Question

Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.

Answer 1

The value of (x, y) is (0, ± √7)

Step-by-step explanation:

Let the point (x, y) be P.

Let the point (-3, 0) be Q.

Let the point (3, 0) be R.

Now, the distance between P and Q is:

PQ = √((x + 3)² - (y - 0)²)

4 = √(x² + 9 + 6x - y²)

On squaring both sides, we get,

16 = x² + 9 + 6x - y²

x² + y² = 16 - 9 - 6x

x² + y² = 7 - 6x → (equation 1)

PR = √((x - 3)² + (y + 0)²)

4 = √(x² + 9 - 6x + y²)

On squaring both sides, we get,

16 = x² + 9 - 6x + y²

x² + y² = 16 - 9 + 6x

x² + y² = 7 + 6x → (equation 2)

On equating equation (1) and (2), we get,

7 - 6x = 7 + 6x

12x = 0

∴ x = 0

On substituting the value of 'x' in equation (1), we get,

0² + y² = 7 - 6(0)

y² = 7

y = √7

∴ y = ± √7