Straight Line and Pair of Straight Lines 3 Question 8
9. Consider the lines given by
$L _1: x+3 y-5=0, \quad L _2: 3 x-k y-1=0$, $L _3: 5 x+2 y-12=0$
| Column I | Column II | ||
|---|---|---|---|
| (A) | $L _1, L _2, L _3$ are concurrent, if | (p) | $k=-9$ |
| (B) | One of $L _1, L _2, L _3$ is parallel to at least one of the other two, if |
(q) | $k=-\frac{6}{5}$ |
| (C) $L _1, L _2, L _3$ form a triangle, if | (r) | $k=\frac{5}{6}$ | |
| (D)$L _1, L _2, L _3$ do not form a triangle, if |
(s) $k=5$ |
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Answer:
Correct Answer: 9. $A \rightarrow s ; B \rightarrow p, q ; C \rightarrow r ; D \rightarrow p, q, s$
Solution:
- (A) Solving equations $L _1$ and $L _3$,
$$ \begin{aligned} \frac{x}{-36+10}=\frac{y}{-25+12} & =\frac{1}{2-15} \\ \therefore \quad x & =2, y=1 \end{aligned} $$
$$ L _1, L _2, L _3 \text { are concurrent, if point }(2,1) \text { lies on } L _2 $$
$$ \therefore \quad 6-k-1=0 \quad \Rightarrow \quad k=5 $$
(B) Either $L _1$ is parallel to $L _2$, or $L _3$ is parallel to $L _2$, then
$$ \begin{aligned} \frac{1}{3} & =\frac{3}{-k} \text { or } \quad \frac{3}{5}=\frac{-k}{2} \quad \Rightarrow \quad k=-9 \\ \text { or } \quad k & =\frac{-6}{5} \end{aligned} $$
(C) $L _1, L _2, L _3$ form a triangle, if they are not concurrent, or not parallel.
$$ \therefore \quad k \neq 5,-9,-\frac{6}{5} \Rightarrow k=\frac{5}{6} $$
(D) $L _1, L _2, L _3$ do not form a triangle, if
$$ k=5,-9,-\frac{6}{5} $$
