the lines `L_1` and `L_2` denoted by `3x^2 + 10xy +8y^2 +14x +22y + 15 = 0` intersect at the point P and have gradients `M_1` and `M_2` respectively.The acute angles between them is `theta` . which of the following relations hold good ?

Options

A. `(-1,-2)`
B. `(1,-2)`
C. `(1,2)`
D. `(-2,1)`

Answer

To solve the problem step by step, we need to analyze the given equation of the lines and find the relationships between the gradients and the angle between them.

### Step 1: Rewrite the given equation
The given equation is:
\[ 3x^2 + 10xy + 8y^2 + 14x + 22y + 15 = 0 \]

### Step 2: Factor the quadratic equation
We can rearrange the equation to isolate the quadratic terms:
\[ 3x^2 + 10xy + 8y^2 = -14x - 22y - 15 \]

### Step 3: Factor the left-hand side
We can factor the left-hand side:
\[ (3x + 4y)(x + 2y) = 0 \]
This gives us two lines:
1. \( 3x + 4y = 0 \) (Line \( L_1 \))
2. \( x + 2y = 0 \) (Line \( L_2 \))

### Step 4: Find the gradients of the lines
From the equations of the lines, we can find their gradients:
- For \( L_1: 3x + 4y = 0 \)
\[ y = -\frac{3}{4}x \]
Thus, the gradient \( M_1 = -\frac{3}{4} \).

- For \( L_2: x + 2y = 0 \)
\[ y = -\frac{1}{2}x \]
Thus, the gradient \( M_2 = -\frac{1}{2} \).

### Step 5: Use the formula for the angle between two lines
The acute angle \( \theta \) between the two lines can be found using the formula:
\[ \tan(\theta) = \left| \frac{M_1 - M_2}{1 + M_1M_2} \right| \]

### Step 6: Substitute the gradients into the formula
Substituting the values of \( M_1 \) and \( M_2 \):
\[ \tan(\theta) = \left| \frac{-\frac{3}{4} - (-\frac{1}{2})}{1 + \left(-\frac{3}{4}\right)\left(-\frac{1}{2}\right)} \right| \]

### Step 7: Simplify the expression
Calculating the numerator:
\[ -\frac{3}{4} + \frac{1}{2} = -\frac{3}{4} + \frac{2}{4} = -\frac{1}{4} \]

Calculating the denominator:
\[ 1 + \left(-\frac{3}{4}\right)\left(-\frac{1}{2}\right) = 1 + \frac{3}{8} = \frac{8}{8} + \frac{3}{8} = \frac{11}{8} \]

Thus, we have:
\[ \tan(\theta) = \left| \frac{-\frac{1}{4}}{\frac{11}{8}} \right| = \frac{1}{4} \cdot \frac{8}{11} = \frac{2}{11} \]

### Step 8: Conclusion
The relationship that holds good between the gradients of the lines and the angle \( \theta \) is:
\[ \tan(\theta) = \frac{2}{11} \]