Question 215352


If you want to find the equation of line with a given a slope of {{{-2/3}}} which goes through the point ({{{6}}},{{{-8}}}), 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--8=(-2/3)(x-6)}}} Plug in {{{m=-2/3}}}, {{{x[1]=6}}}, and {{{y[1]=-8}}} (these values are given)



{{{y+8=(-2/3)(x-6)}}} Rewrite {{{y--8}}} as {{{y+8}}}



{{{y+8=(-2/3)x+(-2/3)(-6)}}} Distribute {{{-2/3}}}


{{{y+8=(-2/3)x+4}}} Multiply {{{-2/3}}} and {{{-6}}} to get {{{4}}}


{{{y=(-2/3)x+4-8}}} Subtract 8 from  both sides to isolate y


{{{y=(-2/3)x-4}}} Combine like terms {{{4}}} and {{{-8}}} to get {{{-4}}} 

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Answer:



So the equation of the line with a slope of {{{-2/3}}} which goes through the point ({{{6}}},{{{-8}}}) is:


{{{y=(-2/3)x-4}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-2/3}}} and the y-intercept is {{{b=-4}}}


Notice if we graph the equation {{{y=(-2/3)x-4}}} and plot the point ({{{6}}},{{{-8}}}),  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, -3, 15, -17, 1,
graph(500, 500, -3, 15, -17, 1,(-2/3)x+-4),
circle(6,-8,0.12),
circle(6,-8,0.12+0.03)
) }}} Graph of {{{y=(-2/3)x-4}}} through the point ({{{6}}},{{{-8}}})

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 ({{{6}}},{{{-8}}}), this verifies our answer.