Let A(1, 2), B(3, 4) be two points and C(x, y) be a point such that area of `DeltaABC` is 3 sq. units and `(x- 1)(x-3)+ (y-2)(y-4)=0`. Then number of positions of C, in the xy plane is

Options

A. 2
B. 4
C. 8
D. 0

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the points A and B
We have two points:
- A(1, 2)
- B(3, 4)

### Step 2: Determine the area of triangle ABC
The area of triangle ABC is given as 3 square units. The formula for the area of a triangle formed by three points (A, B, C) is given by:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of points A and B, we have:

\[ \text{Area} = \frac{1}{2} \left| 1(4 - y) + 3(y - 2) + x(2 - 4) \right| = 3 \]

### Step 3: Simplify the area equation
Expanding the equation:

\[ \frac{1}{2} \left| 4 - y + 3y - 6 + 2x \right| = 3 \]

This simplifies to:

\[ \frac{1}{2} \left| 2x + 2y - 2 \right| = 3 \]

Multiplying both sides by 2 gives:

\[ \left| 2x + 2y - 2 \right| = 6 \]

This results in two equations:

1. \( 2x + 2y - 2 = 6 \) → \( x + y = 4 \)
2. \( 2x + 2y - 2 = -6 \) → \( x + y = -2 \)

### Step 4: Analyze the given equation
The equation given is:

\[ (x - 1)(x - 3) + (y - 2)(y - 4) = 0 \]

This can be rewritten as:

\[ (x - 1)(x - 3) = -(y - 2)(y - 4) \]

This represents a circle with diameter endpoints A(1, 2) and B(3, 4).

### Step 5: Find the radius of the circle
The distance AB (diameter) can be calculated using the distance formula:

\[ AB = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{4 + 4} = 2\sqrt{2} \]

Thus, the radius \( r \) is:

\[ r = \frac{AB}{2} = \sqrt{2} \]

### Step 6: Determine the height from point C to line AB
From the area calculation, we found the altitude (height) from point C to line AB to be:

\[ h = \frac{3}{\sqrt{2}} \]

### Step 7: Compare height with radius
Since \( \frac{3}{\sqrt{2}} \) is greater than the radius \( \sqrt{2} \), the point C cannot lie on the circle if the height from point C to line AB is greater than the radius of the circle.

### Conclusion
Since no such point C exists that satisfies both conditions (area of triangle and lying on the circle), the number of positions of C in the xy-plane is:

\[ \text{Number of positions of C} = 0 \]