Triangle formed by variable lines (a+b)x+(a-b)y-2ab=0 and (a-b)x+(a+b)y-2ab=0 and x+y=0 is (where a, `b in R`)

Options

A. (a) equilateral
B. (b) Isoceles
C. (c) scalene
D. (d) none of these

Explanation

To solve the problem of finding the type of triangle formed by the given lines, we will follow these steps:

### Step 1: Write the equations of the lines
The equations of the lines are given as:
1. \( L_1: (a+b)x + (a-b)y - 2ab = 0 \)
2. \( L_2: (a-b)x + (a+b)y - 2ab = 0 \)
3. \( L_3: x + y = 0 \)

### Step 2: Convert the equations to slope-intercept form
We need to find the slopes of these lines. We can rewrite each line in the form \( y = mx + c \).

**For Line 1 (L1):**
\[ (a-b)y = - (a+b)x + 2ab \implies y = -\frac{(a+b)}{(a-b)}x + \frac{2ab}{(a-b)} \]
Thus, the slope \( m_1 = -\frac{(a+b)}{(a-b)} \).

**For Line 2 (L2):**
\[ (a+b)y = - (a-b)x + 2ab \implies y = -\frac{(a-b)}{(a+b)}x + \frac{2ab}{(a+b)} \]
Thus, the slope \( m_2 = -\frac{(a-b)}{(a+b)} \).

**For Line 3 (L3):**
\[ y = -x \implies m_3 = -1 \]

### Step 3: Find the angles between the lines
We will find the angles between the lines \( L_1 \) and \( L_3 \), and \( L_2 \) and \( L_3 \).

**Angle between L1 and L3:**
Let \( \alpha \) be the angle between \( L_1 \) and \( L_3 \).
Using the formula for the tangent of the angle between two lines:
\[ \tan(\alpha) = \left| \frac{m_1 - m_3}{1 + m_1 m_3} \right| \]
Substituting the values:
\[ \tan(\alpha) = \left| \frac{-\frac{(a+b)}{(a-b)} + 1}{1 + \left(-\frac{(a+b)}{(a-b)}\right)(-1)} \right| \]
This simplifies to:
\[ \tan(\alpha) = \left| \frac{-\frac{(a+b)}{(a-b)} + \frac{(a-b)}{(a-b)}}{1 + \frac{(a+b)}{(a-b)}} \right| = \left| \frac{-2b}{2a} \right| = \frac{b}{a} \]

**Angle between L2 and L3:**
Let \( \beta \) be the angle between \( L_2 \) and \( L_3 \).
Using the same formula:
\[ \tan(\beta) = \left| \frac{m_2 - m_3}{1 + m_2 m_3} \right| \]
Substituting the values:
\[ \tan(\beta) = \left| \frac{-\frac{(a-b)}{(a+b)} + 1}{1 + \left(-\frac{(a-b)}{(a+b)}\right)(-1)} \right| \]
This simplifies to:
\[ \tan(\beta) = \left| \frac{-\frac{(a-b)}{(a+b)} + \frac{(a+b)}{(a+b)}}{1 + \frac{(a-b)}{(a+b)}} \right| = \left| \frac{2b}{2a} \right| = \frac{b}{a} \]

### Step 4: Compare the angles
Since \( \tan(\alpha) = \tan(\beta) \), it follows that \( \alpha = \beta \). Therefore, the triangle formed by the lines is an isosceles triangle.

### Conclusion
The triangle formed by the given lines is an **isosceles triangle**.

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