Question

Find a point on the y-axis such that its distance from the point (8,5) is 10 units.​

Answer 1

Given:

The point is (8,5).

To find:

A point on the y-axis such that its distance from the point (8,5) is 10 units.

Solution:

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

Distance formula:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

The distance between (8,5) and (0,k) is 10 units. Using distance formula, we get

`10=√((0-8)^2+(k-5)^2)`

Taking square on both sides.

`100=64+(k-5)^2`

`100-64=(k-5)^2`

`36=(k-5)^2`

Taking square root on both sides.

`\pm √(36)=k-5`

`\pm 6=k-5`

Now,

`6=k-5` and
`-6=k-5`

`6+5=k` and
`-6+5=k`

`11=k` and
`-1=k`

Therefore, the required point is either (0,11) or (0,-1).