The vertices of a triangle are (5,1), (11,1) and (11,9). Find the co-ordinates of the circumcentre of the triangle.

Answer

To find the coordinates of the circumcenter of the triangle with vertices at (5,1), (11,1), and (11,9), we will follow these steps:

### Step 1: Identify the vertices of the triangle
Let the vertices of the triangle be:
- A(5, 1)
- B(11, 1)
- C(11, 9)

### Step 2: Set up the circumcenter condition
The circumcenter is the point (x, y) that is equidistant from all three vertices of the triangle. Therefore, we can set up the equations based on the distances from the circumcenter to points A and B.

### Step 3: Set up the equation for distances PA and PB
Using the distance formula, we have:
- Distance PA = Distance PB
- \((x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 1)^2\)

### Step 4: Simplify the equation
Since \((y - 1)^2\) appears on both sides, we can cancel it out:
- \((x - 5)^2 = (x - 11)^2\)

### Step 5: Expand both sides
Expanding both sides:
- \(x^2 - 10x + 25 = x^2 - 22x + 121\)

### Step 6: Cancel \(x^2\) and rearrange
Cancelling \(x^2\) from both sides gives:
- \(-10x + 25 = -22x + 121\)

Rearranging gives:
- \(12x = 96\)

### Step 7: Solve for x
Dividing both sides by 12:
- \(x = 8\)

### Step 8: Set up the equation for distances PA and PC
Now we will use the circumcenter condition again for points A and C:
- \((x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 9)^2\)

### Step 9: Substitute x and simplify
Substituting \(x = 8\):
- \((8 - 5)^2 + (y - 1)^2 = (8 - 11)^2 + (y - 9)^2\)
- \(3^2 + (y - 1)^2 = (-3)^2 + (y - 9)^2\)
- \(9 + (y - 1)^2 = 9 + (y - 9)^2\)

### Step 10: Cancel 9 and simplify
Cancelling 9 from both sides gives:
- \((y - 1)^2 = (y - 9)^2\)

### Step 11: Expand both sides
Expanding both sides:
- \(y^2 - 2y + 1 = y^2 - 18y + 81\)

### Step 12: Cancel \(y^2\) and rearrange
Cancelling \(y^2\) from both sides gives:
- \(-2y + 1 = -18y + 81\)

Rearranging gives:
- \(16y = 80\)

### Step 13: Solve for y
Dividing both sides by 16:
- \(y = 5\)

### Final Step: Write the coordinates of the circumcenter
Thus, the coordinates of the circumcenter P are:
- \(P(8, 5)\)