P(x, y) moves such that the area of the triangle with vertices at P(x, y), (1, -2), (-1, 3) is equal to the area of the triangle with vertices at P(x, y), (2, -1), (3, 1). The locus of P is

Answer

To find the locus of the point \( P(x, y) \) such that the area of the triangle formed with the vertices \( P(x, y) \), \( (1, -2) \), and \( (-1, 3) \) is equal to the area of the triangle formed with the vertices \( P(x, y) \), \( (2, -1) \), and \( (3, 1) \), we can follow these steps:

### Step 1: Area of the First Triangle
The area of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For the first triangle with vertices \( P(x, y) \), \( (1, -2) \), and \( (-1, 3) \):
- \( x_1 = x, y_1 = y \)
- \( x_2 = 1, y_2 = -2 \)
- \( x_3 = -1, y_3 = 3 \)

Substituting these values into the area formula gives:
\[ \text{Area}_1 = \frac{1}{2} \left| x(-2 - 3) + 1(3 - y) + (-1)(y + 2) \right| \]
\[ = \frac{1}{2} \left| -5x + 3 - y - 2 \right| \]
\[ = \frac{1}{2} \left| -5x - y + 1 \right| \]

### Step 2: Area of the Second Triangle
For the second triangle with vertices \( P(x, y) \), \( (2, -1) \), and \( (3, 1) \):
- \( x_1 = x, y_1 = y \)
- \( x_2 = 2, y_2 = -1 \)
- \( x_3 = 3, y_3 = 1 \)

Substituting these values into the area formula gives:
\[ \text{Area}_2 = \frac{1}{2} \left| x(-1 - 1) + 2(1 - y) + 3(y + 1) \right| \]
\[ = \frac{1}{2} \left| -2x + 2 - 2y + 3y + 3 \right| \]
\[ = \frac{1}{2} \left| -2x + y + 5 \right| \]

### Step 3: Setting the Areas Equal
Since the areas of the two triangles are equal, we can set the two area expressions equal to each other:
\[ \frac{1}{2} \left| -5x - y + 1 \right| = \frac{1}{2} \left| -2x + y + 5 \right| \]
Removing the \( \frac{1}{2} \) from both sides gives:
\[ \left| -5x - y + 1 \right| = \left| -2x + y + 5 \right| \]

### Step 4: Removing the Absolute Values
This equation can be split into two cases:

**Case 1:**
\[ -5x - y + 1 = -2x + y + 5 \]
Rearranging gives:
\[ -5x + 2x - y - y + 1 - 5 = 0 \implies -3x - 2y - 4 = 0 \implies 3x + 2y = -4 \]

**Case 2:**
\[ -5x - y + 1 = 2x - y - 5 \]
Rearranging gives:
\[ -5x - 2x + 1 + 5 = 0 \implies -7x + 6 = 0 \implies 7x = 6 \implies x = \frac{6}{7} \]

### Step 5: Finding the Locus
The two cases give us two equations:
1. \( 3x + 2y = -4 \) (a line)
2. \( x = \frac{6}{7} \) (a vertical line)

Thus, the locus of point \( P \) is given by the line \( 3x + 2y + 4 = 0 \) and the vertical line \( x = \frac{6}{7} \).