Question

Show that the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n' = 0, mx + ly + n = 0 and mx + ly + n' = 0 include an angle π/2.

Answer 1

The given lines are

lx + my + n = 0 ... (1)

mx + ly + n' = 0 ... (2)

lx + my + n' = 0 ... (3)

mx + ly + n = 0 ... (4)



Solving (1) and (2), we get,

$B\equiv \left(\frac{m{n}^{\text{'}}-ln}{l^2-m^2},\frac{mn-l{n}^{\text{'}}}{l^2-m^2}\right)$

Solving (2) and (3), we get,

$C\equiv \left(-\frac{n^{\text{'}}}{m+l},-\frac{n^{\text{'}}}{m+l}\right)$

Solving (3) and (4), we get,

$D\equiv \left(\frac{mn-l{n}^{\text{'}}}{l^2-m^2},\frac{m{n}^{\text{'}}-ln}{l^2-m^2}\right)$

Solving (1) and (4), we get,

$A\equiv \left(-\frac{n}{m+l},-\frac{n}{m+l}\right)$

Let $m_1\mathrm{and}{m}_2$ be the slope of AC and BD.

$m_1=\frac{-{\displaystyle \frac{n^{\text{'}}}{m+l}}+{\displaystyle \frac{n}{m+l}}}{-{\displaystyle \frac{n^{\text{'}}}{m+l}}+{\displaystyle \frac{n}{m+l}}}=1$



$$\begin{gathered} \mathrm{and}{m}_2=\frac{\frac{mn\text{'}-ln}{l^2-m^2}-\frac{mn-ln\text{'}}{l^2-m^2}}{\frac{mn-ln\text{'}}{l^2-m^2}-\frac{mn\text{'}-ln}{l^2-m^2}} \\ =\frac{mn\text{'}-ln\mathit{-}mn\mathit{+}\mathit{ln}\mathit{\text{'}}}{mn-ln\text{'}-mn\text{'}+ln} \\ =-1 \end{gathered}$$

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

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