If a, c, b are in G.P then the line `ax + by + c= 0`

Options

A. has a fixed direction
B. always passes through a fixed point
C. forms a triangle with the axes whose area is constant
D. always cuts intercepts on the axes such that their sum is zero

Answer

To solve the problem, we need to analyze the given condition that \( a, b, c \) are in Geometric Progression (G.P.) and how it relates to the line equation \( ax + by + c = 0 \).

### Step-by-Step Solution:

1. **Understanding G.P. Condition**:
- If \( a, b, c \) are in G.P., then the condition is given by:
\[ c^2 = ab \]

2. **Line Equation**:
- The line equation is given as:
\[ ax + by + c = 0 \]
- We can rearrange this to express \( y \) in terms of \( x \):
\[ by = -ax - c \quad \Rightarrow \quad y = -\frac{a}{b}x - \frac{c}{b} \]

3. **Finding Intercepts**:
- The x-intercept occurs when \( y = 0 \):
\[ ax + c = 0 \quad \Rightarrow \quad x = -\frac{c}{a} \]
- The y-intercept occurs when \( x = 0 \):
\[ by + c = 0 \quad \Rightarrow \quad y = -\frac{c}{b} \]

4. **Area of Triangle Formed**:
- The area \( A \) of the triangle formed by the line and the axes can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, the base is the x-intercept \( -\frac{c}{a} \) and the height is the y-intercept \( -\frac{c}{b} \):
\[ A = \frac{1}{2} \times \left(-\frac{c}{a}\right) \times \left(-\frac{c}{b}\right) = \frac{1}{2} \times \frac{c^2}{ab} \]

5. **Substituting G.P. Condition**:
- Since \( c^2 = ab \) (from the G.P. condition), we can substitute this into the area formula:
\[ A = \frac{1}{2} \times \frac{ab}{ab} = \frac{1}{2} \]

6. **Conclusion**:
- The area of the triangle formed by the line \( ax + by + c = 0 \) and the coordinate axes is constant and equal to \( \frac{1}{2} \).

### Final Answer:
The area of the triangle formed by the line \( ax + by + c = 0 \) is constant and equal to \( \frac{1}{2} \).

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