Circumcentre of the triangle whose vertices are `(2, -1)`, (3, 2) and (0, 3) is

Options

A. `(1, -1)`
B. `(-1, 1)`
C. (1, 1)
D. None of these

Answer

To find the circumcenter of the triangle with vertices at \( A(2, -1) \), \( B(3, 2) \), and \( C(0, 3) \), we can follow these steps:

### Step 1: Calculate the slopes of the sides of the triangle
1. **Find the slope of line AB**:
\[ \text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-1)}{3 - 2} = \frac{2 + 1}{1} = 3 \]

2. **Find the slope of line BC**:
\[ \text{slope of } BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 2}{0 - 3} = \frac{1}{-3} = -\frac{1}{3} \]

### Step 2: Check if lines AB and BC are perpendicular
To check if the lines are perpendicular, we multiply their slopes:
\[ \text{slope of } AB \times \text{slope of } BC = 3 \times -\frac{1}{3} = -1 \]
Since the product of the slopes is \(-1\), lines AB and BC are perpendicular, indicating that triangle ABC is a right triangle.

### Step 3: Identify the hypotenuse
In a right triangle, the circumcenter is located at the midpoint of the hypotenuse. The hypotenuse is the line segment connecting points A and C.

### Step 4: Calculate the midpoint of the hypotenuse AC
1. **Coordinates of A**: \( (2, -1) \)
2. **Coordinates of C**: \( (0, 3) \)

The midpoint \( M \) of segment AC is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 0}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1) \]

### Conclusion
The circumcenter of the triangle with vertices \( (2, -1) \), \( (3, 2) \), and \( (0, 3) \) is \( (1, 1) \).

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