Question

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

Answer 1

The given lines are

$lx+my+n=0\dots \dots \left(i\right)$

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

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

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

Solving (i) and (ii), we get

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

Solving (ii) and (iii), we get

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

Solving (i) and (iv), we get

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

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

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

and $m_2=\frac{-\frac{m{n}^{{}^{\prime }}-ln}{l^2-m^2}-\frac{mn-l{n}^{{}^{\prime }}}{l^2-m^2}}{\frac{m{n}^{{}^{\prime }}-ln-mn+l{n}^{{}^{\prime }}}{mn-l{n}^{{}^{\prime }}-mn+ln}}=-1$

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

Hence, diagonals of the parallelogram intersect at an angle $\frac{\pi }{2}$.