Question

Which point on the x-axis is equidistant from (5, 9) and (−4, 6)?

Answer 1

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

`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`

Here we are to find out a point on the x−axis which is equidistant from both the points A (5, 9) and B (−4, 6)

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((5 - x)^2 + (9 - y)^2)`

`= sqrt((5 - x)^2 + (9 - 0)^2)`

`AC = sqrt((5 - x)^2 + (9)^2)`

`BC = sqrt((-4 - x)^2 + (6 - y)^2)`

`= sqrt((4 + x)^2 + (6 - 0)^2)`

`BC = sqrt((4 + x)^2 + (6)^2)`

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

AC = BC

`sqrt((5 -x)^2 + (9)^2) = sqrt((4 + x)^2 + (6)^2)`

Squaring on both sides we have,

`(5 - x)^2 + (9)^2 = (4 + x)^2 + (6)^2`

`25 + x^2 - 10x + 81 = 16 + x^2 + 8x + 36`

18x = 54

x = 3

Hence the point on the x-axis which lies at equal distances from the mentioned points is (3,0)