Question 169825


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



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



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



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



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



{{{y=(2/3)x-13/3}}} Combine like terms {{{-4/3}}} and {{{-3}}} to get {{{-13/3}}} (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>)



------------------------------------------------------------------------------------------------------------

Answer:



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


{{{y=(2/3)x-13/3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2/3}}} and the y-intercept is {{{b=-13/3}}}


Notice if we graph the equation {{{y=(2/3)x-13/3}}} and plot the point (2,-3),  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, -7, 11, -12, 6,
graph(500, 500, -7, 11, -12, 6,(2/3)x+-13/3),
circle(2,-3,0.12),
circle(2,-3,0.12+0.03)
) }}} Graph of {{{y=(2/3)x-13/3}}} through the point (2,-3)

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 ({{{2}}},{{{-3}}}), this verifies our answer.