JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 10)
The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle $${\pi \over 4}$$ at the origin, is equal to :
10
$${48 \over 5}$$
$${52 \over 5}$$
3
Explanation
Let $A(\alpha, 2) \quad$ Given $B(2,3)$
$$ \begin{aligned} & m_{O A}=\frac{2}{\alpha} \quad\&\quad m_{O B}=\frac{3}{2} \\\\ & \tan \frac{\pi}{4}=\left|\frac{\frac{2}{\alpha}-\frac{3}{2}}{1+\frac{2}{\alpha} \cdot \frac{3}{2}}\right| \Rightarrow \frac{4-3 \alpha}{2 \alpha+6}=\pm 1 \\\\ & 4-3 \alpha=2 \alpha+6 \quad \& 4-3 \alpha=-2 \alpha-6 \\\\ & \alpha=\frac{-2}{5} \& \alpha=10 \\\\ & A\left(-\frac{2}{5}, 2\right) \& A^{\prime}(10,2) \text { and } B(2,3) \\\\ & A A^{\prime}=10+\frac{2}{5}=\frac{52}{5} \end{aligned} $$
$$ \begin{aligned} & m_{O A}=\frac{2}{\alpha} \quad\&\quad m_{O B}=\frac{3}{2} \\\\ & \tan \frac{\pi}{4}=\left|\frac{\frac{2}{\alpha}-\frac{3}{2}}{1+\frac{2}{\alpha} \cdot \frac{3}{2}}\right| \Rightarrow \frac{4-3 \alpha}{2 \alpha+6}=\pm 1 \\\\ & 4-3 \alpha=2 \alpha+6 \quad \& 4-3 \alpha=-2 \alpha-6 \\\\ & \alpha=\frac{-2}{5} \& \alpha=10 \\\\ & A\left(-\frac{2}{5}, 2\right) \& A^{\prime}(10,2) \text { and } B(2,3) \\\\ & A A^{\prime}=10+\frac{2}{5}=\frac{52}{5} \end{aligned} $$
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