The problem asks to find the point $(0, b)$ on the y-axis that is equidistant from the points $(5, 5)$ and $(5, -4)$.

Step 1: Use the distance formula.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

We need the distance from $(0, b)$ to both points:

1. From $(0, b)$ to $(5, 5)$: $d_1 = \sqrt{(5 - 0)^2 + (5 - b)^2} = \sqrt{25 + (5 - b)^2}$

2. From $(0, b)$ to $(5, -4)$: $d_2 = \sqrt{(5 - 0)^2 + (-4 - b)^2} = \sqrt{25 + (-4 - b)^2}$

Since the distances are equal: $\sqrt{25 + (5 - b)^2} = \sqrt{25 + (-4 - b)^2}$

Step 2: Square both sides to eliminate the square roots:

$25 + (5 - b)^2 = 25 + (-4 - b)^2$

Step 3: Simplify the equation:

$(5 - b)^2 = (-4 - b)^2$

Step 4: Expand both sides:

$(25 - 10b + b^2) = (16 + 8b + b^2)$

Step 5: Cancel out the $b^2$ terms and simplify further:

$25 - 10b = 16 + 8b$

Step 6: Solve for $b$:

$25 - 16 = 10b + 8b$ $9 = 18b$ $b = \frac{9}{18} = \frac{1}{2}$

Final Answer:

The value of $b$ is $\frac{1}{2}$, so the point on the y-axis is $(0, \frac{1}{2})$.

Would you like further details or explanations on any of these steps?

Here are some related questions to explore further:

1. How do you find the midpoint between two points?
2. What are the general steps to solve a quadratic equation?
3. Can we generalize this problem to find points equidistant from two arbitrary points?
4. How does the distance formula change in three dimensions?
5. What are the applications of the distance formula in real-world problems?

Tip: Whenever you have square roots in an equation, squaring both sides is a powerful tool, but always remember to check for extraneous solutions!