Question

Find the point on x-axis which is equidistant from the points (−2, 5) and (2,−3).

Answer 1

The distance d between two points `(x_1, y_1)` and `(x_2,y_2)` is given by the formula

Here we are to find out a point on the x-axis which is equidistant from both the points A(−2,5) and B(2,−3)

Let this point be denoted as C(x, y).

Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words, we have y = 0.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

`AC = sqrt((-2-x)^2 + (5 - y)^2)`

`= sqrt((-2 - x)^2 + (5 - 0))`

`AC = sqrt((-2-x)^2 + (5)^2)`

`BC = sqrt((2 - x)^2 + (-3-0)^2)`

`= sqrt((2 - x)^2 + (-3-0)^2)`

`BC = sqrt((2 - x)^2 + (-3)^2)`

We know that both these distances are the same. So equating both these we get,

AC = BC

`sqrt((-2-x)^2 + (5)^2) = sqrt((2 - x)^2 + (-3)^2)`

Squaring on both sides we have,

`(-2-x)^2 + (5)^2 = (2 - x)^2 + (-3)^2`

`4 + x^2 + 4x + 25 = 4 + x^2 - 4x + 9`

8x = -16

x = -2

Hence the point on the x-axis which lies at equal distances from the mentioned points is (-2, 0)