Find the angle between the pair of straight lines `x^(2) - 3xy +2y^(2) = 0`

Answer

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

### Step 1: Identify the coefficients
The given equation is in the form \(Ax^2 + Bxy + Cy^2 = 0\). Here, we have:
- \(A = 1\)
- \(B = -3\)
- \(C = 2\)

### Step 2: Use the formula for the angle between two lines
The angle \(\theta\) between the two lines represented by the equation can be calculated using the formula:
\[ \tan \theta = \frac{B}{A - C} \]

### Step 3: Substitute the values into the formula
Substituting the values we identified:
\[ \tan \theta = \frac{-3}{1 - 2} = \frac{-3}{-1} = 3 \]

### Step 4: Calculate the angle
Now, we need to find \(\theta\) such that:
\[ \tan \theta = 3 \]
To find \(\theta\), we take the arctangent:
\[ \theta = \tan^{-1}(3) \]

### Step 5: Conclusion
Thus, the angle between the pair of straight lines is:
\[ \theta = \tan^{-1}(3) \]