Question 93643

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



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


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


{{{y=-2x+0+0}}} Add 0 to  both sides to isolate y


{{{y=-2x+0}}} Combine like terms


{{{y=-2x}}} Remove any zero terms

------------------------------------------------------------------------------------------------------------

Answer:



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


{{{y=-2x}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-2}}} and the y-intercept is {{{b=0}}}


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

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}}},{{{0}}}), this verifies our answer.