1. **Problem Statement:** (i) Show that points A(0, 2), B(\sqrt{3}, -1), and C(0, -2) form a right-angled triangle. (ii) Show that points A(3, 1), B(-2, -3), and C(2, 2) form an isosceles triangle. 2. **Step (i) - Right-angled triangle:** - Calculate vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \). \[ \overrightarrow{AB} = (\sqrt{3} - 0, -1 - 2) = (\sqrt{3}, -3) \] \[ \overrightarrow{AC} = (0 - 0, -2 - 2) = (0, -4) \] - Compute their dot product to check orthogonality: \[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (\sqrt{3})(0) + (-3)(-4) = 0 + 12 = 12 \] - Since dot product \( \neq 0 \), check other pairs: - Calculate vector \( \overrightarrow{BC} = (0 - \sqrt{3}, -2 - (-1)) = (-\sqrt{3}, -1) \). - Dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \): \[ (\sqrt{3})(-\sqrt{3}) + (-3)(-1) = -3 + 3 = 0 \] - Dot product is zero, so vectors are perpendicular. - Hence, \( \triangle ABC \) is right-angled at \( B \). 3. **Step (ii) - Isosceles triangle:** - Calculate lengths \( AB \), \( BC \), and \( CA \). - \[ AB = \sqrt{( -2 - 3 )^2 + ( -3 - 1 )^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \] - \[ BC = \sqrt{( 2 - (-2) )^2 + ( 2 - (-3) )^2} = \sqrt{(4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41} \] - \[ CA = \sqrt{( 2 - 3 )^2 + ( 2 - 1 )^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \] - Since \( AB = BC = \sqrt{41} \), two sides are equal. - Therefore, \( \triangle ABC \) is isosceles. **Final answers:** (i) Points A(0, 2), B(\sqrt{3}, -1), and C(0, -2) form a right-angled triangle with right angle at B. (ii) Points A(3, 1), B(-2, -3), and C(2, 2) form an isosceles triangle because \( AB = BC \).