Question

find the coordinates of the point on the y axis which are at the distance of 13 units from the point (12,9)

Answer 1

Answer:

There are two points: (0, 4) and (0, 14).

Step-by-step explanation:

Let the point have coordinates (0, y).  (The x-coordinate is 0 since it is on the y-axis.)

The square of the distance from this point to (12, 9) needs to be 13² = 169 (it's easier and tidier to work with squares of distances).

But the square of the distance from (0, y) to (12, 9) is given by the sum of the squares of the differences in the coordinates, so:

( y - 9 )² + 12² = 13²

=> ( y - 9 )² = 13² - 12² = 169 -144 = 25

=> y - 9 = ±5

=> y = 9 ± 5

=> y = 4 or 14.

Answer 2

Answer:

Step-by-step explanation:

LEt d(AB)=13

D(AB)=√(12-0)^2+(9-y)^2

13=√144+(9-y)^2

(13)^2=144+(9-y)^2......squaring both sides

169-144=(9-y)^2

√25=9-y

5-9=-y

Y=4