Question

Find the points on the x-axis which are at a distance of `2sqrt(5)` from the point (7, – 4). How many such points are there?

Answer 1

We know that, every point on the x-axis in the form (x, 0).

Let P(x, 0) the point on the x-axis have `2sqrt(5)` distance from the point Q(7, – 4).

By given condition,

PQ = `2sqrt(5)`  ...`[∵ "Distance formula" = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`

⇒ (PQ)² = 4 × 5

⇒ (x – 7)² + (0 + 4)² = 20

⇒ x² + 49 – 14x + 16 = 20

⇒ x² – 14x + 65 – 20 = 0

⇒ x² – 14x + 45 = 0

⇒ x² – 9x – 5x + 45 = 0   ...[By factorisation method]

⇒ x(x – 9) – 5(x – 9) = 0

⇒ (x – 9)(x – 5) = 0

∴ x = 5, 9

Hence, there are two points lies on the axis, which are (5, 0) and (9, 0), have `2sqrt(5)` distance from the point (7, – 4).