The combined equation of the pair of straight lines passing through the point of intersection of the pair of lines x2+4xy+3y2−4x−10y+3=0 and having slopes 1/2 and -1/3 is

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The combined equation of the pair of straight lines passing through the point of intersection of the pair of lines $x^{2} + 4 x y + 3 y^{2} - 4 x - 10 y + 3 = 0$ and having slopes 1/2 and -1/3 is

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$x^{2} - y^{2} - 8 x - 2 y + 15 = 0$

$x^{2} + 7 x y + 10 y^{2} - x - 8 y - 2 = 0$

$x^{2} + x y - 6 y^{2} - 7 x - 16 y + 6 = 0$

$x^{2} + 7 x y + 12 y^{2} - x - 4 y = 0$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x^{2} + 4 x y + 3 y^{2} - 4 x - 10 y + 3 = 0 \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 \\ a = 1, h = 2, b = 3, g = - 2, f = - 5, c = 3 \\ P . I = (\frac{h f - b g}{a b - h^{2}}, \frac{g h - a f}{a b - h^{2}}) \\ = (\frac{2 (- 5) - 3 (- 2)}{1 (3) - 4}, \frac{- 2 (2) - 1 (- 5)}{1 (3) - 4}) \Rightarrow (\frac{- 4}{- 1}, \frac{1}{- 1}) \Rightarrow (4, - 1) \\ I f m = \frac{1}{2} a n d p o int (4, - 1) t h e n E q u a t i o n \\ y + 1 = \frac{1}{2} (x - 4) \\ \Rightarrow 2 y + 2 = x - 4 \\ \Rightarrow x - 2 y - 6 = 0 \end{array}$

$\begin{array}{l} I f m = \frac{- 1}{3} a n d p o int (4, - 1) t h e n E q u a t i o n i s \\ y + 1 = \frac{- 1}{3} (x - 4) \\ \Rightarrow 3 y + 3 = - x + 4 \\ \Rightarrow x + 3 y - 1 = 0 \\ C o m b a i n e d E q u a t i o n i s (x - 2 y - 6) (x + 3 y - 1) = 0 \\ \Rightarrow x^{2} + 3 x y - x - 2 x y - 6 y^{2} + 2 y - 6 x - 18 y + 6 = 0 \\ \Rightarrow x^{2} + x y - 6 y^{2} - 7 x - 16 y + 6 = 0 \end{array}$