A point A is at a distance of $\sqrt{10}$ units from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

Let (a, 2a) = (x₁, y₁) and (4, 3) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] ⇒ \sqrt{10} = \sqrt{(4 - a)^2 + (3 - 2a)^2}\\[1em] ⇒ 10 = (4 - a)^2 + (3 - 2a)^2\\[1em] ⇒ 10 = 16 + a^2 - 8a + 9 + 4a^2 - 12a\\[1em] ⇒ 10 = 25 + 5a^2 - 20a\\[1em] ⇒ 25 + 5a^2 - 20a - 10 = 0\\[1em] ⇒ 5a^2 - 20a + 15 = 0\\[1em] ⇒ a^2 - 4a + 3 = 0\\[1em] ⇒ a^2 - (3a + 1a) + 3 = 0\\[1em] ⇒ a^2 - 3a - 1a + 3 = 0\\[1em] ⇒ a(a - 3) - 1(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 1) = 0\\[1em] ⇒ a = 3 \text{ and } 1

For each value of a = x, we can find the corresponding value of y:

• If a = 3, then y = 2a = 6
• If a = 1, then y = 2a = 2

Hence, the co-ordinates of point A are (3, 6) and (1, 2).