Question

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

• A. $(1,-4)$
• B. $(-1,4)$
• C. $(1,-2)$
• D. $(-1,2)$

Answer 1

Answer: (C), (D)

The correct options are
C $(1,-2)$
D $(-1,2)$

$2x^2-3xy-2y^2=0$
$\to (y+2x)(2y-x)=0$

Hence

$2x+y=0$ ...(i)
$-x+2y=0$ ...(ii)

Now $(2,1)$ is a vertex of the square thus formed.

Hence substituting $y=1$ in equation i, we get

$x=\frac{-1}{2}$.

Hence another vertex will be $(\frac{-1}{2}k,k)$ where k is a constant.

Substituting $x=2$ in equation i, we get

$y=-4$

Hence another vertex will be $(2m,-4m)$ where k is a constant.

Hence we get the two vertices lying on $2x+y=0$ as

$(\frac{-1}{2}k,k)$ and $(2m,-4m)$

Let

$k=2$ and $m=\frac{1}{2}$, we get the vertices as

$(-1,2),(1,-2)$.