Question

-------50 points-----

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

Answer 1

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

Given A(2, -5) and B(-2, 9) are equidistant from PThat is PA = PB

 Hence PA2 = PB2  →

(1)Distance between two points is √[(x2 - x1)2+ (y2 - y1)2] PA = √[(2 - x)2 + (-5 - 0)2] PA2 = 4 - 4x +x2 + 25 = x2 - 4x + 29 

Similarly, PB2 = x2 + 4x + 85 Equation (1) becomes x2 - 4x + 29 = x2 + 4x + 85 - 8x = 56 x = -7

 Hence the point on x-axis is (-7, 0)

Answer 2

hope this answer is helpful