If coordinates of the vertices of a triangle are (2,0), (6,0) and (1,5), then distance between its orthocentre and circumcentre is

Options

A. 4
B. 6
C. 5
D. None of these

Answer

To find the distance between the orthocenter and circumcenter of the triangle with vertices at (2,0), (6,0), and (1,5), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- A(2, 0)
- B(6, 0)
- C(1, 5)

### Step 2: Find the circumcenter
The circumcenter is the point where the perpendicular bisectors of the sides of the triangle intersect.

1. **Find the midpoint of AB:**
\[ \text{Midpoint of AB} = \left( \frac{2 + 6}{2}, \frac{0 + 0}{2} \right) = (4, 0) \]

2. **Find the slope of AB:**
Since A and B lie on the x-axis, the slope of AB is:
\[ \text{slope of AB} = \frac{0 - 0}{6 - 2} = 0 \]
The perpendicular slope is undefined (vertical line).

3. **Equation of the perpendicular bisector of AB:**
The equation is \( x = 4 \).

4. **Find the midpoint of AC:**
\[ \text{Midpoint of AC} = \left( \frac{2 + 1}{2}, \frac{0 + 5}{2} \right) = \left( \frac{3}{2}, \frac{5}{2} \right) \]

5. **Find the slope of AC:**
\[ \text{slope of AC} = \frac{5 - 0}{1 - 2} = -5 \]
The perpendicular slope is \( \frac{1}{5} \).

6. **Equation of the perpendicular bisector of AC:**
Using point-slope form:
\[ y - \frac{5}{2} = \frac{1}{5} \left( x - \frac{3}{2} \right) \]
Simplifying, we get:
\[ 5y - 12.5 = x - 1.5 \implies x - 5y + 11 = 0 \quad \text{(Equation 1)} \]

7. **Find the intersection of the two perpendicular bisectors:**
Substitute \( x = 4 \) into Equation 1:
\[ 4 - 5y + 11 = 0 \implies 5y = 15 \implies y = 3 \]
Thus, the circumcenter \( O \) is at \( (4, 3) \).

### Step 3: Find the orthocenter
The orthocenter is the intersection of the altitudes of the triangle.

1. **Find the slope of BC:**
\[ \text{slope of BC} = \frac{5 - 0}{1 - 6} = -1 \]
The slope of the altitude from A (perpendicular to BC) is \( 1 \).

2. **Equation of the altitude from A:**
Using point-slope form:
\[ y - 0 = 1(x - 2) \implies y = x - 2 \quad \text{(Equation 2)} \]

3. **Find the slope of AC:**
(Already calculated as -5, so the altitude from B will have slope \( \frac{1}{5} \)).

4. **Equation of the altitude from B:**
\[ y - 0 = \frac{1}{5}(x - 6) \implies 5y = x - 6 \implies x - 5y - 6 = 0 \quad \text{(Equation 3)} \]

5. **Find the intersection of Equations 2 and 3:**
Substitute \( y = x - 2 \) into Equation 3:
\[ x - 5(x - 2) - 6 = 0 \implies x - 5x + 10 - 6 = 0 \implies -4x + 4 = 0 \implies x = 1 \]
Substitute \( x = 1 \) into Equation 2:
\[ y = 1 - 2 = -1 \]
Thus, the orthocenter \( H \) is at \( (1, -1) \).

### Step 4: Calculate the distance between the orthocenter and circumcenter
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting \( O(4, 3) \) and \( H(1, -1) \):
\[ d = \sqrt{(4 - 1)^2 + (3 - (-1))^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Final Answer
The distance between the orthocenter and circumcenter is \( 5 \) units.

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