In a `Delta ABC ` the equation of the side BC is `2x-y =3` and its circumcentre and orthocentre are `(2,4) and (1,2) ` respetively .
The distnce of orthocentre from vertex A is

Options

A. `1/(sqrt(5))`
B. `6/(sqrt(5))`
C. `3/(sqrt(5))`
D. `2/(sqrt(5))`

Answer

To find the distance of the orthocenter from vertex A in triangle ABC, we can follow these steps:

### Step 1: Understand the given information
We are given:
- The equation of side BC: \(2x - y = 3\)
- The circumcenter (O) at point \(C(2, 4)\)
- The orthocenter (H) at point \(O(1, 2)\)

### Step 2: Find the slope of line BC
The equation of line BC can be rearranged to the slope-intercept form:
\[ y = 2x - 3 \]
From this, we can see that the slope of line BC is \(m_{BC} = 2\).

### Step 3: Find the slope of the altitude from A
The altitude from vertex A to side BC will be perpendicular to BC. The slope of the altitude (let's call it m_A) will be the negative reciprocal of the slope of BC:
\[ m_A = -\frac{1}{m_{BC}} = -\frac{1}{2} \]

### Step 4: Use the circumradius formula
The circumradius \(R\) is related to the distances from the circumcenter to the vertices. The distance from the circumcenter to the orthocenter is given by the formula:
\[ OH^2 = R^2(1 - 8\cos A \cos B \cos C) \]
However, we can use the relationship between the circumradius and the distances to the vertices.

### Step 5: Find the distance from orthocenter to vertex A
The distance from the orthocenter (H) to vertex A can be expressed as:
\[ AH = 2R \cos A \]
We know from the problem that \(R \cos A = \frac{3}{\sqrt{5}}\). Therefore,
\[ AH = 2 \cdot \frac{3}{\sqrt{5}} = \frac{6}{\sqrt{5}} \]

### Step 6: Conclusion
Thus, the distance of the orthocenter from vertex A is:
\[ \boxed{\frac{6}{\sqrt{5}}} \]