1. **Problem 1: Transformation of figure ABCD under ERS** Given points: - A(-2, -2), B(-4, -1), C(-4, -3), D(-2, -3) Transformations: - $$E=\begin{bmatrix}-2 & 0 \\ 0 & -2\end{bmatrix}$$ (scaling by -2) - $$R=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$ (reflection about line y=x) - $$S=\begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}$$ (reflection and rotation) We apply transformations successively in order ERS, meaning first S, then R, then E. 2. **Step 1: Apply S to each point** For a point $$P=(x,y)$$, new point $$P' = S \times \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}-y \\ -x\end{bmatrix}$$ Calculate: - A: $$(-2,-2) \to (-(-2), -(-2)) = (2, 2)$$ - B: $$(-4,-1) \to (-(-1), -(-4)) = (1, 4)$$ - C: $$(-4,-3) \to (-(-3), -(-4)) = (3, 4)$$ - D: $$(-2,-3) \to (-(-3), -(-2)) = (3, 2)$$ 3. **Step 2: Apply R to the results from Step 1** For $$P'=(x', y')$$, new point $$P'' = R \times \begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix}y' \\ x'\end{bmatrix}$$ Calculate: - A: $$(2, 2) \to (2, 2)$$ - B: $$(1, 4) \to (4, 1)$$ - C: $$(3, 4) \to (4, 3)$$ - D: $$(3, 2) \to (2, 3)$$ 4. **Step 3: Apply E to the results from Step 2** For $$P''=(x'', y'')$$, new point $$P''' = E \times \begin{bmatrix}x'' \\ y''\end{bmatrix} = \begin{bmatrix}-2 & 0 \\ 0 & -2\end{bmatrix} \begin{bmatrix}x'' \\ y''\end{bmatrix} = \begin{bmatrix}-2x'' \\ -2y''\end{bmatrix}$$ Calculate: - A: $$(2, 2) \to (-4, -4)$$ - B: $$(4, 1) \to (-8, -2)$$ - C: $$(4, 3) \to (-8, -6)$$ - D: $$(2, 3) \to (-4, -6)$$ **Final image coordinates of ABCD after ERS:** - A'(-4, -4), B'(-8, -2), C'(-8, -6), D'(-4, -6) --- 5. **Problem 2: Shear transformation on triangle ABC** Given triangle ABC with vertices: - A(2, 2), B(2, 0), C(3, 2) Shear factor 3 parallel to the x-axis means the transformation matrix is: $$\text{Shear} = \begin{bmatrix}1 & 3 \\ 0 & 1\end{bmatrix}$$ Apply shear to each point: $$P' = \text{Shear} \times \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}x + 3y \\ y\end{bmatrix}$$ Calculate: - A: $$(2, 2) \to (2 + 3 \times 2, 2) = (8, 2)$$ - B: $$(2, 0) \to (2 + 3 \times 0, 0) = (2, 0)$$ - C: $$(3, 2) \to (3 + 3 \times 2, 2) = (9, 2)$$ **Coordinates of A'B'C' after shear:** - A'(8, 2), B'(2, 0), C'(9, 2)