The orthocentre of the triangle with vertices `(5, 0),(0, 0), ((5)/(2), (5sqrt(3))/(2))` is :

Options

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

Answer

To find the orthocenter of the triangle with vertices \( A(5, 0) \), \( B(0, 0) \), and \( C\left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right) \), we will follow these steps:

### Step 1: Determine the slopes of the sides of the triangle
1. **Calculate the slope of side \( AB \)**:
\[ \text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 0}{0 - 5} = 0 \]
(Since \( A \) and \( B \) are on the x-axis, the slope is 0.)

2. **Calculate the slope of side \( AC \)**:
\[ \text{slope of } AC = \frac{\frac{5\sqrt{3}}{2} - 0}{\frac{5}{2} - 5} = \frac{\frac{5\sqrt{3}}{2}}{-\frac{5}{2}} = -\sqrt{3} \]

3. **Calculate the slope of side \( BC \)**:
\[ \text{slope of } BC = \frac{\frac{5\sqrt{3}}{2} - 0}{\frac{5}{2} - 0} = \frac{\frac{5\sqrt{3}}{2}}{\frac{5}{2}} = \sqrt{3} \]

### Step 2: Find the slopes of the altitudes
1. **The slope of altitude from \( C \) to side \( AB \)** (perpendicular to \( AB \)):
- Since \( AB \) has a slope of 0, the altitude from \( C \) will be vertical, so its equation is \( x = \frac{5}{2} \).

2. **The slope of altitude from \( A \) to side \( BC \)** (perpendicular to \( BC \)):
- The slope of \( BC \) is \( \sqrt{3} \), so the slope of the altitude from \( A \) will be \( -\frac{1}{\sqrt{3}} \).
- The equation of the line through \( A(5, 0) \) with this slope is:
\[ y - 0 = -\frac{1}{\sqrt{3}}(x - 5) \implies y = -\frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}} \]

### Step 3: Find the intersection of the altitudes
1. **Substitute \( x = \frac{5}{2} \) into the equation of the altitude from \( A \)**:
\[ y = -\frac{1}{\sqrt{3}}\left(\frac{5}{2}\right) + \frac{5}{\sqrt{3}} = -\frac{5}{2\sqrt{3}} + \frac{5}{\sqrt{3}} = \frac{5}{2\sqrt{3}} \]

### Step 4: Conclusion
The orthocenter \( H \) of the triangle is:
\[ H\left(\frac{5}{2}, \frac{5}{2\sqrt{3}}\right) \]