If `bx+cy=a`, there `a, b, c` are of the same sign, be a line such that the area enclosed by the line and the axes of reference is `1/8` square units, then : (A) `b,a,c` are in G.P. (B) `b, 2a, c` arein G.P. (C) `b, a/2, c` are in A.P. (D) `b, -2a, c` are in G.P.

Options

A. `b, a, c` and `G.P.`
B. `b, 2a, -c` are in `G.P.`
C. `b, a//2, c` are in `A.P.`
D. `b, +2a, c` are in `G.P.`

Explanation

To solve the problem, we start with the equation of the line given by:

\[ bx + cy = a \]

### Step 1: Find the intercepts
To find the intercepts of the line with the axes, we set \(y = 0\) to find the x-intercept and \(x = 0\) to find the y-intercept.

1. **X-intercept**: Set \(y = 0\):
\[ bx + c(0) = a \implies x = \frac{a}{b} \]
So, the x-intercept is \(\left(\frac{a}{b}, 0\right)\).

2. **Y-intercept**: Set \(x = 0\):
\[ b(0) + cy = a \implies y = \frac{a}{c} \]
So, the y-intercept is \(\left(0, \frac{a}{c}\right)\).

### Step 2: Calculate the area of the triangle formed by the line and the axes
The area \(A\) of the triangle formed by the x-axis, y-axis, and the line can be calculated using the formula for the area of a triangle:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is the x-intercept \(\frac{a}{b}\) and the height is the y-intercept \(\frac{a}{c}\):

\[ A = \frac{1}{2} \times \frac{a}{b} \times \frac{a}{c} = \frac{a^2}{2bc} \]

### Step 3: Set the area equal to \(\frac{1}{8}\)
We know from the problem statement that the area is \(\frac{1}{8}\):

\[ \frac{a^2}{2bc} = \frac{1}{8} \]

### Step 4: Solve for \(bc\)
Cross-multiplying gives:

\[ 8a^2 = 2bc \implies bc = 4a^2 \]

### Step 5: Analyze the relationship between \(b\), \(a\), and \(c\)
We have \(bc = 4a^2\). We can express this relationship in terms of geometric progression (G.P.) and arithmetic progression (A.P.).

1. **For G.P.**: If \(b\), \(2a\), and \(c\) are in G.P., then:
\[ b^2 = 2a \cdot c \]
From \(bc = 4a^2\), we can express \(c\) as \(c = \frac{4a^2}{b}\). Substituting this into the G.P. condition gives:
\[ b^2 = 2a \cdot \frac{4a^2}{b} \implies b^3 = 8a^3 \implies b = 2a \]
Thus, \(b\), \(2a\), and \(c\) can be in G.P.

2. **For G.P.**: If \(b\), \(-2a\), and \(c\) are in G.P., we can similarly check:
\[ b^2 = -2a \cdot c \]
This leads to a valid condition as well.

### Conclusion
Thus, the correct options are:
- (B) \(b, 2a, c\) are in G.P.
- (D) \(b, -2a, c\) are in G.P.