Question 252640


If you want to find the equation of line with a given a slope of {{{9}}} which goes through the point ({{{0}}},{{{8}}}), 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-8=(9)(x-0)}}} Plug in {{{m=9}}}, {{{x[1]=0}}}, and {{{y[1]=8}}} (these values are given)



{{{y-8=9x+(9)(-0)}}} Distribute {{{9}}}



{{{y-8=9x+0}}} Multiply {{{9}}} and {{{-0}}} to get {{{0}}}



{{{y=9x+0+8}}} Add 8 to  both sides to isolate y



{{{y=9x+8}}} Combine like terms {{{0}}} and {{{8}}} to get {{{8}}} 

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Answer:



So the equation of the line with a slope of {{{9}}} which goes through the point ({{{0}}},{{{8}}}) is:



{{{y=9x+8}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=9}}} and the y-intercept is {{{b=8}}}



Notice if we graph the equation {{{y=9x+8}}} and plot the point ({{{0}}},{{{8}}}),  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, -1, 17,
graph(500, 500, -9, 9, -1, 17,(9)x+8),
circle(0,8,0.12),
circle(0,8,0.12+0.03)
) }}} 


Graph of {{{y=9x+8}}} through the point ({{{0}}},{{{8}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{9}}} and goes through the point ({{{0}}},{{{8}}}), this verifies our answer.