Question

Find the point(s) on the x-axis which is at a distance of `sqrt41` units from the point (8, −5).

Answer 1

Let the point on the x-axis be (x, 0).

Using the distance formula, we have

`sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2) = sqrt41`

(x − 8)² + (0 + 5)² = 41

(x − 8)² + (5)² = 41

(x − 8)² + 25 = 41

(x − 8)² = 41 − 25

(x − 8)² = 16

x − 8 = `sqrt16`

x − 8 = ± 4

x − 8 = 4 or x + 8 = −4

x = 4 + 8 or x = −4 + 8

x = 12 or x = 4

The points on the x-axis are (12, 0) and (4, 0).