Question 109124


If you want to find the equation of line with a given a slope of {{{2/3}}} which goes through the point ({{{4}}},{{{-1}}}), 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--1=(2/3)(x-4)}}} Plug in {{{m=2/3}}}, {{{x[1]=4}}}, and {{{y[1]=-1}}} (these values are given)



{{{y+1=(2/3)(x-4)}}} Rewrite {{{y--1}}} as {{{y+1}}}



{{{y+1=(2/3)x+(2/3)(-4)}}} Distribute {{{2/3}}}


{{{y+1=(2/3)x-8/3}}} Multiply {{{2/3}}} and {{{-4}}} to get {{{-8/3}}}


{{{y=(2/3)x-8/3-1}}} Subtract 1 from  both sides to isolate y


{{{y=(2/3)x-11/3}}} Combine like terms {{{-8/3}}} and {{{-1}}} to get {{{-11/3}}} (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 ({{{4}}},{{{-1}}}) is:


{{{y=(2/3)x-11/3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2/3}}} and the y-intercept is {{{b=-11/3}}}


Notice if we graph the equation {{{y=(2/3)x-11/3}}} and plot the point ({{{4}}},{{{-1}}}),  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, -5, 13, -10, 8,
graph(500, 500, -5, 13, -10, 8,(2/3)x+-11/3),
circle(4,-1,0.12),
circle(4,-1,0.12+0.03)
) }}} Graph of {{{y=(2/3)x-11/3}}} through the point ({{{4}}},{{{-1}}})

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 ({{{4}}},{{{-1}}}), this verifies our answer.