Question 92286
If you want to find the equation of line with a given a slope of {{{0}}} which goes through the point ({{{-1}}},{{{4}}}), 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 ({{{x[1]}}},{{{y[1]}}}) is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-4=(0)(x--1)}}} Plug in {{{m=0}}}, {{{x[1]=-1}}}, and {{{y[1]=4}}} (these values are given)


{{{y-4=(0)x-(0)(-1))}}} Distribute {{{0}}}


{{{y-4=(0)x+(0)(-1))}}} Multiply the negatives


{{{y-4=(0)x+0}}} Multiply {{{0}}} and {{{-1}}} to get {{{0}}}


{{{y=(0)x+0+4}}}Add {{{4}}} to both sides


{{{y=0x+4}}} Combine like terms


{{{y=4}}} 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 ({{{-1}}},{{{4}}}) is:


{{{y=4}}}  which is now in {{{y=mx+b}}} form where the slope is {{{m=0}}} and the y-intercept is {{{b=4}}}


Notice if we graph the equation {{{y=4}}} and plot the point ({{{-1}}},{{{4}}}),  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, -10, 8, -5, 13,
graph(500, 500, -10, 8, -5, 13,(0)x+4),
circle(-1,4,0.12),
circle(-1,4,0.12+0.03)
) }}} Graph of {{{y=4}}}  through the point ({{{-1}}},{{{4}}})

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 ({{{-1}}},{{{4}}}), this verifies our answer.