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Q.

If $θ$ is the angle between the pair of lines $x^{2} - 3 xy + 2 y^{2} + 3 x - 5 y + 2 = 0$ the value of ${cosec}^{2} θ$ is

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a

3

b

9

c

10

d

15

answer is C.

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Detailed Solution

$\begin{array}{l} I f θ i s a n g l e b e t w e e n g i v e n p a i r o f l i n e s t h e n \\ \tan θ = \frac{2 \sqrt{(h^{2} - a b)}}{a + b} \\ \Rightarrow \tan^{2} θ = \frac{4 (h^{2} - ab)}{(a + b)^{2}} = \frac{1}{9} \\ ∴ {cosec}^{2} θ = 1 + c o t^{2} θ = 10 \end{array}$

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