Question

Find the distance between the parallel lines $3x–4y+7=0\text{and}3x–4y+5=0.$

Answer 1

Required distance

Given Equation $3x–4y+7=0\text{and}3x–4y+5=0$

Above equations are of the form
$Ax+By+C_1=0$ & $Ax+By+C_2=0$

Where $A=3,B=-4,C_1=7$ and $C_2=5$

We know that distance between two parallel lines $Ax+By+C_1=0$ & $Ax+By+C_2=0$ is

$d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$

$d=\frac{|7-5|}{\sqrt{(3{)}^2+(-4{)}^2}}$

$\Rightarrow d=\frac{|2|}{\sqrt{9+16}}$

$\Rightarrow d=\frac{2}{\sqrt{25}}$

$\Rightarrow d=\frac{2}{\sqrt{5\times 5}}$

$\Rightarrow d=\frac{2}{5}$

$\therefore$ The required distance is $\frac{2}{5}$ units.