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 $O A=a,$ then co-ordinates of $A$ are $(a \cos \alpha$ $a \sin \alpha$ ).

Also $O B \perp O A$, then $O B$ makes an angle $\left(90^{\circ}+\alpha\right)$ with $x$ -axis, then co-ordinates of $B$ are
$\left(a \cos \left(90^{\circ}+\alpha\right),\right. \left.a \sin \left(90^{\circ}+\alpha\right)\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 \quad(\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\left[\sin ^{2} \alpha+\sin \alpha \cos \alpha\right. $
$\left.-\cos \alpha \sin \alpha+\cos ^{2} \alpha\right] $
$\therefore y(\sin \alpha+\cos \alpha)+x(\cos \alpha-\sin \alpha)=a$