If the angle between the two lines represented by `2x^(2)+5xy+3y^(2)+7y+4=0` is `tan^(-1)` m, then m=

Options

A. `1//5`
B. 1
C. `7//5`
D. 7

Answer

To find the value of \( m \) such that the angle between the two lines represented by the equation \( 2x^2 + 5xy + 3y^2 + 7y + 4 = 0 \) is \( \tan^{-1} m \), we can follow these steps:

### Step 1: Identify coefficients
The given equation is in the form of \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). Here, we identify the coefficients:
- \( a = 2 \)
- \( b = 3 \)
- \( h = \frac{5}{2} \)
- \( g = 0 \)
- \( f = \frac{7}{2} \)
- \( c = 4 \)

### Step 2: Use the formula for the angle between two lines
The formula for the tangent of the angle \( \theta \) between two lines represented by the general quadratic equation is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
We need to calculate \( \tan \theta \) using the values we identified.

### Step 3: Calculate \( h^2 - ab \)
First, we calculate \( h^2 \) and \( ab \):
- \( h^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \)
- \( ab = 2 \cdot 3 = 6 \)

Now, substitute these into the formula:
\[ h^2 - ab = \frac{25}{4} - 6 = \frac{25}{4} - \frac{24}{4} = \frac{1}{4} \]

### Step 4: Substitute into the tangent formula
Now substitute \( h^2 - ab \) into the tangent formula:
\[ \tan \theta = \frac{2\sqrt{\frac{1}{4}}}{2 + 3} = \frac{2 \cdot \frac{1}{2}}{5} = \frac{1}{5} \]

### Step 5: Relate to \( m \)
Since we are given that \( \theta = \tan^{-1} m \), we have:
\[ m = \tan \theta = \frac{1}{5} \]

### Final Answer
Thus, the value of \( m \) is:
\[ \boxed{\frac{1}{5}} \]