If r is the geometric mean of p and q, then the line `px+qy+r=0`

Options

A. has a fixed direction
B. passes thorugh a fixed point
C. forms with the axs a triangle of constant area
D. sum of its intercepts on the axes is constant

Answer

To solve the problem, we need to determine the area of the triangle formed by the line \( px + qy + r = 0 \) and the coordinate axes, given that \( r \) is the geometric mean of \( p \) and \( q \).

### Step-by-Step Solution:

1. **Identify the Geometric Mean**:
Since \( r \) is the geometric mean of \( p \) and \( q \), we have:
\[ r = \sqrt{pq} \]
Therefore, squaring both sides gives:
\[ r^2 = pq \]

2. **Find the Intercepts**:
To find the intercepts of the line \( px + qy + r = 0 \):
- **X-intercept**: Set \( y = 0 \):
\[ px + r = 0 \implies x = -\frac{r}{p} \]
Thus, the x-intercept is \( A \left(-\frac{r}{p}, 0\right) \).

- **Y-intercept**: Set \( x = 0 \):
\[ qy + r = 0 \implies y = -\frac{r}{q} \]
Thus, the y-intercept is \( B \left(0, -\frac{r}{q}\right) \).

3. **Calculate the Area of Triangle OAB**:
The area \( A \) of triangle OAB formed by the x-axis, y-axis, and the line can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( -\frac{r}{p} \) and the height is \( -\frac{r}{q} \):
\[ \text{Area} = \frac{1}{2} \times \left(-\frac{r}{p}\right) \times \left(-\frac{r}{q}\right) = \frac{1}{2} \times \frac{r^2}{pq} \]

4. **Substitute \( r^2 \)**:
Now, substitute \( r^2 \) with \( pq \):
\[ \text{Area} = \frac{1}{2} \times \frac{pq}{pq} = \frac{1}{2} \]

5. **Conclusion**:
The area of triangle OAB is a constant value:
\[ \text{Area} = \frac{1}{2} \]

### Final Answer:
The area of the triangle formed by the line \( px + qy + r = 0 \) and the coordinate axes is \( \frac{1}{2} \).