Triangle \( A B C \) has the following vertices: - \( A(1,9) \) - \( B(11,-7) \) - \( C(-9,3) \) Is triangle \( A B C \) a right triangle, and why? Choose 1 answer: (A) Yes, because \( \overline{A B} \perp \overline{A C} \). (B) Yes, because \( \overline{A B} \perp \overline{B C} \). (C) No, because \( \overline{A B} \) is longer than \( \overline{A C} \). (D) No, because \( \triangle A B C \) doesn't have a pair of perpendicular sides.

To determine if triangle \( A B C \) is a right triangle, we can check the slopes of the sides to see if any pair is perpendicular (i.e., has a product of their slopes equal to -1). First, we find the coordinates of the sides: - The slope of \( \overline{A B} \) is \( \frac{-7 - 9}{11 - 1} = \frac{-16}{10} = -\frac{8}{5} \). - The slope of \( \overline{B C} \) is \( \frac{3 - (-7)}{-9 - 11} = \frac{10}{-20} = -\frac{1}{2} \). - The slope of \( \overline{A C} \) is \( \frac{3 - 9}{-9 - 1} = \frac{-6}{-10} = \frac{3}{5} \). Now, we check for perpendicular pairs: - The product of the slopes of \( \overline{A B} \) and \( \overline{B C} \): \( -\frac{8}{5} \times -\frac{1}{2} = \frac{8}{10} = \frac{4}{5} \) (not -1). - The product of the slopes of \( \overline{A B} \) and \( \overline{A C} \): \( -\frac{8}{5} \times \frac{3}{5} = -\frac{24}{25} \) (not -1). - The product of the slopes of \( \overline{A C} \) and \( \overline{B C} \): \( \frac{3}{5} \times -\frac{1}{2} = -\frac{3}{10} \) (not -1). Since no pair of slopes yielded -1, triangle \( A B C \) does not have a pair of perpendicular sides, making it clear that the answer is: (D) No, because \( \triangle A B C \) doesn't have a pair of perpendicular sides.