Question 124533


If you want to find the equation of line with a given a slope of {{{5/6}}} which goes through the point ({{{3}}},{{{4}}}), 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-4=(5/6)(x-3)}}} Plug in {{{m=5/6}}}, {{{x[1]=3}}}, and {{{y[1]=4}}} (these values are given)



{{{y-4=(5/6)x+(5/6)(-3)}}} Distribute {{{5/6}}}


{{{y-4=(5/6)x-5/2}}} Multiply {{{5/6}}} and {{{-3}}} to get {{{-5/2}}}


{{{y=(5/6)x-5/2+4}}} Add 4 to  both sides to isolate y


{{{y=(5/6)x+3/2}}} Combine like terms {{{-5/2}}} and {{{4}}} to get {{{3/2}}} (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 {{{5/6}}} which goes through the point ({{{3}}},{{{4}}}) is:


{{{y=(5/6)x+3/2}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=5/6}}} and the y-intercept is {{{b=3/2}}}


Notice if we graph the equation {{{y=(5/6)x+3/2}}} and plot the point ({{{3}}},{{{4}}}),  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, -6, 12, -5, 13,
graph(500, 500, -6, 12, -5, 13,(5/6)x+3/2),
circle(3,4,0.12),
circle(3,4,0.12+0.03)
) }}} Graph of {{{y=(5/6)x+3/2}}} through the point ({{{3}}},{{{4}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{5/6}}} and goes through the point ({{{3}}},{{{4}}}), this verifies our answer.