If two members of family `(2+lambda)x+(1+2lambda)y-3(1+lambda) = 0` and line x+y=0 make an equilateral triangle, the the incentre of triangle so formed is

Options

A. `((1)/(3), (1)/(3))`
B. `((7)/(6), -(5)/(6))`
C. `((5)/(6), (5)/(6))`
D. `(-(3)/(2), -(3)/(2))`

Explanation

To solve the problem, we need to find the incentre of the equilateral triangle formed by the lines given in the question. Let's go through the steps systematically.

### Step 1: Identify the lines
We have the family of lines given by:
\[ (2 + \lambda)x + (1 + 2\lambda)y - 3(1 + \lambda) = 0 \]
and the line:
\[ x + y = 0 \]

### Step 2: Rewrite the family of lines
Let's rewrite the family of lines in a more manageable form. We can express it as:
\[ (2 + \lambda)x + (1 + 2\lambda)y = 3(1 + \lambda) \]
This represents a family of lines depending on the parameter \(\lambda\).

### Step 3: Find the intersection point
We need to find the intersection point of the two lines. The second line can be rewritten as:
\[ y = -x \]
Substituting \(y = -x\) into the family of lines:
\[ (2 + \lambda)x + (1 + 2\lambda)(-x) = 3(1 + \lambda) \]
This simplifies to:
\[ (2 + \lambda - 1 - 2\lambda)x = 3(1 + \lambda) \]
\[ (1 - \lambda)x = 3(1 + \lambda) \]
From here, we can solve for \(x\):
\[ x = \frac{3(1 + \lambda)}{1 - \lambda} \]
Substituting back to find \(y\):
\[ y = -\frac{3(1 + \lambda)}{1 - \lambda} \]

### Step 4: Find the coordinates of the intersection point
The intersection point \(A\) can be represented as:
\[ A\left(\frac{3(1 + \lambda)}{1 - \lambda}, -\frac{3(1 + \lambda)}{1 - \lambda}\right) \]

### Step 5: Find the other two points of the triangle
To form an equilateral triangle, we need to find two more lines from the family that intersect with the line \(x + y = 0\). We can choose specific values of \(\lambda\) to find these lines.

Let’s take \(\lambda = 0\) and \(\lambda = 1\):
1. For \(\lambda = 0\):
\[ 2x + y - 3 = 0 \quad \Rightarrow \quad y = -2x + 3 \]
2. For \(\lambda = 1\):
\[ 3x + 3y - 12 = 0 \quad \Rightarrow \quad y = -x + 4 \]

### Step 6: Find the intersection points of these lines with \(x + y = 0\)
1. For \(y = -2x + 3\):
\[ -2x + 3 + x = 0 \quad \Rightarrow \quad -x + 3 = 0 \quad \Rightarrow \quad x = 3, y = -3 \]
So, point \(B(3, -3)\).

2. For \(y = -x + 4\):
\[ -x + 4 + x = 0 \quad \Rightarrow \quad 4 = 0 \quad \Rightarrow \quad x = 4, y = -4 \]
So, point \(C(4, -4)\).

### Step 7: Find the incentre of triangle ABC
In an equilateral triangle, the incentre coincides with the centroid. The centroid \(G\) of triangle \(ABC\) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the points \(A(1, 1)\), \(B(3, -3)\), and \(C(4, -4)\):
\[ G\left(\frac{1 + 3 + 4}{3}, \frac{1 - 3 - 4}{3}\right) = G\left(\frac{8}{3}, \frac{-6}{3}\right) = G\left(\frac{8}{3}, -2\right) \]

### Final Answer
The incentre of the triangle formed by the lines is:
\[ \text{Incentre} = \left(\frac{8}{3}, -2\right) \]