1. **State the problem:** We have a triangle and its rotated image on the coordinate plane. We need to find: a) The coordinates of the centre of rotation. b) The direction and angle of rotation, including any possible alternative answers. 2. **Identify the centre of rotation:** The problem states the centre of rotation is approximately at point $(1, -1)$. 3. **Understand rotation properties:** A rotation moves every point of a figure around a fixed point (the centre of rotation) by a certain angle and direction (clockwise or counterclockwise). 4. **Calculate the angle and direction of rotation:** - Consider a point on the original triangle and its image. For example, point $A = (7, -6)$ rotates to $A' = (-3, 4)$. - Find vectors from the centre of rotation $C = (1, -1)$ to these points: $$\vec{CA} = (7 - 1, -6 - (-1)) = (6, -5)$$ $$\vec{CA'} = (-3 - 1, 4 - (-1)) = (-4, 5)$$ 5. **Calculate the angle between vectors $\vec{CA}$ and $\vec{CA'}$:** - Use the dot product formula: $$\vec{CA} \cdot \vec{CA'} = |\vec{CA}| |\vec{CA'}| \cos \theta$$ - Compute dot product: $$6 \times (-4) + (-5) \times 5 = -24 - 25 = -49$$ - Compute magnitudes: $$|\vec{CA}| = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61}$$ $$|\vec{CA'}| = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$$ - Calculate $\cos \theta$: $$\cos \theta = \frac{-49}{\sqrt{61} \times \sqrt{41}} = \frac{-49}{\sqrt{2501}} \approx \frac{-49}{50.01} \approx -0.98$$ - Find $\theta$: $$\theta = \cos^{-1}(-0.98) \approx 168.6^\circ$$ 6. **Determine direction of rotation:** - Use the cross product (2D) to find direction: $$\vec{CA} \times \vec{CA'} = (6)(5) - (-5)(-4) = 30 - 20 = 10 > 0$$ - Positive cross product means rotation is counterclockwise. 7. **Answer:** a) Centre of rotation is at $(1, -1)$. b) The rotation is approximately $168.6^\circ$ counterclockwise. 8. **Alternative answers:** - The same rotation can be described as $191.4^\circ$ clockwise (since $360^\circ - 168.6^\circ = 191.4^\circ$). Thus, the centre is $(1, -1)$ and the rotation is either $168.6^\circ$ counterclockwise or $191.4^\circ$ clockwise.