The distance between the two points A and A ' which lie on y = 2 such that both the line segments A B and A…

Let the slope of $OB$ be $m_1=\frac{3}{2}$

and the slope of $OA$ be $m_2=\frac{2}{x}$

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

i.e. $\left|\frac{{\displaystyle \frac{3}{2}-\frac{2}{x}}}{1+{\displaystyle \frac{6}{2x}}}\right|=1\Rightarrow \left|\frac{3x-4}{2x+6}\right|=1$

$3x-4=\pm \left(2x+6\right)$

$\Rightarrow x_1=10, x_2=-\frac{2}{5}$

$\Rightarrow A{A}^{\text{'}}=10+\frac{2}{5}=\frac{52}{5}$