JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 21)
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 Statement/Expressions in Column I with the Statements/Expressions in Column II.
| 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 atleast one of the other two, if | (Q) | $$K = - {6 \over 5}$$ |
| (C) | L$$_1$$, L$$_2$$, L$$_3$$ form a triangle, if | (R) | $$K = {5 \over 6}$$ |
| (D) | L$$_1$$, L$$_2$$, L$$_3$$ do not form a triangle, if | (S) | $$K = 5$$ |
Explanation
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)
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