Question

Find the 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 P
That 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

Let the point of x-axis be C(x,0)
As points A(2, -5) and B(-2, 9) are equidistant from C
Therefore, PA = PB
∴ √(x-2)² +(0+5)²  = √(x+2)² + (9-0)²
   squaring on both sides.
  x²+ 4 - 4x +25  = x² +4 + 4x + 81
 -4x - 4x = 81- 25
 -8x = 56
 x = -56/8
 x = - 7
hence, pt on the x-axis is (-7,0)