InterviewSolution
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InterviewSolution
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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 \(\frac{2a^2}{7}\) sq. units. |
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Answer» Given: The given lines are 3x − 4y + a = 0 … (1) 3x − 4y + 3a = 0 … (2) 4x − 3y − a = 0 … (3) 4x − 3y − 2a = 0 … (4) To prove: 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 \(\frac{2a^2}{7}\) sq. units. Explanation: From Above solution, We know that Area of the parallelogram = \(\Big|\frac{(c_1-d_1)(c_2-d_2)}{\sqrt{a_1b_2-a_2-b_1}}\Big|\) ⇒ Area of the parallelogram = \(\Big|\frac{(a-3a)(2a-a)}{\sqrt{-9+16}}\Big|\) = \(\frac{2a^2}{7}\) square units Hence proved. |
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