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One side of a square is inclined at an acute angle \alpha with the positive \mathrm{x}-axis and the side of a square is 4 , then the equation of the diagonal of the square which is not passing through the origin is 

Option: 1

(\cos \alpha+\sin \alpha) x+(\cos \alpha-\sin \alpha) y=4


Option: 2

(\cos \alpha+\sin \alpha) x-(\cos \alpha-\sin \alpha) y=4


Option: 3

(\cos \alpha-\sin \alpha) x+(\cos \alpha+\sin \alpha) y=4


Option: 4

(\cos \alpha-\sin \alpha) x-(\cos \alpha+\sin \alpha) y=4 \cos 2 \alpha


Answers (1)

best_answer



  \therefore \: \mathrm{slope \: of\: line}\: \mathrm{AC}=\frac{4 \sin \alpha-4 \cos \alpha}{4 \cos \alpha+4 \sin \alpha}
\therefore \quad \text{equation of line }=\frac{\sin \alpha-\sin \alpha}{\cos \alpha \sin \alpha}

y-\sin \alpha=\frac{\sin \alpha-\sin \alpha}{\cos \alpha \sin \alpha}(x-4 \cos \alpha)
\mathrm{y} \cos \alpha+\mathrm{y} \sin \alpha-4 \sin \alpha \cdot \cos \alpha=\mathrm{x} \sin \alpha-\mathrm{x} \cos \alpha
-4 \sin ^{2} \alpha \quad-\sin \alpha \cos \alpha+4 \cos ^{2} \alpha
\mathrm{x}(\cos \alpha-\sin \alpha)+\mathrm{y}(\cos \alpha+\sin \alpha)=4 \quad \Rightarrow \quad (D)]

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rishi.raj

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