Question 115116


If you want to find the equation of line with a given a slope of {{{5/2}}} which goes through the point ({{{-3}}},{{{-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=(5/2)(x--3)}}} Plug in {{{m=5/2}}}, {{{x[1]=-3}}}, and {{{y[1]=-2}}} (these values are given)



{{{y+2=(5/2)(x--3)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(5/2)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y+2=(5/2)x+(5/2)(3)}}} Distribute {{{5/2}}}


{{{y+2=(5/2)x+15/2}}} Multiply {{{5/2}}} and {{{3}}} to get {{{15/2}}}


{{{y=(5/2)x+15/2-2}}} Subtract 2 from  both sides to isolate y


{{{y=(5/2)x+11/2}}} Combine like terms {{{15/2}}} and {{{-2}}} to get {{{11/2}}} (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 {{{5/2}}} which goes through the point ({{{-3}}},{{{-2}}}) is:


{{{y=(5/2)x+11/2}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=5/2}}} and the y-intercept is {{{b=11/2}}}


Notice if we graph the equation {{{y=(5/2)x+11/2}}} and plot the point ({{{-3}}},{{{-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, -12, 6, -11, 7,
graph(500, 500, -12, 6, -11, 7,(5/2)x+11/2),
circle(-3,-2,0.12),
circle(-3,-2,0.12+0.03)
) }}} Graph of {{{y=(5/2)x+11/2}}} through the point ({{{-3}}},{{{-2}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{5/2}}} and goes through the point ({{{-3}}},{{{-2}}}), this verifies our answer.