Question 112036


If you want to find the equation of line with a given a slope of {{{0}}} which goes through the point ({{{6}}},{{{7}}}), 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-7=(0)(x-6)}}} Plug in {{{m=0}}}, {{{x[1]=6}}}, and {{{y[1]=7}}} (these values are given)



{{{y-7=0x+(0)(-6)}}} Distribute {{{0}}}


{{{y-7=0x+0}}} Multiply {{{0}}} and {{{-6}}} to get {{{0}}}


{{{y=0x+0+7}}} Add 7 to  both sides to isolate y


{{{y=0x+7}}} Combine like terms {{{0}}} and {{{7}}} to get {{{7}}} 


{{{y=7}}} Remove the zero terms

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Answer:



So the equation of the line with a slope of {{{0}}} which goes through the point ({{{6}}},{{{7}}}) is:


{{{y=7}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=0}}} and the y-intercept is {{{b=7}}}


Notice if we graph the equation {{{y=7}}} and plot the point ({{{6}}},{{{7}}}),  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, -3, 15, -2, 16,
graph(500, 500, -3, 15, -2, 16,(0)x+7),
circle(6,7,0.12),
circle(6,7,0.12+0.03)
) }}} Graph of {{{y=7}}} through the point ({{{6}}},{{{7}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{0}}} and goes through the point ({{{6}}},{{{7}}}), this verifies our answer.