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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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Verified Answer
The correct answer is:
parallelogram
Given pair of lines are
$$
x^{2}-3 x y+2 y^{2}=0
$$
and $x^{2}-3 x y+2 y^{2}+x-2=0$
$$
\therefore(x-2 y)(x-y)=0
$$
and $(x-2 y+2)(x-y-1)=0$
$\Rightarrow x-2 y=0, x-y=0$
and $x-2 y+2=0, x-y-1=0$
The lines $x-2 y=0, x-2 y+2=0$ and $x-y=0, x-y-1=0$ are paral lel.
Also, angle between $x-2 y=0$ and $x-y=0$ is not $90^{\circ}$
$\therefore$ It is a parallelogram.
$$
x^{2}-3 x y+2 y^{2}=0
$$
and $x^{2}-3 x y+2 y^{2}+x-2=0$
$$
\therefore(x-2 y)(x-y)=0
$$
and $(x-2 y+2)(x-y-1)=0$
$\Rightarrow x-2 y=0, x-y=0$
and $x-2 y+2=0, x-y-1=0$
The lines $x-2 y=0, x-2 y+2=0$ and $x-y=0, x-y-1=0$ are paral lel.
Also, angle between $x-2 y=0$ and $x-y=0$ is not $90^{\circ}$
$\therefore$ It is a parallelogram.
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