Question 115121


If you want to find the equation of line with a given a slope of {{{-3}}} which goes through the point ({{{3}}},{{{7}}}), 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-7=(-3)(x-3)}}} Plug in {{{m=-3}}}, {{{x[1]=3}}}, and {{{y[1]=7}}} (these values are given)



{{{y-7=-3x+(-3)(-3)}}} Distribute {{{-3}}}


{{{y-7=-3x+9}}} Multiply {{{-3}}} and {{{-3}}} to get {{{9}}}


{{{y=-3x+9+7}}} Add 7 to  both sides to isolate y


{{{y=-3x+16}}} Combine like terms {{{9}}} and {{{7}}} to get {{{16}}} 

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

Answer:



So the equation of the line with a slope of {{{-3}}} which goes through the point ({{{3}}},{{{7}}}) is:


{{{y=-3x+16}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-3}}} and the y-intercept is {{{b=16}}}


Notice if we graph the equation {{{y=-3x+16}}} and plot the point ({{{3}}},{{{7}}}),  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, -6, 12, -2, 16,
graph(500, 500, -6, 12, -2, 16,(-3)x+16),
circle(3,7,0.12),
circle(3,7,0.12+0.03)
) }}} Graph of {{{y=-3x+16}}} through the point ({{{3}}},{{{7}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-3}}} and goes through the point ({{{3}}},{{{7}}}), this verifies our answer.