Question 119887


If you want to find the equation of line with a given a slope of {{{-4}}} 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=(-4)(x-6)}}} Plug in {{{m=-4}}}, {{{x[1]=6}}}, and {{{y[1]=2}}} (these values are given)



{{{y-2=-4x+(-4)(-6)}}} Distribute {{{-4}}}


{{{y-2=-4x+24}}} Multiply {{{-4}}} and {{{-6}}} to get {{{24}}}


{{{y=-4x+24+2}}} Add 2 to  both sides to isolate y


{{{y=-4x+26}}} Combine like terms {{{24}}} and {{{2}}} to get {{{26}}} 

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

Answer:



So the equation of the line with a slope of {{{-4}}} which goes through the point ({{{6}}},{{{2}}}) is:


{{{y=-4x+26}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-4}}} and the y-intercept is {{{b=26}}}


Notice if we graph the equation {{{y=-4x+26}}} and plot the point ({{{6}}},{{{2}}}),  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, -3, 15, -7, 11,
graph(500, 500, -3, 15, -7, 11,(-4)x+26),
circle(6,2,0.12),
circle(6,2,0.12+0.03)
) }}} Graph of {{{y=-4x+26}}} through the point ({{{6}}},{{{2}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-4}}} and goes through the point ({{{6}}},{{{2}}}), this verifies our answer.