Find the point on the x-axis which is equidistant from (2,5) (-2,9)??

1.	Find the point on the x-axis which is equidistant from (2,5) (-2,9)??
Answer» Yogita ingle plz explain about balkan crisis.. (-7,0) Let the point of x-axis be P(x, 0)Given A(2, 5) and B(-2, 9) are equidistant from PThat is PA = PBHence PA2 = PB2 → (1)Distance between two points is\xa0√[(x2 - x1)2 +\xa0(y2\xa0- y1)2]PA =\xa0√[(2\xa0- x)2\xa0+\xa0(5\xa0- 0)2]PA2\xa0= 4 - 4x +x2 + 25 =\xa0x2 - 4x + 29Similarly,\xa0PB2\xa0=\xa0x2\xa0+ 4x + 85Equation (1) becomesx2\xa0- 4x + 29 =\xa0x2\xa0+ 4x + 85- 8x = 56x = -7Hence the point on x-axis is (-7, 0)

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