Let A `(3, 2)` and B `(5,1)`. ABP is an equilateral triangle is constructed one the side of AB remote from the origin then the orthocentre of triangle ABP is:

Options

A. `( 4- (1)/(2)sqrt(3), (3)/(2) - sqrt(3) )`
B. `( 4+(1)/(2)sqrt(3), (3)/(2) + sqrt(3) )`
C. `(4- (1)/(6) sqrt(3) , (3)/(2) - (1)/(3) sqrt(3) )`
D. `(4+(1)/(6) sqrt(3) , (3)/(2) + (1)/(3) sqrt(3) )`

Answer

To find the orthocenter of the equilateral triangle \( ABP \) constructed on the side \( AB \) remote from the origin, we will follow these steps:

### Step 1: Identify the Points A and B
Given points:
- \( A(3, 2) \)
- \( B(5, 1) \)

### Step 2: Calculate the Length of AB
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of A and B:
\[ AB = \sqrt{(5 - 3)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 3: Find the Midpoint M of AB
The midpoint \( M \) of segment \( AB \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of A and B:
\[ M = \left( \frac{3 + 5}{2}, \frac{2 + 1}{2} \right) = \left( 4, \frac{3}{2} \right) \]

### Step 4: Calculate the Slope of AB
The slope \( m_{AB} \) of line segment \( AB \) is given by:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{5 - 3} = \frac{-1}{2} \]

### Step 5: Find the Slope of MP
Since \( MP \) is perpendicular to \( AB \), the slope \( m_{MP} \) is the negative reciprocal of \( m_{AB} \):
\[ m_{MP} = -\frac{1}{m_{AB}} = 2 \]

### Step 6: Determine the Coordinates of Point P
Using the properties of an equilateral triangle, we can find the coordinates of point \( P \) using the midpoint \( M \) and the slope \( m_{MP} \). Let \( P(x_P, y_P) \) be the coordinates of point \( P \):
- The height from \( M \) to \( P \) can be calculated using the slope.

Using the angle of 60 degrees (for an equilateral triangle), we can find the coordinates of \( P \):
\[ x_P = x_M + \frac{\sqrt{5}}{2} \cdot \cos(60^\circ) = 4 + \frac{\sqrt{5}}{2} \cdot \frac{1}{2} = 4 + \frac{\sqrt{5}}{4} \]
\[ y_P = y_M + \frac{\sqrt{5}}{2} \cdot \sin(60^\circ) = \frac{3}{2} + \frac{\sqrt{5}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} + \frac{\sqrt{15}}{4} \]

### Step 7: Calculate the Orthocenter
For an equilateral triangle, the orthocenter coincides with the centroid. The centroid \( G \) can be calculated as:
\[ G = \left( \frac{x_A + x_B + x_P}{3}, \frac{y_A + y_B + y_P}{3} \right) \]
Substituting the coordinates:
\[ G = \left( \frac{3 + 5 + (4 + \frac{\sqrt{5}}{4})}{3}, \frac{2 + 1 + (\frac{3}{2} + \frac{\sqrt{15}}{4})}{3} \right) \]
Calculating:
\[ G = \left( \frac{8 + \frac{\sqrt{5}}{4}}{3}, \frac{3.5 + \frac{\sqrt{15}}{4}}{3} \right) \]

### Final Result
The coordinates of the orthocenter \( O \) are:
\[ O = \left( \frac{8 + \frac{\sqrt{5}}{4}}{3}, \frac{3.5 + \frac{\sqrt{15}}{4}}{3} \right) \]