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+2 votes
74.0k views
in Mathematics by (63.2k points)
edited by
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

1 Answer

+3 votes
by (13.2k points)
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Best answer

Solution:
Let the point on x-axis be P(x, 0)

Let A be (7, 6) and B be (3, 4)

Now PA = image

PA2 = (x – 7)2 + 36

and PB = image

PB2 = (x – 3)2 + 16

Given PA = PB

or PA= PB2

or (x – 7)2 + 36 = (x – 3)2 + 16

or x2 – 14x + 49 + 36 = x2 + 9 – 6x + 16

or -14x + 6x = 25 – 85

or -8x = -60

or 8x = 60

or 2x = 15 ⇒ x = 15/2

Hence the required point is (15/2, 0)

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