Question

The diagonal of the parallelogram whose sides are $\ell x+my+n=0,\ell x+my+n^{\prime }=0,mx+\ell y+n=0,ma+\ell y+n^{\prime }=0$, include an angle-

• A. ${\tan }^{-1}\left(\frac{2\ell m}{{\ell }^2+m^2}\right)$
• B. ${\tan }^{-1}\left(\frac{\ell -m^2}{{\ell }^2+m^2}\right)$
• C. $\pi /2$
• D. $\pi /3$

Answer 1

The correct option is C $\pi /2$

The given lines are $lx+my+n=0\dots \left(1\right)$

$mx+ly+n^{\prime }=0\dots \left(2\right)$

$lx+my+n^{\prime }=0\dots \left(3\right)$

$mx+ly+n=0\dots \left(4\right)$

By solving $\left(1\right)$ and $\left(2\right)$, we get

$B=(\frac{m{n}^{\prime }-ln}{l^2-m^2},\frac{mn-l{n}^{\prime }}{l^2-m^2})$

By solving $\left(2\right)$ and $\left(3\right)$, we get

$C=(-\frac{n^{\prime }}{m+l^{\prime }},-\frac{n^{\prime }}{m+l})$

By solving $\left(3\right)$ and $\left(4\right)$, we get

$D=(\frac{mn-l{n}^{\prime }}{l^2-m^2},\frac{m{n}^{\prime }-ln}{l^2-m^2})$

By solving $\left(1\right)$ and $\left(4\right)$, we get

$A=(-\frac{n}{m+l^{\prime }},-\frac{n}{m+l})$

Let $m_1$ and $m_2$ be the slope of $AC$ and $BD$.

Now, $m_1=\frac{\frac{-n^{\prime }}{m+l}+\frac{n}{m+l}}{\frac{-n^{\prime }}{m+l}+\frac{n}{m+l}}=1$

and, $m_1=\frac{\frac{m{n}^{\prime }-\ln }{l^2-m^2}-\frac{mn-l{n}^{\prime }}{l^2-m^2}}{\frac{mn-l{n}^{\prime }}{l^2-m^2}-\frac{m{n}^{\prime }-\ln }{l^2-m^2}}=-1$

$\therefore m_1{m}_2=-1$

Thus, the diagonals of the parallelogram whose sides are $lx+my+n=0,lx+my+n^{\prime }=0,mx+ly+n=0andmx+ly+n^{\prime }=0$ include an angle $\frac{\pi }{2}$.

So, $\text{C}$ is the correct option.