A(1,1) , B (-2,3) are two points . If a point P forms a triangle of are 2 square units with A, B then find the locus of P.

Answer

To find the locus of point P that forms a triangle of area 2 square units with points A(1,1) and B(-2,3), we can follow these steps:

### Step 1: Set up the area formula
The area \( A \) of a triangle formed by three points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( P(x, y) \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y) + x_2(y - y_1) + x(y_1 - y_2) \right| \]
For our points, \( A(1,1) \) and \( B(-2,3) \), we can substitute these values into the formula.

### Step 2: Substitute the coordinates
Substituting \( A(1,1) \) and \( B(-2,3) \) into the area formula, we have:
\[ A = \frac{1}{2} \left| 1(3 - y) + (-2)(y - 1) + x(1 - 3) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| 3 - y - 2y + 2 - 2x \right| = \frac{1}{2} \left| 5 - 3y - 2x \right| \]

### Step 3: Set the area equal to 2
Since we want the area to be 2 square units, we set up the equation:
\[ \frac{1}{2} \left| 5 - 3y - 2x \right| = 2 \]
Multiplying both sides by 2 gives:
\[ \left| 5 - 3y - 2x \right| = 4 \]

### Step 4: Remove the absolute value
This absolute value equation can be split into two cases:
1. \( 5 - 3y - 2x = 4 \)
2. \( 5 - 3y - 2x = -4 \)

### Step 5: Solve the first case
For the first case:
\[ 5 - 3y - 2x = 4 \implies -3y - 2x = -1 \implies 3y + 2x = 1 \]

### Step 6: Solve the second case
For the second case:
\[ 5 - 3y - 2x = -4 \implies -3y - 2x = -9 \implies 3y + 2x = 9 \]

### Step 7: Write the equations of the locus
Thus, we have two linear equations representing the locus of point P:
1. \( 3y + 2x = 1 \)
2. \( 3y + 2x = 9 \)

### Final Answer
The locus of point P is represented by the two lines:
1. \( 3y + 2x = 1 \)
2. \( 3y + 2x = 9 \)

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