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

Option 1)
$y(\cos \alpha +\sin \alpha )+x(\sin \alpha -\cos \alpha )=a$

Option 2)
$y(\cos \alpha +\sin \alpha )+x(\sin \alpha +\cos \alpha )=a$

Option 3)
$y(\cos \alpha +\sin \alpha )+x(\cos \alpha - \sin \alpha )=a$

Option 4)
$y(\cos \alpha -\sin \alpha )-x(\sin \alpha -\cos \alpha )=a$

Answers (1)

best_answer

As we learnt in

Two – point form of a straight line -

$y-y_{1}=\left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x-x_{1})$

- wherein

The lines passes through $(x_{1}y_{1})$ and $(x_{2}\, y_{2})$

Slope of a line -

If $\Theta$ is the angle at which a straight line is inclined to a positive direction of x-axis, then the slope is defined by

$m= \tan \Theta$ .

- wherein

Equation of diagonal AB is

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

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

Option 1)

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

Incorrect

Option 2)

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

Incorrect

Option 3)

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

Correct

Option 4)

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

Incorrect

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