1. **State the problem:** We have points A(10,10) and B(-2,14). Point A is reflected about the x-axis to get A'. Point B is rotated anticlockwise about the origin by 90° to get B'. We need to find the coordinates of A' and B' and prove that line segment AB is parallel to A'B'. 2. **Find coordinates of A' (reflection about x-axis):** Reflection about the x-axis changes the y-coordinate sign but keeps x the same. So, if A = (x, y), then A' = (x, -y). Given A = (10, 10), $$A' = (10, -10)$$ 3. **Find coordinates of B' (90° anticlockwise rotation about origin):** The formula for 90° anticlockwise rotation of a point $(x, y)$ about the origin is: $$B' = (-y, x)$$ Given B = (-2, 14), $$B' = (-14, -2)$$ 4. **Prove AB is parallel to A'B':** Two lines are parallel if their direction vectors are scalar multiples. Calculate vector $\overrightarrow{AB}$: $$\overrightarrow{AB} = (x_B - x_A, y_B - y_A) = (-2 - 10, 14 - 10) = (-12, 4)$$ Calculate vector $\overrightarrow{A'B'}$: $$\overrightarrow{A'B'} = (x_{B'} - x_{A'}, y_{B'} - y_{A'}) = (-14 - 10, -2 - (-10)) = (-24, 8)$$ Check if $\overrightarrow{A'B'}$ is a scalar multiple of $\overrightarrow{AB}$: $$\frac{-24}{-12} = 2, \quad \frac{8}{4} = 2$$ Both ratios are equal, so: $$\overrightarrow{A'B'} = 2 \times \overrightarrow{AB}$$ Therefore, $AB$ is parallel to $A'B'$. **Final answers:** - Coordinates of $A'$: $(10, -10)$ - Coordinates of $B'$: $(-14, -2)$ - $AB$ is parallel to $A'B'$ because their direction vectors are scalar multiples.