1. **Problem statement:** a. Given triangle ABC with vertices A(3,0), B(6,4), and C(-1,3), show that ABC is a right-angled isosceles triangle. 2. **Formula and rules:** - Distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - A triangle is right-angled if the square of the longest side equals the sum of the squares of the other two sides (Pythagoras theorem). - A triangle is isosceles if at least two sides are equal in length. 3. **Calculate side lengths:** - $AB = \sqrt{(6-3)^2 + (4-0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ - $BC = \sqrt{(-1-6)^2 + (3-4)^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$ - $AC = \sqrt{(-1-3)^2 + (3-0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ 4. **Check for isosceles:** - Sides $AB = 5$ and $AC = 5$ are equal, so triangle ABC is isosceles. 5. **Check for right angle:** - Longest side is $BC = 5\sqrt{2}$. - Check Pythagoras: $AB^2 + AC^2 = 5^2 + 5^2 = 25 + 25 = 50$ - $BC^2 = (5\sqrt{2})^2 = 25 \times 2 = 50$ - Since $AB^2 + AC^2 = BC^2$, triangle ABC is right-angled at vertex A. **Final conclusion:** Triangle ABC is a right-angled isosceles triangle.