Question 139533


If you want to find the equation of line with a given a slope of {{{1/2}}} which goes through the point ({{{-3}}},{{{3}}}), 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-3=(1/2)(x--3)}}} Plug in {{{m=1/2}}}, {{{x[1]=-3}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=(1/2)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-3=(1/2)x+(1/2)(3)}}} Distribute {{{1/2}}}


{{{y-3=(1/2)x+3/2}}} Multiply {{{1/2}}} and {{{3}}} to get {{{3/2}}}


{{{y=(1/2)x+3/2+3}}} Add 3 to  both sides to isolate y


{{{y=(1/2)x+9/2}}} Combine like terms {{{3/2}}} and {{{3}}} to get {{{9/2}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)



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Answer:



So the equation of the line with a slope of {{{1/2}}} which goes through the point ({{{-3}}},{{{3}}}) is:


{{{y=(1/2)x+9/2}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=1/2}}} and the y-intercept is {{{b=9/2}}}


Notice if we graph the equation {{{y=(1/2)x+9/2}}} and plot the point ({{{-3}}},{{{3}}}),  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, -12, 6, -6, 12,
graph(500, 500, -12, 6, -6, 12,(1/2)x+9/2),
circle(-3,3,0.12),
circle(-3,3,0.12+0.03)
) }}} Graph of {{{y=(1/2)x+9/2}}} through the point ({{{-3}}},{{{3}}})

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 ({{{-3}}},{{{3}}}), this verifies our answer.