Question:
Find the distance between the parallel lines 4x – 3y + 5 = 0 and 4x – 3y + 7 = 0
Solution:
Given: parallel lines $4 x-3 y+5=0$ and $4 x-3 y+7=0$
To find : distance between the parallel lines
Formula used: The distance between the parallel lines $a x+b y+c=0$ and $a x+b y+d$ $=0$ is,
$D=\frac{|d-c|}{\sqrt{a^{2}+b^{2}}}$
Here $a=4, b=-3, c=5, d=7$
$D=\frac{|7-5|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{|2|}{\sqrt{16+9}}=\frac{2}{\sqrt{25}}=\frac{2}{5}$
The distance between the parallel lines $4 x-3 y+5=0$ and $4 x-3 y+7=0$ is $\frac{2}{5}$ units