SOLUTION: a. The point on the x-axis that is equidistant (equal distance) from (8,4) and (-4,2) b. The distance between the points (8,2) and (16,2). 8 c. The distance between the points

Algebra ->  Rational-functions -> SOLUTION: a. The point on the x-axis that is equidistant (equal distance) from (8,4) and (-4,2) b. The distance between the points (8,2) and (16,2). 8 c. The distance between the points       Log On


   



Question 387349: a. The point on the x-axis that is equidistant (equal distance) from (8,4) and (-4,2)
b. The distance between the points (8,2) and (16,2). 8
c. The distance between the points (2,6) and (12,8).2
d. The midpoint between (8,2) and (10,6).
e. The midpoint between (-4, 2) and (-5, 3).

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a. The point on the x-axis that is equidistant (equal distance) from (8,4) and (-4,2)
The point will be (x,0)
d%5E2+=+diffy%5E2+%2B+diffx%5E2 Use d squared, since it will be equal if d is =
d%5E2+=+4%5E2+%2B+%288-x%29%5E2+=+2%5E2+%2B+%28-4-x%29%5E2
16+%2B+%288-x%29%5E2+=+4+%2B+%28-4-x%29%5E2
12+%2B+64+-+16x+%2B+x%5E2+=+16+%2B+8x+%2B+x%5E2
76 - 16x = 16 + 8x
60 = 24x
x = 2.5
---> (2.5,0)
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b. The distance between the points (8,2) and (16,2). 8
8 is right.
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c. The distance between the points (2,6) and (12,8).2
How did you get 2?
d+=+sqrt%28diffy%5E2+%2B+diffx%5E2%29+=+sqrt%2810%5E2+%2B+2%5E2%29+=+sqrt%28104%29
d=~ 10.198 units
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d. The midpoint between (8,2) and (10,6)
Find the average of x and y separately
x: (8 + 10)/2 = 0
y: (2 + 6)/2 = 4
--> (0,4)
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e. The midpoint between (-4, 2) and (-5, 3)
Same method as d --> (-4.5,2.5)