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  • The diagonals of the parallelogram whose sides are <img alt="\mathrm{I x+m y+n=0, l x+m y+n^{\prime}=0, m x+l y+\mathrm{n}=0, \mathrm{mx}+\mathrm{ly}+\mathrm{n}^{\prime}=0}" src="https://entranceco

The diagonals of the parallelogram whose sides are \mathrm{I x+m y+n=0, l x+m y+n^{\prime}=0, m x+l y+\mathrm{n}=0, \mathrm{mx}+\mathrm{ly}+\mathrm{n}^{\prime}=0} include an angle

Option: 1

\pi / 2


Option: 2

\pi / 3


Option: 3

\mathrm{\tan ^{-1}\left(\frac{e^{2}-m^{2}}{e^{2}+m^{2}}\right)}


Option: 4

\mathrm{\tan ^{-1}\left(\frac{2 \mathrm{Im}}{\mathrm{I}^{2}+\mathrm{m}^{2}}\right)}


Answers (1)

best_answer

Since the distance between the parallel lines \mathrm{I x+m y+n=0 \: and \: I x+m y+n ' =0} is same as the distance between the parallel lines \mathrm{m x+l y+n=0\, and \, I x+m y+n^{\prime}=0}. Therefore, the parallelogram is rhombus. Since the diagonals of a rhombus are at right angles, therefore the required angle is \mathrm{\pi / 2}
 

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sudhir.kumar

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