Question 143644
First let's find the midpoint of the segment through (-1, 6) and (-7, -4)



In order to find the midpoint between the points (-1, 6) and (-7, -4), we need to average each corresponding coordinate. In other words, we need to add up the corresponding coordinates and divide the sum by 2.



So lets find the averages between the two points




To find *[Tex \Large  \textrm{x_{mid}}], average the x-coordinates between the two points

{{{x[mid]=(-1+-7)/2=(-8)/2=-4}}}



So the x-coordinate of the midpoint is -4 (i.e. x=-4)

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To find *[Tex \Large  \textrm{y_{mid}}], average the y-coordinates between the two points

{{{y[mid]=( 6+ -4)/2=(2)/2=1}}}



So the y-coordinate of the midpoint is 1 (i.e. y=1)



So  the midpoint is (-4,1)





Now let's find the equation of the line with a slope of {{{2/3}}} and goes  through the point (-4,1)







If you want to find the equation of line with a given a slope of {{{2/3}}} which goes through the point ({{{-4}}},{{{1}}}), 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-1=(2/3)(x--4)}}} Plug in {{{m=2/3}}}, {{{x[1]=-4}}}, and {{{y[1]=1}}} (these values are given)



{{{y-1=(2/3)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y-1=(2/3)x+(2/3)(4)}}} Distribute {{{2/3}}}


{{{y-1=(2/3)x+8/3}}} Multiply {{{2/3}}} and {{{4}}} to get {{{8/3}}}


{{{y=(2/3)x+8/3+1}}} Add 1 to  both sides to isolate y


{{{y=(2/3)x+11/3}}} Combine like terms {{{8/3}}} and {{{1}}} to get {{{11/3}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)



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Answer:



So the equation of the line with a slope of {{{2/3}}} which goes through the point ({{{-4}}},{{{1}}}) is:


{{{y=(2/3)x+11/3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2/3}}} and the y-intercept is {{{b=11/3}}}


Notice if we graph the equation {{{y=(2/3)x+11/3}}} and plot the point ({{{-4}}},{{{1}}}),  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, -8, 10,
graph(500, 500, -13, 5, -8, 10,(2/3)x+11/3),
circle(-4,1,0.12),
circle(-4,1,0.12+0.03)
) }}} Graph of {{{y=(2/3)x+11/3}}} through the point ({{{-4}}},{{{1}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{2/3}}} and goes through the point ({{{-4}}},{{{1}}}), this verifies our answer.