1. **State the problem:** Determine if triangle ABC with vertices A(0,4), B(1,2), and C(4,6) is a right triangle. 2. **Recall the formula:** To check if a triangle is right-angled, use the distance formula to find the lengths of the sides and then apply the Pythagorean theorem. Distance formula between points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate side lengths:** - Side AB: $$AB = \sqrt{(1-0)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ - Side BC: $$BC = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ - Side AC: $$AC = \sqrt{(4-0)^2 + (6-4)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20}$$ 4. **Check the Pythagorean theorem:** The triangle is right-angled if the square of the longest side equals the sum of the squares of the other two sides. Longest side is $BC = 5$. Check: $$BC^2 = 5^2 = 25$$ $$AB^2 + AC^2 = (\sqrt{5})^2 + (\sqrt{20})^2 = 5 + 20 = 25$$ Since $$BC^2 = AB^2 + AC^2$$, triangle ABC is a right triangle. **Final answer:** Triangle ABC is a right triangle because the Pythagorean theorem holds for its side lengths.