Question

The distance between the lines \( 3x + 4y + 1 = 0 \) and \( 6x + 8y - 1 = 0 \) is

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Hint:

For parallel lines, the distance formula simplifies to \( \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} \).

Answer 1

The Correct Option is C

The general formula for the distance between two parallel lines \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \) is: \[ \text{Distance} = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}. \] For the lines \( 3x + 4y + 1 = 0 \) and \( 6x + 8y - 1 = 0 \), the coefficients of \( x \) and \( y \) are proportional. The lines are parallel, so we can use the distance formula for parallel lines. Thus, the distance is: \[ \frac{|(-1) - 1|}{\sqrt{3^2 + 4^2}} = \frac{2}{\sqrt{9 + 16}} = \frac{2}{5} = 0.(3) \]