Question

A square of side length $a$ units lies above the $x-$axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha (0<\alpha <\pi /4)$ with the positive direction of $x-$axis. The equation of its diagonal not passing through the origin is

• A. $y(\cos \alpha +\sin \alpha )+x(\sin \alpha -\cos \alpha )=a$
• B. $y(\cos \alpha +\sin \alpha )+x(\sin \alpha +\cos \alpha )=a$
• C. $y(\cos \alpha +\sin \alpha )+x(\cos \alpha -\sin \alpha )=a$
• D. $y(\cos \alpha -\sin \alpha )-x(\sin \alpha -\cos \alpha )=a$

Answer 1

The correct option is C $y(\cos \alpha +\sin \alpha )+x(\cos \alpha -\sin \alpha )=a$

The parametric coordinate of one vertex of square is $(0+a\cos \alpha ,0+a\sin \alpha )$


Inclination of diagonal not passing through origin is $135°+\alpha$
$\therefore$equation of the diagonal which is not passing through origin is
$y-a\sin \alpha =\tan (135°+\alpha )[x-a\cos \alpha ]\Rightarrow y-a\sin \alpha =-\tan (45°-\alpha )[x-a\cos \alpha ]\Rightarrow y-a\sin \alpha =-\left(\frac{\cos \alpha -\sin \alpha }{\sin \alpha +\cos \alpha }\right)[x-a\cos \alpha ]\Rightarrow (y-a\sin \alpha )(\sin \alpha +\cos \alpha )+(\cos \alpha -\sin \alpha )(x-a\cos \alpha )=0\Rightarrow y(\sin \alpha +\cos \alpha )+x(\cos \alpha -\sin \alpha )=a$