Orthocentre of the triangle formed by the lines `xy-3x-5y+15=0 and 3x+5y=15` is

Options

A. (-5,-3)
B. (5,3)
C. (-3,-5)
D. (3,5)

Answer

To find the orthocenter of the triangle formed by the lines \( xy - 3x - 5y + 15 = 0 \) and \( 3x + 5y = 15 \), we can follow these steps:

### Step 1: Understand the lines
We have two equations:
1. \( xy - 3x - 5y + 15 = 0 \)
2. \( 3x + 5y = 15 \)

### Step 2: Simplify the first equation
We can rewrite the first equation in a more manageable form. Rearranging gives:
\[ xy = 3x + 5y - 15 \]

### Step 3: Solve the second equation for \( y \)
From the second equation \( 3x + 5y = 15 \), we can express \( y \) in terms of \( x \):
\[ 5y = 15 - 3x \]
\[ y = \frac{15 - 3x}{5} \]

### Step 4: Substitute \( y \) into the first equation
Now, substitute \( y \) from the second equation into the first equation:
\[ x\left(\frac{15 - 3x}{5}\right) - 3x - 5\left(\frac{15 - 3x}{5}\right) + 15 = 0 \]

### Step 5: Simplify the equation
Multiply through by 5 to eliminate the fraction:
\[ x(15 - 3x) - 15x - (15 - 3x) + 75 = 0 \]
Expanding gives:
\[ 15x - 3x^2 - 15x - 15 + 3x + 75 = 0 \]
This simplifies to:
\[ -3x^2 + 3x + 60 = 0 \]

### Step 6: Factor or use the quadratic formula
We can factor out -3:
\[ -3(x^2 - x - 20) = 0 \]
Now we can factor the quadratic:
\[ (x - 5)(x + 4) = 0 \]
Thus, \( x = 5 \) or \( x = -4 \).

### Step 7: Find corresponding \( y \) values
For \( x = 5 \):
\[ y = \frac{15 - 3(5)}{5} = \frac{0}{5} = 0 \]
So one vertex is \( (5, 0) \).

For \( x = -4 \):
\[ y = \frac{15 - 3(-4)}{5} = \frac{15 + 12}{5} = \frac{27}{5} \]
So the other vertex is \( (-4, \frac{27}{5}) \).

### Step 8: Find the orthocenter
The orthocenter of a right triangle is the vertex at the right angle. In this case, the right angle is at the origin \( (0, 0) \).

### Step 9: Conclusion
The orthocenter of the triangle formed by the given lines is \( (0, 0) \).