Question 117105


If you want to find the equation of line with a given a slope of {{{9/7}}} which goes through the point ({{{0}}},{{{5}}}), 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-5=(9/7)(x-0)}}} Plug in {{{m=9/7}}}, {{{x[1]=0}}}, and {{{y[1]=5}}} (these values are given)



{{{y-5=(9/7)x+(9/7)(-0)}}} Distribute {{{9/7}}}


{{{y-5=(9/7)x+0}}} Multiply {{{9/7}}} and {{{-0}}} to get {{{0}}}


{{{y=(9/7)x+0+5}}} Add 5 to  both sides to isolate y


{{{y=(9/7)x+5}}} Combine like terms {{{0}}} and {{{5}}} to get {{{5}}} 

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Answer:



So the equation of the line with a slope of {{{9/7}}} which goes through the point ({{{0}}},{{{5}}}) is:


{{{y=(9/7)x+5}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=9/7}}} and the y-intercept is {{{b=5}}}


Notice if we graph the equation {{{y=(9/7)x+5}}} and plot the point ({{{0}}},{{{5}}}),  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, -4, 14,
graph(500, 500, -9, 9, -4, 14,(9/7)x+5),
circle(0,5,0.12),
circle(0,5,0.12+0.03)
) }}} Graph of {{{y=(9/7)x+5}}} through the point ({{{0}}},{{{5}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{9/7}}} and goes through the point ({{{0}}},{{{5}}}), this verifies our answer.