Question 118783

If you want to find the equation of line with a given a slope of {{{5}}} 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=(5)(x-0)}}} Plug in {{{m=5}}}, {{{x[1]=0}}}, and {{{y[1]=-2}}} (these values are given)

{{{y+2=(5)(x-0)}}} Rewrite {{{y--2}}} as {{{y+2}}}

{{{y+2=5x+(5)(-0)}}} Distribute {{{5}}}

{{{y+2=5x+0}}} Multiply {{{5}}} and {{{-0}}} to get {{{0}}}

{{{y=5x+0-2}}} Subtract 2 from  both sides to isolate y

{{{y=5x-2}}} Combine like terms {{{0}}} and {{{-2}}} to get {{{-2}}}

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So the equation of the line with a slope of {{{5}}} which goes through the point ({{{0}}},{{{-2}}}) is:

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

Notice if we graph the equation {{{y=5x-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, -11, 7,
graph(500, 500, -9, 9, -11, 7,(5)x+-2),
circle(0,-2,0.12),
circle(0,-2,0.12+0.03)
) }}} Graph of {{{y=5x-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 {{{5}}} and goes through the point ({{{0}}},{{{-2}}}), this verifies our answer.