In mathematics, we are often asked to find the shortest distance between two lines. When the lines are intersecting, the shortest distance between them is eventually zero. For two non-intersecting lines lying in the same plane, the shortest distance is the distance that is shortest of all the distances between two points lying on both lines. For two parallel lines, this distance would be the length of perpendicular drawn from one point on a line to the other line.** In this page, we will study about the shortest distance between two parallel lines in detail.**

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Parallel lines have uniform shortest distance between them at any point. The parallel lines are shown in the following diagram :

Finding shortest distance between two parallel lines is equal to determining how far apart they are. This can be done by measuring the length of a line that is perpendicular to both of them. We may derive a formula using this approach and use this formula directly to find the shortest distance between two parallel lines.

(the slopes two parallel lines are equal)

y = $\frac{-x}{m}$ ____(2)

$x\ (m\ +$ $\frac{1}{m})$ = $-c_{1}$

$x$ = $\frac{-c_{1} m}{1\ +\ m^{2}}$

y = $\frac{c_{1} }{1\ +\ m^{2}}$

**Distance between these two points is :**

= $\sqrt{\frac{(c_{1} -c_{2})^{2}}{1\ +\ m^{2}}}$

= $\frac{|c_{1} -c_{2}|} {\sqrt{1\ +\ m^{2}}}$

**Example 1 :** Find the distance between two parallel lines y = x + 6 and y = x - 2.

Solution : This is the form of y = mx + c

Here, m = 1, c$_{1}$ = 6, c$_{2}$ = -2

d = $\frac{|c_{2} - c_{1}|}{\sqrt{1 + m^{2}}}$

d = $\frac{|-2 - 6|}{\sqrt{1 + 1^{2}}}$

d = $\frac{8}{\sqrt{2}}$ = 5.657 units**Example 2 :** Calculate the distance between parallel lines 15x + 8y - 34 = 0 and 15x + 8y + 31 = 0.

**Solution :** a = 15, b = 8, d$_{1}$ = -34, d$_{2}$ = 31

The formula is given by :

d = $\frac{|d_{2} - d_{1}|}{\sqrt{a^{2} + b^{2}}}$

d = $\frac{|31 + 34|}{\sqrt{15^{2} + 8^{2}}}$

d = $\frac{|65|}{\sqrt{225 + 64}}$

d = $\frac{65}{\sqrt{289}}$

d = $\frac{65}{17}$ = 3.82 units

Finding shortest distance between two parallel lines is equal to determining how far apart they are. This can be done by measuring the length of a line that is perpendicular to both of them. We may derive a formula using this approach and use this formula directly to find the shortest distance between two parallel lines.

Let us suppose that two parallel lines are represented in the following form :

y = mx + c$_{1}$

y = mx + c$_{2}$

(the slopes two parallel lines are equal)

Then, the formula for shortest distance can be written as under :

d = $\frac{|c_{2} - c_{1}|}{\sqrt{1 + m^{2}}}$

If the equations of two parallel lines are expressed in the following way :

ax + by + d$_{1}$ = 0

ax + by + d$_{2}$ = 0

then there is a little change in the formula.

d = $\frac{|d_{2} - d_{1}|}{\sqrt{a^{2} + b^{2}}}$

Two given equations are

y = mx + c$_{1}$

y = mx + c$_{2}$

To find the distance between these two, we are required to calculate the difference of two intercepts of these lines through the perpendicular line given by

- y = $\frac{-x}{m}$

- y = $\frac{-x}{m}$

(as slope of perpendicular line would be $\frac{-1}{m}$)

Intersecting point of the first line and perpendicular line can be found by solving following linear system

y = mx + c$_{1}$ _____(1)

y = $\frac{-x}{m}$ ____(2)

We have

$mx + c_{1}$ = $\frac{-x}{m}$

$x\ (m\ +$ $\frac{1}{m})$ = $-c_{1}$

$x$ = $\frac{-c_{1} m}{1\ +\ m^{2}}$

From equation (2)

y = $\frac{c_{1} }{1\ +\ m^{2}}$

Thus, we got $(\frac{-c_{1} m}{1\ +\ m^{2}}, \frac{c_{1} }{1\ +\ m^{2}})$ as the intersecting point of perpendicular line and first line.

Similarly, on solving y = mx + c$_{2}$ and y = $- \frac{x}{m}$, we obtain the following $(\frac{-c_{2} m}{1\ +\ m^{2}}, \frac{c_{2} }{1 + m^{2}})$.

$d$ = $\sqrt{(\frac{c_{1} m-c_{2} m}{1 + m^{2}})^{2} + (\frac{c_{1} -c_{2} }{1\ +\

m^{2}})^{2}}$

m^{2}})^{2}}$

= $\sqrt{\frac{(c_{1} -c_{2})^{2}}{1\ +\ m^{2}}}$

= $\frac{|c_{1} -c_{2}|} {\sqrt{1\ +\ m^{2}}}$

If we perform similar process for the set of equations

ax + by + d$_{1}$ = 0

ax + by + d$_{2}$ = 0

We get the following formula

$d$ = $\frac{|d_{2} - d_{1}|}{\sqrt{a^{2} + b^{2}}}$

$d$ = $\frac{|d_{2} - d_{1}|}{\sqrt{a^{2} + b^{2}}}$

The examples on shortest distance between two lines are discussed below.

Solution :

Here, m = 1, c$_{1}$ = 6, c$_{2}$ = -2

Formula is :

d = $\frac{|c_{2} - c_{1}|}{\sqrt{1 + m^{2}}}$

d = $\frac{|-2 - 6|}{\sqrt{1 + 1^{2}}}$

d = $\frac{8}{\sqrt{2}}$ = 5.657 units

The formula is given by :

d = $\frac{|d_{2} - d_{1}|}{\sqrt{a^{2} + b^{2}}}$

d = $\frac{|31 + 34|}{\sqrt{15^{2} + 8^{2}}}$

d = $\frac{|65|}{\sqrt{225 + 64}}$

d = $\frac{65}{\sqrt{289}}$

d = $\frac{65}{17}$ = 3.82 units

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