Question

Find the distance between parallel lines
(i)$15x+8y-34$ $=0$ and $15x+8y+31=0$
(ii)$(x+y)+p$ $=0$ and $(x+y)-r=0$

Answer 1

(i) The given parallel lines are $15x+8y-34=0$ and $15x+8y+31=0$
Here, $A=15,B=8$ ,$C_1=-34,$ and $C_2=31$
Thus distance between them is given by,
$d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}=\frac{|-34-31|}{\sqrt{(15{)}^2+(8{)}^2}}units=\frac{|-65|}{17}units=\frac{65}{17}units$
(ii)The given parallel lines are $l(x+y)+p=0$ and $l(x+y)-r=0.$
Here $A=l,B=l,C_1=p,$ and $C_2=-r$
Therefore, the distance between these parallel lines is
$d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}=\frac{|p+r|}{\sqrt{l^2+l^2}}units=\frac{|p+r|}{\sqrt{2l^2}}units$$=\frac{|p+r|}{l\sqrt{2}}units=\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right|units$