To show that the points $(4, 0)$, $(2, 1)$, and $(-1, -5)$ are vertices of a right triangle, we need to verify that the triangle formed by these points satisfies the Pythagorean theorem. We'll also find the area of the triangle.

Step 1: Calculate the distances between the points

We use the distance formula to find the lengths of the sides of the triangle:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

1. Distance between $(4, 0)$ and $(2, 1)$:

$d_1 = \sqrt{(2 - 4)^2 + (1 - 0)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$

2. Distance between $(2, 1)$ and $(-1, -5)$:

$d_2 = \sqrt{(-1 - 2)^2 + (-5 - 1)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$

3. Distance between $(4, 0)$ and $(-1, -5)$:

$d_3 = \sqrt{(-1 - 4)^2 + (-5 - 0)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$

Step 2: Verify the Pythagorean theorem

The Pythagorean theorem states that for a right triangle:

$\text{Hypotenuse}^2 = \text{Side}_1^2 + \text{Side}_2^2$

Here, we compare the squares of the distances:

• $\text{Hypotenuse} = d_3 = 5\sqrt{2}$
• $\text{Side}_1 = d_2 = 3\sqrt{5}$
• $\text{Side}_2 = d_1 = \sqrt{5}$

Now, check if:

$(5\sqrt{2})^2 = (3\sqrt{5})^2 + (\sqrt{5})^2$

Calculate each side:

• $(5\sqrt{2})^2 = 50$
• $(3\sqrt{5})^2 = 45$
• $(\sqrt{5})^2 = 5$

Since:

$50 = 45 + 5$

The equation holds true, confirming that the points form a right triangle.

Step 3: Calculate the area of the right triangle

The area $A$ of a right triangle is given by:

$A = \frac{1}{2} \times \text{base} \times \text{height}$

Here, we can take (d_1 = \