Q. The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is:
Straight Lines
Solution:
Let $d_1$ and $d_2$ be the distances of two lines 3x + 4y - 9 = 0 and 6x + 8y - 15 = 0 respectively from origin.
$\therefore d_{1} = \frac{\left|3\left(0\right)+4\left(0\right)-9\right|}{\sqrt{3^{2}+4^{2}}} \Rightarrow d_{1} = \frac{9}{5}$
and $ d_{2} = \frac{\left|6\left(0\right)+8\left(0\right)-15\right|}{\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} $
