To find the point on the $y$-axis that is equidistant from the points $(-5, -2)$ and $(3, 2)$, we can follow these steps:

1. Identify the coordinates of the point on the $y$-axis:
   Any point on the $y$-axis has coordinates of the form $(0, y)$.

2. Set up the distance formulas:
   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's compute the distances from the point $(0, y)$ to each of the two given points:

   • Distance to $(-5, -2)$: $d_1 = \sqrt{(0 - (-5))^2 + (y - (-2))^2} = \sqrt{(5)^2 + (y + 2)^2}$

   • Distance to $(3, 2)$: $d_2 = \sqrt{(0 - 3)^2 + (y - 2)^2} = \sqrt{(-3)^2 + (y - 2)^2}$

3. Set the distances equal to each other (since the point $(0, y)$ is equidistant from both points): $\sqrt{5^2 + (y + 2)^2} = \sqrt{(-3)^2 + (y - 2)^2}$

4. Square both sides to eliminate the square roots: $5^2 + (y + 2)^2 = (-3)^2 + (y - 2)^2$

5. Simplify the equation: $25 + (y + 2)^2 = 9 + (y - 2)^2$

6. Expand the squares: $25 + (y^2 + 4y + 4) = 9 + (y^2 - 4y + 4)$

   $25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4$

7. Combine like terms: $29 + y^2 + 4y = 13 + y^2 - 4y$

   $29 + 4y = 13 - 4y$

8. Solve for $y$: $4y + 4y = 13 - 29$

   $8y = -16$

   $y = -2$

9. Conclusion:
   The point on the $y$-axis that is equidistant from $(-5, -2)$ and $(3, 2)$ is $(0, -2)$.

Would you like more details or have any questions about this solution?

Related Questions

1. How do you find a point equidistant from three points in a plane?
2. What is the formula for the distance between a point and a line?
3. How do you find the perpendicular bisector of a line segment in coordinate geometry?
4. What is the shortest distance from a point to a line?
5. How do you calculate the midpoint of a line segment in a coordinate plane?

Tip

When solving geometric problems involving distances, sketching the situation on a graph can provide a visual aid and help clarify the relationships between points and lines.