QJEEAdv08Mad767: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Mad767 Solution: Consider the lines given by: Match the Statement/Expressions in …

We have,

(A)

${L_1}:x + 3y - 5 = 0$

${L_2}:3x - ky - 1 = 0$

${L_3}:5x + 2y - 12 = 0$

Point of intersection of L $_1$ and L $_2$ is (2, 1)

If lines are concurrent, then (2, 1) will satisfy L $_2$

$\Rightarrow 3(2)-k(1)-1=0$

$k=5$

(A) $\to$ (iv)

(B) If L $_1$ and L $_2$ are parallel

${3 \over 1} = {{ - k} \over 3} \ne {{ - 1} \over { - 5}} \Rightarrow k - 9$

If L $_2$ and L $_3$ are parallel

${3 \over 5} = {{ - k} \over 2}$

$k = {{ - 6} \over 5}$

(B) $\to$ (i, ii)

(C) As lines form a triangle, they cannot be concurrent and no two of them are parallel.

$\Rightarrow k\ne5,-9,\frac{-6}{5}$

$\Rightarrow k=\frac{5}{6}$

(C) $\to$ (iii)

(D) If the lines do not form a triangle then

$k=5,-9,\frac{-6}{5}$

(D) $\to$ (i, ii, iv)