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

Options

A. |a| = |b|
B. |a| = |c|
C. |b| = |c|
D. None of these

Answer

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: Rewrite the Line Equation
We start with the line equation:
\[ ax + by + c = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ by = -ax - c \quad \Rightarrow \quad y = -\frac{a}{b}x - \frac{c}{b} \]

### Step 2: Find the Intercepts
To find the x-intercept and y-intercept of the line, we set \( y = 0 \) for the x-intercept and \( x = 0 \) for the y-intercept.

- **X-Intercept**: Set \( y = 0 \):
\[ ax + c = 0 \quad \Rightarrow \quad x = -\frac{c}{a} \]

- **Y-Intercept**: Set \( x = 0 \):
\[ by + c = 0 \quad \Rightarrow \quad y = -\frac{c}{b} \]

### Step 3: Form the Triangle
The x-intercept and y-intercept form a triangle with the origin (0,0). The vertices of the triangle are:
- A (x-intercept, 0) = \(\left(-\frac{c}{a}, 0\right)\)
- B (0, y-intercept) = \(\left(0, -\frac{c}{b}\right)\)
- O (origin) = (0, 0)

### Step 4: Condition for Isosceles Triangle
For the triangle AOB to be isosceles, the lengths OA and OB must be equal:
\[ OA = OB \]
Where:
- \( OA = \left| -\frac{c}{a} \right| \)
- \( OB = \left| -\frac{c}{b} \right| \)

Setting these equal gives:
\[ \left| -\frac{c}{a} \right| = \left| -\frac{c}{b} \right| \]
This simplifies to:
\[ \frac{|c|}{|a|} = \frac{|c|}{|b|} \]

Assuming \( c \neq 0 \) (to avoid division by zero), we can cancel \( |c| \) from both sides:
\[ \frac{1}{|a|} = \frac{1}{|b|} \quad \Rightarrow \quad |a| = |b| \]

### Conclusion
Thus, the condition for the triangle formed by the line \( ax + by + c = 0 \) and the coordinate axes to be isosceles is:
\[ \boxed{|a| = |b|} \]