Question 154338


If you want to find the equation of line with a given a slope of {{{8}}} which goes through the point ({{{0}}},{{{6}}}), 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-6=(8)(x-0)}}} Plug in {{{m=8}}}, {{{x[1]=0}}}, and {{{y[1]=6}}} (these values are given)



{{{y-6=8x+(8)(0)}}} Distribute {{{8}}}



{{{y-6=8x+0}}} Multiply {{{8}}} and {{{0}}} to get {{{0}}}



{{{y=8x+0+6}}} Add 6 to  both sides to isolate y



{{{y=8x+6}}} Combine like terms {{{0}}} and {{{6}}} to get {{{6}}} 

------------------------------------------------------------------------------------------------------------

Answer:



So the equation of the line with a slope of {{{8}}} which goes through the point ({{{0}}},{{{6}}}) is:



{{{y=8x+6}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=8}}} and the y-intercept is {{{b=6}}}



Notice if we graph the equation {{{y=8x+6}}} and plot the point ({{{0}}},{{{6}}}),  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, -3, 15,
graph(500, 500, -9, 9, -3, 15,(8)x+6),
circle(0,6,0.12),
circle(0,6,0.12+0.03)
) }}} Graph of {{{y=8x+6}}} through the point ({{{0}}},{{{6}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{8}}} and goes through the point ({{{0}}},{{{6}}}), this verifies our answer.