Prove that the area of the parallelogram contained by the lines `4y-3x-a=0,3y-4x+a=0,4y-3x-3a=0,` and `3y-4x+2a=0` is `(2/7)a^2dot`

Explanation

Rewriting the equations of the sides of the parallelogram, we have `3x - 4y + a = 0`

`4x - 3y - a = 0 `

`3x - 4y + 3a = 0`

and `4x - 3y - 2a = 0`

Here, `c_1 = a`,` d_1 = 3a`, `c_2 = −a`,` d2 = −2a`, `a1 = 3`, `b_1 = −4`, `a_2 = 4` and `b_2 = −3`.

Therefore, by Problem

`Area=[(d_1-c_1)(d_2-c_2)]/(a_1b_2-a_2b_1)`

`area=[(3a-a)(-2a+a)]/(3(-3)-4(-4))`

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