1. Problem: Find the image of point A(−1, 3) after reflection about the line $x=4$ followed by reflection about the Y-axis. Step 1: Reflect A about $x=4$. The formula for reflection about line $x=k$ is $(x,y) \to (2k - x, y)$. Here $k=4$. $$A' = (2\times4 - (-1), 3) = (8+1, 3) = (9, 3)$$ Step 2: Reflect $A'$ about Y-axis: $(x,y) \to (-x,y)$, $$A'' = (-9, 3)$$ Answer: B. $(-9, 3)$ 2. Problem: Find the image of point A(5, −2) after reflection about the X-axis, followed by a 90° rotation about origin O. Step 1: Reflection about X-axis: $(x,y) \to (x,-y)$, $$A' = (5, 2)$$ Step 2: Rotation 90° CCW about origin: $(x,y) \to (-y,x)$, $$A'' = (-2, 5)$$ Answer: B. $(-2, 5)$ 3. Problem: Find image of P(1,4) after reflection about $x=3$ then reflection about $y=1$. Step 1: Reflect about $x=3$: $(x,y) \to (2\times3 - x, y) = (6-1,4)=(5,4)$ Step 2: Reflect about $y=1$: $(x,y) \to (x, 2\times1 - y) = (5, 2-4)=(5,-2)$ Answer: C. $(5, -2)$ 4. Problem: Image of S(2,4) after 90° CCW rotation about origin, then reflection about $y=x$. Step 1: Rotate 90° CCW: $(x,y) \to (-y,x)$, $$S' = (-4, 2)$$ Step 2: Reflect about $y=x$: $(x,y) \to (y,x)$, $$S'' = (2,-4)$$ Answer: A. $S''(2,-4)$ 5. Problem: Point A(3, −2), translated by $T=(1,-2)$, then rotated 90° CCW about origin. Step 1: Translate: $(x,y) \to (x+1,y-2)$, $$A'=(3+1, -2-2) = (4, -4)$$ Step 2: Rotate 90° CCW: $(x,y) \to (-y,x)$, $$A'' = (4,4)$$ Answer: A. $(4, 4)$ 6. Problem: P(−3,1), rotated 90° CCW about origin, then translated by $T=(3,4)$. Step 1: Rotate 90° CCW: $(x,y) \to (-y,x)$, $$P' = (-1, -3)$$ Step 2: Translate: $(x,y) \to (x+3, y+4)$, $$P'' = (2, 1)$$ Answer: A. $P''(2, 1)$ 7. Problem: A(8,-12), dilated from origin with scale 2, then rotated 180° about origin. Step 1: Dilate: $(x,y) \to (2x, 2y)$, $$A' = (16, -24)$$ Step 2: Rotate 180°: $(x,y) \to (-x,-y)$, $$A'' = (-16, 24)$$ Answer: E. $(-16, 24)$ 8. Problem: $T_1$ rotate 90°, $T_2$ reflect about $y=-x$, and $T_1 \circ T_2 (A) = (8, -6)$. Find $A$. Step 1: Let $B = T_2(A)$, then $T_1(B) = (8,-6)$ Step 2: Rotation 90° CCW inverse to $T_1$: rotate 90° CW $(x,y) \to (y,-x)$, $$B = (y, -x) = (8,-6) \to B = (-6, -8)$$ Step 3: $B = T_2(A)$ is reflection about $y = -x$: $(x,y) \to (-y, -x)$ so $$B = (-y_A, -x_A) = (-6, -8) \Rightarrow (x_A,y_A) = (8,6)$$ Answer: d. $(8, 6)$ 9. Problem: Reflect A(2, -8) about $y=-x$, then transform by matrix $\begin{pmatrix} 2 & 3 \\ 0 & -1 \end{pmatrix}$. Step 1: Reflect about $y=-x$: $(x,y) \to (-y, -x)$, $$M(A) = (8, -2)$$ Step 2: Multiply matrix: $$\begin{pmatrix}2 & 3\\0 & -1\end{pmatrix} \times \begin{pmatrix}8 \\ -2\end{pmatrix} = \begin{pmatrix}2\times8 + 3\times(-2) \\ 0\times8 + (-1)\times(-2)\end{pmatrix} = \begin{pmatrix}16 - 6 \\ 2\end{pmatrix} = (10, 2)$$ Answer: C. $(10, 2)$ 10. Problem: P(4,3) reflected about Y-axis then transformed by matrix $\begin{pmatrix}a & 4 \\ 2 & a+1\end{pmatrix}$ yielding P'(4,1). Find the image of K(7,2) via the same composition. Step 1: Reflect P about Y-axis: $(4,3) \to (-4,3)$. Step 2: Transformation: $\begin{pmatrix}a & 4 \\ 2 & a+1\end{pmatrix} \times \begin{pmatrix}-4 \\ 3\end{pmatrix} = \begin{pmatrix} -4a + 12 \\ -8 + 3(a+1) \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ So, equations: $$-4a + 12 = 4 \Rightarrow -4a = -8 \Rightarrow a=2$$ $$-8 + 3(a+1) = 1 \Rightarrow -8 + 3(3) =1 \Rightarrow -8 + 9 = 1$$ Step 3: Reflect K(7,2): $(-7, 2)$ Step 4: Transform K: $$\begin{pmatrix}2 & 4 \\ 2 & 3\end{pmatrix} \times \begin{pmatrix}-7 \\ 2\end{pmatrix} = \begin{pmatrix}2(-7) + 4(2) \\ 2(-7) + 3(2)\end{pmatrix} = \begin{pmatrix}-14 + 8 \\ -14 + 6\end{pmatrix} = (-6, -8)$$ Answer: b. $(-6, -8)$