# Find the distance between parallel lines

Question:

Find the distance between parallel lines

(i) $15 x+8 y-34=0$ and $15 x+8 y+31=0$

(ii) $I(x+y)+p=0$ and $I(x+y)-r=0$

Solution:

It is known that the distance $(d)$ between parallel lines $A x+B y+C_{1}=0$ and $A x+B y+C_{2}=0$ is given by $d=\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}$.

(i) The given parallel lines are $15 x+8 y-34=0$ and $15 x+8 y+31=0$.

Here, $A=15, B=8, C_{1}=-34$, and $C_{2}=31$.

Therefore, the distance between the parallel lines is

$d=\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}=\frac{|-34-31|}{\sqrt{(15)^{2}+(8)^{2}}}$ units $=\frac{|-65|}{17}$ units $=\frac{65}{17}$ units

(ii) The given parallel lines are $I(x+y)+p=0$ and $I(x+y)-r=0$.

$I x+l y+p=0$ and $l x+l y-r=0$

Here, $A=l, B=l, C_{1}=p$, and $C_{2}=-r$.

Therefore, the distance between the parallel lines is

$d=\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}=\frac{|p+r|}{\sqrt{l^{2}+l^{2}}}$ units $=\frac{|p+r|}{\sqrt{2 l^{2}}}$ units $=\frac{|p+r|}{I \sqrt{2}}$ units $=\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right|$ units