Two points A and B with co-ordinates (5, 3), (3, -2) are given. A point P moves so that the area of `DeltaPAB` is constant and equal to 9 square units. Find the equation to the locus of the point P.

Answer

To find the equation of the locus of point P such that the area of triangle PAB is constant and equal to 9 square units, we can follow these steps:

### Step 1: Identify the coordinates of points A and B
Given points are:
- A(5, 3)
- B(3, -2)

### Step 2: Use the formula for the area of a triangle
The area of triangle formed by points P(x, y), A(x1, y1), and B(x2, y2) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
In our case, we can substitute:
- \( P(x, y) \) as \( (x, y) \)
- \( A(5, 3) \) as \( (x_1, y_1) \)
- \( B(3, -2) \) as \( (x_2, y_2) \)

### Step 3: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula gives:
\[ \text{Area} = \frac{1}{2} \left| 5(-2 - y) + 3(y - 3) + x(3 - (-2)) \right| \]
This simplifies to:
\[ \text{Area} = \frac{1}{2} \left| 5(-2 - y) + 3(y - 3) + x(5) \right| \]

### Step 4: Set the area equal to 9
Since the area is given as 9 square units, we set up the equation:
\[ \frac{1}{2} \left| 5(-2 - y) + 3(y - 3) + 5x \right| = 9 \]
Multiplying both sides by 2:
\[ \left| 5(-2 - y) + 3(y - 3) + 5x \right| = 18 \]

### Step 5: Simplify the expression inside the absolute value
Now, simplify the expression:
\[ 5(-2 - y) + 3(y - 3) + 5x = -10 - 5y + 3y - 9 + 5x = 5x - 2y - 19 \]
Thus, we have:
\[ \left| 5x - 2y - 19 \right| = 18 \]

### Step 6: Solve the absolute value equation
This gives us two cases:
1. \( 5x - 2y - 19 = 18 \)
2. \( 5x - 2y - 19 = -18 \)

### Step 7: Solve each case for y
**Case 1:**
\[ 5x - 2y - 19 = 18 \implies 5x - 2y = 37 \implies 2y = 5x - 37 \implies y = \frac{5}{2}x - \frac{37}{2} \]

**Case 2:**
\[ 5x - 2y - 19 = -18 \implies 5x - 2y = 1 \implies 2y = 5x - 1 \implies y = \frac{5}{2}x - \frac{1}{2} \]

### Step 8: Write the final equations of the locus
Thus, the equations of the locus of point P are:
1. \( 5x - 2y - 37 = 0 \)
2. \( 5x - 2y - 1 = 0 \)