To find:
The coordinates of a point P such that PA = PB.
Solution:
Given that A(4, 0) and B(0, 9) are the coordinates.
Let the point P is (x,0) because the point is on x-axis, and it is given that |PA| = |PB|.
So,

Now, squaring both the sides:

Thus, the coordinates of point P are (-65/8, 0).