Type of quadrilateral formed by the two pairs of lines $6 x^{2} - 5 x y - 6 y^{2} = 0$ and $6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0$ is

kite

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Rhombus

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Parallelogram

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Rectangle

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Solution

The correct option is D Rectangle
$6 x^{2} - 5 x y - 6 y^{2} = 0$

$x = \frac{5 y \pm \sqrt{25 y^{2} + 144 y^{2}}}{12}$

$12 x = 5 y \pm 13 y$

$12 x = 18 y$ and $12 x = - 8 y$

$2 x - 3 y = 0; 3 x + 2 y = 0$

These lines are mutually perpendicular. Since these are adjacent sides., hence the angle included within the sides is $90$

$6 x^{2} + x (- 5 y + 1) - 6 y^{2} + 5 y - 1 = 0$

$x = \frac{5 y - 1 \pm \sqrt{25 y^{2} + 1 - 10 y - 24 (- 6 y^{2} + 5 y - 1)}}{12}$

$12 x = 5 y - 1 \pm \sqrt{169 y^{2} - 130 y + 25}$

$12 x = 18 y - 6$ and $12 x = - 8 y + 4$

Thus $2 x - 3 y = - 1$ and $3 x + 2 y = 1$

Again these lines are perpendicular and angle is $90$

Thus the respective sides are mutually parallel and perpendicular.
Hence it is a rectangle

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