Step-by-step Solution:

Let the vertices be \( A(4, 0) \), \( B(-1, -1) \), and \( C(3, 5) \). - Slope of AB: \[ \text{Slope of AB} = \frac{-1 - 0}{-1 - 4} = \frac{-1}{-5} = \frac{1}{5} \] - Slope of BC: \[ \text{Slope of BC} = \frac{5 - (-1)}{3 - (-1)} = \frac{6}{4} = \frac{3}{2} \] - Slope of CA: \[ \text{Slope of CA} = \frac{0 - 5}{4 - 3} = \frac{-5}{1} = -5 \] Since the product of the slopes of AB and CA is: \[ \frac{1}{5} \times (-5) = -1 \] we conclude that \( AB \perp CA \). - Lengths of the sides: \[ AB = \sqrt{(-1 - 4)^2 + (-1 - 0)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \] \[ BC = \sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \] \[ CA = \sqrt{(4 - 3)^2 + (0 - 5)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \] Since \( AB = CA \) and \( AB \perp CA \), the triangle \( ABC \) is a right-angled isosceles triangle with the right angle at \( A \).