Quadrilateral \( A B C D \) has the following vertices: - \( A(-3,0) \) - \( B(7,2) \) - \( C(1,-4) \) - \( D(-9,-6) \) Is quadrilateral \( A B C D \) a parallelogram, and why? Choose 1 answer: (A) Yes, because \( \overline{A B} \| \overline{C D} \), and \( \overline{B C} \perp \overline{A B} \). (B) Yes, because \( \overline{A B} \| \overline{C D} \), and \( \overline{B C} \| \overline{A D} \). (C) No, because \( \overline{A B} \) is longer than \( \overline{C D} \). (D) No, because \( \overline{B C} \) is not parallel to \( \overline{A D} \).

To determine if quadrilateral \( A B C D \) is a parallelogram, we need to analyze the slopes of the opposite sides. The slope of line segment \( \overline{A B} \) is given by the formula \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{7 - (-3)} = \frac{2}{10} = \frac{1}{5} \). For \( \overline{C D} \), the slope calculation yields \( \frac{-6 - (-4)}{-9 - 1} = \frac{-2}{-10} = \frac{1}{5} \), indicating that \( \overline{A B} \parallel \overline{C D} \). Next, we find the slopes for \( \overline{B C} \) and \( \overline{A D} \). The slope of \( \overline{B C} \) is \( \frac{-4 - 2}{1 - 7} = \frac{-6}{-6} = 1 \), and the slope for \( \overline{A D} \) is \( \frac{-6 - 0}{-9 - (-3)} = \frac{-6}{-6} = 1 \), indicating that \( \overline{B C} \parallel \overline{A D} \) as well. Since both pairs of opposite sides are parallel, we can conclude that \( A B C D \) is indeed a parallelogram, which makes option (B) the correct choice: *Yes, because \( \overline{A B} \| \overline{C D} \), and \( \overline{B C} \| \overline{A D} \).*