Question

Find a point on the x-axis which is equidistant from the points (7, 6) and (−3, 4).

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(7,6) and B(−3,4).

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((7 - x)^2 + (6 - y)^2)`

`= sqrt((7 - x)^2 + (6 - 0)^2)`

`AC = sqrt((7-x)^2 + (6)^2)`

`BC= sqrt((-3-x)^2 + (4- y)^2)`

`= sqrt((-3-x)^2 + (4 - 0)^2)`

`BC = sqrt((-3-x)^2 + (4)^2)`

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

AC = BC

`sqrt((7 - x)^2 + (6)^2) = sqrt((-3-x)^2 + (4)^2)` 

Squaring on both sides we have,

`(7 -x)^2 + (6)^2 = (-3-x)^2+ (4)^2`

`49 + x^2 -14x + 36 = 9 + x^2 + 6x + 16`

20x = 60

x = 3

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