Question

If the lines x + 2y = 3, 3x - y = 4 and $\lambda$x + 4y = 0 are concurrent, then $\lambda$ is equal to

• A. -5
• B. $\frac{-15}{11}$
• C. $\frac{-20}{11}$
• D. $\frac{-5}{11}$

Answer 1

Answer: (C)

$\frac{-20}{11}$

Answer 2

For three lines $a_1x + b_1y + c_1 = 0$, $a_2x + b_2y + c_2 = 0$, and $a_3x + b_3y + c_3 = 0$ to be concurrent, the determinant of their coefficients must be zero. The given lines are $x + 2y - 3 = 0$, $3x - y - 4 = 0$, and $\lambda x + 4y + 0 = 0$. The condition for concurrency is: $\begin{vmatrix} 1 & 2 & -3 \\ 3 & -1 & -4 \\ \lambda & 4 & 0 \end{vmatrix} = 0$ Expanding this determinant: $16 - 8\lambda - 36 - 3\lambda = 0$ $-20 - 11\lambda = 0$ Solving for $\lambda$: $\lambda = -\frac{20}{11}$