Q73 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Mathematics
The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is equal to units
- A\(\frac{1}{10}\)
- Correct answerB\(\frac{3}{10}\)
- C\(\frac{5}{10}\)
- D\(\frac{7}{10}\)
Solution
The distance between given parallel lines
\(\begin{aligned}
3 x+4 y-9 & =0 \\
6 x+8 y-15 & =0 \\
\text{is } \frac{18-15}{\sqrt{36+64}} & =\frac{3}{10} \text { unit }
\end{aligned}\)
Hence, option (b) is correct.
Hence, the correct answer is option: 'B'.