1. **Problem Statement:** Determine the type of triangle (scalene, isosceles, or equilateral) formed by the given vertices. 2. **Formula Used:** To classify triangles by side lengths, calculate the distance between each pair of vertices using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Important Rules:** - Equilateral triangle: all three sides equal. - Isosceles triangle: exactly two sides equal. - Scalene triangle: all sides different. 4. **Calculate side lengths for triangle (a) J(-8,3), K(-1,0), L(-1,6):** - $JK = \sqrt{(-1 + 8)^2 + (0 - 3)^2} = \sqrt{7^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58}$ - $KL = \sqrt{(-1 + 1)^2 + (6 - 0)^2} = \sqrt{0^2 + 6^2} = 6$ - $JL = \sqrt{(-1 + 8)^2 + (6 - 3)^2} = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58}$ 5. **Check equality:** $JK = JL = \sqrt{58}$, $KL = 6$; two sides equal, so triangle (a) is **isosceles**. 6. **Calculate side lengths for triangle (b) A(3,1), B(-3,1), C(0,6):** - $AB = \sqrt{(-3 - 3)^2 + (1 - 1)^2} = \sqrt{(-6)^2 + 0} = 6$ - $BC = \sqrt{(0 + 3)^2 + (6 - 1)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$ - $AC = \sqrt{(0 - 3)^2 + (6 - 1)^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$ 7. **Check equality:** $BC = AC = \sqrt{34}$, $AB = 6$; two sides equal, so triangle (b) is **isosceles**. 8. **Calculate side lengths for triangle (c) D(2,0), E(-1,-2), F(2,6):** - $DE = \sqrt{(-1 - 2)^2 + (-2 - 0)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$ - $EF = \sqrt{(2 + 1)^2 + (6 + 2)^2} = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}$ - $DF = \sqrt{(2 - 2)^2 + (6 - 0)^2} = \sqrt{0 + 6^2} = 6$ 9. **Check equality:** $DE = \sqrt{13}$, $EF = \sqrt{73}$, $DF = 6$; all sides different, so triangle (c) is **scalene**. **Final classification:** - (a) Isosceles - (b) Isosceles - (c) Scalene