If the coordinates of the orthocentre of the triangle formed by the lines `y=0,37x-36y+37xx36=0` and `64x-63y+64xx63=0` is (a, b). Then `a+b` is equal to

Answer

To find the value of \( a + b \) where \( (a, b) \) are the coordinates of the orthocenter of the triangle formed by the given lines, we will follow these steps:

### Step 1: Identify the lines forming the triangle
The lines given are:
1. \( y = 0 \) (the x-axis)
2. \( 37x - 36y + 37 \cdot 36 = 0 \)
3. \( 64x - 63y + 64 \cdot 63 = 0 \)

### Step 2: Find the intersection points (vertices of the triangle)
- **Vertex B**: Intersection of \( y = 0 \) and \( 37x - 36y + 37 \cdot 36 = 0 \):
\[ 37x + 37 \cdot 36 = 0 \implies x = -36 \quad \text{(since \( y = 0 \))} \]
Thus, \( B(-36, 0) \).

- **Vertex C**: Intersection of \( y = 0 \) and \( 64x - 63y + 64 \cdot 63 = 0 \):
\[ 64x + 64 \cdot 63 = 0 \implies x = -63 \quad \text{(since \( y = 0 \))} \]
Thus, \( C(-63, 0) \).

### Step 3: Find the coordinates of point A
To find point A, we need to find the intersection of the two lines:
1. \( 37x - 36y + 1332 = 0 \) (where \( 1332 = 37 \cdot 36 \))
2. \( 64x - 63y + 4032 = 0 \) (where \( 4032 = 64 \cdot 63 \))

We can solve these two equations simultaneously.

From the first equation:
\[ 36y = 37x + 1332 \implies y = \frac{37}{36}x + 37 \]

Substituting this expression for \( y \) into the second equation:
\[ 64x - 63\left(\frac{37}{36}x + 37\right) + 4032 = 0 \]

Distributing \( -63 \):
\[ 64x - \frac{63 \cdot 37}{36}x - 63 \cdot 37 + 4032 = 0 \]

Calculating \( 63 \cdot 37 = 2331 \):
\[ 64x - \frac{2331}{36}x - 2331 + 4032 = 0 \]

Combining the constants:
\[ 64x - \frac{2331}{36}x + 1701 = 0 \]

Finding a common denominator (36):
\[ \frac{64 \cdot 36}{36}x - \frac{2331}{36}x + 1701 = 0 \]
\[ \frac{2304 - 2331}{36}x + 1701 = 0 \]
\[ \frac{-27}{36}x + 1701 = 0 \]
\[ -27x + 1701 \cdot 36 = 0 \implies x = \frac{1701 \cdot 36}{27} \]
Calculating \( 1701 \div 27 = 63 \):
\[ x = 63 \cdot 36 = 2268 \]

Substituting back to find \( y \):
\[ y = \frac{37}{36}(63) + 37 = 63 + 37 = 100 \]
Thus, point A is \( (63, 100) \).

### Step 4: Find the orthocenter
The orthocenter is found by determining the intersection of the altitudes from points B and C.

- **Slope of line AC**:
\[ \text{slope of AC} = \frac{100 - 0}{63 - (-63)} = \frac{100}{126} = \frac{50}{63} \]
Thus, the slope of the altitude from B (perpendicular to AC) is:
\[ -\frac{63}{50} \]
The equation of the altitude from B is:
\[ y - 0 = -\frac{63}{50}(x + 36) \]

- **Slope of line BC**:
\[ \text{slope of BC} = 0 \quad \text{(horizontal line)} \]
Thus, the altitude from C is vertical, and its equation is:
\[ x = -63 \]

### Step 5: Solve for the intersection of the altitudes
Substituting \( x = -63 \) into the altitude equation from B:
\[ y = -\frac{63}{50}(-63 + 36) = -\frac{63}{50}(-27) = \frac{1701}{50} \]

### Step 6: Find \( a + b \)
The orthocenter coordinates are \( (-63, \frac{1701}{50}) \):
\[ a + b = -63 + \frac{1701}{50} \]
Converting \( -63 \) to a fraction:
\[ -63 = -\frac{3150}{50} \]
Thus:
\[ a + b = -\frac{3150}{50} + \frac{1701}{50} = -\frac{1449}{50} \]

### Final Answer
The value of \( a + b \) is \( -\frac{1449}{50} \).