Question

Find the point on x-axis which is equi-distance 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)