Question 190702


If you want to find the equation of line with a given a slope of {{{6}}} which goes through the point ({{{0}}},{{{9}}}), 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-9=(6)(x-0)}}} Plug in {{{m=6}}}, {{{x[1]=0}}}, and {{{y[1]=9}}} (these values are given)



{{{y-9=6x+(6)(-0)}}} Distribute {{{6}}}


{{{y-9=6x+0}}} Multiply {{{6}}} and {{{-0}}} to get {{{0}}}


{{{y=6x+0+9}}} Add 9 to  both sides to isolate y


{{{y=6x+9}}} Combine like terms {{{0}}} and {{{9}}} to get {{{9}}} 

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Answer:



So the equation of the line with a slope of {{{6}}} which goes through the point ({{{0}}},{{{9}}}) is:


{{{y=6x+9}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=6}}} and the y-intercept is {{{b=9}}}


Notice if we graph the equation {{{y=6x+9}}} and plot the point ({{{0}}},{{{9}}}),  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, 0, 18,
graph(500, 500, -9, 9, 0, 18,(6)x+9),
circle(0,9,0.12),
circle(0,9,0.12+0.03)
) }}} Graph of {{{y=6x+9}}} through the point ({{{0}}},{{{9}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{6}}} and goes through the point ({{{0}}},{{{9}}}), this verifies our answer.