1. **Problem (i):** Show that A(0,2), B($\sqrt{3}$,-1), C(0,-2) form a right-angled triangle. 2. Calculate lengths of sides using distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$: - $AB=\sqrt{(\sqrt{3}-0)^2+(-1-2)^2}=\sqrt{3+( -3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3}$ - $BC=\sqrt{(0-\sqrt{3})^2+(-2+1)^2}=\sqrt{(-\sqrt{3})^2+(-1)^2}=\sqrt{3+1}=2$ - $AC=\sqrt{(0-0)^2+(-2-2)^2}=\sqrt{0+(-4)^2}=4$ 3. Check Pythagorean theorem for right angle: - $AB^2+BC^2=(2\sqrt{3})^2+2^2=4\times3+4=12+4=16$ - $AC^2=4^2=16$ Since $AB^2+BC^2=AC^2$, triangle ABC is right-angled at B. 4. **Problem (ii):** Show that A(3,1), B(-2,-3), C(2,2) form an isosceles triangle. 5. Calculate side lengths: - $AB=\sqrt{(-2-3)^2+(-3-1)^2}=\sqrt{(-5)^2+(-4)^2}=\sqrt{25+16}=\sqrt{41}$ - $BC=\sqrt{(2+2)^2+(2+3)^2}=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}$ - $AC=\sqrt{(2-3)^2+(2-1)^2}=\sqrt{(-1)^2+1^2}=\sqrt{1+1}=\sqrt{2}$ 6. Since $AB=BC=\sqrt{41}$ and $AC=\sqrt{2}$, triangle ABC has two equal sides and is isosceles. **Final answers:** - (i) Triangle ABC is right-angled at vertex B. - (ii) Triangle ABC is isosceles with equal sides AB and BC.