One side of a square is inclined to the axis of $x$ at an angle $α$ and one of its extremities is at the origin ; prove that the equations to its diagonals are
$y (cos α - sin α) = x (sin α + cos α)$
and $y (sin α + cos α) + x (cos α - sin α) = a$ .

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Solution

$∠ X O A = α$ and $O A = a$

So the coordinates of $A$ are $(a cos α, a sin α)$

$∠ X O B = 45 + α$ and $O B = O B = \sqrt{2} a$

So the coordiantes of $B$ are $(\sqrt{2} a cos (α + 45^{\circ}), \sqrt{2} a sin (α + 45^{\circ}))$

$\Rightarrow B (a cos α - a sin α, a cos α + a sin α)$

$∠ X O C = 90 + α$ and $O C = a$

So the coordinates of $C$ are $(a cos (α + 90^{\circ}), a sin (α + 90^{\circ}))$

$\Rightarrow C (- a sin α, a cos α)$

Equation of $O B$ is

$y - 0 = \frac{a cos α + a sin α - 0}{a cos α - a sin α - 0} (x - 0) (cos α - sin α) y = (cos α + sin α) x$

Equation of $A C$ is

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

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