The straight line `ax+by+c=0` and the coordinate axes form an isosceles triangle under which one of the following consitions ?

Options

A. a,b,c are in G.P.
B. a,c,b are in G.P.
C. c,a,b are in G.P.
D. none of these

Explanation

To determine the condition under which the straight line \( ax + by + c = 0 \) and the coordinate axes form an isosceles triangle, we can follow these steps:

### Step 1: Find the intercepts of the line with the axes
To find the x-intercept, set \( y = 0 \):
\[ ax + c = 0 \implies x = -\frac{c}{a} \]
Thus, the x-intercept is \( A\left(-\frac{c}{a}, 0\right) \).

To find the y-intercept, set \( x = 0 \):
\[ by + c = 0 \implies y = -\frac{c}{b} \]
Thus, the y-intercept is \( B\left(0, -\frac{c}{b}\right) \).

### Step 2: Determine the lengths of the sides of the triangle
The triangle formed by the line and the coordinate axes has vertices at the origin \( O(0, 0) \), \( A\left(-\frac{c}{a}, 0\right) \), and \( B\left(0, -\frac{c}{b}\right) \).

The lengths of the sides from the origin to the intercepts are:
- Length \( OA = \left| -\frac{c}{a} \right| = \frac{|c|}{|a|} \)
- Length \( OB = \left| -\frac{c}{b} \right| = \frac{|c|}{|b|} \)

### Step 3: Set the condition for the triangle to be isosceles
For the triangle \( OAB \) to be isosceles, the lengths \( OA \) and \( OB \) must be equal:
\[ \frac{|c|}{|a|} = \frac{|c|}{|b|} \]

### Step 4: Simplify the condition
Assuming \( |c| \neq 0 \) (as the line must not pass through the origin), we can cancel \( |c| \) from both sides:
\[ \frac{1}{|a|} = \frac{1}{|b|} \implies |a| = |b| \]

### Step 5: Relate this to the geometric progression
The condition \( |a| = |b| \) can be expressed in terms of the coefficients \( a, b, c \). For the triangle to be isosceles, we can also consider the relationship among \( a, b, c \) in terms of geometric progression.

If \( c^2 = ab \), then \( a, c, b \) are in geometric progression.

### Conclusion
Thus, the condition under which the straight line \( ax + by + c = 0 \) and the coordinate axes form an isosceles triangle is:
\[ c^2 = ab \]