How to Find the Distance Between Two Parallel Lines

The distance between two parallel lines in the Cartesian plane is a fundamental concept in coordinate geometry, involving the determination of the unique perpendicular segment that joins them and quantifies their constant separation at every point.

General Formulation of Parallel Lines and Their Distance

Consider two lines given by the equations $a x + b y + c_1 = 0$ and $a x + b y + c_2 = 0$. The requirement for parallelism is that the coefficients of $x$ and $y$ must be identical, ensuring both lines have the same slope. Accordingly, their separation is entirely dictated by the differing constant terms $c_1$ and $c_2$.

The perpendicular, or shortest, distance between these two lines is determined by identifying any point on one line and measuring its perpendicular distance to the other. Owing to parallelism, this value is invariant regardless of the chosen point.

Derivation of the Distance Formula for Two Parallel Lines

Let $L_1 : a x + b y + c_1 = 0$ and $L_2 : a x + b y + c_2 = 0$ denote the parallel lines. To derive the distance, select an arbitrary point $(x_0, y_0)$ on $L_1$. Therefore, $a x_0 + b y_0 + c_1 = 0$ holds.

The distance $d$ from $(x_0, y_0)$ to $L_2$ is given by the formula:

$d = \frac{|a x_0 + b y_0 + c_2|}{\sqrt{a^2 + b^2}}$

Since $a x_0 + b y_0 = -c_1$, substitute to obtain:

$d = \frac{|-c_1 + c_2|}{\sqrt{a^2 + b^2}}$

$d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$

This is the canonical formula for the perpendicular distance between two parallel lines in the form $a x + b y + c_1=0$ and $a x + b y + c_2=0$, with $a$ and $b$ not simultaneously zero.

Explanation of the Perpendicular Segment and Its Behaviour

The perpendicular drawn from any point on one of the parallel lines to the other line always has the same length, as parallel lines never intersect and their separation is uniform across the plane. This property mathematically guarantees the invariance of the calculated distance regardless of location along either line.

Parallel lines in the plane extend indefinitely without intersection, and the value $d$ thus quantifies the minimal, constant spacing between them. For vector generalisation in three-dimensional geometry, the computation requires the direction vector of the lines and a point on each line, but the two-dimensional result remains concise and direct, as stated above.

Distance Between Two Parallel Lines in Slope-Intercept Form

If the lines are provided in the form $y = m x + c_1$ and $y = m x + c_2$, convert each to the standard form. The general slope-intercept equation, $y = m x + c$, can be rewritten as $-m x + y - c = 0$. Applying the distance formula:

$d = \frac{|c_2 - c_1|}{\sqrt{1 + m^2}}$

This result is derived by substituting $a = -m$, $b = 1$, and applying the previously established arrangement for $|c_2-c_1|$ in the numerator and $\sqrt{a^2+b^2}$ in the denominator.

For further distinction between two and three-dimensional contexts, or between various forms such as vector and parametric representations, see Distance Between Two Parallel Lines.

Structural Conditions for Parallelism

Two lines are parallel if, and only if, their direction vectors (or, equivalently, their slopes in the plane) are equal. For lines written as $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$, parallelism requires that there exists a non-zero scalar $\lambda$ so that $a_1 = \lambda a_2$ and $b_1 = \lambda b_2$. For lines in $y = m x + c$ form, parallelism demands exactly equality of slopes ($m_1 = m_2$).

Worked Example 1: Calculation for Lines with Numerical Coefficients

Given: Determine the distance between $3x + 4y = 9$ and $9x + 12y + 28 = 0$.

Begin by rewriting both equations in comparable forms. Express $9x + 12y + 28 = 0$ as $3x + 4y + \frac{28}{3}=0$ by dividing all terms by $3$.

Let $a=3$, $b=4$, $c_1=-9$, $c_2= \frac{28}{3}$.

Apply the distance formula:

$d = \dfrac{\left|c_2 - c_1\right|}{\sqrt{a^2 + b^2}}$

$d = \dfrac{\left| \dfrac{28}{3} - (-9) \right|}{\sqrt{3^2 + 4^2}}$

Calculate numerator: $ \dfrac{28}{3} + 9 = \dfrac{28 + 27}{3} = \dfrac{55}{3} $.

Calculate denominator: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

So $d = \dfrac{ \dfrac{55}{3} }{5 } = \dfrac{55}{15} = \dfrac{11}{3}$ units.

Result: The distance between $3x + 4y = 9$ and $9x + 12y + 28 = 0$ is $\dfrac{11}{3}$ units.

Worked Example 2: Parallel Lines in Slope-Intercept Form

Given: Find the distance between $y = 2x + 4$ and $y = 2x - 1$.

Identify the slope and intercepts: $m = 2$, $c_1 = 4$, $c_2 = -1$.

Apply the slope-intercept form of the distance formula:

$d = \dfrac{|c_1 - c_2|}{\sqrt{1 + m^2}}$

$d = \dfrac{|4 - (-1)|}{\sqrt{1 + 2^2}} = \dfrac{5}{\sqrt{5}}$

$d = \sqrt{5} \approx 2.236$ units.

Result: The perpendicular distance between the lines $y = 2x + 4$ and $y = 2x - 1$ is approximately $2.236$ units.

Worked Example 3: Standard Form Parallel Lines with Integer Coefficients

Given: Compute the distance between $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

Here, $a = 3$, $b = -4$, $c_1 = 7$, $c_2 = 5$.

Distance is $d = \dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.

$d = \dfrac{|7 - 5|}{\sqrt{3^2 + (-4)^2}}$

$d = \dfrac{2}{\sqrt{9 + 16}} = \dfrac{2}{5} = 0.4$ units.

Result: The perpendicular distance between $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$ is $0.4$ units.

For practice and further exploration of related geometric concepts, consult Difference Between Distance and Displacement.

Interpretation and Limitations of the Distance Formula

The derived formula and procedures remain applicable only when the lines are truly parallel, as guaranteed by identical direction ratios or slopes. For non-parallel or coincident lines, the interpretation and resulting values lose geometric meaning, and separate analysis is necessary.

Extension to vector equations and three-dimensional contexts requires adaptation of the concepts described. For additional worked examples and topic links connected to coordinate geometry, see Difference Between 2D and 3D Shapes.