To solve the problem step by step, we will follow the outlined process:
### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- A(2a, a)
- B(a, a)
- C(a, 2a)
### Step 2: Use the formula for the area of a triangle
The area \( A \) of a triangle with 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:
\[
A = \frac{1}{2} \left| 2a(a - 2a) + a(2a - a) + a(a - a) \right|
\]
This simplifies to:
\[
A = \frac{1}{2} \left| 2a(-a) + a(a) + 0 \right| = \frac{1}{2} \left| -2a^2 + a^2 \right| = \frac{1}{2} \left| -a^2 \right| = \frac{a^2}{2}
\]
### Step 3: Set the area equal to 18 square units
We know from the problem statement that the area is 18 square units:
\[
\frac{a^2}{2} = 18
\]
Multiplying both sides by 2 gives:
\[
a^2 = 36
\]
Taking the square root of both sides, we find:
\[
a = 6 \quad \text{or} \quad a = -6
\]
### Step 4: Determine the circumcenter of the triangle
For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse. We need to first identify the lengths of the sides to find the hypotenuse.
#### Step 4.1: Calculate the lengths of the sides
Using the distance formula:
1. Length \( AB \):
\[
AB = \sqrt{(2a - a)^2 + (a - a)^2} = \sqrt{a^2} = a
\]
2. Length \( BC \):
\[
BC = \sqrt{(a - a)^2 + (a - 2a)^2} = \sqrt{(-a)^2} = a
\]
3. Length \( AC \):
\[
AC = \sqrt{(2a - a)^2 + (a - 2a)^2} = \sqrt{a^2 + (-a)^2} = \sqrt{2a^2} = a\sqrt{2}
\]
Since \( AC \) is the longest side, it is the hypotenuse.
#### Step 4.2: Calculate the midpoint of the hypotenuse \( AC \)
The midpoint \( M \) of segment \( AC \) can be calculated using the midpoint formula:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
Substituting the coordinates of points A and C:
\[
M = \left( \frac{2a + a}{2}, \frac{a + 2a}{2} \right) = \left( \frac{3a}{2}, \frac{3a}{2} \right)
\]
### Step 5: Substitute the values of \( a \)
Now substituting \( a = 6 \) and \( a = -6 \):
1. For \( a = 6 \):
\[
M = \left( \frac{3 \times 6}{2}, \frac{3 \times 6}{2} \right) = (9, 9)
\]
2. For \( a = -6 \):
\[
M = \left( \frac{3 \times -6}{2}, \frac{3 \times -6}{2} \right) = (-9, -9)
\]
### Final Answer
The circumcenter of the triangle can be either \( (9, 9) \) or \( (-9, -9) \).
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