1. **Problem:** Find the coordinates of triangle DEF after each transformation. Given vertices: D(4, -1), E(5, 2), F(1, 2). 2. **Reflection in x-axis:** Formula: Reflecting a point $(x,y)$ in the x-axis gives $(x,-y)$. - $D' = (4, \cancel{-1}^{+1}) = (4,1)$ - $E' = (5, \cancel{2}^{-2}) = (5,-2)$ - $F' = (1, \cancel{2}^{-2}) = (1,-2)$ 3. **Reflection in y-axis:** Formula: Reflecting a point $(x,y)$ in the y-axis gives $(-x,y)$. - $D' = (\cancel{4}^{-4}, -1) = (-4,-1)$ - $E' = (\cancel{5}^{-5}, 2) = (-5,2)$ - $F' = (\cancel{1}^{-1}, 2) = (-1,2)$ 4. **Translation along vector $\langle 1, 0 \rangle$:** Formula: Translate $(x,y)$ by adding vector components: $(x+1, y+0)$. - $D' = (4+1, -1+0) = (5,-1)$ - $E' = (5+1, 2+0) = (6,2)$ - $F' = (1+1, 2+0) = (2,2)$ 5. **Translation along vector $\langle -3, 1 \rangle$:** Formula: Translate $(x,y)$ by $(x-3, y+1)$. - $D' = (4-3, -1+1) = (1,0)$ - $E' = (5-3, 2+1) = (2,3)$ - $F' = (1-3, 2+1) = (-2,3)$ 6. **Rotation 180° about the origin:** Formula: Rotating $(x,y)$ by 180° about origin gives $(-x,-y)$. - $D' = (\cancel{4}^{-4}, \cancel{-1}^{1}) = (-4,1)$ - $E' = (\cancel{5}^{-5}, \cancel{2}^{-2}) = (-5,-2)$ - $F' = (\cancel{1}^{-1}, \cancel{2}^{-2}) = (-1,-2)$ 7. **Rotation 270° counterclockwise about the origin:** Formula: Rotating $(x,y)$ by 270° CCW is equivalent to rotating 90° clockwise, giving $(y,-x)$. - $D' = (-1, \cancel{4}^{-4}) = (-1,-4)$ - $E' = (2, \cancel{5}^{-5}) = (2,-5)$ - $F' = (2, \cancel{1}^{-1}) = (2,-1)$ 8. **AIR SHOW transformation:** Planes start at $(-20,-15)$ and $(10,-15)$ and end at $(-30,20)$ and $(0,20)$. Calculate translation vector: - For first plane: $(-30 - (-20), 20 - (-15)) = (-10, 35)$ - For second plane: $(0 - 10, 20 - (-15)) = (-10, 35)$ Both planes translate by vector $\langle -10, 35 \rangle$. **Final answer:** The transformation is a translation by vector $\langle -10, 35 \rangle$.