Question

If \( a, c, b \) are in GP, then the area of the triangle formed by the lines \( ax + by + c = 0 \) with the coordinate axes is equal to:

• A. 1
• B. 2
• C. \( \frac{1}{2} \)
• D. None of these

Hint:

When given a triangle formed by the coordinate axes and a line, you can use the formula for the area of a triangle to calculate it. If the coefficients of the line are in geometric progression, use the relationship between the coefficients to simplify the area expression.

Answer 1

The Correct Option is C

Given \( a, c, b \) are in GP, so \( c^2 = ab \).
The area of the triangle formed by the line \( ax + by + c = 0 \) and the coordinate axes can be found using the formula for the area of a triangle formed by two lines intersecting the axes at \( x = \frac{-c}{a} \) and \( y = \frac{-c}{b} \). The area of the triangle \( AOB \) is: \[ {Area} = \frac{1}{2} \times \left( \frac{-c}{b} \right) \times \left( \frac{-c}{a} \right) = \frac{1}{2} \times \frac{c^2}{ab} = \frac{1}{2} \times \frac{c^2}{ab} = \frac{1}{2} { (using } c^2 = ab). \] Thus, the area is \( \frac{1}{2} \).

Answer 2

Step 1: Given the lines
We are given the lines in the form \( ax + by + c = 0 \), where \( a \), \( b \), and \( c \) are in geometric progression (GP). The equation of the line is: \[ ax + by + c = 0 \]
Step 2: Find the intercepts
The intercepts of the line with the coordinate axes can be found by setting \( x = 0 \) and \( y = 0 \) respectively. - x-intercept: Set \( y = 0 \) in the equation \( ax + by + c = 0 \): \[ ax + c = 0 \quad \Rightarrow \quad x = -\frac{c}{a} \] So, the x-intercept is \( \left( -\frac{c}{a}, 0 \right) \). - y-intercept: Set \( x = 0 \) in the equation \( ax + by + c = 0 \): \[ by + c = 0 \quad \Rightarrow \quad y = -\frac{c}{b} \] So, the y-intercept is \( \left( 0, -\frac{c}{b} \right) \).
Step 3: Area of the Triangle
The area \( A \) of the triangle formed by the line and the coordinate axes is given by the formula: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] Here, the base is the x-intercept \( \left( -\frac{c}{a}, 0 \right) \), and the height is the y-intercept \( \left( 0, -\frac{c}{b} \right) \). Thus, the area becomes: \[ A = \frac{1}{2} \times \left( -\frac{c}{a} \right) \times \left( -\frac{c}{b} \right) \] \[ A = \frac{1}{2} \times \frac{c^2}{ab} \]
Step 4: Use the Given Condition (GP Condition)
Since \( a \), \( b \), and \( c \) are in geometric progression, we have: \[ \frac{b}{a} = \frac{c}{b} \quad \Rightarrow \quad b^2 = ac \] Now substitute \( b^2 = ac \) into the area formula: \[ A = \frac{1}{2} \times \frac{c^2}{ab} = \frac{1}{2} \times \frac{c^2}{\sqrt{ac} \cdot \sqrt{ac}} = \frac{1}{2} \]
Step 5: Conclusion
Thus, the area of the triangle formed by the lines is: \[ \boxed{\frac{1}{2}} \]