Question 173358

If you want to find the equation of line with a given a slope of {{{0}}} which goes through the point ({{{-6}}},{{{-2}}}), 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--2=(0)(x--6)}}} Plug in {{{m=0}}}, {{{x[1]=-6}}}, and {{{y[1]=-2}}} (these values are given)



{{{y+2=(0)(x--6)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(0)(x+6)}}} Rewrite {{{x--6}}} as {{{x+6}}}



{{{y+2=0x+(0)(6)}}} Distribute {{{0}}}



{{{y+2=0+0}}} Multiply {{{0}}} and {{{x}}} to get {{{0}}}



{{{y+2=0x+0}}} Multiply {{{0}}} and {{{6}}} to get {{{0}}}



{{{y=0+0-2}}} Subtract 2 from  both sides to isolate y



{{{y=-2}}} Combine like terms and simplify 

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Answer:



So the equation of the line with a slope of {{{0}}} which goes through the point (-6,-2) is:


{{{y=-2}}} 



The graph looks like this:



{{{drawing(500, 500, -15, 3, -11, 7,
graph(500, 500, -15, 3, -11, 7,0,-2),
circle(-6,-2,0.12),
circle(-6,-2,0.12+0.03)
) }}} Graph of {{{y=-2}}} (green)