If the square ABCD, where `A(0,0),B(2,0)C(2,2) and D(0,2)` undergoes the following three transformations successively
(i) `f_(1)(x,y) to (y,x)`
(ii) `f_(2)(x,y) to (x+3y,y)`
(iii) `f_(3) (x,y) to ((x-y)/2,(x+y)/2)`
then the final figure is a

Options

A. square
B. parallelogram
C. rhombus
D. None of these

Answer

To solve the problem, we need to apply the three transformations to the vertices of the square ABCD and determine the final shape formed by the transformed points.

### Step 1: Identify the Initial Points
The vertices of the square ABCD are given as:
- A(0, 0)
- B(2, 0)
- C(2, 2)
- D(0, 2)

### Step 2: Apply the First Transformation
The first transformation is \( f_1(x, y) = (y, x) \). We will apply this transformation to each of the points.

- \( A(0, 0) \) transforms to \( (0, 0) \)
- \( B(2, 0) \) transforms to \( (0, 2) \)
- \( C(2, 2) \) transforms to \( (2, 2) \)
- \( D(0, 2) \) transforms to \( (2, 0) \)

After the first transformation, the new points are:
- A'(0, 0)
- B'(0, 2)
- C'(2, 2)
- D'(2, 0)

### Step 3: Apply the Second Transformation
The second transformation is \( f_2(x, y) = (x + 3y, y) \). We will apply this transformation to the new points.

- \( A'(0, 0) \) transforms to \( (0 + 3*0, 0) = (0, 0) \)
- \( B'(0, 2) \) transforms to \( (0 + 3*2, 2) = (6, 2) \)
- \( C'(2, 2) \) transforms to \( (2 + 3*2, 2) = (8, 2) \)
- \( D'(2, 0) \) transforms to \( (2 + 3*0, 0) = (2, 0) \)

After the second transformation, the new points are:
- A''(0, 0)
- B''(6, 2)
- C''(8, 2)
- D''(2, 0)

### Step 4: Apply the Third Transformation
The third transformation is \( f_3(x, y) = \left( \frac{x - y}{2}, \frac{x + y}{2} \right) \). We will apply this transformation to the new points.

- \( A''(0, 0) \) transforms to \( \left( \frac{0 - 0}{2}, \frac{0 + 0}{2} \right) = (0, 0) \)
- \( B''(6, 2) \) transforms to \( \left( \frac{6 - 2}{2}, \frac{6 + 2}{2} \right) = \left( \frac{4}{2}, \frac{8}{2} \right) = (2, 4) \)
- \( C''(8, 2) \) transforms to \( \left( \frac{8 - 2}{2}, \frac{8 + 2}{2} \right) = \left( \frac{6}{2}, \frac{10}{2} \right) = (3, 5) \)
- \( D''(2, 0) \) transforms to \( \left( \frac{2 - 0}{2}, \frac{2 + 0}{2} \right) = (1, 1) \)

After the third transformation, the new points are:
- A'''(0, 0)
- B'''(2, 4)
- C'''(3, 5)
- D'''(1, 1)

### Step 5: Determine the Final Figure
Now we have the transformed points:
- A'''(0, 0)
- B'''(2, 4)
- C'''(3, 5)
- D'''(1, 1)

To determine the shape formed by these points, we can analyze the slopes of the segments formed by these points:

1. **Slope of A'''B''':**
\[ \text{slope} = \frac{4 - 0}{2 - 0} = 2 \]

2. **Slope of B'''C''':**
\[ \text{slope} = \frac{5 - 4}{3 - 2} = 1 \]

3. **Slope of C'''D''':**
\[ \text{slope} = \frac{1 - 5}{1 - 3} = \frac{-4}{-2} = 2 \]

4. **Slope of D'''A''':**
\[ \text{slope} = \frac{0 - 1}{0 - 1} = 1 \]

From the slopes, we can see that:
- The slopes of segments A'''B''' and C'''D''' are both 2, indicating that these two segments are parallel.
- The slopes of segments B'''C''' and D'''A''' are both 1, indicating that these two segments are also parallel.

Since both pairs of opposite sides are parallel, the final figure formed by the points A'''(0, 0), B'''(2, 4), C'''(3, 5), and D'''(1, 1) is a **parallelogram**.

### Final Answer
The final figure is a **parallelogram**.

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