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:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

$10=\sqrt{(0-8)^2+(k-5)^2}$

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.

$\pm \sqrt{36}=k-5$

$\pm 6=k-5$

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).