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Question
If the diagonals of the quadrilateral formed by the lines l1x + m1y + n1 = 0, l2x + m2y + n2 = 0, l1x + m1y + n1' = 0 and l2x + m2y + n2' = 0 are perpendicular, then write the value of l12 − l22 + m12 − m22.
Solution
The given lines are
l1x + m1y + n1 = 0 ... (1)
l2x + m2y + n2 = 0 ... (2)
l1x + m1y + n1' = 0 ... (3)
l2x + m2y + n2' = 0 ... (4)
Let (1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.
The equation of diagonal AC passing through the intersection of (2) and (3) is given by
l1x + m1y + n1' + \[\lambda\] (l2x + m2y + n2) = 0
\[\Rightarrow \left( l_1 + \lambda l_2 \right)x + \left( m_1 + \lambda m_2 \right)y + \left( n_1 ' + \lambda n_2 \right) = 0\]
\[ \Rightarrow \text { Slope of diagonal AC } = - \left( \frac{l_1 + \lambda l_2}{m_1 + \lambda m_2} \right)\]
Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by
l1x + m1y + n1 + \[\mu\] (l2x + m2y + n2) = 0
\[\Rightarrow \left( l_1 + \mu l_2 \right)x + \left( m_1 + \mu m_2 \right)y + \left( n_1 + \mu n_2 \right) = 0\]
\[ \Rightarrow \text { Slope of diagonal BD }= - \left( \frac{l_1 + \mu l_2}{m_1 + \mu m_2} \right)\]
The diagonals are perpendicular to each other.
∴ \[\left( \frac{l_1 + \lambda l_2}{m_1 + \lambda m_2} \right)\left( \frac{l_1 + \mu l_2}{m_1 + \mu m_2} \right) = - 1\]
\[\Rightarrow \left( l_1 + \lambda l_2 \right)\left( l_1 + \mu l_2 \right) = - \left( m_1 + \lambda m_2 \right)\left( m_1 + \mu m_2 \right)\]
\[\text { Let }\lambda = - 1, \mu = 1\]
\[ \Rightarrow \left( l_1 - l_2 \right)\left( l_1 + l_2 \right) = - \left( m_1 - m_2 \right)\left( m_1 + m_2 \right)\]
\[ \Rightarrow \left( {l_1}^2 - {l_2}^2 \right) = - \left( {m_1}^2 - {m_2}^2 \right)\]
\[ \Rightarrow \left( {l_1}^2 - {l_2}^2 \right) + \left( {m_1}^2 - {m_2}^2 \right) = 0\]
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