If $θ$ is an angle between the lines given by the equation $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ , then equation of the line passing through the point of intersection of these lines and making an angle $θ$ with the positive x - axis is

$2 x + 11 y + 13 = 0$

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$11 x - 2 y + 13 = 0$

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$2 x - 11 y + 2 = 0$

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$11 x + 2 y - 11 = 0$

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Solution

The correct option is B
$11 x - 2 y + 13 = 0$

Writing the given equation as a quadratic in x we have $6 x^{2} + (5 y + 7) x - (4 y^{2} - 13 y + 3) = 0$
$\Rightarrow x = \frac{- (5 y + 7) \pm \sqrt{(5 y + 7)^{2} + 24 (4 y^{2} - 13 y + 3)}}{12} = \frac{- (5 y + 7) \pm \sqrt{121 y^{2} - 242 y + 121}}{12} = \frac{- (5 y + 7) \pm 11 (y - 1)}{12} = \frac{6 y - 18}{12}, \frac{- 16 y + 4}{12} \Rightarrow 2 x - y + 3 = 0 a n d 2 x + 4 y - 1 = 0$
which are the two lines represented by the given equation and the point of intersection is (-1, 1), obtained by solving these equations.
Also $t a n θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$ where a = 6, b = -4, $h = \frac{5}{2}$
$= \frac{2 \sqrt{{(\frac{5}{2})}^{2} - 6 (- 4)}}{6 - 4} = \sqrt{\frac{121}{4}} = \frac{11}{2}$
So the equation of the required line is
$y - 1 = \frac{11}{2} (x + 1) \Rightarrow 11 x - 2 y + 13 = 0$

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If $θ$ is an angle between the lines given by the equation $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ , then equation of the line passing through the point of intersection of these lines and making an angle $θ$ with the positive x - axis is

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If θ is an angle between the lines given by the equation 6x2+5xy−4y2+7x+13y−3=0, then equation of the line passing through the point of intersection of these lines and making an angle θ with the positive x - axis is

If $θ$ is an angle between the lines given by the equation $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ , then equation of the line passing through the point of intersection of these lines and making an angle $θ$ with the positive x - axis is