Question

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

Answer 1

It is known that the distance (d) between parallel lines $Ax + By + C1 = 0$ and $Ax + By + C2 = 0$ is given by
$d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$

(i) The given parallel lines are $15x + 8y \- 34 = 0$ and $15x + 8y + 31 = 0.$
Here, $A = 15, B = 8, C1 = -34$, and $C2 = 31$.
Therefore, the distance between the parallel lines is
$d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$

$=\frac{|-34-31|}{\sqrt{(15)^2+(8)^2}}$ units

$=\frac{|-65|}{17}$ units

$=\frac{65}{17}$ units.

(ii) The given parallel lines are $l (x + y) + p = 0$ and $l (x + y) \- r = 0.$
$lx + ly + p = 0$ and $lx + ly \- r = 0$
Here, $A = l, B = l, C_1 = p$, and $C_2 = - r.$
Therefore, the distance between the 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

$=\frac{1}{\sqrt{2}}|\frac{p+r}{l}|$ units.