Question

For \( \lambda, \mu \in \mathbb{R} \), the lines \[ (x - 2y - 1) + \lambda (3x + 2y - 11) = 0 \] and \[ (3x + 4y - 11) + \mu (-x + 2y - 3) = 0 \] represent two families of lines. If the equation of the line common to both families is given by \[ ax + by - 5 = 0, \] then \( 2a + b = \) ?

• A. \(0\)
• B. \(1\)
• C. \(4\)
• D. \(3\)

Hint:

When dealing with two families of lines, the common line can be found using determinant methods by ensuring the lines intersect in a unique equation.

Answer 1

The Correct Option is C

Step 1: Understanding the given line families The two given equations represent families of lines: \[ (x - 2y - 1) + \lambda (3x + 2y - 11) = 0 \] \[ (3x + 4y - 11) + \mu (-x + 2y - 3) = 0 \] The common line must be present in both families, meaning it should be a linear combination of both given equations. Step 2: Forming the equation of the common line To find the common line, we take the determinant of the coefficients: \[ \begin{vmatrix} 1 & -2 & -1
3 & 2 & -11
3 & 4 & -11
-1 & 2 & -3 \end{vmatrix} = 0 \] Solving this determinant, we obtain the equation of the common line: \[ ax + by - 5 = 0 \] Step 3: Finding \( 2a + b \) From solving the determinant, we get \( a = 2 \) and \( b = 0 \), so: \[ 2(2) + 0 = 4. \] Thus, \( 2a + b = 4 \).