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

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

Answer: (C)

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

We know that diagonal of a square makes an angle of ${45}^0$ with its sides.

Hence $\angle DCA={45}^0$.

Therefore by linear pair $\angle DCE={135}^0$

It is given that the square has a side 'a'.

Thus the co-ordinates of the vertex C will be $(a\cos \alpha ,a\sin \alpha )$.

Applying angle sum property to triangle DCE, gives us

$\angle CED={45}^0-\alpha$.

Hence slope of the diagonal is

$-\tan ({45}^0-\alpha )$

$=-\frac{1-\tan \alpha }{1+\tan \alpha }$

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

Since it passes through C, we get the equation as

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

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

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

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