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,

\[B \equiv \left( \frac{m n^{\prime}- ln}{l^2 - m^2},\frac{mn - ln^{\prime}}{l^2 - m^2} \right) \]

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

\[C \equiv \left( - \frac{n^{\prime}}{m + l}, - \frac{n^{\prime}}{m + l} \right)\]

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

\[D \equiv \left( \frac{mn - l n^{\prime}}{l^2 - m^2}, \frac{m n^{\prime} - ln}{l^2 - m^2} \right)\]

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

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

Let   \[m_1\text {  and } m_2 \] be the slope of AC and BD.

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

\[\text { and }m_2 = \frac{\frac{mn' - ln}{l^2 - m^2} - \frac{mn - ln'}{l^2 - m^2}}{\frac{mn - ln'}{l^2 - m^2} - \frac{mn' - ln}{l^2 - m^2}}\]

\[ = \frac{mn' - ln - mn + \ln'}{mn - ln' - mn' + ln}\]

\[ = - 1\]

\[\therefore m_1 m_2 = - 1\]

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