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Prove that the area of the parallelogram...

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`

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` tan theta =|((4//3)-(3//4))/(1+(4//3) xx (3//4))| = (7)/(24)`
` or "cosec" theta = (25)/(7)`
` "Now " d_(1) = (|-a+3a|)/(5) = |(2a)/(5)`
`"and " d_(2) = (|2a-a|)/(5) = |(a)/(5)|`
`therefore " Area" = 2("Area of" Delta ABD") = 2 ((1)/(2)d_(1) * AB)`
` = d_(1)d_(2) "cosec " theta = (|(a)/(5)|)(|(2a)/(5)|)(25)/(7)= (2a^(2))/(7) `
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