Question 146190


If you want to find the equation of line with a given a slope of {{{-2}}} which goes through the point (-3,6), 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-6=(-2)(x--3)}}} Plug in {{{m=-2}}}, {{{x[1]=-3}}}, and {{{y[1]=6}}} (these values are given)



{{{y-6=(-2)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-6=-2x+(-2)(3)}}} Distribute {{{-2}}}


{{{y-6=-2x-6}}} Multiply {{{-2}}} and {{{3}}} to get {{{-6}}}


{{{y=-2x-6+6}}} Add 6 to  both sides to isolate y


{{{y=-2x+0}}} Combine like terms {{{-6}}} and {{{6}}} to get {{{0}}} 


{{{y=-2x}}} Remove the zero term


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Answer:



So the equation of the line with a slope of {{{-2}}} which goes through the point ({{{-3}}},{{{6}}}) 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 ({{{-3}}},{{{6}}}),  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, -12, 6, -3, 15,
graph(500, 500, -12, 6, -3, 15,(-2)x+0),
circle(-3,6,0.12),
circle(-3,6,0.12+0.03)
) }}} Graph of {{{y=-2x}}} through the point (-3,6)

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 (-3,6), this verifies our answer.