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Q.

The figure formed by the four lines $3 x^{2} + 10 x y + 3 y^{2} = 0$ and $3 x^{2} + 10 x y + 3 y^{2} - 28 x - 28 y + 49 = 0$

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a

rectangle

b

square

c

parallelogram

d

rhombus

answer is B.

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Detailed Solution

$a = 3, 2 h = 10, b = 3$

$2 g = - 28, 2 f = - 28, c = 49$

$a + b \neq 0, (a - b) f g + h (f^{2} - g^{2}) = 0$

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