The distance between the lines $3 x+4 y=9$ and $6 x+8 y=15$ is
The distance between the lines $3 x+4 y=9$ and $6 x+8 y=15$ is
$\frac{3}{2}$
$\frac{3}{10}$
$6$
$\frac{3}{5}$
Solution
To find, distance between the lines $3 x+4 y=9$ and $6 x+8 y=15$, since we know that distance between the lines $a x+b y+c_1=0$ and $a x+b y+c_2=0$
$\Rightarrow \quad d=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$
Here lines are $3 x+4 y-9=0$
and $3 x+4 y-\frac{15}{2}=0$
$\therefore \quad d=\left|\frac{15 / 2-9}{\sqrt{3^2+4^2}}\right|$
$\begin{aligned} & d=\left|\frac{15-18}{2 \times 5}\right| \\ & d=\frac{3}{10} \text { units }\end{aligned}$