A square with each side equal to 'a' above the $x$-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle $\alpha\left(0 \lt \alpha \lt \frac{\pi}{4}\right)$ with the positive direction of the axis. Equation of the diagonals of the square.

• Correct answer
  A

  $y(\cos \alpha-\sin \alpha)=x(\sin \alpha+\cos \alpha)$
• B

  $y(\cos \alpha+\sin \alpha)=x(\cos \alpha-\sin \alpha)$
• Correct answer
  C

  $y(\sin \alpha+\cos \alpha)+x(\cos \alpha-\sin \alpha)=a$
• D

  $y(\cos \alpha-\sin \alpha)+x(\cos \alpha+\sin \alpha)=a$

Solution

The parametric coordinate of one vertex of square is \((0+a \cos a, 0+a \sin a)\) Inclination of diagonal not passing through origin is \(135^{\circ}+\mathrm{a}\) \(\therefore\) equation of the diagonal which is not passing through origin is \(\begin{aligned} & y-a \sin \alpha=\tan \left(135^{\circ}+\alpha\right)[x-a \cos \alpha] \\ & \Rightarrow y-a \sin a=-\tan \left(45^{\circ}-\alpha\right)[x-a \cos \alpha] \\ & \Rightarrow y-a \sin a=-\left(\frac{\cos a-\sin }{\sin \alpha+\cos \alpha}\right)[x-a \cos \alpha] \\ & \Rightarrow(y-a \sin \alpha)(\sin \alpha+\cos \alpha)+(\cos \alpha-\sin a)(x-a \cos \alpha)=0 \\ & \Rightarrow y(\sin a+\cos \alpha)+x(\cos \alpha-\sin \alpha)=a \end{aligned}\)

Hence, the correct answer is options: 'A', 'C'.