SATHEE: Straight Line and Pair of Straight Lines 3 Question 8

Straight Line and Pair of Straight Lines 3 Question 8

9.

Consider the lines given by

$L_{1} : x + 3 y - 5 = 0, L_{2} : 3 x - k y - 1 = 0$ , $L_{3} : 5 x + 2 y - 12 = 0$

	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)	$k = 5$

Show Answer

Answer:

Correct Answer: 9. $A \to s; B \to p, q; C \to r; D \to p, q, s$

Solution:

(A) Solving equations $L_{1}$ and $L_{3}$ ,

$\frac{x}{- 36 + 10} = \frac{y}{- 25 + 12} = \frac{1}{2 - 15}$

$∴ x = 2, y = 1$

$L_{1}, L_{2}, L_{3} are concurrent, if point (2, 1) lies on L_{2}$

$∴ 6 - k - 1 = 0 \Rightarrow k = 5$

(B) Either $L_{1}$ is parallel to $L_{2}$ , or $L_{3}$ is parallel to $L_{2}$ , then

$\begin{aligned} \frac{1}{3} & = \frac{3}{- k} or \frac{3}{5} = \frac{- k}{2} \Rightarrow k = - 9 \\ or k & = \frac{- 6}{5} \end{aligned}$

(C) $L_{1}, L_{2}, L_{3}$ form a triangle, if they are not concurrent, or not parallel.

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

(D) $L_{1}, L_{2}, L_{3}$ do not form a triangle, if

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

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