Question

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

• A. $y(cosa+sina)-x(sina-cosa)=a$
• B. $y(cosa+sina)+x(sina-cosa)=a$
• C. $y(cosa+sina)+x(sina+cosa)=a$
• D. $y(cosa+sina)-x(cosa-sina)=a$

Answer 1

Answer: (A)

$y(cosa+sina)-x(sina-cosa)=a$

According to the given problem square lies above $x-$axis.

Now, equation of $AB$ using two point form, we get

$y-y_1=m(x-x_1)$

$\Rightarrow y-a\sin \alpha =\frac{-a(\cos \alpha -\sin \alpha )}{a(\cos \alpha +\sin \alpha )}[x-a\cos \alpha ]$

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

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

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

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

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

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

$\Rightarrow y(\sin \alpha +\cos \alpha )-x(\sin \alpha -\cos \alpha )=a$ since ${\sin }^2\alpha +{\cos }^2\alpha =1$