a moving point p is always equidistant from the points (-2, 3) and (4, -1). Find the equation of the point p

To find the equation of the locus of a point \( P(x, y) \) that is equidistant from the points \((-2, 3)\) and \( (4, -1) \), we need to set up an equation based on the definition of equidistance. 

### Step-by-step Solution

1. **Define the Distances:**

   The distance between the point \( P(x, y) \) and the point \((-2, 3)\) is given by the distance formula:
   \[
   d_1 = \sqrt{(x + 2)^2 + (y - 3)^2}
   \]

   The distance between the point \( P(x, y) \) and the point \( (4, -1) \) is:
   \[
   d_2 = \sqrt{(x - 4)^2 + (y + 1)^2}
   \]

2. **Set the Distances Equal:**

   Since \( P(x, y) \) is equidistant from both points, set \( d_1 = d_2 \):
   \[
   \sqrt{(x + 2)^2 + (y - 3)^2} = \sqrt{(x - 4)^2 + (y + 1)^2}
   \]

3. **Square Both Sides to Eliminate the Square Roots:**

   \[
   (x + 2)^2 + (y - 3)^2 = (x - 4)^2 + (y + 1)^2
   \]

4. **Expand Both Sides:**

   \[
   (x^2 + 4x + 4) + (y^2 - 6y + 9) = (x^2 - 8x + 16) + (y^2 + 2y + 1)
   \]

5. **Simplify:**

   Combine like terms:

   \[
   x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 + 2y + 1
   \]

6. **Cancel \(x^2\) and \(y^2\) from both sides:**

   \[
   4x - 6y + 13 = -8x + 2y + 17
   \]

7. **Combine Like Terms:**

   \[
   4x + 8x - 6y - 2y = 17 - 13
   \]

   \[
   12x - 8y = 4
   \]

8. **Simplify the Equation:**

   Divide every term by 4:

   \[
   3x - 2y = 1
   \]

Thus, the equation of the locus of the point \( P(x, y) \), which is equidistant from the points \((-2, 3)\) and \( (4, -1)\), is:

\[
3x - 2y = 1
\]

### Conclusion

The locus of points equidistant from two given points is a straight line called the perpendicular bisector of the segment connecting those two points. The equation \( 3x - 2y = 1 \) represents this line.

### Questions

1. What is the definition of equidistant in terms of a point and two other points?
2. How do you derive the distance formula between two points?
3. What geometric shape does the locus of points equidistant from two fixed points form?
4. How can we verify that the derived equation is correct?
5. How do you find the midpoint of the segment between two points?
6. What is the significance of squaring both sides of the equation in this context?
7. Can this method be used to find a locus for three or more points? How?
8. What happens if the points are the same, or the distance is zero?

### Tip

When solving problems involving loci of points, consider geometric interpretations such as symmetry and perpendicular bisectors to simplify your work and verify your results.

Learn how to find the equation of the locus of a point (x, y) that is equidistant from the points (-2, 3) and (4, -1). This problem involves using the distance formula and the perpendicular bisector theorem to derive a straight line equation. Suitable for Grades 9-12.

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