## Distance Between Two Parallel Lines

If two lines are parallel, then they have the same distance between them throughout,

Therefore the distance between two parallel lines \(ax + by + c_1\) and \(ax + by + c_2\) is given by :

D = \(|c_1 – c_2|\over \sqrt{a^2 + b^2}\)

**Note** – Both equation must be in the given form \(ax + by + c_1\) and \(ax + by + c_2\), if it is not in the given form reduce them to the given form as shown in the example below.

Example : Find the the distance between two parallel lines 3x – 4y + 9 and 6x – 8y – 15 = 0.

Solution : Given lines are 3x – 4y + 9 and 6x – 8y – 15 = 0.

Divide line 6x – 8y – 15 = 0 by 2

we get, 3x – 4y – 15/2 = 0.

Now both the equation are reduced to given form.

Hence, we can find the distance using above formula

D = \(|c_1 – c_2|\over \sqrt{a^2 + b^2}\)

Required distance D = \(|9 – (-15/2)|\over \sqrt{3^2 + (-4)^2}\)

D = \(9 + {15\over 2}\over 5\) = \(33\over 10\)

Example : Find the equation of lines parallel to 3x – 4y – 5 = 0 at a unit distance from it.

Solution : Equation of any line parallel to 3x – 4y – 5 = 0 is

3x – 4y + \(\lambda\) = 0 …..(i)

It is given that the distance between the line 3x – 4y – 5 = 0 and line (i) is 1 unit.

\(\therefore\) \(|\lambda – (-5)|\over \sqrt{3^2 + (-4)^2}\) = 1

\(\implies\) \(|\lambda + 5|\over 5\) = 1

\(|\lambda + 5|\) = 5 \(\implies\) \(\lambda + 5\) = \(\pm 5\)

\(\implies\) \(\lambda\) = 0 , -10

Substituting the values of \(\lambda\) in (i), we get

3x – 4y = 0 and 3x – 4y – 10 = 0

as the equations of required lines.