The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is equal to units

The distance between the lines \(3 x+4 y=9\) and \(6 x+8 y=15\) is equal to units
  1. \(\frac{1}{10}\)
  2. \(\frac{3}{10}\)
  3. \(\frac{5}{10}\)
  4. \(\frac{7}{10}\)

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.

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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