To solve the problem, we will follow these steps:
### Step 1: Identify the Points
We have two fixed points A(3, 0) and B(6, 0). The variable point U is represented as U(x₁, y₁).
### Step 2: Find the Coordinates of Points C and D
1. **Finding Point C (intersection of AU with the y-axis)**:
- The line AU can be expressed using the two-point form of the equation of a line:
\[
y - y_1 = \frac{0 - y_1}{3 - x_1}(x - x_1)
\]
- To find C, we set \(x = 0\):
\[
y - y_1 = \frac{-y_1}{3 - x_1}(0 - x_1)
\]
\[
y - y_1 = \frac{y_1 x_1}{3 - x_1}
\]
\[
y = y_1 + \frac{y_1 x_1}{3 - x_1}
\]
\[
y = \frac{y_1 (3 - x_1 + x_1)}{3 - x_1} = \frac{3y_1}{3 - x_1}
\]
- Thus, the coordinates of C are:
\[
C(0, \frac{3y_1}{3 - x_1})
\]
2. **Finding Point D (intersection of BU with the y-axis)**:
- The line BU can be expressed similarly:
\[
y - y_1 = \frac{0 - y_1}{6 - x_1}(x - x_1)
\]
- Setting \(x = 0\) gives:
\[
y - y_1 = \frac{-y_1}{6 - x_1}(0 - x_1)
\]
\[
y - y_1 = \frac{y_1 x_1}{6 - x_1}
\]
\[
y = y_1 + \frac{y_1 x_1}{6 - x_1}
\]
\[
y = \frac{y_1 (6 - x_1 + x_1)}{6 - x_1} = \frac{6y_1}{6 - x_1}
\]
- Thus, the coordinates of D are:
\[
D(0, \frac{6y_1}{6 - x_1})
\]
### Step 3: Find the Equation of Line AD
- Using points A(3, 0) and D(0, \frac{6y_1}{6 - x_1}), the equation of line AD can be derived:
\[
\frac{y - 0}{\frac{6y_1}{6 - x_1} - 0} = \frac{x - 3}{0 - 3}
\]
This simplifies to:
\[
\frac{y}{\frac{6y_1}{6 - x_1}} = \frac{x - 3}{-3}
\]
Cross-multiplying gives:
\[
y(6 - x_1) = -2(x - 3)
\]
This is the equation of line AD.
### Step 4: Find the Equation of Line OU
- The line OU can be expressed as:
\[
y_1 x = x_1 y
\]
or
\[
y = \frac{y_1}{x_1} x
\]
### Step 5: Find Intersection Point V of Lines AD and OU
- Solving the equations of lines AD and OU simultaneously will yield the coordinates of point V.
### Step 6: Find the Equation of Line CV
- Using points C(0, \frac{3y_1}{3 - x_1}) and V, we can find the equation of line CV.
### Step 7: Identify the Fixed Point (p, q)
- By analyzing the equation of line CV, we find that it passes through a fixed point (2, 0) regardless of the position of U.
### Step 8: Calculate the Distance from the Origin
- The distance from the origin (0, 0) to the point (2, 0) is:
\[
\sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{4} = 2 \text{ units}
\]
### Conclusion
The distance from the origin to the fixed point (p, q) is **2 units**.
