1. **Problem Statement:** Describe the transformations J and K on parallelogram EFGH and find images of points P and Q under these transformations. 2. **Transformation J:** Given matrix $J = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$, this is the identity transformation which maps every point to itself. 3. **Transformation K:** Given matrix $K = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$, this represents a rotation by 90 degrees counterclockwise about the origin. 4. **Coordinates of P' under J:** Since J is identity, $P' = (6, 2)$. 5. **Coordinates of Q' under K:** Multiply $K$ by $Q = (5, -4)$: $$Q' = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}5 \\ -4\end{pmatrix} = \begin{pmatrix}0 \times 5 + (-1) \times (-4) \\ 1 \times 5 + 0 \times (-4)\end{pmatrix} = \begin{pmatrix}4 \\ 5\end{pmatrix}$$ 6. **Final answers:** - Transformation J is the identity (no change). - Transformation K is a 90° counterclockwise rotation. - $P' = (6, 2)$ - $Q' = (4, 5)$