Question 117502


If you want to find the equation of line with a given a slope of {{{-7/3}}} which goes through the point ({{{6}}},{{{-6}}}), 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--6=(-7/3)(x-6)}}} Plug in {{{m=-7/3}}}, {{{x[1]=6}}}, and {{{y[1]=-6}}} (these values are given)



{{{y+6=(-7/3)(x-6)}}} Rewrite {{{y--6}}} as {{{y+6}}}



{{{y+6=(-7/3)x+(-7/3)(-6)}}} Distribute {{{-7/3}}}


{{{y+6=(-7/3)x+14}}} Multiply {{{-7/3}}} and {{{-6}}} to get {{{14}}}


{{{y=(-7/3)x+14-6}}} Subtract 6 from  both sides to isolate y


{{{y=(-7/3)x+8}}} Combine like terms {{{14}}} and {{{-6}}} to get {{{8}}} 

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Answer:



So the equation of the line with a slope of {{{-7/3}}} which goes through the point ({{{6}}},{{{-6}}}) is:


{{{y=(-7/3)x+8}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-7/3}}} and the y-intercept is {{{b=8}}}


Notice if we graph the equation {{{y=(-7/3)x+8}}} and plot the point ({{{6}}},{{{-6}}}),  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, -15, 3,
graph(500, 500, -3, 15, -15, 3,(-7/3)x+8),
circle(6,-6,0.12),
circle(6,-6,0.12+0.03)
) }}} Graph of {{{y=(-7/3)x+8}}} through the point ({{{6}}},{{{-6}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-7/3}}} and goes through the point ({{{6}}},{{{-6}}}), this verifies our answer.