1. **Problem Statement:** Given triangle ABC with vertices A(-4,6), B(-1,2), and C(0,5), we need to find the images of ABC under various transformations: central symmetry about (1,-1), translation by vector T(-1,4), reflection about line $x=2$, and rotation by 90° in the negative direction around (1,4). 2. **Central Symmetry (Point Reflection) about center $O(1,-1)$:** The formula for central symmetry of a point $P(x,y)$ about $O(x_0,y_0)$ is: $$P' = (2x_0 - x, 2y_0 - y)$$ Calculate: - $A' = (2\cdot1 - (-4), 2\cdot(-1) - 6) = (2+4, -2-6) = (6, -8)$ - $B' = (2 - (-1), -2 - 2) = (3, -4)$ - $C' = (2 - 0, -2 - 5) = (2, -7)$ 3. **Translation by vector $T(-1,4)$:** The formula for translation is: $$P'' = (x + t_x, y + t_y)$$ Calculate: - $A'' = (-4 - 1, 6 + 4) = (-5, 10)$ - $B'' = (-1 - 1, 2 + 4) = (-2, 6)$ - $C'' = (0 - 1, 5 + 4) = (-1, 9)$ 4. **Reflection about line $L: x=2$:** Reflection about vertical line $x=a$ is: $$P''' = (2a - x, y)$$ Calculate: - $A''' = (2\cdot2 - (-4), 6) = (4 + 4, 6) = (8, 6)$ - $B''' = (4 - (-1), 2) = (5, 2)$ - $C''' = (4 - 0, 5) = (4, 5)$ 5. **Rotation by 90° in negative (clockwise) direction around center $R(1,4)$:** The rotation formula for angle $\theta$ around $R(x_0,y_0)$ is: $$P'''' = \left(x_0 + (x - x_0)\cos\theta + (y - y_0)\sin\theta, y_0 - (x - x_0)\sin\theta + (y - y_0)\cos\theta\right)$$ For $\theta = -90^\circ$, $\cos(-90^\circ) = 0$, $\sin(-90^\circ) = -1$: $$P'''' = \left(x_0 + (y - y_0)(-1), y_0 - (x - x_0)(-1)\right) = (x_0 - (y - y_0), y_0 + (x - x_0))$$ Calculate: - $A'''' = (1 - (6 - 4), 4 + (-4 - 1)) = (1 - 2, 4 - 5) = (-1, -1)$ - $B'''' = (1 - (2 - 4), 4 + (-1 - 1)) = (1 - (-2), 4 - 2) = (3, 2)$ - $C'''' = (1 - (5 - 4), 4 + (0 - 1)) = (1 - 1, 4 - 1) = (0, 3)$ **Final answers:** - $A' = (6, -8), B' = (3, -4), C' = (2, -7)$ - $A'' = (-5, 10), B'' = (-2, 6), C'' = (-1, 9)$ - $A''' = (8, 6), B''' = (5, 2), C''' = (4, 5)$ - $A'''' = (-1, -1), B'''' = (3, 2), C'''' = (0, 3)$