Step-by-step Solution:

Correct Answer: B ((2,0))

Here's the step-by-step solution to find the centroid of the equilateral triangle inscribed in the parabola.

Step 1: Define the Vertices of the Triangle

First, we identify the properties of the parabola and the vertices of the triangle.

• The equation of the parabola is $y^2 = x$. This is a standard parabola that opens to the right, and its vertex is at the origin, A = (0, 0).
• One vertex of the equilateral triangle is at the vertex of the parabola, A.
• Since the parabola is symmetric about the x-axis, and one vertex of the triangle lies on this axis, the other two vertices (let's call them B and C) must also be symmetric with respect to the x-axis.
• Let the coordinates of the other two vertices be B = $(x_1, y_1)$ and C = $(x_1, -y_1)$, with $y_1 > 0$.
• Since points B and C lie on the parabola $y^2 = x$, we have the relation $x_1 = y_1^2$.

Step 2: Use the Equilateral Property to Find Coordinates

For an equilateral triangle, all side lengths are equal. Therefore, the square of the distance from A to B must equal the square of the distance from B to C ($|AB|^2 = |BC|^2$).

• Calculate $|AB|^2$: $|AB|^2 = (x_1 - 0)^2 + (y_1 - 0)^2 = x_1^2 + y_1^2$
• Calculate $|BC|^2$: $|BC|^2 = (x_1 - x_1)^2 + (-y_1 - y_1)^2 = 0^2 + (-2y_1)^2 = 4y_1^2$

Now, set the squared lengths equal:

$x_1^2 + y_1^2 = 4y_1^2 \implies x_1^2 = 3y_1^2$

We know from Step 1 that $x_1 = y_1^2$. Substitute this into the equation:

$(y_1^2)^2 = 3y_1^2 \implies y_1^4 - 3y_1^2 = 0$

$y_1^2(y_1^2 - 3) = 0$

Since B is a distinct vertex, $y_1 \neq 0$, so we must have $y_1^2 = 3$. This gives $x_1 = 3$. The coordinates of the vertices are:

• A = (0, 0)
• B = (3, $\sqrt{3}$)
• C = (3, $-\sqrt{3}$)

Step 3: Calculate the Centroid

The centroid of a triangle with vertices $(x_A, y_A)$, $(x_B, y_B)$, and $(x_C, y_C)$ is found by averaging the coordinates:

$$ \text{Centroid} = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) $$

$$ \text{Centroid} = \left( \frac{0 + 3 + 3}{3}, \frac{0 + \sqrt{3} + (-\sqrt{3})}{3} \right) = \left( \frac{6}{3}, \frac{0}{3} \right) = (2, 0) $$

Thus, the centroid of the triangle is (2, 0).