Question 169805


If you want to find the equation of line with a given a slope of {{{2/3}}} which goes through the point ({{{-3}}},{{{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=(2/3)(x--3)}}} Plug in {{{m=2/3}}}, {{{x[1]=-3}}}, and {{{y[1]=5}}} (these values are given)



{{{y-5=(2/3)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-5=(2/3)x+(2/3)(3)}}} Distribute {{{2/3}}}


{{{y-5=(2/3)x+2}}} Multiply {{{2/3}}} and {{{3}}} to get {{{2}}}


{{{y=(2/3)x+2+5}}} Add 5 to  both sides to isolate y


{{{y=(2/3)x+7}}} Combine like terms {{{2}}} and {{{5}}} to get {{{7}}} 

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Answer:



So the equation of the line with a slope of {{{2/3}}} which goes through the point ({{{-3}}},{{{5}}}) is:


{{{y=(2/3)x+7}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2/3}}} and the y-intercept is {{{b=7}}}


Notice if we graph the equation {{{y=(2/3)x+7}}} and plot the point ({{{-3}}},{{{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, -12, 6, -4, 14,
graph(500, 500, -12, 6, -4, 14,(2/3)x+7),
circle(-3,5,0.12),
circle(-3,5,0.12+0.03)
) }}} Graph of {{{y=(2/3)x+7}}} through the point ({{{-3}}},{{{5}}})

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 ({{{-3}}},{{{5}}}), this verifies our answer.