Question 113889


If you want to find the equation of line with a given a slope of {{{3}}} which goes through the point ({{{-7}}},{{{-15}}}), 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--15=(3)(x--7)}}} Plug in {{{m=3}}}, {{{x[1]=-7}}}, and {{{y[1]=-15}}} (these values are given)



{{{y+15=(3)(x--7)}}} Rewrite {{{y--15}}} as {{{y+15}}}



{{{y+15=(3)(x+7)}}} Rewrite {{{x--7}}} as {{{x+7}}}



{{{y+15=3x+(3)(7)}}} Distribute {{{3}}}


{{{y+15=3x+21}}} Multiply {{{3}}} and {{{7}}} to get {{{21}}}


{{{y=3x+21-15}}} Subtract 15 from  both sides to isolate y


{{{y=3x+6}}} Combine like terms {{{21}}} and {{{-15}}} to get {{{6}}} 

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Answer:



So the equation of the line with a slope of {{{3}}} which goes through the point ({{{-7}}},{{{-15}}}) is:


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


Notice if we graph the equation {{{y=3x+6}}} and plot the point ({{{-7}}},{{{-15}}}),  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, -16, 2, -24, -6,
graph(500, 500, -16, 2, -24, -6,(3)x+6),
circle(-7,-15,0.12),
circle(-7,-15,0.12+0.03)
) }}} Graph of {{{y=3x+6}}} through the point ({{{-7}}},{{{-15}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{3}}} and goes through the point ({{{-7}}},{{{-15}}}), this verifies our answer.