SOLUTION: What is the equation of the line with slope 2/3 and passes through the midpoint of the segment joining points (-1, 6) and (-7, -4)?

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Question 143644: What is the equation of the line with slope 2/3 and passes through the midpoint of the segment joining points (-1, 6) and (-7, -4)?
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
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 , average the x-coordinates between the two points
x%5Bmid%5D=%28-1%2B-7%29%2F2=%28-8%29%2F2=-4


So the x-coordinate of the midpoint is -4 (i.e. x=-4)
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To find , average the y-coordinates between the two points
y%5Bmid%5D=%28+6%2B+-4%29%2F2=%282%29%2F2=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%2F3 and goes through the point (-4,1)






If you want to find the equation of line with a given a slope of 2%2F3 which goes through the point (-4,1), you can simply use the point-slope formula to find the equation:


---Point-Slope Formula---
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point

So lets use the Point-Slope Formula to find the equation of the line

y-1=%282%2F3%29%28x--4%29 Plug in m=2%2F3, x%5B1%5D=-4, and y%5B1%5D=1 (these values are given)


y-1=%282%2F3%29%28x%2B4%29 Rewrite x--4 as x%2B4


y-1=%282%2F3%29x%2B%282%2F3%29%284%29 Distribute 2%2F3

y-1=%282%2F3%29x%2B8%2F3 Multiply 2%2F3 and 4 to get 8%2F3

y=%282%2F3%29x%2B8%2F3%2B1 Add 1 to both sides to isolate y

y=%282%2F3%29x%2B11%2F3 Combine like terms 8%2F3 and 1 to get 11%2F3 (note: if you need help with combining fractions, check out this solver)


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Answer:


So the equation of the line with a slope of 2%2F3 which goes through the point (-4,1) is:

y=%282%2F3%29x%2B11%2F3 which is now in y=mx%2Bb form where the slope is m=2%2F3 and the y-intercept is b=11%2F3

Notice if we graph the equation y=%282%2F3%29x%2B11%2F3 and plot the point (-4,1), we get (note: if you need help with graphing, check out this solver)

drawing%28500%2C+500%2C+-13%2C+5%2C+-8%2C+10%2C%0D%0Agraph%28500%2C+500%2C+-13%2C+5%2C+-8%2C+10%2C%282%2F3%29x%2B11%2F3%29%2C%0D%0Acircle%28-4%2C1%2C0.12%29%2C%0D%0Acircle%28-4%2C1%2C0.12%2B0.03%29%0D%0A%29+ Graph of y=%282%2F3%29x%2B11%2F3 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%2F3 and goes through the point (-4,1), this verifies our answer.