If (2,0) and (-2, 0) are the two vertices of an equilateral triangle , then the third vertex can be `"_______"`.

Options

A. (0,0)
B. `( 2 , -2)`
C. `(0 , 2sqrt3)`
D. `(sqrt3 , sqrt3)`

Explanation

To find the third vertex of the equilateral triangle with the given vertices (2, 0) and (-2, 0), we can follow these steps:

### Step 1: Identify the given points
Let the two given vertices be:
- A = (2, 0)
- B = (-2, 0)

### Step 2: Determine the distance between A and B
The distance \( AB \) can be calculated using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of A and B:
\[ AB = \sqrt{((-2) - 2)^2 + (0 - 0)^2} = \sqrt{(-4)^2} = \sqrt{16} = 4 \]

### Step 3: Set up the coordinates for the third vertex
Let the coordinates of the third vertex C be \( (x, y) \).

### Step 4: Use the property of equilateral triangles
In an equilateral triangle, all sides are equal. Therefore, we need to set up two equations based on the distances \( AC \) and \( BC \), which should also equal 4.

1. Distance \( AC \):
\[ AC = \sqrt{(x - 2)^2 + (y - 0)^2} = 4 \]
Squaring both sides:
\[ (x - 2)^2 + y^2 = 16 \quad \text{(Equation 1)} \]

2. Distance \( BC \):
\[ BC = \sqrt{(x + 2)^2 + (y - 0)^2} = 4 \]
Squaring both sides:
\[ (x + 2)^2 + y^2 = 16 \quad \text{(Equation 2)} \]

### Step 5: Expand both equations
Expanding Equation 1:
\[ (x - 2)^2 + y^2 = 16 \implies x^2 - 4x + 4 + y^2 = 16 \implies x^2 + y^2 - 4x - 12 = 0 \quad \text{(Equation 1)} \]

Expanding Equation 2:
\[ (x + 2)^2 + y^2 = 16 \implies x^2 + 4x + 4 + y^2 = 16 \implies x^2 + y^2 + 4x - 12 = 0 \quad \text{(Equation 2)} \]

### Step 6: Set the two equations equal
From Equation 1 and Equation 2, we can eliminate \( x^2 + y^2 \):
\[ x^2 + y^2 - 4x - 12 = 0 \]
\[ x^2 + y^2 + 4x - 12 = 0 \]
Subtracting the first from the second:
\[ (x^2 + y^2 + 4x - 12) - (x^2 + y^2 - 4x - 12) = 0 \]
This simplifies to:
\[ 8x = 0 \implies x = 0 \]

### Step 7: Substitute \( x \) back into one of the equations
Substituting \( x = 0 \) into Equation 1:
\[ 0^2 + y^2 - 4(0) - 12 = 0 \implies y^2 - 12 = 0 \implies y^2 = 12 \implies y = \pm 2\sqrt{3} \]

### Step 8: Final coordinates of the third vertex
Thus, the third vertex C can be:
\[ (0, 2\sqrt{3}) \quad \text{or} \quad (0, -2\sqrt{3}) \]

### Conclusion
The third vertex of the equilateral triangle can be either \( (0, 2\sqrt{3}) \) or \( (0, -2\sqrt{3}) \).

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