Question 151613


If you want to find the equation of line with a given a slope of {{{3/2}}} which goes through the point (0,2), 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=(3/2)(x-0)}}} Plug in {{{m=3/2}}}, {{{x[1]=0}}}, and {{{y[1]=2}}} (these values are given)



{{{y-2=(3/2)x+(3/2)(-0)}}} Distribute {{{3/2}}}


{{{y-2=(3/2)x+0}}} Multiply {{{3/2}}} and {{{-0}}} to get {{{0}}}


{{{y=(3/2)x+0+2}}} Add 2 to  both sides to isolate y


{{{y=(3/2)x+2}}} Combine like terms {{{0}}} and {{{2}}} to get {{{2}}} 

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Answer:



So the equation of the line with a slope of {{{3/2}}} which goes through the point (0,2) is:


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


Notice if we graph the equation {{{y=(3/2)x+2}}} and plot the point (0,2),  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, -9, 9, -7, 11,
graph(500, 500, -9, 9, -7, 11,(3/2)x+2),
circle(0,2,0.12),
circle(0,2,0.12+0.03)
) }}} Graph of {{{y=(3/2)x+2}}} through the point (0,2)

and we can see that the point lies on the line. Since we know the equation has a slope of {{{3/2}}} and goes through the point (0,2), this verifies our answer.