In the equilateral triangle ABC, the equation of the side BC is `x+y-2=0` and the centroid of `DeltaABC` is (0, 0). If the points A, B, C are in anticlockwise oder, then the midpoint of the line segment joining A and C is

Options

A. `((-sqrt3+1)/(2),(sqrt3+1)/(2))`
B. `((-sqrt3-1)/(2),(sqrt3-1)/(2))`
C. `((-sqrt3-1)/(2),(sqrt3+1)/(2))`
D. `((-sqrt3+1)/(2),(sqrt3-1)/(2))`

Answer

To solve the problem step by step, we will find the midpoint of the line segment joining points A and C in the equilateral triangle ABC, given the equation of side BC and the centroid of the triangle.

### Step 1: Understand the given information
We have an equilateral triangle ABC with the following properties:
- The equation of side BC is given by \(x + y - 2 = 0\).
- The centroid of triangle ABC is at the origin, which is (0, 0).
- The points A, B, and C are arranged in an anticlockwise order.

### Step 2: Find the coordinates of points B and C
The equation \(x + y - 2 = 0\) can be rewritten as \(y = 2 - x\). This line will represent the side BC of the triangle.

To find points B and C, we can select two points on this line. For example:
- Let \(x = 0\), then \(y = 2\). So, point B can be \(B(0, 2)\).
- Let \(x = 2\), then \(y = 0\). So, point C can be \(C(2, 0)\).

### Step 3: Find the coordinates of point A
Since the centroid G of triangle ABC is at (0, 0), we can use the formula for the centroid:
\[ G = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) \]
Given \(G(0, 0)\), we have:
\[ \frac{x_A + 0 + 2}{3} = 0 \quad \Rightarrow \quad x_A + 2 = 0 \quad \Rightarrow \quad x_A = -2 \]
\[ \frac{y_A + 2 + 0}{3} = 0 \quad \Rightarrow \quad y_A + 2 = 0 \quad \Rightarrow \quad y_A = -2 \]
Thus, point A is \(A(-2, -2)\).

### Step 4: Find the midpoint of line segment AC
The coordinates of points A and C are:
- A(-2, -2)
- C(2, 0)

The formula for the midpoint M of a line segment joining points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the coordinates of A and C:
\[ M = \left(\frac{-2 + 2}{2}, \frac{-2 + 0}{2}\right) = \left(\frac{0}{2}, \frac{-2}{2}\right) = (0, -1) \]

### Conclusion
The midpoint of the line segment joining points A and C is \(M(0, -1)\).