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 coefficients
The given equation is in the form \(Ax^2 + 2Hxy + By^2 = 0\). Here, we can identify:
- \(A = 1\)
- \(2H = -3\) (thus \(H = -\frac{3}{2}\))
- \(B = 2\)

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

### Step 3: Calculate \(H^2 - AB\)
First, we need to calculate \(H^2\) and \(AB\):
- \(H^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4}\)
- \(AB = 1 \cdot 2 = 2\)

Now, we can find \(H^2 - AB\):
\[ H^2 - AB = \frac{9}{4} - 2 = \frac{9}{4} - \frac{8}{4} = \frac{1}{4} \]

### Step 4: Substitute into the formula
Now, substituting \(H^2 - AB\) into the formula for \(\tan \theta\):
\[ \tan \theta = \frac{2\sqrt{\frac{1}{4}}}{1 + 2} = \frac{2 \cdot \frac{1}{2}}{3} = \frac{1}{3} \]

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

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

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