Question

The abscissa of a point A is twice its ordinate and B ≡ (10, 0). Find the coordinates of A if AB = 5 units.

Answer 1

Given

The abscissa (x-coordinate) of point A is twice its ordinate (y-coordinate).

Point B = (10, 0)

The distance AB = 5 units.

Let the coordinates of A be (x, y).

Since the abscissa of A is twice its ordinate, we have: x = 2y

∴ coordinates of A are: A = (2y, y)

Using the distance formula between points A and B: \[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substitute A = (2y, y)  and (B = (10, 0): \[5 = \sqrt{(10 - 2y)^2 + (0 - y)^2}\]

Square both sides:

25 = (10 − 2y)² + (−y)²

25 = (10 − 2y)² + y²

25 = (10)² − 2 × 10 × 2y + (2y)² + y²

25 = 100 − 40y + 4y² + y²

25 = 100 − 40y + 5y²

= 5y² − 40y + 100 = 25   ... [Rearrange to form a quadratic equation]

5y² − 40y + 75 = 0

y² − 8y + 15 = 0     ...[Divide through by 5 to simplify]

y² − 8y + 15 = (y − 3) (y − 5) = 0     ...[Factorize the quadratic equation]

y = 3 or y = 5

x = 2y For (y = 3), (x = 6)     ...[Find corresponding values of (x)]

For (y = 5), (x = 10)

A = (6, 3) or A = (10, 5)

A = (10, 5) and B = (10, 0)

\[ AB = \sqrt{(10-10)^2 + (5-0)^2} = \sqrt{0 + 25} = 5 \]

The coordinates of point A are (6, 3) or (10, 5)