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The distance between the two points $\mathrm{{A \: and \:A^{\prime}}$ which lie on $\mathrm{y=2}$ such that both the line segments $\mathrm{AB}$ and $\mathrm{A}^{\prime} \mathrm{B}$ (where $\mathrm{B}$ is the point $(2,3)$ ) subtend angle $\frac{\pi}{4}$ at the origin, is equal to :
Option: 1
$10$

Option: 2
$\frac{48}{5}$

Option: 3
$\frac{52}{5}$

Option: 4
$3$

Answers (1)

best_answer

$\mathrm{Let \: A \: be (p, 2) }\\$

$\mathrm{Slope \: of \: O A=m_{1}=\frac{2}{p}}\\$

$\mathrm{Slope\: of \: O B=m_{2}=\frac{3}{2}}$

$\mathrm{\text{Angle between these} =\frac{\pi}{4}}\\$

$\mathrm{\therefore \tan \left(\frac{\pi}{4}\right)=\left|\frac{\frac{3}{2}-\frac{2}{p}}{1+\frac{3}{2} \cdot \frac{2}{p}}\right| }$

$\mathrm{\Rightarrow 1=\left|\frac{3 p-4}{2 p+6}\right|} \\$

$\mathrm{\Rightarrow \frac{3 p-4}{2 p+6}=1 \text { or } \frac{3 p-4}{2 p+6}=-1 }$

$\mathrm{\Rightarrow 3 p-4=2 p+6 \text { or } 3 p-4=-2 p-6} \\$

$\mathrm{\Rightarrow p=10 \quad \text { or } 5 p=-2} \\$

$\mathrm{\Rightarrow p=10 \text { or } p=-\frac{2}{5}} \\$

$\mathrm{\therefore A A^{\prime}=10+\frac{2}{5}=\frac{52}{5} }$

Hence the correct answer is option 3.

Posted by

shivangi.shekhar

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