Findthe angle between the pair of straight lines `x^2 -4y^2+3xy = 0 `

Answer

To find the angle between the pair of straight lines given by the equation \(x^2 - 4y^2 + 3xy = 0\), we can follow these steps:

### Step 1: Identify coefficients
The general form of the equation of a pair of straight lines is given by:
\[ ax^2 + by^2 + 2hxy = 0 \]
From the given equation \(x^2 - 4y^2 + 3xy = 0\), we can identify:
- \(a = 1\)
- \(b = -4\)
- \(h = \frac{3}{2}\) (since \(2h = 3\))

### Step 2: Use the formula for the angle between the lines
The formula for the tangent of the angle \(\alpha\) between the two lines is given by:
\[ \tan \alpha = \frac{2\sqrt{h^2 - ab}}{a + b} \]

### Step 3: Calculate \(h^2\), \(ab\), and \(a + b\)
1. Calculate \(h^2\):
\[ h^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

2. Calculate \(ab\):
\[ ab = 1 \cdot (-4) = -4 \]

3. Calculate \(a + b\):
\[ a + b = 1 + (-4) = -3 \]

### Step 4: Substitute values into the formula
Now substitute \(h^2\), \(ab\), and \(a + b\) into the formula:
\[ \tan \alpha = \frac{2\sqrt{\frac{9}{4} - (-4)}}{-3} \]

### Step 5: Simplify the expression
1. Calculate \(h^2 - ab\):
\[ h^2 - ab = \frac{9}{4} + 4 = \frac{9}{4} + \frac{16}{4} = \frac{25}{4} \]

2. Substitute this back into the formula:
\[ \tan \alpha = \frac{2\sqrt{\frac{25}{4}}}{-3} = \frac{2 \cdot \frac{5}{2}}{-3} = \frac{5}{-3} = -\frac{5}{3} \]

### Step 6: Find \(\alpha\)
To find the angle \(\alpha\), we take the arctangent:
\[ \alpha = \tan^{-1}\left(-\frac{5}{3}\right) \]

### Final Answer
Thus, the angle between the pair of straight lines is:
\[ \alpha = \tan^{-1}\left(-\frac{5}{3}\right) \]