What is the area of rectangle $$ A B C D $$ with vertices $$ A(-6,3), B(2,5), C(3,1) $$, and $$ D(-5,-1) $$ $$ A=\square $$ units $$ ^{2} $$

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To find the area of rectangle $$ ABCD $$ with vertices $$ A(-6,3) $$, $$ B(2,5) $$, $$ C(3,1) $$, and $$ D(-5,-1) $$, we can use the **shoelace formula** or verify using vector properties. Here's a step-by-step approach:

### Using the Shoelace Formula

1. **List the vertices in order** (either clockwise or counterclockwise):
   $$
   A(-6, 3),\ B(2, 5),\ C(3, 1),\ D(-5, -1),\ A(-6, 3)
   $$

2. **Apply the Shoelace Formula**:
   $$
   \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1}) - \sum_{i=1}^{n} (y_i x_{i+1}) \right|
   $$
   
   Plugging in the coordinates:
   $$
   \begin{align*}
   \sum (x_i y_{i+1}) &= (-6)(5) + (2)(1) + (3)(-1) + (-5)(3) = -30 + 2 - 3 - 15 = -46 \
   \sum (y_i x_{i+1}) &= (3)(2) + (5)(3) + (1)(-5) + (-1)(-6) = 6 + 15 - 5 + 6 = 22 \
   \end{align*}
   $$
   
   $$
   \text{Area} = \frac{1}{2} | -46 - 22 | = \frac{1}{2} \times 68 = 34
   $$

### Verifying with Vectors

1. **Find vectors for adjacent sides**:
   - $$ \vec{AB} = B - A = (2 - (-6), 5 - 3) = (8, 2) $$
   - $$ \vec{AD} = D - A = (-5 - (-6), -1 - 3) = (1, -4) $$

2. **Check perpendicularity**:
   $$
   \vec{AB} \cdot \vec{AD} = (8)(1) + (2)(-4) = 8 - 8 = 0 \quad \text{(Perpendicular)}
   $$

3. **Calculate lengths**:
   - $$ |\vec{AB}| = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} $$
   - $$ |\vec{AD}| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} $$
   
4. **Compute area**:
   $$
   \text{Area} = |\vec{AB}| \times |\vec{AD}| = \sqrt{68} \times \sqrt{17} = \sqrt{68 \times 17} = \sqrt{1156} = 34
   $$

Both methods confirm that the area of rectangle $$ ABCD $$ is **34 square units**.

**Answer:**  
$$ A = 34 $$ units$$^2$$
The area of rectangle $$ ABCD $$ is 34 square units.

What is the area of rectangle with vertices , and
units

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Solución

Using the Shoelace Formula

Verifying with Vectors

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What is the area of rectangle with vertices , and units