Question 175489


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



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


{{{y-2=(1/2)x-1}}} Multiply {{{1/2}}} and {{{-2}}} to get {{{-1}}}


{{{y=(1/2)x-1+2}}} Add 2 to  both sides to isolate y


{{{y=(1/2)x+1}}} Combine like terms {{{-1}}} and {{{2}}} to get {{{1}}} 

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Answer:



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


{{{y=(1/2)x+1}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=1/2}}} and the y-intercept is {{{b=1}}}


Notice if we graph the equation {{{y=(1/2)x+1}}} and plot the point ({{{2}}},{{{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, -7, 11, -7, 11,
graph(500, 500, -7, 11, -7, 11,(1/2)x+1),
circle(2,2,0.12),
circle(2,2,0.12+0.03)
) }}} Graph of {{{y=(1/2)x+1}}} through the point ({{{2}}},{{{2}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{1/2}}} and goes through the point ({{{2}}},{{{2}}}), this verifies our answer.