If the vertices of a triangle ABC are `A (-4, -1), B (1,2)` and C (4, -3), then the coordinates of the circumcentre of the triangle are,

Options

A. `(1/3, -2//3)`
B. `(0, -4)`
C. `(0, -2)`
D. `(-3//2, 1//2)`

Answer

To find the coordinates of the circumcentre of triangle ABC with vertices A(-4, -1), B(1, 2), and C(4, -3), we will follow these steps:

### Step 1: Set up the equations for the circumcentre
The circumcentre (O) of triangle ABC is equidistant from all three vertices A, B, and C. We denote the coordinates of the circumcentre as O(x, y).

The distances OA, OB, and OC can be expressed as:
- OA² = (x + 4)² + (y + 1)²
- OB² = (x - 1)² + (y - 2)²
- OC² = (x - 4)² + (y + 3)²

### Step 2: Equate OA² and OB²
We will first equate OA² and OB²:
\[ (x + 4)² + (y + 1)² = (x - 1)² + (y - 2)² \]
Expanding both sides:
\[ (x² + 8x + 16 + y² + 2y + 1) = (x² - 2x + 1 + y² - 4y + 4) \]
Simplifying:
\[ x² + 8x + 16 + y² + 2y + 1 = x² - 2x + 1 + y² - 4y + 4 \]
Cancelling \(x²\) and \(y²\) from both sides:
\[ 8x + 17 + 2y = -2x + 5 - 4y \]
Rearranging gives:
\[ 10x + 6y + 12 = 0 \]
Dividing through by 2:
\[ 5x + 3y + 6 = 0 \quad \text{(Equation 1)} \]

### Step 3: Equate OB² and OC²
Next, we equate OB² and OC²:
\[ (x - 1)² + (y - 2)² = (x - 4)² + (y + 3)² \]
Expanding both sides:
\[ (x² - 2x + 1 + y² - 4y + 4) = (x² - 8x + 16 + y² + 6y + 9) \]
Simplifying:
\[ x² - 2x + 5 - 4y = x² - 8x + 25 + 6y \]
Cancelling \(x²\) and \(y²\):
\[ -2x + 5 - 4y = -8x + 25 + 6y \]
Rearranging gives:
\[ 6x - 10y - 20 = 0 \]
Dividing through by 2:
\[ 3x - 5y - 10 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we will solve the two equations:
1. \(5x + 3y + 6 = 0\)
2. \(3x - 5y - 10 = 0\)

From Equation 1, we can express \(y\) in terms of \(x\):
\[ 3y = -5x - 6 \implies y = -\frac{5}{3}x - 2 \]

Substituting this expression for \(y\) into Equation 2:
\[ 3x - 5(-\frac{5}{3}x - 2) - 10 = 0 \]
Expanding:
\[ 3x + \frac{25}{3}x + 10 - 10 = 0 \]
Combining like terms:
\[ 3x + \frac{25}{3}x = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 9x + 25x = 0 \implies 34x = 0 \implies x = 0 \]

### Step 5: Find y-coordinate
Substituting \(x = 0\) back into the equation for \(y\):
\[ y = -\frac{5}{3}(0) - 2 = -2 \]

### Conclusion
Thus, the coordinates of the circumcentre O are:
\[ \boxed{(0, -2)} \]