SOLUTION: Find the point on the x-axis that is equidistant from (-4,-3) and (-1,5).

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Question 941945: Find the point on the x-axis that is equidistant from (-4,-3) and (-1,5).
Found 2 solutions by Edwin McCravy, MathLover1:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find the point on the x-axis that is equidistant from (-4,-3) and (-1,5)



Every point on the x-axis has its y-coordinate as 0.  So we label
the point we are looking for (x,0).

Then we use the distance formula to set the distances from (x,0) to
the two given points equal to each other:

sqrt%28%28x%5E%22%22-%28-1%29%29%5E2%2B%280-5%29%5E2%29%22%22=%22%22sqrt%28%28x%5E%22%22-%28-4%29%29%5E2%2B%280-%28-3%29%29%5E2%29

sqrt%28%28x%2B1%29%5E2%2B%28-5%29%5E2%29%22%22=%22%22sqrt%28%28x%2B4%29%5E2%2B%280%2B3%29%5E2%29

sqrt%28%28x%2B1%29%5E2%2B25%29%22%22=%22%22sqrt%28%28x%2B4%29%5E2%2B%283%29%5E2%29

sqrt%28%28x%2B1%29%5E2%2B25%29%22%22=%22%22sqrt%28%28x%2B4%29%5E2%2B9%29

%28x%2B1%29%5E2%2B25%22%22=%22%22%28x%2B4%29%5E2%2B9

x%5E2%2B2x%2B1%2B25%22%22=%22%22x%5E2%2B8x%2B16%2B9

x%5E2%2B2x%2B26%22%22=%22%22x%5E2%2B8x%2B25

2x%2B26%22%22=%22%228x%2B25

1%22%22=%22%226x

1%2F6%22%22=%22%22x

So the point on the x-axis is %28matrix%281%2C3%2C1%2F6%2C%22%2C%22%2C0%29%29

Edwin

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Find the point on the x-axis that is equidistant from (-4,-3) and (-1,5) .
Since this unknown point is on the x axis, then its y coordinate = 0.
So this unknown point is (x, 0).
The distance from(x, 0) to (-4,-3) must equal the distance from(x, 0) (x, 0) to (-1,5).
The formula for the distance between points (x%5B1%5D, y%5B1%5D) and (x%5B2%5D, y%5B2%5D) is:
%28distance%29%5E2+=+%28x%5B2%5D+-+x%5B1%5D%29%5E2+%2B+%28y%5B2%5D+-+y%5B1%5D%29%5E2
So the distance between (x, 0) and (-4,-3) is
%28distance%29%5E2+=+%28-4+-+x%29%5E2+%2B+%28-3+-+0%29%5E2
%28distance%29%5E2+=+%28-4+-+x%29%5E2+%2B+%28-3+%29%5E2
%28distance%29%5E2+=+%28-4+-+x%29%5E2+%2B+9
And the distance between (x, 0) and (-1,5) is
%28distance%29%5E2+=+%28-1+-+x%29%5E2+%2B+%285+-+0%29%5E2
%28distance%29%5E2+=+%28-1-+x%29%5E2+%2B+%285%29%5E2
%28distance%29%5E2+=+%28-1-+x%29%5E2+%2B+25
Remembering that these two distances must be equal, we set both equations equal to each other.
%28-4+-+x%29%5E2+%2B+9=+%28-1-+x%29%5E2+%2B+25
16+%2B8x%2Bx%5E2+%2B+9=+1%2B2x%2Bx%5E2+%2B+25
cross%2825%29+%2B8x%2Bcross%28x%5E2+%29=+1%2B2x%2Bcross%28x%5E2%29+%2B+cross%2825%29
8x=+1%2B2x
8x-2x=+1
6x=+1
x=+1%2F6
so, (x, 0) =(1%2F6, 0)
and (1%2F6, 0) is equidistant from points (-4,-3) and (-1,5)