Question

A square of side $a$ lies above the x-axis and has one vertex at the origin. The side passing through it makes an angle $\alpha (0<\alpha <\frac{\pi }{4})$ with the positive direction of x-axis. The equation of its diagonal not through the origin

• 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

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

Consider the diagram shown below.

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

Also, $OB\perp OA$. Therefore, $OB$ makes an angle $({90}^{\circ }+\alpha )$ with $x-$axis, then coordinates of $B$ are,

$\Rightarrow \left[a\cos \right({90}^{\circ }+\alpha ),a\sin ({90}^{\circ }+\alpha \left)\right]$

Here, the diagonal which is not passing through the origin is $AB$. Therefore, the equation of this diagonal is,

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

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

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

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

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

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

This is the required equation of the diagonal.