Orthocentre of the triangle formed by the lines whose combined equation is `(y ^(2) - 6xy -x ^(2)) ( 4x -y +7) =0` is

Options

A. `(1,2)`
B. `(2,3)`
C. `(0,1)`
D. none of these

Answer

To find the orthocenter of the triangle formed by the lines given in the combined equation \((y^2 - 6xy - x^2)(4x - y + 7) = 0\), we can follow these steps:

### Step 1: Identify the Lines
The combined equation consists of two parts:
1. \(y^2 - 6xy - x^2 = 0\)
2. \(4x - y + 7 = 0\)

### Step 2: Solve the First Equation
We will first solve the quadratic equation \(y^2 - 6xy - x^2 = 0\) for \(y\):
This can be rearranged as:
\[ y^2 - 6xy - x^2 = 0 \]
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6x\), and \(c = -x^2\):
\[ y = \frac{6x \pm \sqrt{(-6x)^2 - 4 \cdot 1 \cdot (-x^2)}}{2 \cdot 1} \]
\[ y = \frac{6x \pm \sqrt{36x^2 + 4x^2}}{2} \]
\[ y = \frac{6x \pm \sqrt{40x^2}}{2} \]
\[ y = \frac{6x \pm 2\sqrt{10}x}{2} \]
\[ y = 3x \pm \sqrt{10}x \]
Thus, we have two lines:
1. \(y = (3 + \sqrt{10})x\)
2. \(y = (3 - \sqrt{10})x\)

### Step 3: Solve the Second Equation
The second line is given by:
\[ 4x - y + 7 = 0 \implies y = 4x + 7 \]

### Step 4: Find the Intersection Points
Now we need to find the intersection points of these lines.

**Intersection of \(y = (3 + \sqrt{10})x\) and \(y = 4x + 7\):**
\[ (3 + \sqrt{10})x = 4x + 7 \]
\[ (3 + \sqrt{10} - 4)x = 7 \]
\[ (\sqrt{10} - 1)x = 7 \implies x = \frac{7}{\sqrt{10} - 1} \]
Substituting \(x\) back to find \(y\):
\[ y = 4\left(\frac{7}{\sqrt{10} - 1}\right) + 7 \]

**Intersection of \(y = (3 - \sqrt{10})x\) and \(y = 4x + 7\):**
\[ (3 - \sqrt{10})x = 4x + 7 \]
\[ (3 - \sqrt{10} - 4)x = 7 \]
\[ (-\sqrt{10} - 1)x = 7 \implies x = \frac{-7}{\sqrt{10} + 1} \]
Substituting \(x\) back to find \(y\):
\[ y = 4\left(\frac{-7}{\sqrt{10} + 1}\right) + 7 \]

### Step 5: Find the Orthocenter
For a right triangle, the orthocenter is at the vertex where the right angle is formed. Since the two lines from the first equation are perpendicular, the orthocenter will be at the intersection of these two lines.

### Step 6: Conclusion
The orthocenter of the triangle formed by these lines is at the intersection point of the two lines derived from the first equation.