Question 175490


If you want to find the equation of line with a given a slope of {{{5}}} which goes through the point ({{{-1}}},{{{-3}}}), 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--3=(5)(x--1)}}} Plug in {{{m=5}}}, {{{x[1]=-1}}}, and {{{y[1]=-3}}} (these values are given)



{{{y+3=(5)(x--1)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=(5)(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y+3=5x+(5)(1)}}} Distribute {{{5}}}


{{{y+3=5x+5}}} Multiply {{{5}}} and {{{1}}} to get {{{5}}}


{{{y=5x+5-3}}} Subtract 3 from  both sides to isolate y


{{{y=5x+2}}} Combine like terms {{{5}}} and {{{-3}}} to get {{{2}}} 

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Answer:



So the equation of the line with a slope of {{{5}}} which goes through the point ({{{-1}}},{{{-3}}}) 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 ({{{-1}}},{{{-3}}}),  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, -10, 8, -12, 6,
graph(500, 500, -10, 8, -12, 6,(5)x+2),
circle(-1,-3,0.12),
circle(-1,-3,0.12+0.03)
) }}} Graph of {{{y=5x+2}}} through the point ({{{-1}}},{{{-3}}})

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 ({{{-1}}},{{{-3}}}), this verifies our answer.