1. **Problem Statement:** We have a square ABCD with vertices A(3,1), B(3,6), C(8,1), and D(8,6). We will perform the following transformations: (a) Translate 8 units left and 5 units down. (b) Rotate 90 degrees about the vertical line $x=1$. (c) Reflect across the vertical line $x=-2$. (d) Dilate by a factor of 2 from the origin. 2. **Translation (a):** Translation moves each point by subtracting 8 from the x-coordinate and 5 from the y-coordinate. Formula: $A'(x',y') = (x - 8, y - 5)$ Calculate new vertices: - $A'(3 - 8, 1 - 5) = (-5, -4)$ - $B'(3 - 8, 6 - 5) = (-5, 1)$ - $C'(8 - 8, 1 - 5) = (0, -4)$ - $D'(8 - 8, 6 - 5) = (0, 1)$ 3. **Rotation (b):** Rotate 90 degrees about the vertical line $x=1$. Rotation about a vertical line $x = h$ by 90 degrees clockwise transforms a point $(x,y)$ to $(h + (y - k), k - (x - h))$ if rotating about $(h,k)$, but here rotation is about the line $x=1$ (vertical line), so we treat it as rotation around each point's relative position to $x=1$. For 90 degrees clockwise rotation about $x=1$: - Translate point so line $x=1$ becomes $x=0$: $(x - 1, y)$ - Rotate 90 degrees clockwise: $(x', y') = (y, -x)$ - Translate back: $(x' + 1, y')$ Apply to each vertex: - $A(3,1)$: translate $(2,1)$, rotate $(1,-2)$, translate back $(2,-2)$ - $B(3,6)$: translate $(2,6)$, rotate $(6,-2)$, translate back $(7,-2)$ - $C(8,1)$: translate $(7,1)$, rotate $(1,-7)$, translate back $(2,-7)$ - $D(8,6)$: translate $(7,6)$, rotate $(6,-7)$, translate back $(7,-7)$ So new vertices: - $A' = (2,-2)$ - $B' = (7,-2)$ - $C' = (2,-7)$ - $D' = (7,-7)$ 4. **Reflection (c):** Reflect across vertical line $x = -2$. Reflection formula about vertical line $x = h$: $$x' = 2h - x, \quad y' = y$$ Apply to each vertex: - $A(3,1)$: $x' = 2(-2) - 3 = -4 - 3 = -7$, $y' = 1$ - $B(3,6)$: $x' = -7$, $y' = 6$ - $C(8,1)$: $x' = 2(-2) - 8 = -4 - 8 = -12$, $y' = 1$ - $D(8,6)$: $x' = -12$, $y' = 6$ New vertices: - $A' = (-7,1)$ - $B' = (-7,6)$ - $C' = (-12,1)$ - $D' = (-12,6)$ 5. **Dilation (d):** Dilate by factor 2 from origin (0,0). Dilation formula: $$x' = 2x, \quad y' = 2y$$ Apply to each vertex: - $A(3,1)$: $(6,2)$ - $B(3,6)$: $(6,12)$ - $C(8,1)$: $(16,2)$ - $D(8,6)$: $(16,12)$ **Final answers:** - (a) $A'(-5,-4), B'(-5,1), C'(0,-4), D'(0,1)$ - (b) $A'(2,-2), B'(7,-2), C'(2,-7), D'(7,-7)$ - (c) $A'(-7,1), B'(-7,6), C'(-12,1), D'(-12,6)$ - (d) $A'(6,2), B'(6,12), C'(16,2), D'(16,12)$