Find the point on the y-axis that is equidistant from (-5,-2) and (3, 2).


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 \((0, -2)\).

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### 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.

Learn how to find the point on the y-axis equidistant from the points (-5, -2) and (3, 2) using coordinate geometry and the distance formula. This problem involves calculating distances and solving equations to determine the exact coordinates. Suitable for students in Grades 9-12.

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