Question

A square with each side equal to \( a \) lies above the \( x \)-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle \( \alpha \) (\( 0 < \alpha < \frac{\pi}{4} \)) with the positive direction of the \( x \)-axis. The equation of the diagonals of the square is:

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

Hint:

For problems involving geometric shapes, use symmetry and coordinate geometry to derive the required equations.

Answer 1

The Correct Option is A, C

1. The vertices of the square are: - At the origin: (0,0), - Along the side making an angle \(\alpha\):

\((a \cos \alpha, a \sin \alpha),\)

- Opposite vertex: \((a(\cos \alpha - \sin \alpha), a(\sin \alpha + \cos \alpha)),\)

- The fourth vertex: \((a(-\sin \alpha), a(\cos \alpha)).\)

2. The diagonals of the square intersect at their midpoints. The equation of a diagonal passing through (0,0) and \((a(\cos \alpha - \sin \alpha), a(\sin \alpha + \cos \alpha))\) can be derived as:

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

3. Similarly, the second diagonal has the same form but shifted by symmetry.

Thus, the equation of the diagonals is:

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