Q. 45.0( 3 Votes )

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Find a point on the x-axis, which is equidistant from the points(7, 6) and (3,4).

Answer :

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

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

Now PA =

PA^{2} = (x – 7)^{2} + 36

and PB =

PB^{2} = (x – 3)^{2} + 16

Given PA = PB

or PA^{2 }= PB^{2}

or (x – 7)^{2} + 36 = (x – 3)^{2} + 16

or x^{2} – 14x + 49 + 36 = x^{2} + 9 – 6x + 16

or -14x + 6x = 25 – 85

or -8x = -60

or 8x = 60

or 2x = 15 ⇒ x =

Hence the required point is

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