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Question: A square of side a lies above the *x*-axis and has one vertex at the origin. The side passing throug...

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 is

A

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

Explanation

Solution

Co-ordinates of A=(acos⁡α,asin⁡α)A = ( a \cos \alpha , a \sin \alpha ); Equation of OBO B y=tan⁡(π4+α)xy = \tan \left( \frac { \pi } { 4 } + \alpha \right) x

CA⊥C A \perp to OB ; ∴\therefore slope of CA=−cot⁡(π4+α)C A = - \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) ( x - a \cos \alpha ) ⇒

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