Question 243697
# 1


Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(5,3\right)]. So this means that {{{x[1]=5}}} and {{{y[1]=3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(8,-4\right)].  So this means that {{{x[2]=8}}} and {{{y[2]=-4}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(-4-3)/(8-5)}}} Plug in {{{y[2]=-4}}}, {{{y[1]=3}}}, {{{x[2]=8}}}, and {{{x[1]=5}}}



{{{m=(-7)/(8-5)}}} Subtract {{{3}}} from {{{-4}}} to get {{{-7}}}



{{{m=(-7)/(3)}}} Subtract {{{5}}} from {{{8}}} to get {{{3}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(5,3\right)] and *[Tex \LARGE \left(8,-4\right)] is {{{m=-7/3}}}



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# 2


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



{{{y+2=(-1/5)(x-0)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(-1/5)x+(-1/5)(-0)}}} Distribute {{{-1/5}}}



{{{y+2=(-1/5)x+0}}} Multiply {{{-1/5}}} and {{{-0}}} to get {{{0}}}



{{{y=(-1/5)x+0-2}}} Subtract 2 from  both sides to isolate y



{{{y=(-1/5)x-2}}} Combine like terms

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Answer:



So the equation of the line with a slope of {{{-1/5}}} which goes through the point (0,-2) is:


{{{y=(-1/5)x-2}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-1/5}}} and the y-intercept is {{{b=-2}}}


Notice if we graph the equation {{{y=(-1/5)x-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,(-1/5)x+-2),
circle(0,-2,0.12),
circle(0,-2,0.12+0.03)
) }}} 

Graph of {{{y=(-1/5)x-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 {{{-1/5}}} and goes through the point (0,-2), this verifies our answer.