Решебник по геометрии 9 класс Мерзляк Задание 302

\[\mathrm{\Delta}ABC - прямоугольный:\]

\[A( - 1;\ 2),\ \ \ B(7;\ 4).\]

\[1)\ C(7;\ 2):\]

\[AB = \sqrt{( - 1 - 7)^{2} + (2 - 4)^{2}} =\]

\[= \sqrt{64 + 4} = \sqrt{68};\]

\[BC = \sqrt{(7 - 7)^{2} + (4 - 2)^{2}} =\]

\[= \sqrt{0 + 4} = 2;\]

\[AC = \sqrt{(7 + 1)^{2} + (2 - 2)^{2}} =\]

\[= \sqrt{64 + 0} = 8.\]

\[\cos{\angle C} = \frac{AC^{2} + BC^{2} - AB^{2}}{2AC \bullet BC} =\]

\[= \frac{64 + 4 - 68}{2 \bullet 8 \bullet 2} = 0;\]

\[\angle C = \arccos 0 = 90{^\circ}.\]

\[Ответ:\ \ да.\]

\[2)\ C(2;\ - 3):\]

\[AB = \sqrt{( - 1 - 7)^{2} + (2 - 4)^{2}} =\]

\[= \sqrt{64 + 4} = \sqrt{68};\]

\[BC = \sqrt{(7 - 2)^{2} + (4 + 3)^{2}} =\]

\[= \sqrt{25 + 49} = \sqrt{74};\]

\[AC = \sqrt{( - 1 - 2)^{2} + (2 + 3)^{2}} =\]

\[= \sqrt{9 + 25} = \sqrt{34};\]

\[\cos{\angle A} = \frac{AB^{2} + AC^{2} - BC^{2}}{2AB \bullet AC} =\]

\[= \frac{68 + 34 - 74}{2\sqrt{68} \bullet \sqrt{34}} \neq 0;\]

\[\angle A \neq \arccos 0 = 90{^\circ}.\]

\[Ответ:\ \ нет.\]