A point P moves such that its distances from (1 , 2) and (-2 , 3) are equal . Then the locus of P is

Options

A. Straight line
B. Parabola
C. Ellipse
D. Hyperbola

Answer

To find the locus of point P that moves such that its distances from the points (1, 2) and (-2, 3) are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Points**: We have two fixed points: A(1, 2) and B(-2, 3).

2. **Set Up the Distance Formula**: The distance from point P(x, y) to point A(1, 2) is given by the formula:
\[ d_A = \sqrt{(x - 1)^2 + (y - 2)^2} \]
The distance from point P(x, y) to point B(-2, 3) is:
\[ d_B = \sqrt{(x + 2)^2 + (y - 3)^2} \]

3. **Set the Distances Equal**: Since the distances are equal, we can set the two distance equations equal to each other:
\[ \sqrt{(x - 1)^2 + (y - 2)^2} = \sqrt{(x + 2)^2 + (y - 3)^2} \]

4. **Square Both Sides**: To eliminate the square roots, we square both sides of the equation:
\[ (x - 1)^2 + (y - 2)^2 = (x + 2)^2 + (y - 3)^2 \]

5. **Expand Both Sides**:
- Left Side:
\[ (x - 1)^2 + (y - 2)^2 = (x^2 - 2x + 1) + (y^2 - 4y + 4) = x^2 + y^2 - 2x - 4y + 5 \]
- Right Side:
\[ (x + 2)^2 + (y - 3)^2 = (x^2 + 4x + 4) + (y^2 - 6y + 9) = x^2 + y^2 + 4x - 6y + 13 \]

6. **Set the Expanded Equations Equal**:
\[ x^2 + y^2 - 2x - 4y + 5 = x^2 + y^2 + 4x - 6y + 13 \]

7. **Cancel Common Terms**: Subtract \(x^2 + y^2\) from both sides:
\[ -2x - 4y + 5 = 4x - 6y + 13 \]

8. **Rearrange the Equation**: Move all terms to one side:
\[ -2x - 4y - 4x + 6y + 5 - 13 = 0 \]
Simplifying gives:
\[ -6x + 2y - 8 = 0 \]

9. **Final Equation**: Multiply through by -1 to get a standard form:
\[ 6x - 2y + 8 = 0 \]

10. **Identify the Locus**: The equation \(6x - 2y + 8 = 0\) represents a straight line.

### Conclusion:
The locus of point P is a straight line given by the equation \(6x - 2y + 8 = 0\).