Question

Consider the lines given by
$L_1:x+3y-5=0$
$L_2:3x-ky-1=0$
$L_3:5x+2y-12=0$

Match the Statements/Exp[ressions in Column I with the Statements/Expressions in Column II.

Column I	Column II
$L_1,L_2,L_3$ are concurrent, if	k = -9
$Oneof{L}_1,L_2,L_3$ is parallel to at least one of the other two, if	$k=\frac{6}{5}$
$L_1,L_2,L_3$ form a triangle, if	$k=\frac{5}{6}$
$L_1,L_2,L_3$ do not form a triangle, if	k = 5

• A. (a) $\to$ (s), (b) $\to$ (p, q), (c) $\to$ (r), (d) $\to$ (p, q, s)
• B. (a) $\to$ (p), (b) $\to$ (p, q, s), (c) $\to$ (r), (d) $\to$ (p, q)
• C. (a) $\to$ (s), (b) $\to$ (p, q, s), (c) $\to$ (s, r), (d) $\to$ (p, q, s)
• D. none of these

Answer 1

The correct option is A (a) $\to$ (s), (b) $\to$ (p, q), (c) $\to$ (r), (d) $\to$ (p, q, s)

$\left(a\right)\to \left(s\right),\left(b\right)\to (p,q),\left(c\right)\to \left(r\right),\left(d\right)\to (p,q,s)$
$\left(a\right)L_{1}and{L}_{3}intersectat(1,2)anditwilllieon{L}_{2}ifk=5$
(b) Condition of parallelism gives
$\frac{3}{1}=-\frac{k}{3}=\frac{1}{5}\therefore k=-9\therefore \left(b\right)\to \left(p\right)$
Also $\frac{3}{5}=-\frac{k}{2}=\frac{1}{12}\therefore k=\frac{6}{5}\therefore \left(b\right)\to \left(p\right)\therefore \left(b\right)\to (p,q)$

(c) If $k\ne 5,-9,-\frac{6}{5}$ i.e., (a) and (b) i.e., k = 5, -9 -6/5, they will not form a triangle.
$\therefore$ (d) $\to$ (r)

(d) Exclude the cases of (a) and (b) i.e., k=5, -9, -6/5, they will not form a triangle.
$\therefore$ (d) $\to$ (p, q, s)