ABCD is a square with side AB = 2. A point P moves such that its distance from A equals its distance from the line BD. The locus of P meets the line AC at `T_1` and the line through A parallel to BD at `T_2` and `T_3`. The area of the triangle `T_1 T_2 T_3` is :

Options

A. `1/2"sq. unit"`
B. `2/3"sq. unit"`
C. `1"sq. unit"`
D. `2"sq. unit"`

Explanation

To solve the problem step by step, we will follow the given information and derive the necessary equations and points.

### Step 1: Define the Square and Points
Let’s define the square ABCD with vertices:
- A(0, 0)
- B(2, 0)
- C(2, 2)
- D(0, 2)

### Step 2: Identify the Directrix and Focus
The problem states that point P moves such that its distance from point A equals its distance from the line BD. The line BD can be represented by the equation \( x + y - 2 = 0 \).

- The focus of the parabola is point A(0, 0).
- The directrix is the line BD, given by \( x + y - 2 = 0 \).

### Step 3: Equation of the Parabola
Using the definition of a parabola, we can derive its equation. The distance from point P(x, y) to the focus A(0, 0) is equal to the distance from point P to the directrix.

1. Distance from P to A:
\[ \sqrt{x^2 + y^2} \]

2. Distance from P to the line \( x + y - 2 = 0 \):
\[ \frac{|x + y - 2|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 2|}{\sqrt{2}} \]

Setting these two distances equal gives:
\[ \sqrt{x^2 + y^2} = \frac{|x + y - 2|}{\sqrt{2}} \]

### Step 4: Squaring Both Sides
Squaring both sides results in:
\[ x^2 + y^2 = \frac{(x + y - 2)^2}{2} \]

Multiplying through by 2 to eliminate the fraction:
\[ 2(x^2 + y^2) = (x + y - 2)^2 \]

### Step 5: Expanding the Right Side
Expanding the right side:
\[ 2x^2 + 2y^2 = x^2 + 2xy + y^2 - 4x - 4y + 4 \]
Rearranging gives:
\[ x^2 + y^2 - 2xy + 4x + 4y - 4 = 0 \]

### Step 6: Finding Points T1, T2, and T3
1. **Finding T1**: The locus of P meets the line AC (which is \( y = x \)) at T1.
Substitute \( y = x \) into the parabola equation:
\[ x^2 + x^2 - 2x^2 + 4x + 4x - 4 = 0 \implies 8x - 4 = 0 \implies x = \frac{1}{2} \]
Thus, T1 is \( \left(\frac{1}{2}, \frac{1}{2}\right) \).

2. **Finding T2 and T3**: The line through A parallel to BD has the equation \( y = -x \).
Substitute \( y = -x \) into the parabola equation:
\[ x^2 + (-x)^2 - 2(-x)x + 4x + 4(-x) - 4 = 0 \implies 2x^2 + 2x - 4 = 0 \implies x^2 + x - 2 = 0 \]
Factoring gives:
\[ (x - 1)(x + 2) = 0 \implies x = 1 \text{ or } x = -2 \]
Thus, T2 is \( (1, -1) \) and T3 is \( (-2, 2) \).

### Step 7: Area of Triangle T1 T2 T3
Using the formula for the area of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the points:
- T1 \( \left(\frac{1}{2}, \frac{1}{2}\right) \)
- T2 \( (1, -1) \)
- T3 \( (-2, 2) \)

Calculating the area:
\[ \text{Area} = \frac{1}{2} \left| \frac{1}{2}(-1 - 2) + 1(2 - \frac{1}{2}) + (-2)(\frac{1}{2} + 1) \right| \]
\[ = \frac{1}{2} \left| \frac{1}{2}(-3) + 1(\frac{3}{2}) - 2(\frac{3}{2}) \right| \]
\[ = \frac{1}{2} \left| -\frac{3}{2} + \frac{3}{2} - 3 \right| = \frac{1}{2} \left| -3 \right| = \frac{3}{2} \]

Thus, the area of triangle T1 T2 T3 is \( \frac{3}{2} \).

### Final Answer
The area of triangle T1 T2 T3 is \( 2 \) square units.