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If $2 x^2+3 x y-2 y^2=0$ represents two sides of a parallelogram and $3 x+y+1=0$ is one of its diagonals, then the other diagonal is
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Verified Answer
The correct answer is:
$x-3 y=0$
As we know that,
If the lines $a x^2+2 h x y+b y^2=0$ be two sides of a parallelogram and the line $l x+m y=1$ be one of its diagonals, then other diagonal is
$$
y(b l-h m)=x(a m-h l)
$$
Here, $a=2, b=-2, h=\frac{3}{2}$
$$
l=-3, m=-1
$$
Putting all values in Eq. (i), we get
$$
\begin{aligned}
\therefore y\left(6+\frac{3}{2}\right) & =x\left(-2+\frac{9}{2}\right) \\
\Rightarrow y\left(\frac{15}{2}\right) & =x\left(\frac{5}{2}\right) \\
15 y & =5 x \\
3 y & =x \Rightarrow x-3 y=0
\end{aligned}
$$
If the lines $a x^2+2 h x y+b y^2=0$ be two sides of a parallelogram and the line $l x+m y=1$ be one of its diagonals, then other diagonal is
$$
y(b l-h m)=x(a m-h l)
$$
Here, $a=2, b=-2, h=\frac{3}{2}$
$$
l=-3, m=-1
$$
Putting all values in Eq. (i), we get
$$
\begin{aligned}
\therefore y\left(6+\frac{3}{2}\right) & =x\left(-2+\frac{9}{2}\right) \\
\Rightarrow y\left(\frac{15}{2}\right) & =x\left(\frac{5}{2}\right) \\
15 y & =5 x \\
3 y & =x \Rightarrow x-3 y=0
\end{aligned}
$$
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