Orthocentre of triangle whose vertices are `(0, 0), (3, 4), (4, 0)` is

Options

A. (3,5/2)
B. (3,12)
C. (3,3/4)
D. (3,9)

Explanation

To find the orthocenter of the triangle with vertices at \( A(0, 0) \), \( B(3, 4) \), and \( C(4, 0) \), we will follow these steps:

### Step 1: Find the slopes of the sides of the triangle
1. **Slope of side BC**:
- Points: \( B(3, 4) \) and \( C(4, 0) \)
- Slope \( m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{4 - 3} = \frac{-4}{1} = -4 \)

2. **Slope of side AC**:
- Points: \( A(0, 0) \) and \( C(4, 0) \)
- Slope \( m_{AC} = \frac{0 - 0}{4 - 0} = 0 \)

3. **Slope of side AB**:
- Points: \( A(0, 0) \) and \( B(3, 4) \)
- Slope \( m_{AB} = \frac{4 - 0}{3 - 0} = \frac{4}{3} \)

### Step 2: Find the equations of the altitudes
1. **Altitude from A (perpendicular to BC)**:
- Slope of altitude \( m_{A} = -\frac{1}{m_{BC}} = \frac{1}{4} \)
- Using point-slope form: \( y - y_1 = m(x - x_1) \)
- Equation: \( y - 0 = \frac{1}{4}(x - 0) \) → \( y = \frac{1}{4}x \)

2. **Altitude from B (perpendicular to AC)**:
- Slope of altitude \( m_{B} = -\frac{1}{m_{AC}} \) (since slope of AC is 0, the altitude is vertical)
- Equation: \( x = 3 \)

### Step 3: Find the intersection of the altitudes
To find the orthocenter, we need to find the intersection of the two altitudes we calculated:
- From altitude at A: \( y = \frac{1}{4}x \)
- From altitude at B: \( x = 3 \)

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

### Step 4: Conclusion
The orthocenter of the triangle is at the point \( (3, \frac{3}{4}) \).

### Final Answer
The orthocenter of the triangle with vertices \( (0, 0) \), \( (3, 4) \), and \( (4, 0) \) is \( \left(3, \frac{3}{4}\right) \).

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