Orthocentre of the triangle whose vertices are (1, 1) (3, 5) and (3, 0), is

Options

A. `(-1, 1)`
B. `(1, -1)`
C. (1, 1)
D. None of these

Answer

To find the orthocenter of the triangle with vertices A(1, 1), B(3, 5), and C(3, 0), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are:
- A(1, 1)
- B(3, 5)
- C(3, 0)

### Step 2: Find the slopes of the sides of the triangle
1. **Slope of BC**:
\[ \text{slope of } BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 0}{3 - 3} = \text{undefined (vertical line)} \]

2. **Slope of AC**:
\[ \text{slope of } AC = \frac{0 - 1}{3 - 1} = \frac{-1}{2} \]

3. **Slope of AB**:
\[ \text{slope of } AB = \frac{5 - 1}{3 - 1} = \frac{4}{2} = 2 \]

### Step 3: Find the equations of the altitudes
1. **Altitude from A (perpendicular to BC)**:
Since BC is a vertical line (undefined slope), the altitude from A will be a horizontal line through A(1, 1).
\[ \text{Equation of altitude AD: } y = 1 \]

2. **Altitude from B (perpendicular to AC)**:
The slope of AC is \(-\frac{1}{2}\), so the slope of the altitude from B will be the negative reciprocal:
\[ \text{slope of BE} = 2 \]
Using point-slope form (y - y1 = m(x - x1)):
\[ y - 5 = 2(x - 3) \]
Simplifying:
\[ y - 5 = 2x - 6 \implies 2x - y = 1 \quad \text{(Equation of BE)} \]

### Step 4: Find the intersection of the altitudes
To find the orthocenter, we need to find the intersection of the lines \(y = 1\) (AD) and \(2x - y = 1\) (BE).

Substituting \(y = 1\) into the equation of BE:
\[ 2x - 1 = 1 \implies 2x = 2 \implies x = 1 \]

### Step 5: Determine the coordinates of the orthocenter
Thus, the coordinates of the orthocenter are:
\[ (1, 1) \]

### Final Answer
The orthocenter of the triangle is at the point **(1, 1)**.