Question

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

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

Answer 1

Answer: (A), (B)

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

Let the side $OA$ make an angle $\alpha$ with the $x$-axis. Then the coordinates of $A$ are $(a\cos \alpha ,a\sin \alpha )$. Also, the diagonal $OB$ makes an angle $\alpha +\frac{\pi }{4}$ with the $x$-axis, so that its equation is
$y=\tan (\alpha +\frac{\pi }{4})x$ or $y(\cos \alpha -\sin \alpha )=x(\sin \alpha +\cos \alpha )$

Since $AC$ is perpendicular to $OB$, its slope is $-\cot (\alpha +\frac{\pi }{4})$, and as it passes through $A(a\cos \alpha ,a\sin \alpha )$, its equation is
$y-a\sin \alpha =-\cot (\frac{\pi }{4}+\alpha )(x-a\cos \alpha )$ or $y(\sin \alpha +\cos \alpha )+x(\cos \alpha -\sin \alpha )=a$