Verified answer

Answer:

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

Given that:-

A(3, 2) and B(-5, -20) are equidistant from P

PA = PB

=> PA² = PB²

Distance between two points is √[(x2 - x1)² + (y2 - y1)²]

=> PA = √[3 - x)² + (2 - 0)²]

=> PA² = 9 + X² + 6X + 4

=> PA² = 13 + X² + 6X

AND,

PB = √[-5 - x)² + (-20 - 0)²]

=> PB² = [25 + X² + 10X + 400]

=> PB² = 425 + X² + 10X

NOW,

PA² = PB²

=> 13 + X² + 6X = 425 + x² + 10x

=> x² - x² = 10x - 6x + 425 - 13

=> 4x = -412

=> x = -412/4

=> x = -103

Hence the point on x-axis is (-103, 0)