Question 124397


If you want to find the equation of line with a given a slope of {{{2}}} which goes through the point ({{{1}}},{{{5}}}), 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-5=(2)(x-1)}}} Plug in {{{m=2}}}, {{{x[1]=1}}}, and {{{y[1]=5}}} (these values are given)



{{{y-5=2x+(2)(-1)}}} Distribute {{{2}}}


{{{y-5=2x-2}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}


{{{y=2x-2+5}}} Add 5 to  both sides to isolate y


{{{y=2x+3}}} Combine like terms {{{-2}}} and {{{5}}} to get {{{3}}} 

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Answer:



So the equation of the line with a slope of {{{2}}} which goes through the point ({{{1}}},{{{5}}}) is:


{{{y=2x+3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} and the y-intercept is {{{b=3}}}


Notice if we graph the equation {{{y=2x+3}}} and plot the point ({{{1}}},{{{5}}}),  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, -8, 10, -4, 14,
graph(500, 500, -8, 10, -4, 14,(2)x+3),
circle(1,5,0.12),
circle(1,5,0.12+0.03)
) }}} Graph of {{{y=2x+3}}} through the point ({{{1}}},{{{5}}})

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 ({{{1}}},{{{5}}}), this verifies our answer.