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Q.

Consider the lines:
$L_{1} : 2 x - y + 5 = 0$ ,
$L_{2} : x + k y - 1 = 0$ ,
$L_{3} : k x + 2 y + 4 = 0$ .
If the lines $L_{1}, L_{2} and L_{3}$ do not form a triangle, then absolute value of sum of all possible values of k is

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answer is 2.7.

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Detailed Solution

Triangle will not be formed if
$L_{1} | | L_{2} \Rightarrow 2 = - \frac{1}{k} \Rightarrow k = - \frac{1}{2}$
or $L_{1} | | L_{3} \Rightarrow 2 = - \frac{k}{2} \Rightarrow k = - 4$
or $L_{2} | | L_{3} \Rightarrow \frac{- 1}{k} = \frac{- k}{2} \Rightarrow k^{2} = 2 \Rightarrow k = \pm \sqrt{2}$
or $L_{1}, L_{2}, L_{3} are concurrent$
$\Rightarrow k = - \frac{6}{5}, 3$
sum $= - \frac{1}{2} - 4 - \frac{6}{5} + 3 = - 2.7$

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Consider the lines:L1:2x−y+5=0 ,L2:x+ky−1=0 ,L3:kx+2y+4=0 .If the lines L1,L2 and L3 do not form a triangle, then absolute value of sum of all possible values of k is