1. **State the problem:** Determine if points X(3,1), Y(3,7), and Z(11,1) can form a triangle by checking the side lengths and triangle inequality. 2. **Formula for distance between two points:** $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate side lengths:** - $XY = \sqrt{(3-3)^2 + (7-1)^2} = \sqrt{0 + 6^2} = \sqrt{36} = 6$ - $YZ = \sqrt{(11-3)^2 + (1-7)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ - $XZ = \sqrt{(11-3)^2 + (1-1)^2} = \sqrt{8^2 + 0} = \sqrt{64} = 8$ 4. **Check triangle inequality:** - $XY + YZ = 6 + 10 = 16 > 8 = XZ$ - $XY + XZ = 6 + 8 = 14 > 10 = YZ$ - $YZ + XZ = 10 + 8 = 18 > 6 = XY$ Since all sums of two sides are greater than the third side, the points can form a triangle. 5. **Classify the triangle:** Use the Pythagorean theorem to check if the triangle is right, acute, or obtuse. - Largest side is $YZ = 10$ - Check $YZ^2$ vs $XY^2 + XZ^2$ - $YZ^2 = 10^2 = 100$ - $XY^2 + XZ^2 = 6^2 + 8^2 = 36 + 64 = 100$ Since $YZ^2 = XY^2 + XZ^2$, the triangle is a **right triangle**. **Final answer:** The points X, Y, and Z can form a right triangle.