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Mathematics Question on Straight lines

A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle α(0<α<π4)\alpha\left(0 < \alpha < \frac{\pi}{4}\right) with the positive direction of x-axis. The equation of its diagonal not passing through the origin :

A

y(cos⁡α+sin⁡ α)+x(cos⁡ α−sin⁡ α)=ay(\cos \alpha + \sin \, \alpha) + x(\cos \, \alpha - \sin\, \alpha ) = a

B

y(cos⁡α−sin⁡α)−x(sin⁡α−cos α)=ay(\cos \alpha - \sin \alpha) - x(\sin \alpha - cos \,\alpha) = a

C

y(cos⁡α+sin⁡α)+x(sin⁡α−cos⁡α)=ay(\cos \alpha + \sin \alpha ) + x(\sin \alpha - \cos \alpha) = a

D

y(cos⁡α+sin⁡α)+x(sin⁡α+cos⁡α)=ay(\cos \alpha + \sin \alpha) + x(\sin \alpha + \cos \alpha) = a

Answer

y(cos⁡α+sin⁡ α)+x(cos⁡ α−sin⁡ α)=ay(\cos \alpha + \sin \, \alpha) + x(\cos \, \alpha - \sin\, \alpha ) = a

Explanation

Solution

Co-ordinates of A=(acos⁡α,asin⁡α)A = (a \cos \alpha , a \sin \alpha ) Equation of OB, y=tan⁡(π4+α)x y = \tan \left( \frac{\pi}{4} + \alpha \right) x CA⊥rCA \bot^r to OB ∴\therefore slope of CA =−cot⁡(π4+α) = - \cot \left( \frac{\pi}{4} + \alpha \right) Equation of CA y−asin⁡α=−cot⁡(π4+α)(x−acos⁡α)y -a \sin\alpha =- \cot\left(\frac{\pi}{4} + \alpha\right) \left(x-a \cos\alpha\right) ⇒(y−asin⁡α)(tan⁡(π4+α))=(acos⁡α−x)\Rightarrow\left(y -a \sin\alpha\right) \left(\tan \left(\frac{\pi}{4} + \alpha\right)\right) = \left(a \cos\alpha-x\right) ⇒(y−asin⁡α)(tan⁡π4+tan⁡α1−tan⁡π4tan⁡α)(acos⁡α−x)\Rightarrow \left(y -a \sin\alpha\right) \left(\frac{\tan \frac{\pi}{4}+ \tan\alpha}{1- \tan \frac{\pi}{4} \tan\alpha}\right) \left(a \cos\alpha-x\right) ⇒(y−asin⁡α)(1+tan⁡α)=(acos⁡α−x)(1−tan⁡α)\Rightarrow \left(y -a\sin \alpha\right) \left(1+\tan\alpha\right) = \left(a\cos\alpha-x\right)\left(1 -\tan\alpha\right) ⇒(y−asin⁡α)(cos⁡α+sin⁡α)=(acos⁡α−x)(cos⁡α−sin⁡ α) \Rightarrow \left(y -a \sin\alpha\right)\left(\cos\alpha+\sin\alpha\right)=\left(a \cos\alpha - x\right)\left(\cos\alpha - \sin\,\alpha\right) ⇒y(cos⁡+sin⁡α)−asin⁡αcos⁡α−asin⁡2α\Rightarrow y \left(\cos+\sin\alpha\right) -a \sin\alpha \cos\alpha -a \sin^{2} \alpha =acos⁡2α−acos⁡αsin⁡α−x(cos⁡α−sin⁡α)= a \cos^{2} \alpha - a \cos\alpha \sin\alpha - x \left(\cos\alpha - \sin\alpha\right) ⇒y(cos⁡α+sin⁡α)+x(cos⁡α−sin⁡α)=a\Rightarrow y \left(\cos\alpha + \sin\alpha\right) + x\left(\cos\alpha - \sin\alpha\right) = a y(sin⁡α+cos⁡α)+x(cos⁡α−sin⁡α)=ay \left(\sin\alpha + \cos\alpha\right) + x \left(\cos\alpha - \sin\alpha\right) = a.