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,
[tex]\sqrt{(4-x)^2+(0-0)^2}=\sqrt{(x-0)^2+(0-9)^2}[/tex]
Now, squaring both the sides:
[tex]\begin{gathered} (4-x)^2=x^2+9^2 \\ 16+x^2-8x=x^2+81 \\ 8x=-65 \\ x=\frac{-65}{8} \end{gathered}[/tex]
Thus, the coordinates of point P are (-65/8, 0).
