Question 150897
If an equation has a y-intercept of 17, this means that the equation goes through the point (0,17)



If you want to find the equation of line with a given a slope of {{{-2/3}}} which goes through the point (0,17), 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-17=(-2/3)(x-0)}}} Plug in {{{m=-2/3}}}, {{{x[1]=0}}}, and {{{y[1]=17}}} (these values are given)



{{{y-17=(-2/3)x+(-2/3)(-0)}}} Distribute {{{-2/3}}}


{{{y-17=(-2/3)x+0}}} Multiply {{{-2/3}}} and {{{-0}}} to get {{{0}}}


{{{y=(-2/3)x+0+17}}} Add 17 to  both sides to isolate y


{{{y=(-2/3)x+17}}} Combine like terms {{{0}}} and {{{17}}} to get {{{17}}} 

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Answer:



So the equation of the line with a slope of {{{-2/3}}} which goes through the point (0,17) is:


{{{y=(-2/3)x+17}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-2/3}}} and the y-intercept is {{{b=17}}}


Notice if we graph the equation {{{y=(-2/3)x+17}}} and plot the point (0,17),  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, -9, 9, -5, 26,
graph(500, 500, -9, 9, -5, 26,(-2/3)x+17),
circle(0,17,0.12),
circle(0,17,0.12+0.03)
) }}} Graph of {{{y=(-2/3)x+17}}} through the point (0,17)

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 (0,17), this verifies our answer.