Of the three lines `x+sqrt(3)y=0,x+y=1` and `x-sqrt(3)y=0` two are equations of two altitudes of an equilateral triangle. The centroid of `Delta` is the point

Options

A. `(0,0)`
B. `((sqrt(3))/(sqrt(3)-1),1/(sqrt(3)-1))`
C. `((sqrt(3))/(sqrt(3)+1),1/(sqrt(3)+1))`
D. None of these

Answer

To find the centroid of the equilateral triangle formed by the given lines, we will first identify the points of intersection of the lines, which will form the vertices of the triangle. The lines given are:

1. \( x + \sqrt{3}y = 0 \)
2. \( x + y = 1 \)
3. \( x - \sqrt{3}y = 0 \)

### Step 1: Find the intersection of the first two lines

We will solve the equations \( x + \sqrt{3}y = 0 \) and \( x + y = 1 \).

From the first equation, we can express \( x \) in terms of \( y \):
\[ x = -\sqrt{3}y \]

Now, substitute \( x \) into the second equation:
\[ -\sqrt{3}y + y = 1 \]
\[ y(1 - \sqrt{3}) = 1 \]
\[ y = \frac{1}{1 - \sqrt{3}} \]

Now, substitute \( y \) back into \( x = -\sqrt{3}y \):
\[ x = -\sqrt{3} \cdot \frac{1}{1 - \sqrt{3}} = \frac{-\sqrt{3}}{1 - \sqrt{3}} \]

### Step 2: Find the intersection of the first and third lines

Next, we will solve \( x + \sqrt{3}y = 0 \) and \( x - \sqrt{3}y = 0 \).

From the second equation, we can express \( x \) in terms of \( y \):
\[ x = \sqrt{3}y \]

Now, substitute \( x \) into the first equation:
\[ \sqrt{3}y + \sqrt{3}y = 0 \]
\[ 2\sqrt{3}y = 0 \implies y = 0 \]

Substituting \( y = 0 \) back into \( x = \sqrt{3}y \):
\[ x = 0 \]

Thus, the intersection point is \( (0, 0) \).

### Step 3: Find the intersection of the second and third lines

Now we will solve \( x + y = 1 \) and \( x - \sqrt{3}y = 0 \).

From the second equation, we can express \( x \) in terms of \( y \):
\[ x = \sqrt{3}y \]

Now, substitute \( x \) into the first equation:
\[ \sqrt{3}y + y = 1 \]
\[ y(\sqrt{3} + 1) = 1 \]
\[ y = \frac{1}{\sqrt{3} + 1} \]

Substituting \( y \) back into \( x = \sqrt{3}y \):
\[ x = \sqrt{3} \cdot \frac{1}{\sqrt{3} + 1} = \frac{\sqrt{3}}{\sqrt{3} + 1} \]

### Step 4: Find the centroid of the triangle

The vertices of the triangle are:
1. \( A = \left( \frac{-\sqrt{3}}{1 - \sqrt{3}}, \frac{1}{1 - \sqrt{3}} \right) \)
2. \( B = (0, 0) \)
3. \( C = \left( \frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1} \right) \)

The centroid \( G \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Substituting the coordinates of the vertices into the formula, we can find the coordinates of the centroid.