Question 124709


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



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


{{{y-2=4x-16}}} Multiply {{{4}}} and {{{-4}}} to get {{{-16}}}


{{{y=4x-16+2}}} Add 2 to  both sides to isolate y


{{{y=4x-14}}} Combine like terms {{{-16}}} and {{{2}}} to get {{{-14}}} 

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Answer:



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


{{{y=4x-14}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=4}}} and the y-intercept is {{{b=-14}}}


Notice if we graph the equation {{{y=4x-14}}} and plot the point ({{{4}}},{{{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, -5, 13, -7, 11,
graph(500, 500, -5, 13, -7, 11,(4)x+-14),
circle(4,2,0.12),
circle(4,2,0.12+0.03)
) }}} Graph of {{{y=4x-14}}} through the point ({{{4}}},{{{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 ({{{4}}},{{{2}}}), this verifies our answer.