Question

If the point A(x, y) is equidistant from the points B(–2, 0) and C(2, 0), prove that the point A lies on y-axis. Also, find the coordinates of the point A, if ΔABC is an equilateral triangle.

Answer 1

1. Prove A lies on the y-axis

Since point A(x, y) is equidistant from B(–2, 0) and C(2, 0), the distances AB and AC are equal.

Using the distance formula `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`:

`AB = sqrt((x - (-2))^2 + (y - 0)^2)`

= `sqrt((x + 2)^2 + y^2)`

`AC = sqrt((x - 2)^2 + (y - 0)^2)`

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

Since AB = AC, their squares are also equal:

(x + 2)² + y² = (x – 2)² + y²

Expanding both sides:

x² + 4x + 4 + y² = x² – 4x + 4 + y²

4x = –4x

⇒ 8x = 0

⇒ x = 0

Any point with an x-coordinate of 0 lies on the y-axis.

2. Find coordinates for an equilateral triangle

In an equilateral triangle, all sides are equal.

First, find the length of the base BC:

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

= `sqrt(4^2)`

= 4 units

Since the triangle is equilateral, AB = BC = 4.

Using the distance AB with x = 0:

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

2² + y² = 4²

4 + y² = 16

⇒ y² = 12

`y = ± sqrt(12)`

= `± 2sqrt(3)`

The coordinates of point A are either `(0, 2sqrt(3))` or `(0, -2sqrt(3))`.

The point A lies on the y-axis because its x-coordinate must be 0 to be equidistant from B(–2, 0) and C(2, 0) and its coordinates in an equilateral triangle are `(0, 2sqrt(3))` or `(0, -2sqrt(3))`.