If the lines `3x^(2)-4xy +y^(2) +8x - 2y- 3 = 0` and `2x - 3y +lambda = 0` are concurrent, then the value of `lambda` is

Options

A. `4pi - 11`
B. `-11`
C. `(1)/(11)`
D. 11

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the lines represented by the equations \( 3x^2 - 4xy + y^2 + 8x - 2y - 3 = 0 \) and \( 2x - 3y + \lambda = 0 \) are concurrent.

### Step 1: Differentiate the first equation with respect to \( x \)
We start with the equation of the pair of straight lines:
\[ 3x^2 - 4xy + y^2 + 8x - 2y - 3 = 0 \]

Differentiating with respect to \( x \) while treating \( y \) as a constant gives:
\[ \frac{d}{dx}(3x^2) - \frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(8x) - \frac{d}{dx}(2y) - \frac{d}{dx}(3) = 0 \]

This simplifies to:
\[ 6x - 4y + 8 = 0 \]

### Step 2: Differentiate the first equation with respect to \( y \)
Next, we differentiate the same equation with respect to \( y \) while treating \( x \) as a constant:
\[ \frac{d}{dy}(3x^2) - \frac{d}{dy}(4xy) + \frac{d}{dy}(y^2) + \frac{d}{dy}(8x) - \frac{d}{dy}(2y) - \frac{d}{dy}(3) = 0 \]

This simplifies to:
\[ -4x + 2y - 2 = 0 \]

### Step 3: Solve the two equations
Now we have two equations:
1. \( 6x - 4y + 8 = 0 \) (Equation 1)
2. \( -4x + 2y - 2 = 0 \) (Equation 2)

From Equation 1, we can express \( x \) in terms of \( y \):
\[ 6x = 4y - 8 \]
\[ x = \frac{4y - 8}{6} = \frac{2y - 4}{3} \]

Substituting this expression for \( x \) into Equation 2:
\[ -4\left(\frac{2y - 4}{3}\right) + 2y - 2 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ -4(2y - 4) + 6y - 6 = 0 \]
\[ -8y + 16 + 6y - 6 = 0 \]
Combining like terms:
\[ -2y + 10 = 0 \]
Thus, we find:
\[ 2y = 10 \]
\[ y = 5 \]

### Step 4: Substitute \( y \) back to find \( x \)
Now substituting \( y = 5 \) back into the expression for \( x \):
\[ x = \frac{2(5) - 4}{3} = \frac{10 - 4}{3} = \frac{6}{3} = 2 \]

### Step 5: Substitute \( x \) and \( y \) into the second line equation
Now we have \( x = 2 \) and \( y = 5 \). We substitute these values into the second line equation:
\[ 2x - 3y + \lambda = 0 \]
Substituting \( x \) and \( y \):
\[ 2(2) - 3(5) + \lambda = 0 \]
\[ 4 - 15 + \lambda = 0 \]
\[ \lambda - 11 = 0 \]
Thus, we find:
\[ \lambda = 11 \]

### Final Answer
The value of \( \lambda \) is \( 11 \).

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