Question 92287


If you want to find the equation of line with a given a slope of {{{6}}} which goes through the point ({{{0}}},{{{-9}}}), 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 ({{{x[1]}}},{{{y[1]}}}) is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y--9=(6)(x-0)}}} Plug in {{{m=6}}}, {{{x[1]=0}}}, and {{{y[1]=-9}}} (these values are given)


{{{y--9=(6)x-(6)(0))}}} Distribute {{{6}}}


{{{y--9=(6)x+(-6)(0))}}} Multiply the negatives


{{{y+9=(6)x+0}}} Multiply {{{-6}}} and {{{0}}} to get {{{0}}}


{{{y=(6)x+0+-9}}}Subtract {{{-9}}} from both sides



{{{y=6x-9}}} Combine like terms

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Answer:



So the equation of the line with a slope of {{{6}}} which goes through the point ({{{0}}},{{{-9}}}) is:


{{{y=6x-9}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=6}}} and the y-intercept is {{{b=-9}}}


Notice if we graph the equation {{{y=6x-9}}} and plot the point ({{{0}}},{{{-9}}}),  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, -18, 0,
graph(500, 500, -9, 9, -18, 0,(6)x+-9),
circle(0,-9,0.12),
circle(0,-9,0.12+0.03)
) }}} Graph of {{{y=6x-9}}} through the point ({{{0}}},{{{-9}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{6}}} and goes through the point ({{{0}}},{{{-9}}}), this verifies our answer.