Question

Find the coordinates of a point on x-axis which is equidistant from A(2, −4) and B(8, 4).

Answer 1

Step 1: General form of a point on the x-axis P(x, 0)

Since the point lies on the x-axis, its y-coordinate is 0.

Step 2: Use the distance formula

We use the distance formula to equate the distances from P to A and P to B.

PA = `sqrt((x − 2)^2 + (0 + 4)^2)`

PA = `sqrt((x − 2)^2 + 16)`

PB = `sqrt((x − 8)^2 + (0 − 4)^2)`

PB = `sqrt((x − 8)^2 + 16)`

Step 3: Equating the distances

= `sqrt((x −2)^2 + 16)`

= `sqrt((x − 8)^2 + 16)`

Step 4: Square both sides

= `(x − 2)^2 + 16 = (x − 8)^2 + 16`

Subtract 16 from both sides:

(x − 2)² = (x − 8)²

x² − 4x + 4 = x² − 16x + 64

−4x + 4 = −16x + 64

12x = 60

x = 5

The point on the x-axis that is equidistant from A(2, −4) and (8, 4) is:

(5, 0).