Find the co-ordinates of the circumcentre of a triangle whose vertices are `(7,5)`, `(6,6)` and `(-2,2)`.

Answer

To find the coordinates of the circumcentre of the triangle with vertices at \( A(7, 5) \), \( B(6, 6) \), and \( C(-2, 2) \), we can follow these steps:

### Step 1: Understand the Circumcentre
The circumcentre of a triangle is the point where the perpendicular bisectors of the sides of the triangle intersect. It is also the center of the circumcircle that passes through all three vertices of the triangle.

### Step 2: Find the Midpoints of Two Sides
Let's find the midpoints of two sides of the triangle. We can choose sides \( AB \) and \( AC \).

1. **Midpoint of AB**:
\[ M_{AB} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{7 + 6}{2}, \frac{5 + 6}{2} \right) = \left( \frac{13}{2}, \frac{11}{2} \right) \]

2. **Midpoint of AC**:
\[ M_{AC} = \left( \frac{7 + (-2)}{2}, \frac{5 + 2}{2} \right) = \left( \frac{5}{2}, \frac{7}{2} \right) \]

### Step 3: Find the Slopes of the Sides
Next, we need to find the slopes of sides \( AB \) and \( AC \).

1. **Slope of AB**:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 5}{6 - 7} = \frac{1}{-1} = -1 \]

2. **Slope of AC**:
\[ m_{AC} = \frac{2 - 5}{-2 - 7} = \frac{-3}{-9} = \frac{1}{3} \]

### Step 4: Find the Slopes of the Perpendicular Bisectors
The slopes of the perpendicular bisectors are the negative reciprocals of the slopes of the sides.

1. **Slope of the perpendicular bisector of AB**:
\[ m_{pAB} = -\frac{1}{m_{AB}} = 1 \]

2. **Slope of the perpendicular bisector of AC**:
\[ m_{pAC} = -\frac{1}{m_{AC}} = -3 \]

### Step 5: Write the Equations of the Perpendicular Bisectors
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \):

1. **Equation of the perpendicular bisector of AB**:
\[ y - \frac{11}{2} = 1 \left( x - \frac{13}{2} \right) \]
Simplifying this:
\[ y = x - \frac{13}{2} + \frac{11}{2} \implies y = x - 1 \]

2. **Equation of the perpendicular bisector of AC**:
\[ y - \frac{7}{2} = -3 \left( x - \frac{5}{2} \right) \]
Simplifying this:
\[ y - \frac{7}{2} = -3x + \frac{15}{2} \implies y = -3x + \frac{15}{2} + \frac{7}{2} \implies y = -3x + 11 \]

### Step 6: Solve the System of Equations
Now, we need to solve the equations:
1. \( y = x - 1 \)
2. \( y = -3x + 11 \)

Setting them equal:
\[ x - 1 = -3x + 11 \]
Solving for \( x \):
\[ x + 3x = 11 + 1 \implies 4x = 12 \implies x = 3 \]

Now substitute \( x = 3 \) back into one of the equations to find \( y \):
\[ y = 3 - 1 = 2 \]

### Final Answer
The coordinates of the circumcentre are \( (3, 2) \).

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