Show that the points A(2, 1), B(5, 2). C(6, 4) and D(3, 3) are vertices of a parallelogram. Is this figure a rectangle?

Answer

To show that the points A(2, 1), B(5, 2), C(6, 4), and D(3, 3) are vertices of a parallelogram, we need to prove that the opposite sides are equal in length. We will use the distance formula to calculate the lengths of the sides.

### Step 1: Calculate the length of AB
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For points A(2, 1) and B(5, 2):
\[ AB = \sqrt{(5 - 2)^2 + (2 - 1)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 2: Calculate the length of CD
For points C(6, 4) and D(3, 3):
\[ CD = \sqrt{(3 - 6)^2 + (3 - 4)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 3: Calculate the length of AD
For points A(2, 1) and D(3, 3):
\[ AD = \sqrt{(3 - 2)^2 + (3 - 1)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 4: Calculate the length of BC
For points B(5, 2) and C(6, 4):
\[ BC = \sqrt{(6 - 5)^2 + (4 - 2)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 5: Compare the lengths
From our calculations:
- \(AB = CD = \sqrt{10}\)
- \(AD = BC = \sqrt{5}\)

Since \(AB = CD\) and \(AD = BC\), we can conclude that ABCD is a parallelogram.

### Step 6: Check if ABCD is a rectangle
To check if ABCD is a rectangle, we need to see if the diagonals AC and BD are equal.

### Step 7: Calculate the length of AC
For points A(2, 1) and C(6, 4):
\[ AC = \sqrt{(6 - 2)^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 8: Calculate the length of BD
For points B(5, 2) and D(3, 3):
\[ BD = \sqrt{(3 - 5)^2 + (3 - 2)^2} = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 9: Compare the lengths of diagonals
- \(AC = 5\)
- \(BD = \sqrt{5}\)

Since \(AC \neq BD\), we conclude that ABCD is not a rectangle.

### Conclusion
The points A(2, 1), B(5, 2), C(6, 4), and D(3, 3) form a parallelogram, but it is not a rectangle.

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