The acute angle between the line represented by `x^(2)-6xy+5y^(2)+10x-4y+9=0` is

Options

A. `tan^(-1)((3)/(2))`
B. `tan^(-1)((3)/(4))`
C. `tan^(-1)((2)/(3))`
D. `tan^(-1)((4)/(3))`

Answer

To find the acute angle between the lines represented by the equation \(x^2 - 6xy + 5y^2 + 10x - 4y + 9 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The general form of a pair of straight lines is given by:
\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]
From the given equation, we can identify the coefficients:
- \(A = 1\)
- \(H = -3\) (since \(2H = -6\))
- \(B = 5\)
- \(G = 5\) (since \(2G = 10\))
- \(F = -2\) (since \(2F = -4\))
- \(C = 9\)

### Step 2: Use the formula for the angle between the lines
The acute angle \(\theta\) between the two lines can be calculated using the formula:
\[ \tan \theta = \frac{2\sqrt{H^2 - AB}}{A + B} \]
We need to calculate \(H^2 - AB\) and \(A + B\).

### Step 3: Calculate \(H^2 - AB\)
Calculate \(H^2\):
\[ H^2 = (-3)^2 = 9 \]
Calculate \(AB\):
\[ AB = 1 \cdot 5 = 5 \]
Now, calculate \(H^2 - AB\):
\[ H^2 - AB = 9 - 5 = 4 \]

### Step 4: Calculate \(A + B\)
Now, calculate \(A + B\):
\[ A + B = 1 + 5 = 6 \]

### Step 5: Substitute into the formula
Now substitute \(H^2 - AB\) and \(A + B\) into the formula for \(\tan \theta\):
\[ \tan \theta = \frac{2\sqrt{4}}{6} = \frac{2 \cdot 2}{6} = \frac{4}{6} = \frac{2}{3} \]

### Step 6: Calculate the angle \(\theta\)
To find the angle \(\theta\), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

### Step 7: Find the acute angle
The acute angle between the lines is:
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

### Final Result
The acute angle between the lines represented by the equation is \(\theta = \tan^{-1}\left(\frac{2}{3}\right)\).

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