The distance between the lines given by $3 x+4 \mathrm{y}=9$ and $6 x+8 \mathrm{y}=15$ is
The distance between the lines given by $3 x+4 \mathrm{y}=9$ and $6 x+8 \mathrm{y}=15$ is
$5$ units
3 units
$0\cdot 5$ units
$0\cdot 3$ units
Solution
Given parallel lines are
$3 x+4 y=9 \Rightarrow 6 x+8 y=18$ and $6 x+8 y=15$
Distance between them is
$d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|=\frac{|18-15|}{\sqrt{36+64}}=\frac{3}{10}=0.3$