$I f 3 x^{2} - 11 x y + 10 y^{2} - 7 x + 13 y + k = 0 d e n o t e s a p a i r o f s t r a i g h t l i n e s, t h e n t h e p o int o f int e r \sec t i o n o f t h e l i n e s i s$

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$(3, 1)$

$(1, - 3)$

$(1, 3)$

$(- 3, - 1)$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 3 x^{2} - 11 x y + 10 y^{2} - 7 x + 13 y + k = 0 - - (1) \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 \\ a = 3, h = \frac{- 11}{2}, b = 10, g = \frac{- 7}{2}, f = \frac{13}{2}, c = k \\ P . I = (\frac{h f - b y}{a b - h^{2}}, \frac{g h - a f}{a b - h^{2}}) \\ \Rightarrow (\frac{(\frac{- 11}{2}) (\frac{13}{2}) - 10 (\frac{- 7}{2})}{3 (10) - {(\frac{- 11}{2})}^{2}}, \frac{(\frac{- 7}{2}) (\frac{- 11}{2}) - 3 (\frac{13}{2})}{3 (10) - {(\frac{- 11}{2})}^{2}}) \\ \Rightarrow (\frac{143 + 140}{120 - 121}, \frac{77 - 78}{120 - 121}) = (3, 1) \end{array}$