Find the Point on the x_axis which is equidistant from (2,-5) and (-2,

1.	Find the Point on the x_axis which is equidistant from (2,-5) and (-2,9).
Answer» 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\xa0= PB2 → (1)Distance between two points is\xa0{tex}\\sqrt{[(x_2\xa0- x_1)^2\xa0+\xa0(y_2\xa0- y_1)^2]}{/tex}PA =\xa0{tex}\\sqrt{[(2\xa0- x)^2\xa0+\xa0(-5\xa0- 0)^2]}{/tex}PA2\xa0= 4 - 4x +x2\xa0+ 25=\xa0x2\xa0- 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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