Key Concepts and Formulas

• Slope of a line: The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Angle between two lines: If two lines have slopes $m_1$ and $m_2$, the tangent of the angle $\theta$ between them is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
• Distance formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Define the coordinates of the relevant points. Let the coordinates of point A be $(\alpha, 2)$, since it lies on the line $y=2$. The given point B is $(2, 3)$. The origin O is $(0, 0)$.

Step 2: Calculate the slopes of the lines OA and OB.

• Slope of line OA ($m_{OA}$): This line passes through $O(0,0)$ and $A(\alpha, 2)$. $m_{OA} = \frac{2 - 0}{\alpha - 0} = \frac{2}{\alpha}$ Explanation: We apply the slope formula.

• Slope of line OB ($m_{OB}$): This line passes through $O(0,0)$ and $B(2, 3)$. $m_{OB} = \frac{3 - 0}{2 - 0} = \frac{3}{2}$ Explanation: We apply the slope formula.

Step 3: Apply the angle formula between lines OA and OB. We are given that the angle $\theta$ between line OA and line OB is $\frac{\pi}{4}$. Using the angle formula: $\tan \theta = \left| \frac{m_{OA} - m_{OB}}{1 + m_{OA} m_{OB}} \right|$ Substitute $\theta = \frac{\pi}{4}$, $m_{OA} = \frac{2}{\alpha}$, and $m_{OB} = \frac{3}{2}$: $\tan \left( \frac{\pi}{4} \right) = \left| \frac{\frac{2}{\alpha} - \frac{3}{2}}{1 + \left(\frac{2}{\alpha}\right) \cdot \left(\frac{3}{2}\right)} \right|$ Since $\tan \left( \frac{\pi}{4} \right) = 1$: $1 = \left| \frac{\frac{4 - 3\alpha}{2\alpha}}{1 + \frac{3}{\alpha}} \right| = \left| \frac{4 - 3\alpha}{2\alpha + 6} \right|$ Explanation: We substitute the known values into the tangent formula.

Step 4: Solve for $\alpha$. We have two cases: Case 1: $\frac{4 - 3\alpha}{2\alpha + 6} = 1$ $4 - 3\alpha = 2\alpha + 6$ $-2 = 5\alpha$ $\alpha = -\frac{2}{5}$

Case 2: $\frac{4 - 3\alpha}{2\alpha + 6} = -1$ $4 - 3\alpha = -2\alpha - 6$ $10 = \alpha$ $\alpha = 10$ Explanation: Because of the absolute value, we have two possible equations to solve.

Step 5: Determine the coordinates of A and A'. We have two possible x-coordinates for the points on the line $y = 2$. Thus, the points are $A = (-\frac{2}{5}, 2)$ and $A' = (10, 2)$.

Step 6: Calculate the distance between A and A'. Using the distance formula: $AA' = \sqrt{\left(10 - \left(-\frac{2}{5}\right)\right)^2 + (2 - 2)^2} = \sqrt{\left(10 + \frac{2}{5}\right)^2 + 0^2} = \sqrt{\left(\frac{52}{5}\right)^2} = \frac{52}{5}$ Explanation: Applying the distance formula to find the distance between the two possible locations for A.

Step 7: Check if B can also be A or A'. If B was also equal to A or A', the problem statement would be invalid. If A = B, then $(\alpha, 2) = (2,3)$, implying $\alpha = 2$ and $2 = 3$, which is impossible. If A' = B, then $(\alpha', 2) = (2,3)$, implying $\alpha' = 2$ and $2 = 3$, which is impossible. Therefore, we do not have to worry about the possibility of A or A' being identical to B.

Common Mistakes & Tips

• Remember the absolute value in the angle formula. This leads to two cases to solve.
• Be careful with algebraic manipulations, especially when dealing with fractions.
• Always check if the solution makes sense in the context of the problem.

Summary

We found the coordinates of points A and A' on the line y=2 such that the line segments AB and A'B subtend an angle of $\frac{\pi}{4}$ at the origin. We used the formula for the tangent of the angle between two lines to solve for the x-coordinates of A and A'. Finally, we calculated the distance between A and A' using the distance formula, which resulted in $\frac{52}{5}$.

The final answer is $\boxed{\frac{52}{5}}$, which corresponds to option (C).