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

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

Answer 1

Answer: (D)

Line OA makes an angle $\alpha$ with $x$ -axis and $OA=a,$ then co-ordinates of $A$ are $(a\cos \alpha$ $a\sin \alpha$ ).

Also $OB\perp OA$, then $OB$ makes an angle $\left(90^{\circ }+\alpha \right)$ with $x$ -axis, then co-ordinates of $B$ are
$(a\cos \left(90^{\circ }+\alpha \right),a\sin \left(90^{\circ }+\alpha \right))$
$=(-a\sin \alpha ,a\cos \alpha )$
Equation of the diagonal not passing through the origin is
$(y-a\sin \alpha )=\frac{a\cos \alpha -a\sin \alpha }{-a\sin \alpha -a\cos \alpha }(x-a\cos \alpha )$
$\Rightarrow (y-a\sin \alpha )=\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }(x-a\cos \alpha )$
$\Rightarrow (\sin \alpha +\cos \alpha )(y-a\sin \alpha )$
$=(\sin \alpha -\cos \alpha )(x-a\cos \alpha )$
$\Rightarrow (\sin \alpha +\cos \alpha )y-a\sin \alpha (\sin \alpha +\cos \alpha )$
$=(\sin \alpha -\cos \alpha )x-a\cos \alpha (\sin \alpha -\cos \alpha )$
$\Rightarrow y(\sin \alpha +\cos \alpha )+x(\cos \alpha -\sin \alpha )$
$=a\sin \alpha (\sin \alpha +\cos \alpha )$
$-a\cos \alpha (\sin \alpha -\cos \alpha )$
$=a[{\sin }^2\alpha +\sin \alpha \cos \alpha$
$-\cos \alpha \sin \alpha +{\cos }^2\alpha ]$
$\therefore y(\sin \alpha +\cos \alpha )+x(\cos \alpha -\sin \alpha )=a$