1. **Problem statement:** We have triangle $\triangle LMN$ with vertices $L(2,2)$, $M(6,6)$, $N(6,2)$ and its image $\triangle L'M'N'$ with vertices $L'(11,2)$, $M'(7,6)$, $N'(7,2)$. We need to: (a) Describe the transformation mapping $\triangle LMN$ onto $\triangle L'M'N'$. (b) Draw $\triangle L''M''N''$ after translating $\triangle LMN$ by vector $\begin{pmatrix}-2 \\ 3\end{pmatrix}$. (c) Name and describe two transformations mapping $\triangle L'M'N'$ onto $\triangle L''M''N''$. 2. **Part (a): Describe the transformation from $\triangle LMN$ to $\triangle L'M'N'$.** - Check coordinates: $L(2,2) \to L'(11,2)$ $M(6,6) \to M'(7,6)$ $N(6,2) \to N'(7,2)$ - Notice $L$ moves from $x=2$ to $x=11$ (shift right 9), but $M$ and $N$ move left from $x=6$ to $x=7$ (shift right 1). So a simple translation is not the full transformation. - Check if reflection is involved. The $y$-coordinates of $L$ and $L'$ are the same (2), and $M$ and $M'$ are the same (6), $N$ and $N'$ are the same (2). So vertical positions are preserved. - Check if horizontal reflection about a vertical line is possible. Reflecting $M(6,6)$ to $M'(7,6)$ means reflection about $x=6.5$ would map $6$ to $7$? No, reflection about $x=6.5$ maps $6$ to $7$? Actually, reflection about $x=6.5$ maps $x$ to $2*6.5 - x = 13 - x$. So $6$ maps to $7$ because $13-6=7$. Similarly, $N(6,2)$ maps to $N'(7,2)$. - Check $L(2,2)$: reflection about $x=6.5$ maps $2$ to $13-2=11$, which matches $L'(11,2)$. - So the transformation is a reflection about the vertical line $x=6.5$. 3. **Part (b): Translate $\triangle LMN$ by vector $\begin{pmatrix}-2 \\ 3\end{pmatrix}$.** - Translation formula: $$\begin{pmatrix}x' \\ y'\end{pmatrix} = \begin{pmatrix}x \\ y\end{pmatrix} + \begin{pmatrix}-2 \\ 3\end{pmatrix}$$ - Calculate new vertices: $L'' = (2 - 2, 2 + 3) = (0, 5)$ $M'' = (6 - 2, 6 + 3) = (4, 9)$ $N'' = (6 - 2, 2 + 3) = (4, 5)$ 4. **Part (c): Describe two transformations mapping $\triangle L'M'N'$ onto $\triangle L''M''N''$.** - We want to map $L'(11,2)$ to $L''(0,5)$, $M'(7,6)$ to $M''(4,9)$, $N'(7,2)$ to $N''(4,5)$. - Step 1: Translate $\triangle L'M'N'$ by vector $\begin{pmatrix}-7 \\ 3\end{pmatrix}$ to move $L'(11,2)$ to $(4,5)$, but this does not match $L''(0,5)$. - Instead, consider first a translation by $\begin{pmatrix}-7 \\ 0\end{pmatrix}$: $L'(11,2) \to (4,2)$ $M'(7,6) \to (0,6)$ $N'(7,2) \to (0,2)$ - Then a reflection about the horizontal line $y=3.5$ maps $y$ to $2*3.5 - y = 7 - y$: $L'' = (4, 7-2) = (4,5)$ $M'' = (0, 7-6) = (0,1)$ (does not match $M''(4,9)$) - This does not match $M''$ and $N''$. - Try reflection about $y=5$: $L'(11,2) \to (11, 8)$ (no) - Alternatively, consider translation by $\begin{pmatrix}-7 \\ 3\end{pmatrix}$: $L'(11,2) \to (4,5)$ $M'(7,6) \to (0,9)$ $N'(7,2) \to (0,5)$ - Then reflect about vertical line $x=2$: $x \to 2*2 - x = 4 - x$ $L'' = (4 - 4, 5) = (0,5)$ $M'' = (4 - 0, 9) = (4,9)$ $N'' = (4 - 0, 5) = (4,5)$ - This matches $L''(0,5)$, $M''(4,9)$, $N''(4,5)$. - So the two transformations are: 1) Translation by vector $\begin{pmatrix}-7 \\ 3\end{pmatrix}$ 2) Reflection about the vertical line $x=2$. **Final answers:** (a) Reflection about the vertical line $x=6.5$. (b) $L''(0,5)$, $M''(4,9)$, $N''(4,5)$ after translation by $\begin{pmatrix}-2 \\ 3\end{pmatrix}$. (c) Translation by $\begin{pmatrix}-7 \\ 3\end{pmatrix}$ followed by reflection about $x=2$.