A square of side a lies above the x-axis and has one vertex at the origin.

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

Accepted Answer

Co-ordinate of  $$A = \left( {a\,\cos \,\alpha ,\,\,a\,\sin \,\alpha } \right)$$

Equation of $OB,$

$$y = \tan \left( {{\pi  \over 4} + \alpha } \right)x$$

$$CA{ \bot ^r}$$ to $OB$

$$\therefore$$ slope of $CA=-$ $$\cot \left( {{\pi  \over 4} + \alpha } \right)$$

Equation of $CA$

$$y - a\sin \alpha  =  - cot\left( {{\pi  \over 4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\tan \left( {{\pi  \over 4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha  - x} \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {{{\tan {\pi  \over 4} + \tan \alpha } \over {1 - \tan {\pi  \over 4}\tan \alpha }}} \right)\left( {a\,\cos \,\alpha  - x} \right)$$

$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {1 + \tan \alpha } \right) = \left( {a\cos \alpha  - x} \right)\left( {1 - \tan \alpha } \right)$$

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

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

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a{\cos ^2}\alpha  - a\cos \alpha \sin \alpha  - x\left( {\cos \alpha  - \sin \alpha } \right)$$

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

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

Solutions · 1