A triangle with vertices (4, 0), (−1, −1), and (3, 5) is:

Vertices of the triangle are
A(4, 0), B(−1, −1), and C(3, 5)

Formula used:
Distance between two points (x₁, y₁) and (x₂, y₂):
d = $$\sqrt{[x₂ − x₁)² + (y₂ − y₁)²]}$$​​

AB $$= \sqrt{(-1 - 4)^2 + (-1 - 0)^2} \\$$​​
​$$= \sqrt{(-5)^2 + (-1)^2} \\= \sqrt{25 + 1} = \sqrt{26}$$​​

BC $$= \sqrt{(3 - (-1))^2 + (5 - (-1))^2} \\$$​​
​$$= \sqrt{(4)^2 + (6)^2} \\= \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$​​

CA $$= \sqrt{(3 - 4)^2 + (5 - 0)^2} \\$$​​
​$$= \sqrt{(-1)^2 + (5)^2} \\= \sqrt{1 + 25} = \sqrt{26}$$​​

Thus, AB = CA = √26 => triangle is isosceles.

Now check for right angle:
For a right triangle, (hypotenuse)² = (side₁)² + (side₂)²

(2√13)² = (√26)² + (√26)²
52 = 26 + 26 = 52

Hence, the triangle is isosceles right triangle.