To find the locus of the point \( P(x, y) \) such that the area of the quadrilateral \( PABC \) is 10 square units, we will follow these steps:
### Step 1: Define the points
Let the points be:
- \( A(5, 3) \)
- \( B(3, -2) \)
- \( C(2, -1) \)
- \( P(x, y) \)
### Step 2: Use the area formula for a quadrilateral
The area \( A \) of a quadrilateral formed by points \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) can be calculated using the formula:
\[
A = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|
\]
For our points \( A, B, C, P \), we substitute:
- \( (x_1, y_1) = (5, 3) \)
- \( (x_2, y_2) = (3, -2) \)
- \( (x_3, y_3) = (2, -1) \)
- \( (x_4, y_4) = (x, y) \)
### Step 3: Set up the area equation
Substituting into the area formula, we have:
\[
\text{Area} = \frac{1}{2} \left| 5(-2) + 3(-1) + 2y + x(3) - (3 \cdot 3 + (-2) \cdot 2 + (-1)x + y \cdot 5) \right|
\]
This simplifies to:
\[
\text{Area} = \frac{1}{2} \left| -10 - 3 + 2y + 3x - (9 - 4 - x + 5y) \right|
\]
\[
= \frac{1}{2} \left| -13 + 2y + 3x - 9 + 4 + x - 5y \right|
\]
\[
= \frac{1}{2} \left| 4x - 3y - 18 \right|
\]
### Step 4: Set the area equal to 10
Since we want the area to be 10 square units, we set up the equation:
\[
\frac{1}{2} \left| 4x - 3y - 18 \right| = 10
\]
Multiplying both sides by 2 gives:
\[
\left| 4x - 3y - 18 \right| = 20
\]
### Step 5: Remove the absolute value
This leads to two equations:
1. \( 4x - 3y - 18 = 20 \)
2. \( 4x - 3y - 18 = -20 \)
### Step 6: Solve the equations
For the first equation:
\[
4x - 3y = 38 \quad \text{(1)}
\]
For the second equation:
\[
4x - 3y = -2 \quad \text{(2)}
\]
### Step 7: Write the equations of the lines
These equations represent two lines in the coordinate plane:
1. \( 4x - 3y = 38 \)
2. \( 4x - 3y = -2 \)
### Step 8: Conclusion
The locus of point \( P \) is the union of the two lines given by the equations:
- \( 4x - 3y = 38 \)
- \( 4x - 3y = -2 \)
