To solve this, let us analyze the geometry of the problem step by step:

---

1. Points on the line $y = 2$

Any point on the line $y = 2$ can be represented as: $A(x, 2) \quad \text{and} \quad A'(x', 2)$

---

2. Conditions for subtending $\frac{\pi}{4}$ at the origin

The line segments $AB$ and $A'B$, where $B(2, 3)$, subtend an angle of $\frac{\pi}{4}$ at the origin. This means the angle between the position vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$, as well as $\overrightarrow{OA'}$ and $\overrightarrow{OB}$, is $\frac{\pi}{4}$.

Using the dot product formula, for $\overrightarrow{OA}$ and $\overrightarrow{OB}$: $\cos \theta = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}| |\overrightarrow{OB}|}, \quad \text{where } \theta = \frac{\pi}{4}.$ Thus: $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \implies \overrightarrow{OA} \cdot \overrightarrow{OB} = \frac{|\overrightarrow{OA}| |\overrightarrow{OB}|}{\sqrt{2}}.$

Let $A(x, 2)$. Then: $\overrightarrow{OA} = (x, 2), \quad \overrightarrow{OB} = (2, 3).$

---

3. Dot Product Condition

Using $\overrightarrow{OA} \cdot \overrightarrow{OB} = \frac{|\overrightarrow{OA}| |\overrightarrow{OB}|}{\sqrt{2}}$: $x \cdot 2 + 2 \cdot 3 = \frac{\sqrt{x^2 + 2^2} \cdot \sqrt{2^2 + 3^2}}{\sqrt{2}}.$

Simplify: $2x + 6 = \frac{\sqrt{x^2 + 4} \cdot \sqrt{13}}{\sqrt{2}}.$

Squaring both sides: $(2x + 6)^2 = \frac{13(x^2 + 4)}{2}.$

Expanding: $4x^2 + 24x + 36 = \frac{13x^2 + 52}{2}.$

Multiply through by 2: $8x^2 + 48x + 72 = 13x^2 + 52.$

Rearrange: $0 = 5x^2 - 48x - 20.$

---

4. Solving the Quadratic Equation

The quadratic equation is: $5x^2 - 48x - 20 = 0.$

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 5$, $b = -48$, and $c = -20$: $x = \frac{-(-48) \pm \sqrt{(-48)^2 - 4(5)(-20)}}{2(5)}.$

Simplify: $x = \frac{48 \pm \sqrt{2304 + 400}}{10}.$ $x = \frac{48 \pm \sqrt{2704}}{10}.$ $x = \frac{48 \pm 52}{10}.$

Thus: $x = \frac{48 + 52}{10} = 10, \quad x = \frac{48 - 52}{10} = -0.4.$

---

5. Distance Between $A$ and $A'$

The points $A$ and $A'$ are: $A(10, 2), \quad A'(-0.4, 2).$

The distance between them is: $\text{Distance} = \sqrt{(10 - (-0.4))^2 + (2 - 2)^2}.$ $\text{Distance} = \sqrt{(10.4)^2}.$ $\text{Distance} = 10.4.$

---

Final Answer:

The distance between $A$ and $A'$ is: $\boxed{10.4}.$

Would you like a detailed explanation of any specific step?

---

5 Related Questions:

1. What is the geometric interpretation of subtending an angle at a point?
2. How does the dot product help in finding angles between vectors?
3. Can we solve this problem using a coordinate transformation?
4. How is the quadratic formula derived?
5. What are some practical applications of subtended angles in geometry?

Tip:

When working with subtended angles, always carefully verify the geometry of the problem to avoid missing constraints.