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

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

Answer 1

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

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

Also $OB\perp OA$

Then, OB makes an angle $(90+\alpha )$ with x-axis, then coordinates of B are $(-a\sin\alpha,a\cos\alpha)

Equation of the diagonal not passing through origin is,

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

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

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

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

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

$\therefore y(\sin \alpha +\cos \alpha )+x(\cos \alpha -\sin \alpha )=0$