Vertices of a triangle are A (6,0), B (0,6) and C(6,6). The distance between its circumcentre and orthocentre is

Options

A. `2 sqrt(2)`
B. 2
C. `3sqrt(2)`
D. 1

Answer

To find the distance between the circumcentre and orthocentre of triangle ABC with vertices A(6,0), B(0,6), and C(6,6), we can follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- A(6, 0)
- B(0, 6)
- C(6, 6)

### Step 2: Determine the type of triangle
Since angle C is 90 degrees (as A and B lie on the axes and C is at (6,6)), triangle ABC is a right-angled triangle with C as the right angle.

### Step 3: Find the orthocentre (H)
For a right-angled triangle, the orthocentre is located at the vertex where the right angle is formed. Therefore, the orthocentre H is at point C:
- H = C(6, 6)

### Step 4: Find the circumcentre (O)
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse. The hypotenuse is the line segment AB.

To find the midpoint O of AB:
- Coordinates of A = (6, 0)
- Coordinates of B = (0, 6)

Using the midpoint formula:
\[ O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{6 + 0}{2}, \frac{0 + 6}{2} \right) = \left( 3, 3 \right) \]

### Step 5: Calculate the distance between the circumcentre O and orthocentre H
Now we need to find the distance between points O(3, 3) and H(6, 6) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(6 - 3)^2 + (6 - 3)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Final Answer
The distance between the circumcentre and orthocentre is \(3\sqrt{2}\) units.

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