Question

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 $\alpha$ where $0<\alpha <\frac{\pi }{4}$ with the positive direction of x- axis. The equation of its diagonal not passing through the origin is

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

Answer 1

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

Equation of $AB\Rightarrow y=\tan \alpha x$

& equation of $AD\Rightarrow y=\tan (90+\alpha )x$

Point $B=(a\cos \alpha ,a\sin \alpha )$

And point $D=(-a\cos (90-x),a\sin (90-\alpha \left)\right)=(-a\sin \alpha ,a\cos \alpha )$

Equation of $DB$ is

$y-a\sin \alpha =\left(\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\right)(x-a\cos \alpha )$

$(\sin \alpha +\cos \alpha )y-x(\sin \alpha -\cos \alpha )=a({\sin }^2\alpha +{\cos }^2\alpha )$

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