Question

Find the points on the y-axis which are at a distance of 2 root 5 units from the point (-4,7).how many such points are there.

Answer 1

let the point be Y(0,y)
using the distance formula,
$\sqrt{16+(7-y)^{2} } =2 \sqrt{5} \\ 16+(7-y) ^{2} =20 \\ (7-y) ^{2} =4 \\ 7-y=2 \\ y=5$
so the coordinates are Y(0,5)

Answer 2

Point on Y axis be P(0, y)
Distance between two points = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
⇒$\sqrt{(-4-0)^2 + (7-y)^2} = 2\sqrt{5}$
⇒$\sqrt{16 + (7-y)^2} = 2\sqrt{5}$
Squaring on both sides
16 +  (7 - y)² = 20
⇒ (7 - y)² = 4
⇒$7 - y = \pm 2$
7 - y = 2      and      7 - y = - 2

y = 5       and y = 9

The points are  (0,5) and (0, 9)

So, there are two points.