The diagonals of a square are along the pair of lines whose equation is $2 x^{2} - 3 x y - 2 y^{2} = 0$ . If $(2, 1)$ is a vertex of the square then another vertex consecutive to it can be

(1, - 2)

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(1, 4)

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(- 1, 2)

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(- 1, - 4)

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Solution

The correct options are
A $(1, - 2)$
C $(- 1, 2)$
Substituting $x = 2$ in the above equation, we get
$8 - 6 y - 2 y^{2} = 0$
$y^{2} + 3 y - 4 = 0$
$(y + 4) (y - 1) = 0$
Hence $y = - 4, 1$
Therefore two vertices are $(2, 1)$ and $(2, - 4)$ .
Now consider $y = 1$ . We get
$2 x^{2} - 3 x - 2 = 0$
$2 x^{2} - 4 x + x - 2 = 0$
$2 x (x - 2) + 1 (x - 2) = 0$
$(x - 2) (2 x + 1) = 0$
Hence $x = 2, \frac{- 1}{2}$
Thus other two vertices are $(2, 1), (\frac{- 1}{2}, 1)$ .
From the above we get two uncommon vertices as
$(2, - 4)$ and $(\frac{- 1}{2}, 1)$ .
Now decreasing first vertex by a factor of 2 and increasing the second by the same, we get
$(1, - 2)$ and $(- 1, 2)$ .
Hence these are possible vertices of the square.

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The diagonals of a square are along the pair of lines whose equation is 2x2−3xy−2y2=0. If (2,1) is a vertex of the square then another vertex consecutive to it can be

The diagonals of a square are along the pair of lines whose equation is $2 x^{2} - 3 x y - 2 y^{2} = 0$ . If $(2, 1)$ is a vertex of the square then another vertex consecutive to it can be