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 \(\frac{2a^2}{7}\) sq. units.

Answer 1

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.