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Question
Prove that the area of the parallelogram formed by the lines 3x − 4y + a = 0, 3x − 4y + 3a = 0, 4x − 3y− a = 0 and 4x − 3y − 2a = 0 is \[\frac{2}{7} a^2\] sq. units..
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Solution
The given lines are
3x − 4y + a = 0 ... (1)
3x − 4y + 3a = 0 ... (2)
4x − 3y − a = 0 ... (3)
4x − 3y − 2a = 0 ... (4)
\[\text { Area of the parallelogram }= \left| \frac{\left( c_1 - d_1 \right)\left( c_2 - d_2 \right)}{a_1 b_2 - a_2 b_1} \right|\]
\[ \Rightarrow \text { Area of the parallelogram } = \left| \frac{\left( a - 3a \right)\left( 2a - a \right)}{- 9 + 16} \right| = \frac{2 a^2}{7}\text { square units }\]
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