Question

Which point on the y-axis is equidistant from (2, 3)  and (−4, 1)?

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 y-axis which is equidistant from both the points A (2, 3) and B (−4, 1).

Let this point be denoted as C(x, y).

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

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

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

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

`AC = sqrt((2)^2 + (3 - y)^2)`

`BC = sqrt((-4-x)^2 + (1 - y)^2)`

`= sqrt((-4-0)^2 + (1 - y)^2)`

`BC = sqrt((-4)^2 + (1 - y)^2)`

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

AC = BC

`sqrt((2)^2 + (3 - y)^2) = sqrt((-4)^2 + (1 - y)^2)`

`(2)^2 + (3 - y)^2 = (-4)^2 + (1 - y)^2`

`4 + 9 + y^2 - 6y = 16 + 1 + y^2 - 2y`

4y = -4

y = -1

Hence the point on the y-axis which lies at equal distances from the mentioned points is (0, -1)