Question

The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is:

• A. $\frac{3}{2}$
• B. $\frac{3}{10}$
• C. 6
• D. $\frac{9}{4}$

Answer 1

Answer: (B)

Let $d_1$ and $d_2$ be the distances of two lines 3x + 4y - 9 = 0 and 6x + 8y - 15 = 0 respectively from origin.
$\therefore d_1=\frac{|3\left(0\right)+4\left(0\right)-9|}{\sqrt{3^2+4^2}}\Rightarrow d_1=\frac{9}{5}$
and $d_2=\frac{|6\left(0\right)+8\left(0\right)-15|}{\sqrt{36+64}}=\frac{15}{10}=\frac{3}{2}$
$\therefore$ distance between these lines is, $d=d_1-d_2$
$\Rightarrow d=\frac{9}{5}-\frac{3}{2}=\frac{18-15}{10}=\frac{3}{10}$