Formula for distance between two perpendicular lines

Finding the distance between two parallel lines is to determine how far apart the lines are. This can be done by measuring the perpendicular distance between them. We may derive a formula using this approach and use this formula directly to find the shortest distance between two parallel lines. 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. In this page, we will study the shortest distance between two lines in detail.

Table of Content:

Distance between two Straight Lines

The distance between two straight lines in a plane is the minimum distance between any two points lying on the lines. In geometry, we often deal with different sets of lines such as parallel lines, intersecting lines or skew lines.

Related Articles:

Straight Lines

3D Geometry

Distance Formula

Shortest Distance Between Two Parallel Lines

Formula to find distance between two parallel line:

Consider two parallel lines are represented in the following form :

y = mx + c1 …(i)

y = mx + c2 ….(ii)

Where m = slope of line

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

\(\begin{array}{l}d = \frac{|c_2 – c_1|}{\sqrt{1+m^2}}\end{array} \)

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

ax + by + d1 = 0

ax + by + d2 = 0

then there is a small change in the formula.

\(\begin{array}{l}d=\frac{\left | d_{2} -d_{1}\right |}{\sqrt{a^{2}+b^{2}}}\end{array} \)

Remark: The perpendicular distance between parallel lines is always a constant, so we can pick any point to measure the distance.

Proof

Consider two parallel lines given by

y = mx + c1 ..(i)

y = mx + c2 ..(ii)

Here line (i) intersects the x axis at A. So y = 0 at that point.

We can write (i) as 0 = mx + c1

So mx = -c1

x = -c1/m

The point A will be (-c1/m, 0). 

The perpendicular distance from A to line (ii) is the distance between line (i) and (ii).

Equation of line (ii) can be written as mx – y + c2 = 0

Comparing with general equation Ax+ By + C = 0

We get A = m, B = -1, C = c2

Here (x1, y1) = (-c1/m, 0)

The distance d = |(Ax1 + By1+C)/√(A2 + B2)|

= |(m(-c1/m) + -1(0) + c2)/√(m2 + 1)|

= |(-c1 + 0 + c2)/√(m2 + 1)|

= |(c2-c1)/√(1 + m2)|

In vector Form:

If

\(\begin{array}{l}\text{If}\ \vec{r}=\vec{a_1} + \lambda \vec{b}\ \text{and}\ \vec{r}=\vec{a_2} + \mu \vec{b}\end{array} \)

Then,

\(\begin{array}{l}d = |\frac{\vec{b} \times (\vec{a_2}-\vec{a_1})}{|\vec{b}|}|\end{array} \)

Shortest Distance Between Skew Lines

A set of lines that do not intersect each other at any point and are not parallel are called skew lines (also known as agonic lines). Such a set of lines mostly exist in three or more dimensions.

For Example: In the below diagram, RY and PS are skew lines among the given pairs.

Distance formula:

The distance between two lines of the form,

\(\begin{array}{l}\vec{l_{1}}= \vec{a_{1}}+t\vec{b_{1}}\end{array} \)

\(\begin{array}{l}\vec{l_{2}}= \vec{a_{2}}+t\vec{b_{2}}\end{array} \)

is

\(\begin{array}{l}d=\frac{(\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})}{\left | \vec{b_{1}} \times \vec{b_{2}}\right |}\end{array} \)

Solved Examples

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

Solution: Given equations are of the form, y = mx + c

Here, m = 1, c1 = 6, c2 = -2

Formula: d = |c1 – c2|/√(1 + m2)

Therefore, d = 8/√2 or 5.65 Units.

Example 2: Find the shortest distance between lines

\(\begin{array}{l}\vec{r} = \hat{i} + 2\hat{j} + \hat{k} + \lambda ( 2\hat{i} + \hat{j} + 2\hat{k})\end{array} \)

Solution:

Using formula,

\(\begin{array}{l}d = |\frac{\vec{b} \times (\vec{a_2}-\vec{a_1})}{|\vec{b}|}|\end{array} \)

Here,

\(\begin{array}{l}|\vec{b} \times (\vec{a_2}-\vec{a_1})| = \begin{vmatrix} i & j & k\\ 2 & 1 &2 \\ 1 & -3 &-2 \end{vmatrix}\end{array} \)

= |4i + 6j – 7k|

= √101

And

\(\begin{array}{l}|\vec{b}|= 3\end{array} \)

Therefore, d = √101/3