The distance between the two points $AandA\text{'}$  which lie on $y=2$ such that both the line segment $ABandAB\text{'}$ (where B is the point (2,3) ) subtend angle $\frac{\pi }{4}$ at the origin, is equal to

Given : $AandA\text{'}$  lies on $y=2$ 
Let coordinates of  $A\equiv (x_1,2)andA\text{'}\equiv (x_2,2)$
As we know slope of line passing through $(x_1,y_1)and(x_2,y_2)$  is given by  $m=\frac{y_2-y_1}{x_2-x_1}$

Also, we know that if angle between two lines having slope  $m_{1}and{m}_{2}is\theta ,$ then 
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$$\Rightarrow \tan \frac{\pi }{4}=\left|\frac{\frac{2}{x_1}-\frac{3}{2}}{1+\left(\frac{2}{x_1}\right)\left(\frac{3}{2}\right)}\right|\Rightarrow \left|\frac{\frac{2}{x_1}-\frac{3}{2}}{1+\frac{3}{x_1}}\right|\Rightarrow 1=\left|\frac{(4-3x_1)}{2(x_1+3)}\right|\Rightarrow \pm 1=\frac{(4-3x_1)}{2(x_1+3)}$ 
$\Rightarrow 4-3x_1=\pm (2x_1+6)\Rightarrow 4-3x_1=2x_1+6and4-3x=-2x_1-6$ 
 $\Rightarrow 5x_1=-2and{x}_1=10\Rightarrow x_1=\frac{-2}{5}and{x}_1=10$
$\therefore x_1=10and{x}_2=\frac{-2}{5}\Rightarrow A\equiv (10,2)\&A\text{'}\equiv (\frac{-2}{5},2)$ 
$\therefore$ by distance formula,  $AA\text{'}=\sqrt{{(10+\frac{2}{5})}^2+{(2-2)}^2}$
 $\Rightarrow AA\text{'}=\frac{52}{5}units$