Let A(3,4) B(5,0), C(0,5) be the vertices of a triangle ABC then orthocentre of `Delta ABC` is `(alpha, beta)` Find `alpha + beta`

Explanation

To find the orthocenter of triangle ABC with vertices A(3,4), B(5,0), and C(0,5), we will follow these steps:

### Step 1: Find the slopes of the sides of the triangle.
- **Slope of line AB**:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{5 - 3} = \frac{-4}{2} = -2 \]

- **Slope of line BC**:
\[ m_{BC} = \frac{5 - 0}{0 - 5} = \frac{5}{-5} = -1 \]

- **Slope of line AC**:
\[ m_{AC} = \frac{5 - 4}{0 - 3} = \frac{1}{-3} = -\frac{1}{3} \]

### Step 2: Find the slopes of the altitudes.
- The slope of the altitude from A (which is perpendicular to BC) is the negative reciprocal of \( m_{BC} \):
\[ m_{AD} = \frac{1}{1} = 1 \]

- The slope of the altitude from B (which is perpendicular to AC) is the negative reciprocal of \( m_{AC} \):
\[ m_{BE} = 3 \]

### Step 3: Write the equations of the altitudes.
- **Equation of altitude AD** (from A(3,4)):
Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 4 = 1(x - 3) \implies y - 4 = x - 3 \implies x - y + 1 = 0 \]

- **Equation of altitude BE** (from B(5,0)):
\[ y - 0 = 3(x - 5) \implies y = 3x - 15 \implies 3x - y - 15 = 0 \]

### Step 4: Solve the equations of the altitudes to find the orthocenter.
We have the two equations:
1. \( x - y + 1 = 0 \) (1)
2. \( 3x - y - 15 = 0 \) (2)

Now, we can solve these equations simultaneously.

From equation (1):
\[ y = x + 1 \]

Substituting \( y \) in equation (2):
\[ 3x - (x + 1) - 15 = 0 \implies 3x - x - 1 - 15 = 0 \implies 2x - 16 = 0 \implies x = 8 \]

Now substituting \( x = 8 \) back into equation (1) to find \( y \):
\[ y = 8 + 1 = 9 \]

Thus, the orthocenter \( ( \alpha, \beta ) \) is \( (8, 9) \).

### Step 5: Find \( \alpha + \beta \).
\[ \alpha + \beta = 8 + 9 = 17 \]

### Final Answer:
The value of \( \alpha + \beta \) is \( 17 \).

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