The distance between the two points A and A ^ ' which lie on y = 2...

Let the slope of $OB$ be ${m}_{1} = \dfrac{3}{2}$

and the slope of $OA$ be ${m}_{2} = \dfrac{2}{x}$

Now angle between two lines, $tan\dfrac{\pi}{4} = \left|\dfrac{\dfrac{3}{2} - \dfrac{2}{x}}{1 + \dfrac{6}{2x}}\right|$

i.e. \left|\dfrac{\dfrac{3}{2} - \dfrac{2}{x}}{1 + \dfrac{6}{2x}}\right| = 1 ⇒ \left|\dfrac{3x - 4}{2x + 6}\right| = 1$</p><p>$3x - 4 = \pm \left(2x + 6\right)$</p><p>$⇒ {x}_{1} = 10, {x}_{2} = - \dfrac{2}{5}$</p><p>$⇒ A{A}^{'} = 10 + \dfrac{2}{5} = \dfrac{52}{5}