Question

find a point on the x axis which is equidistant from 2, -5 and - 2, 5​

Answer 1

Step-by-step explanation:

the required point on x axis will be

as it is on x axis so y is zero

both points are equal from it

by distance formula..

underroot ( x2 - x1)^2 +( y2 - y1 )^2

underroot (x2 -2)^2 + (0-(-5))^2

underroot x^2+4 -4x2 + 25

underroot x^2 -4x +29 ------(1)

2nd condition

underroot (x-(-2))^2 + (0 -5)^2

underroot (x+2)^2 + 25

underroot x^2 + 4 + 4x +25

underroot x^2 +4x +29 _____(2)

equating both eq 1 &2

underroot x^2 -4x +29 =underroot x^2 +4x +29

x^2 -4x +29 = x^2+4x +29

-4x = 4x

-8x = 0

x =0

required coordinate is (0,0)