The purpose of this post is to briefly discuss parallelism and perpendicularity of lines in both a geometric and algebraic setting.

## Lines

What is a line? In Euclidean Geometry we usually don’t define a line and instead call it a primitive object (the properties of lines are then determined by the axioms which refer to them). If instead points and line segments – defined by pairs of points $P,Q$ $[PQ]$ are taken as the primitive objects, the following might define lines:

Geometric Definition Candidate

line, $\ell$, is a set of points with the property that for each pair of points in the line, $P,Q\in \ell$,

$[PQ]\cap \ell=[PQ]$.

In terms of a picture this just says that when you have a line, that if you take two points in the line (the language in comes from set theory), that the line segment is a subset of the line:

### Exercise:

Why is this objectively not a good definition of a line.

Once we move into Cartesian\Coordinate Geometry we can perhaps do a similar trick. We can use line segments, and their lengths to define slope, (slope = rise over run) and then define a line as follows:

Algebraic Definition Candidate

A line, $\ell$, is a set of points such that for all pairs of distinct points $P,Q\in\ell$, the slope is a constant.

This means that if you take two pairs of distinct points in a line $\ell$, and then calculate the slopes between them, you get the same answer, and therefore it makes sense to talk about the slope of a line, $m$.

This definition, however, has exactly the same problem as the previous. The definition we use isn’t too important but I do want to use a definition that considers the line a set of points.

## The Equation of a Line

We can use such a definition to derive the equation of a line ‘formula’ for a line of slope $m$ containing a point $(x_1,y_1)$.

Suppose first of all that we have an $x\text{-}y$ axis and a point $P(x_1,y_1)$ in the line. What does it take for a second point $Q(x,y)$ to be in the line?

The point $Q(x,y)$ is on the line if and only if the slope between $P$ and $Q$ is equal to $m$:

$m_{[PQ]}=m$

$\displaystyle \Rightarrow \frac{y-y_1}{x-x_1}=m$

$\Rightarrow y-y_1=m(x-x_1)$.

For any line at all this can be rewritten as $y=mx+c$, with $c$ the slope and $c$ the $y$-coordinate when $x=0$: i.e. the $y$-intercept:

Parallel and perpendicular are then relations on the set of lines $\mathcal{L}$. This gives yet another definition of a line (that does not, by the way, include vertical lines. Again, we can include them with small tweaks to this presentation).

Algebraic Definition

line, $\ell$, of slope $m$ and $y$-intercept $c$, is the set of solutions (in the plane) to the equation

$y=mx+c$.

A point $(x_0,y_0)$ is on the line $y=mx+c$ if and only if it satisfies the equation:

$y_0=mx_0+c$.

## Parallel

It is probably more natural to come at these concepts first from a geometric angle (pun not intended). There are various definitions. For example, The Penguin Dictionary of Mathematics gives the following description:

Describing lines, curves, planes, or surfaces that are always equidistant, and that will never meet no matter how far they are produced. Parallel lines and curves must both lie in the same plane.

Whatever the definition, we know a pair of parallel lines when we see them.

One implication of this definition is that a line is not parallel to itself. This illustrates a feature of mathematics that is perhaps not appreciated at the primary and secondary levels, namely that

We have seen a similar thing above: we usually take lines themselves as primitive objects and define line segments in terms of lines. It seems to be possible to define lines in terms of line segments (although the two definitions above do not achieve this).

There is a choice here and the choice made dictates the detail of the sequel but not the essence… it is going to be slightly easier for me to present things if I say that a line is not parallel to itself (although if we do say this we get that ‘parallel’ is an equivalence relation). There are other choices but here is the definition that I will use is as follows.

Geometric Definition
A line $\ell_1$ is parallel to a line $\ell_2$, written $\ell_1\parallel\ell_2$, if $\ell_1\cap\ell_2=\emptyset$, the empty set.

So rather than explicitly including the equi-distance I am just going to say that a pair of lines are parallel if they do not intersect. I am almost completely sure that equi-distance is a consequence of the above definition (perhaps with an additional axiom from Euclidean Geometry).

What we will show now is that parallel lines have the same slope.

Suppose that $\ell_1$ and $\ell_2$ are two parallel lines with equations:

$y=m_1x+c_1$ and $y=m_2x+c_2$.

By the Geometric Definition, these lines do not intersect. That means there is no point $(x_0,y_0)$ that satisfies both equations, so no simultaneous solution to the equations. This means there is no solution to:

$m_1x+c_1=m_2x+c_2$

$\Rightarrow m_1x-m_2x=c_2-c_1$

$\Rightarrow (m_1-m_2)x=c_2-c_1$     (*)

$\displaystyle\Rightarrow x=\frac{c_2-c_1}{m_1-m_2}$.

This clearly is a solution unless there is a division by zero, that is unless $m_1-m_2=0\Rightarrow m_1=m_2$.

That is, parallel lines have the same slope. Note that (*) has a solution if $m_1-m_2=0=c_2-c_1\Rightarrow \ell_1=\ell_2$ but this is not allowed by the definition of parallel that I am using.

This suggests the following algebraic definition for parallel.

Algebraic Definition

A line $\ell_1$ is parallel to a different line $\ell_2$, written $\ell_1\parallel\ell_2$ ,if the lines have the same slope.

If we think of slope as measuring steepness/direction, this definition is quite natural.

If we take this as the definition then we can derive the geometric definition. Suppose that $\ell_1$ and $\ell_2$ are parallel. Therefore their equations are

$y=mx+c_1$ and $y=mx+c_2$,

with $c_1\neq c_2$.

To find the intersection $\ell_1\cap\ell_2$ we solve the simultaneous equations. Doing this leads to $c_1=c_2$ which is false (the two lines must be different according to the definition). Therefore there is no intersection.

We have shown (with the restriction of lines to non-vertical lines) that

Geometric Definition $\Rightarrow$ Algebraic Definition, and

Algebraic Definition $\Rightarrow$ Geometric Definition

This implies that the definitions are equivalent. This post introduces the idea of Duality: in this case between Geometry and Algebra. The power of this duality is that sometimes it is easier to consider ideas/questions in the geometry picture, and sometimes it is easier to consider the algebra picture. The discussion here shows that when convenient we can think of parallel lines as being lines that do not intersect (geometry (or maybe even a set theory picture?)), and when convenient we can think of them as (distinct) lines that have the same slope; and that these are equivalent pictures.

Now we try the same kind of argument with perpendicularity.

## Perpendicular

Again, we should know perpendicular lines when we see them:

Lines are perpendicular when the angle between them is a right angle, $90^\circ$. We don’t have to write down a good definition of angle: let us just say that an ordered triple $(P,O,Q)$ defines an angle, with $O$, the ‘common’ point:

$\ell$ is perpendicular to $k$ because $|\angle POQ|=90^\circ$.

As there are four angles involved, with two of them equal, and they add up to $360^\circ$, we can define the angle between two lines as the smallest of the four and it is automatically less than or equal to $90^\circ$.

This time, the candidate for an algebraic definition, using slopes, does not seem natural (although it is possible to use vectors and the dot product to generate a natural algebraic definition (that has the great property of being generalisable to more abstract settings)). Therefore we will simply write down the Geometric Definition and then prove an Algebraic Condition that holds for perpendicular lines. Again we restrict to non-vertical lines.

Geometric Definition

Lines $\ell_1$ and $\ell_2$ are perpendicular, written $\ell_1\perp \ell_2$, if the angle between $\ell_1$ and $\ell_2$ is $90^\circ$.

Now consider perpendicular lines $\ell_1$ and $\ell_2$ with slopes $m_1$ and $m_2$:

Without affecting the slopes of $\ell_1$ and $\ell_2$ we can consider a new $x'\text{-}y'$-axis that has the same scale as the old but has as origin at $\ell_1\cap\ell_2=\{O\}$:

So we can assume if we want that the perpendicular lines both go through the origin. Now go to $x=1$ and drop perpendiculars up and down to $P$ and $Q$ as shown:

Now if we call the point $(1,0)$ by $R$, note that $OPR$ and $OQR$ are both right-angled triangles and so Pythagoras Theorem applies:

$|OP|^2=1^2+|RP|^2$ and $|OQ|^2=1^2+|RQ|^2$.

Calculate the length of $|RP|$ using slope.

$\displaystyle m_1=\frac{|RP|}{1}\Rightarrow |RP|=m_1$.

Similarly $|RQ|=-m_2$.

Now consider the right-angled triangle $POQ$ with hypotenuse $|PQ|=|RP|+|RQ|=m_1+(-m_2)=m_1-m_2$ and other sides such that $|OP|^2=1^2+|RP|^2$ and $|OQ|^2=1^2+|RQ|^2$.

### Exercise

Apply Pythagoras Theorem to $\triangle POQ$ to show that:

Algebraic Condition

Lines $\ell_1$ and $\ell_2$ are perpendicular, written $\ell_1\perp \ell_2$, if $m_1\cdot m_2=-1$.