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Question -

Prove that the area of the parallelogramformed by the lines 3x – 4y + a = 0, 3x –4y + 3a = 0, 4x – 3y – a = 0 and 4x –3y – 2a = 0 is 2a2/7 sq. units.



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)

Let us prove, the areaof the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x– 3y – a = 0 and 4x – 3y – 2a = 0 is 2a2/7 sq. units.

From above solution,we know that

Hence proved.

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