1. **State the problem:** We need to prove that points $A(1,2)$, $B(3,4)$, and $C(0,-1)$ form a scalene triangle. 2. **Calculate the lengths of each side using the distance formula**: The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$ 3. **Side AB:** $$ AB = \sqrt{(3-1)^2 + (4-2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ 4. **Side BC:** $$ BC = \sqrt{(0-3)^2 + (-1-4)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} $$ 5. **Side CA:** $$ CA = \sqrt{(0-1)^2 + (-1-2)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} $$ 6. **Check if all side lengths are different:** $$ AB = 2\sqrt{2} \approx 2.828, \quad BC = \sqrt{34} \approx 5.831, \quad CA = \sqrt{10} \approx 3.162 $$ Clearly, $AB \neq BC$, $BC \neq CA$, and $CA \neq AB$. 7. **Conclusion:** Since all sides have different lengths, triangle $ABC$ is scalene (a triangle with all sides of different lengths).