Question 89182
If you want to find the equation of line with a given a slope of {{{3}}} which goes through the point ({{{-4}}},{{{-17}}}), 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--17=(3)(x--4)}}} Plug in {{{m=3}}}, {{{x[1]=-4}}}, and {{{y[1]=-17}}} (these values are given)


{{{y--17=(3)x-(3)(-4))}}} Distribute {{{3}}}


{{{y+17=(3)x+12}}} Multiply {{{-3}}} and {{{-4}}} to get {{{12}}}


{{{y=(3)x+12+-17}}}Subtract {{{-17}}} from both sides



{{{y=3x-5}}} Combine like terms

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Answer:



So the equation of the line with a slope of {{{3}}} which goes through the point ({{{-4}}},{{{-17}}}) is:


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


Notice if we graph the equation {{{y=3x-5}}} and plot the point ({{{-4}}},{{{-17}}}),  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, -13, 5, -26, -8,
graph(500, 500, -13, 5, -26, -8,(3)x+-5),
circle(-4,-17,0.12),
circle(-4,-17,0.12+0.03)
) }}} Graph of {{{y=3x-5}}} through the point ({{{-4}}},{{{-17}}})

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 ({{{-4}}},{{{-17}}}), this verifies our answer.