If the points `(2k, k), (k, 2k)` and (k, k) with `kgt0` enclose in a triangle of area 18 sq units, then the centroid of triangle is equal to

Options

A. (8,8)
B. (4,4)
C. (-4,-4)
D. `(4sqrt2,4sqrt2)`

Answer

To solve the problem step by step, we need to find the centroid of the triangle formed by the points (2k, k), (k, 2k), and (k, k) given that the area of the triangle is 18 square units.

### Step 1: Identify the vertices of the triangle
Let the points be:
- A = (2k, k)
- B = (k, 2k)
- C = (k, k)

### Step 2: Calculate the area of the triangle
The area \( A \) of a triangle formed by the vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points A, B, and C:
\[ A = \frac{1}{2} \left| 2k(2k - k) + k(k - k) + k(k - 2k) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| 2k(k) + 0 + k(-k) \right| \]
\[ A = \frac{1}{2} \left| 2k^2 - k^2 \right| = \frac{1}{2} \left| k^2 \right| = \frac{k^2}{2} \]
Given that the area is 18 square units, we have:
\[ \frac{k^2}{2} = 18 \]

### Step 3: Solve for \( k \)
Multiplying both sides by 2:
\[ k^2 = 36 \]
Taking the square root:
\[ k = 6 \quad (\text{since } k > 0) \]

### Step 4: Find the coordinates of the vertices
Now substituting \( k = 6 \) into the coordinates of the points:
- A = (2k, k) = (12, 6)
- B = (k, 2k) = (6, 12)
- C = (k, k) = (6, 6)

### Step 5: Calculate the centroid of the triangle
The centroid \( G \) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) 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 points A, B, and C:
\[ G = \left( \frac{12 + 6 + 6}{3}, \frac{6 + 12 + 6}{3} \right) \]
Calculating the x-coordinate:
\[ G_x = \frac{24}{3} = 8 \]
Calculating the y-coordinate:
\[ G_y = \frac{24}{3} = 8 \]
Thus, the centroid \( G \) is:
\[ G = (8, 8) \]

### Final Answer
The centroid of the triangle is \( (8, 8) \).

---