Find the points on the y-axis which are at a distance of $2{\sqrt5}$ units from the point (-4, 7).

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

Let (0, y) = (x₁, y₁) and (-4, 7) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4 - 0)^2 + (7 - y)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4)^2 + (7 - y)^2}\\[1em] ⇒ (2 \sqrt5)^2 = (-4)^2 + (7 - y)^2\\[1em] ⇒ 20 = 16 + 49 + y^2 - 14y\\[1em] ⇒ 20 = 65 + y^2 - 14y\\[1em] ⇒ 65 + y^2 - 14y - 20 = 0\\[1em] ⇒ y^2 - 14y + 45 = 0\\[1em] ⇒ y^2 - (9y + 5y) + 45 = 0\\[1em] ⇒ y^2 - 9y - 5y + 45 = 0\\[1em] ⇒ (y^2 - 9y) - (5y - 45) = 0\\[1em] ⇒ y(y - 9) - 5(y - 9) = 0\\[1em] ⇒ (y - 9)(y - 5) = 0\\[1em] ⇒ y = 9 \text{ or } 5

Hence, the the points on the y-axis are (0, 9) or (0, 5).