Analyzing the Equilateral Triangle Vertices and Area

The problem asks us to consider an equilateral triangle with two given vertices at (0, 0) and (2, 2). We need to evaluate two statements about the third vertex and the area of this triangle.

Step-by-Step Solution to Find the Third Vertex and Area

First, let's find the side length of the equilateral triangle. The distance between the two given vertices (0, 0) and (2, 2) will be the side length.

Let the two vertices be $A(0, 0)$ and $B(2, 2)$. The distance $AB$ is calculated using the distance formula:

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

$$ s = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

So, the side length of the equilateral triangle is $2\sqrt{2}$.

Next, let the third vertex be $C(x, y)$. Since it's an equilateral triangle, the distance from $C$ to $A$ must be equal to the side length, and the distance from $C$ to $B$ must also be equal to the side length.

Distance $AC = s$:

$$ \sqrt{(x - 0)^2 + (y - 0)^2} = 2\sqrt{2} $$

$$ x^2 + y^2 = (2\sqrt{2})^2 = 8 \quad \text{(Equation 1)} $$

Distance $BC = s$:

$$ \sqrt{(x - 2)^2 + (y - 2)^2} = 2\sqrt{2} $$

$$ (x - 2)^2 + (y - 2)^2 = (2\sqrt{2})^2 = 8 \quad \text{(Equation 2)} $$

Now, we solve the system of equations. Expand Equation 2:

$$ x^2 - 4x + 4 + y^2 - 4y + 4 = 8 $$

$$ x^2 + y^2 - 4x - 4y + 8 = 8 $$

Substitute $x^2 + y^2 = 8$ from Equation 1 into the expanded Equation 2:

$$ 8 - 4x - 4y + 8 = 8 $$

$$ 16 - 4x - 4y = 8 $$

$$ 4x + 4y = 16 - 8 $$

$$ 4x + 4y = 8 $$

Divide by 4:

$$ x + y = 2 \quad \text{or} \quad y = 2 - x $$

Substitute $y = 2 - x$ into Equation 1:

$$ x^2 + (2 - x)^2 = 8 $$

$$ x^2 + (4 - 4x + x^2) = 8 $$

$$ 2x^2 - 4x + 4 = 8 $$

$$ 2x^2 - 4x - 4 = 0 $$

Divide by 2:

$$ x^2 - 2x - 2 = 0 $$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $x$:

$$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} $$

$$ x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} $$

$$ x = 1 \pm \sqrt{3} $$

Now find the corresponding $y$ values using $y = 2 - x$:

Case 1: $x = 1 + \sqrt{3}$

$$ y = 2 - (1 + \sqrt{3}) = 2 - 1 - \sqrt{3} = 1 - \sqrt{3} $$

Third vertex 1: $(1 + \sqrt{3}, 1 - \sqrt{3})$

Case 2: $x = 1 - \sqrt{3}$

$$ y = 2 - (1 - \sqrt{3}) = 2 - 1 + \sqrt{3} = 1 + \sqrt{3} $$

Third vertex 2: $(1 - \sqrt{3}, 1 + \sqrt{3})$

There are two possible locations for the third vertex, both resulting in an equilateral triangle with the given two points. In both cases, the coordinates involve $\sqrt{3}$, which is an irrational number. Thus, the third vertex has at least one irrational coordinate.

Next, let's calculate the area of the equilateral triangle. The area of an equilateral triangle with side length $s$ is given by the formula:

$$ \text{Area} = \frac{\sqrt{3}}{4} s^2 $$

We found the side length $s = 2\sqrt{2}$. Substitute this into the area formula:

$$ \text{Area} = \frac{\sqrt{3}}{4} (2\sqrt{2})^2 = \frac{\sqrt{3}}{4} (4 \times 2) = \frac{\sqrt{3}}{4} (8) = 2\sqrt{3} $$

The area of the triangle is $2\sqrt{3}$. Since $\sqrt{3}$ is irrational, $2\sqrt{3}$ is also an irrational number.

Analyzing the Statements

Let's evaluate the given statements based on our calculations for the equilateral triangle.

1. The third vertex has at least one irrational coordinate.
2. The area is irrational.

Statement 1: We found the possible third vertices are $(1 + \sqrt{3}, 1 - \sqrt{3})$ and $(1 - \sqrt{3}, 1 + \sqrt{3})$. Both coordinates in each case are irrational because they involve $\sqrt{3}$. Therefore, the third vertex indeed has at least one irrational coordinate.

Statement 1 is Correct.

Statement 2: We calculated the area of the triangle to be $2\sqrt{3}$. Since $\sqrt{3}$ is an irrational number, the product $2\sqrt{3}$ is also irrational.

Statement 2 is Correct.

Both statements 1 and 2 are correct for the equilateral triangle with the given vertices.

Based on the analysis, both Statement 1 and Statement 2 are correct.

Revision Table: Key Equilateral Triangle Concepts

Additional Information on Irrational Coordinates and Area

In coordinate geometry problems involving equilateral triangles, if the initial vertices have rational coordinates, the third vertex coordinates often involve square roots of non-perfect squares, leading to irrational numbers. This is because the geometry of the triangle (specifically the 60-degree angles) relates rational distances to irrational side lengths through trigonometric ratios or distance formulas that introduce square roots.

For an equilateral triangle with side length $s$, the area formula $\frac{\sqrt{3}}{4} s^2$ inherently includes the irrational number $\sqrt{3}$. Unless $s^2$ is a specific value that cancels out $\sqrt{3}$ (which isn't possible if $s$ is a real number derived from distances between rational points, as $s^2$ will be rational), the area will typically involve $\sqrt{3}$ and thus be irrational.

In this case, the side length squared was $s^2 = 8$, which is rational. The area is $\frac{\sqrt{3}}{4} \times 8 = 2\sqrt{3}$, which is irrational. The coordinates of the third vertex $(1 \pm \sqrt{3}, 1 \mp \sqrt{3})$ also clearly contain irrational components.