If the circumcentre of a triangle lies at the point (a, a) and the centroid is the mid - point of the line joining the points `(2a+3, a+4)` and `(a-4, 2a-3)`, then the orthocentre of the triangle lies on the line

Options

A. y = x
B. `(a-1)x+(a+1)y=0`
C. `(a-1)x-(a+1)y=0`
D. `(a+1)x-(a-1)y=2a`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the coordinates of the centroid
The centroid \( G \) of a triangle formed by points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by the formula:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

In this case, we are given two points:
- Point 1: \( (2a + 3, a + 4) \)
- Point 2: \( (a - 4, 2a - 3) \)

The centroid \( G \) is the midpoint of these two points. Therefore, we calculate the coordinates of the centroid \( G \):
\[ G = \left( \frac{(2a + 3) + (a - 4)}{2}, \frac{(a + 4) + (2a - 3)}{2} \right) \]

### Step 2: Simplify the coordinates of the centroid
Calculating the x-coordinate:
\[ \text{x-coordinate} = \frac{(2a + 3) + (a - 4)}{2} = \frac{3a - 1}{2} \]

Calculating the y-coordinate:
\[ \text{y-coordinate} = \frac{(a + 4) + (2a - 3)}{2} = \frac{3a + 1}{2} \]

Thus, the coordinates of the centroid \( G \) are:
\[ G = \left( \frac{3a - 1}{2}, \frac{3a + 1}{2} \right) \]

### Step 3: Set the coordinates of the centroid equal to the given point
We know that the centroid is also given to be at the point \( (a, a) \). Therefore, we set the coordinates equal:
\[ \frac{3a - 1}{2} = a \quad \text{and} \quad \frac{3a + 1}{2} = a \]

### Step 4: Solve the equations
Solving the first equation:
\[ 3a - 1 = 2a \implies a = 1 \]

Solving the second equation:
\[ 3a + 1 = 2a \implies a = -1 \]

### Step 5: Find the orthocenter
The circumcenter \( O \) is at \( (a, a) \) which gives us two potential circumcenters based on the values of \( a \):
1. If \( a = 1 \), then \( O = (1, 1) \).
2. If \( a = -1 \), then \( O = (-1, -1) \).

### Step 6: Use the relationship between circumcenter, centroid, and orthocenter
For any triangle, the circumcenter \( O \), centroid \( G \), and orthocenter \( H \) are collinear. The line on which they lie can be expressed as:
\[ y - y_G = m(x - x_G) \]
where \( m \) is the slope between \( O \) and \( G \).

### Step 7: Calculate the slope and derive the line equation
Using the coordinates of \( O \) and \( G \):
- For \( a = 1 \):
\[ O = (1, 1), \quad G = \left( \frac{3(1) - 1}{2}, \frac{3(1) + 1}{2} \right) = \left( 1, 2 \right) \]
The slope \( m \) is:
\[ m = \frac{2 - 1}{1 - 1} \text{ (undefined, vertical line)} \]

- For \( a = -1 \):
\[ O = (-1, -1), \quad G = \left( \frac{3(-1) - 1}{2}, \frac{3(-1) + 1}{2} \right) = \left( -2, -1 \right) \]
The slope \( m \) is:
\[ m = \frac{-1 + 1}{-2 + 1} = 0 \text{ (horizontal line)} \]

### Conclusion
The orthocenter lies on the line defined by the relationship between the circumcenter and the centroid. The final equation for the line can be derived based on the values of \( a \).