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The pairs of straight lines $x^2-3 x y+2 y^2=0$ and $x^2-3 x y+2 y^2+x-2=0$ form a
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The correct answer is:
parallelogram
Given pair of lines are $x^2-3 x y+2 y^2=0$ and
$\begin{array}{ll}
& x^2-3 x y+2 y^2+x-2=0 . \\
\therefore & (x-2 y)(x-y)=0
\end{array}$
$\begin{aligned}
& \text { and } \quad(x-2 y+2)(x-y-1)=0 \\
& \Rightarrow x-2 y=0, \quad x-y=0 \text { and } x-2 y+2=0, \\
& \quad x-y-1=0
\end{aligned}$
Since, the lines $x-2 y=0, x-2 y+2=0$ and $x-y=0, x-y-1=0$ are parallel.
Also, angle between $x-2 y=0$ and $x-y=0$ is not $90^{\circ}$.
$\therefore$ It is a parallelogram.
$\begin{array}{ll}
& x^2-3 x y+2 y^2+x-2=0 . \\
\therefore & (x-2 y)(x-y)=0
\end{array}$
$\begin{aligned}
& \text { and } \quad(x-2 y+2)(x-y-1)=0 \\
& \Rightarrow x-2 y=0, \quad x-y=0 \text { and } x-2 y+2=0, \\
& \quad x-y-1=0
\end{aligned}$
Since, the lines $x-2 y=0, x-2 y+2=0$ and $x-y=0, x-y-1=0$ are parallel.
Also, angle between $x-2 y=0$ and $x-y=0$ is not $90^{\circ}$.
$\therefore$ It is a parallelogram.
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