Triangle A'B'C' is obtained from triangle ABC by a rotation and a translation. First, reflect ΔABC across the x-axis. Then translate the reflected triangle 5 units to the left.
The reflection across the x-axis is denoted by \(r_{x-axis}\). The translation 5 units to the left is denoted by \(T_{\langle -5, 0 \rangle}\).
The composition of transformations is performed from right to left. So, the transformation is \(T_{\langle -5, 0 \rangle} \circ r_{x-axis}\).
The coordinates of P are (1, 1). First, reflect P across line m (y=2). The y-coordinate of the reflected point is 2 + (2-1) = 3. The x-coordinate remains the same. So, the reflected point is (1, 3).
Next, translate the point (1,3) by <-2, 0>. This means moving the point 2 units to the left. The new coordinates are (1 - 2, 3) = (-1, 3).
The coordinates of P are (1, 1). First, reflect P across line n (x=-1). The x-coordinate of the reflected point is -1 - (1 - (-1)) = -1 - 2 = -3. The y-coordinate remains the same. So the reflected point is (-3, 1).
Next, translate the point (-3, 1) by <0, -5>. This means moving the point 5 units down. The new coordinates are (-3, 1 - 5) = (-3, -4).
- \( \boxed{T_{\langle -5, 0 \rangle} \circ r_{x-axis}} \)
- \( \boxed{(-1, 3)} \)
- \( \boxed{(-3, -4)} \)