Question

The value of λ for which the lines, x-y+1=0, 2x+y+λ=0 and 2x-2y+1=0 are concurrent, is

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Answer 1

Step-by-step explanation:

• The given lines are  x − 2y + 3 = 0 … (1)  λx + 3y + 1 = 0 … (2)  4x − λy + 2 = 0 … (3)  It is given that (1), (2) and (3) are concurrent.
• ∴  ⇒ (6 + λ) + 2(2λ – 4) + 3(-λ2 – 12) = 0  ⇒ 6 + λ + 4λ – 8 – 3λ2 – 36 = 0  ⇒ 5λ – 3λ2 – 38 = 0  ⇒ 3λ2 – 5λ + 38 = 0  The discriminant of this equation is 25 - 4 × 3 × 38 = - 431  Hence, there is no real value of λ for which the lines x − 2y + 3 = 0, λx + 3y + 1 = 0 and 4x − λy + 2 = 0 are concurrent

Answer 2

Given,

$\[\begin{array}{l}x - y + 1 = 0\\2x + y + \lambda = 0\\2x - 2y + 1 = 0\end{array}\]$

Solution,

Calculate the value of $\[\lambda \]$.

It is given that the lines are concurrent so the determinant of coefficient matrix will be zero.

$\[\begin{array}{*{20}{c}}\vline& 1&{ - 1}&1\vline& \\\vline& 2&1&\lambda \vline& \\\vline& 2&{ - 2}&1\vline& \end{array} = 0\]$

$\[ \Rightarrow 1\left( {1 \times 1 + 2\lambda } \right) + 1\left( {2 \times 1 - 2 \times \lambda } \right) + \left( { - 2 \times 2 - 2 \times 1} \right) = 0\]$

$\[\begin{array}{l} \Rightarrow 1 + 2\lambda + 2 - 2\lambda - 6 = 0\\ \Rightarrow - 3 \ne 0\end{array}\]$

Here determinant value is independent of $\[\lambda \]$ value.

Hence for concurrency it is independent from $\[\lambda \]$ value.