Question 125744

If you want to find the equation of line with a given a slope of {{{-2}}} which goes through the point ({{{0}}},{{{-3}}}), 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--3=(-2)(x-0)}}} Plug in {{{m=-2}}}, {{{x[1]=0}}}, and {{{y[1]=-3}}} (these values are given)



{{{y+3=(-2)(x-0)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=-2x+(-2)(0)}}} Distribute {{{-2}}}


{{{y+3=-2x+0}}} Multiply {{{-2}}} and {{{0}}} to get {{{0}}}


{{{y=-2x+0-3}}} Subtract 3 from  both sides to isolate y


{{{y=-2x-3}}} Combine like terms {{{0}}} and {{{-3}}} to get {{{-3}}} 

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Answer:



So the equation of the line with a slope of {{{-2}}} which goes through the point ({{{0}}},{{{-3}}}) is:


{{{y=-2x-3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-2}}} and the y-intercept is {{{b=-3}}}


Notice if we graph the equation {{{y=-2x-3}}} and plot the point ({{{0}}},{{{-3}}}),  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, -12, 6,
graph(500, 500, -9, 9, -12, 6,(-2)x+-3),
circle(0,-3,0.12),
circle(0,-3,0.12+0.03)
) }}} Graph of {{{y=-2x-3}}} through the point ({{{0}}},{{{-3}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-2}}} and goes through the point ({{{0}}},{{{-3}}}), this verifies our answer.