{{-8}}},{{{-2}}}), 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--2=(-2/3)(x--8)}}} Plug in {{{m=-2/3}}}, {{{x[1]=-8}}}, and {{{y[1]=-2}}} (these values are given)



{{{y+2=(-2/3)(x--8)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(-2/3)(x+8)}}} Rewrite {{{x--8}}} as {{{x+8}}}



{{{y+2=(-2/3)x+(-2/3)(8)}}} Distribute {{{-2/3}}}


{{{y+2=(-2/3)x-16/3}}} Multiply {{{-2/3}}} and {{{8}}} to get {{{-16/3}}}


{{{y=(-2/3)x-16/3-2}}} Subtract 2 from  both sides to isolate y


{{{y=(-2/3)x-22/3}}} Combine like terms {{{-16/3}}} and {{{-2}}} to get {{{-22/3}}} (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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Answer:



So the equation of the line with a slope of {{{-2/3}}} which goes through the point ({{{-8}}},{{{-2}}}) is:


{{{y=(-2/3)x-22/3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-2/3}}} and the y-intercept is {{{b=-22/3}}}


Notice if we graph the equation {{{y=(-2/3)x-22/3}}} and plot the point ({{{-8}}},{{{-2}}}),  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, -17, 1, -11, 7,
graph(500, 500, -17, 1, -11, 7,(-2/3)x+-22/3),
circle(-8,-2,0.12),
circle(-8,-2,0.12+0.03)
) }}} Graph of {{{y=(-2/3)x-22/3}}} through the point ({{{-8}}},{{{-2}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-2/3}}} and goes through the point ({{{-8}}},{{{-2}}}), this verifies our answer.