The orthocentre of the triangle whose vertices are `(1, 1), (5, 1) and (4, 5)` is

Options

A. `((9)/(4), -(1)/(3))`
B. `(4, 13)`
C. `(4, (9)/(4))`
D. `(4, (7)/(4))`

Answer

To find the orthocenter of the triangle with vertices \( A(1, 1) \), \( B(5, 1) \), and \( C(4, 5) \), we will follow these steps:

### Step 1: Find the slopes of the sides of the triangle
1. **Calculate the slope of side \( AB \)**:
\[ \text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 1}{5 - 1} = \frac{0}{4} = 0 \]
This means line \( AB \) is horizontal.

2. **Calculate the slope of side \( BC \)**:
\[ \text{slope of } BC = \frac{5 - 1}{4 - 5} = \frac{4}{-1} = -4 \]

3. **Calculate the slope of side \( CA \)**:
\[ \text{slope of } CA = \frac{1 - 5}{1 - 4} = \frac{-4}{-3} = \frac{4}{3} \]

### Step 2: Find the slopes of the altitudes
1. **The altitude from vertex \( C \) (to side \( AB \))**:
Since \( AB \) is horizontal (slope = 0), the altitude from \( C \) will be vertical (slope = undefined).

2. **The altitude from vertex \( A \) (to side \( BC \))**:
The slope of \( BC \) is \(-4\), so the slope of the altitude from \( A \) is the negative reciprocal:
\[ m_{altitude \, A} = \frac{1}{4} \]

3. **The altitude from vertex \( B \) (to side \( CA \))**:
The slope of \( CA \) is \( \frac{4}{3} \), so the slope of the altitude from \( B \) is:
\[ m_{altitude \, B} = -\frac{3}{4} \]

### Step 3: Find the equations of the altitudes
1. **Equation of the altitude from \( C(4, 5) \)**:
Since this altitude is vertical, its equation is:
\[ x = 4 \]

2. **Equation of the altitude from \( A(1, 1) \)**:
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 1 = \frac{1}{4}(x - 1) \]
Simplifying:
\[ y - 1 = \frac{1}{4}x - \frac{1}{4} \implies y = \frac{1}{4}x + \frac{3}{4} \]

3. **Equation of the altitude from \( B(5, 1) \)**:
Using the point-slope form:
\[ y - 1 = -\frac{3}{4}(x - 5) \]
Simplifying:
\[ y - 1 = -\frac{3}{4}x + \frac{15}{4} \implies y = -\frac{3}{4}x + \frac{19}{4} \]

### Step 4: Find the intersection of the altitudes
To find the orthocenter, we need to find the intersection of the altitude from \( A \) and the altitude from \( B \).

Set the equations equal:
\[ \frac{1}{4}x + \frac{3}{4} = -\frac{3}{4}x + \frac{19}{4} \]

Multiplying through by 4 to eliminate fractions:
\[ x + 3 = -3x + 19 \]
\[ 4x = 16 \implies x = 4 \]

Now substitute \( x = 4 \) into the equation of the altitude from \( A \):
\[ y = \frac{1}{4}(4) + \frac{3}{4} = 1 + \frac{3}{4} = \frac{7}{4} \]

### Final Result
Thus, the orthocenter of the triangle is:
\[ \boxed{\left(4, \frac{7}{4}\right)} \]