The lines given by the equation `(y^(2)-4xy-x^(2))(x+y-1)=0` form a triangle which is

Options

A. equilateral
B. isosceles
C. right angled
D. obtuse angled

Answer

To solve the problem, we need to analyze the given equation of lines and determine the type of triangle formed by these lines. The equation provided is:

\[ (y^2 - 4xy - x^2)(x + y - 1) = 0 \]

This equation indicates that there are two components: \(y^2 - 4xy - x^2 = 0\) and \(x + y - 1 = 0\). We will first focus on the quadratic equation \(y^2 - 4xy - x^2 = 0\).

### Step 1: Identify the lines from the quadratic equation

The quadratic equation \(y^2 - 4xy - x^2 = 0\) can be rewritten in standard form as:

\[ y^2 - 4xy - x^2 = 0 \]

This is a quadratic in \(y\). We can use the quadratic formula to find the roots (lines) in terms of \(x\):

\[ y = \frac{-(-4x) \pm \sqrt{(-4x)^2 - 4 \cdot 1 \cdot (-x^2)}}{2 \cdot 1} \]

### Step 2: Simplify the expression

Calculating the discriminant:

\[ (-4x)^2 - 4 \cdot 1 \cdot (-x^2) = 16x^2 + 4x^2 = 20x^2 \]

Thus, the roots become:

\[ y = \frac{4x \pm \sqrt{20x^2}}{2} \]

This simplifies to:

\[ y = \frac{4x \pm 2\sqrt{5}x}{2} = 2x \pm \sqrt{5}x \]

So, we have two lines:

\[ y = (2 + \sqrt{5})x \quad \text{and} \quad y = (2 - \sqrt{5})x \]

### Step 3: Identify the third line

The second part of the equation \(x + y - 1 = 0\) can be rearranged to give:

\[ y = 1 - x \]

### Step 4: Determine the angles between the lines

Now we have three lines:

1. \(y = (2 + \sqrt{5})x\)
2. \(y = (2 - \sqrt{5})x\)
3. \(y = 1 - x\)

To find the angles between the lines, we can use the slopes:

- Slope of the first line \(m_1 = 2 + \sqrt{5}\)
- Slope of the second line \(m_2 = 2 - \sqrt{5}\)
- Slope of the third line \(m_3 = -1\)

### Step 5: Calculate the angle between the lines

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) can be calculated using the formula:

\[ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]

Calculating the angle between the first two lines:

\[ \tan(\theta) = \left|\frac{(2 + \sqrt{5}) - (2 - \sqrt{5})}{1 + (2 + \sqrt{5})(2 - \sqrt{5})}\right| \]

Simplifying gives:

\[ \tan(\theta) = \left|\frac{2\sqrt{5}}{1 + (4 - 5)}\right| = \left|\frac{2\sqrt{5}}{0}\right| \]

This indicates that the angle is \(90^\circ\) (or \(\frac{\pi}{2}\) radians), meaning the lines are perpendicular.

### Conclusion: Type of Triangle

Since one of the angles formed by these lines is \(90^\circ\), the triangle formed by these lines is a right triangle.

Thus, the answer is:

**The triangle formed by the lines is a right-angled triangle.**