Question 111348


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



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


{{{y-4=2x-12}}} Multiply {{{2}}} and {{{-6}}} to get {{{-12}}}


{{{y=2x-12+4}}} Add 4 to  both sides to isolate y


{{{y=2x-8}}} Combine like terms {{{-12}}} and {{{4}}} to get {{{-8}}} 

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Answer:



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


{{{y=2x-8}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} and the y-intercept is {{{b=-8}}}


Notice if we graph the equation {{{y=2x-8}}} and plot the point ({{{6}}},{{{4}}}),  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, -5, 13,
graph(500, 500, -3, 15, -5, 13,(2)x+-8),
circle(6,4,0.12),
circle(6,4,0.12+0.03)
) }}} Graph of {{{y=2x-8}}} through the point ({{{6}}},{{{4}}})

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 ({{{6}}},{{{4}}}), this verifies our answer.