To find the point on the y-axis that is equidistant from the points $(2, 3)$ and $(-1, 2)$, follow these steps:

Step 1: Use the distance formula

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

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

Let the point on the y-axis be $(0, y)$, since any point on the y-axis has an $x$-coordinate of 0. The distance from $(0, y)$ to $(2, 3)$ is:

$d_1 = \sqrt{(2 - 0)^2 + (3 - y)^2} = \sqrt{4 + (3 - y)^2}$

The distance from $(0, y)$ to $(-1, 2)$ is:

$d_2 = \sqrt{(-1 - 0)^2 + (2 - y)^2} = \sqrt{1 + (2 - y)^2}$

Step 2: Set the distances equal

Since the point on the y-axis is equidistant from both points, set $d_1 = d_2$:

$\sqrt{4 + (3 - y)^2} = \sqrt{1 + (2 - y)^2}$

Step 3: Square both sides

Square both sides to eliminate the square roots:

$4 + (3 - y)^2 = 1 + (2 - y)^2$

Step 4: Expand both sides

First, expand the squared terms:

$(3 - y)^2 = 9 - 6y + y^2$ $(2 - y)^2 = 4 - 4y + y^2$

Now substitute these into the equation:

$4 + 9 - 6y + y^2 = 1 + 4 - 4y + y^2$

Step 5: Simplify the equation

Combine like terms:

$13 - 6y + y^2 = 5 - 4y + y^2$

Now cancel the $y^2$ terms from both sides:

$13 - 6y = 5 - 4y$

Step 6: Solve for $y$

Move terms involving $y$ to one side and constants to the other side:

$13 - 5 = -4y + 6y$ $8 = 2y$ $y = 4$

Final Answer:

The point on the y-axis that is equidistant from $(2, 3)$ and $(-1, 2)$ is $(0, 4)$.

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Would you like more details on any part of the solution? Feel free to ask!

Here are some related questions you might find interesting:

1. How would the solution change if the point was on the x-axis instead of the y-axis?
2. How can you generalize this method for points not on the axes?
3. Can we apply this method to points in three-dimensional space?
4. How do you calculate the midpoint of two points geometrically?
5. What is the relationship between the distance formula and the Pythagorean theorem?

Tip: When solving distance-related problems, always be mindful of the distance formula and the geometry of the problem—especially how symmetry (like being equidistant) can simplify the algebra.