Lines that are **parallel**, in the sense of **Euclidean geometry** (geometry of the plane) are lines in the same plane that never intersect. In coordinate geometry (the algebra of lines and other figures), they are lines with the same slope.

**Parallel lines** exist in the same plane but do not intersect. Parallel lines have the same **slope**.

In diagrams, we usually indicate that two or more lines are parallel by placing an arrow symbol on each line, as shown. There are two sets of parallel lines (one consisting of three parallel lines, the other of two), and we use different arrow symbols to differentiate them, much as we'd use hash marks for congruent line segments.

We also use this written notation to match the diagram:

$$a \parallel b, \: \: a \parallel c, \: \: b \parallel c, \: \text{and} \: d \parallel e.$$

A **transversal** is a line or segment that cuts across two or more parallel lines, forming eight or more angles that we'll name below.

In this figure, the transversal, **t**, cuts parallel lines **a** and **b**.

In this figure, the pairs of angles formed by a transversal and parallel lines are shown. They are:

- alternate interior angles (2 pairs)
- alternate exterior angles (2 pairs)
- corresponding angles (4 pairs)
- interior angles on the same side (2 pairs)
- exterior angles on the same side (2 pairs)
- vertical angles (4 pairs)

Those pairs are listed in the table below.

Angle pairs | Description |
---|---|

Alternate interior angles: 4‒6, 3‒5 | Angles between || lines and on opposite sides of the transversal |

Alternate exterior angles: 1‒7, 2‒8 | Angles on the outside of || lines and on opposite sides of the transversal |

Corresponding angles: 1‒5, 4‒8, 2‒6, 3‒7 | Angles in corresponding positions of their respective intersections, e.g. upper-left to upper left (1‒5) |

Interior angles on the same side: 4‒5, 3‒6 | Angles between || lines and on the same side of the transversal |

Exterior angles on the same side: 1‒8, 2‒7 | Angles on the outside of || lines and on the same side of the transversal |

Vertical angles: 1‒3, 2‒4, 5‒7, 6‒8 | Angles opposite one-another at intersections, e.g. 1‒3 |

This one is a **postulate**, something that's **self-evident**, and not possible to actually prove:

When two parallel lines are cut by the same line, any pair of corresponding angles formed are congruent. It makes sense, doesn't it? If the lines cut by the transversal are parallel, then the intersections produced are "clones" of one another.

Pairs of corresponding angles are labeled with the same letter, A, B, C, D, in the figure above.

The figure below shows two intersecting lines with four labeled angles. **∠A** is supplementary to **∠B**, and **∠D** supp. **∠B**.

If $\angle A + \angle B = 180˚,$ then $\angle A = 180˚ - \angle B,$ and

if $\angle D + \angle B = 180˚,$ then $\angle D = 180˚ - \angle B,$

so $\angle A \cong \angle D:$ **Vertical angles are congruent.**

(Figure ↑) $\angle 1 \cong \angle 5$ by the corresponding angle postulate, and $\angle 1 \cong \angle 3$ because vertical angles are congruent, therefore $\angle 3 \cong \angle 5$ by substitution of $\angle 3$ for $\angle 1$ in the first expression. Therefore **alternate interior angles are congruent**.

(Figure ↑) $\angle 1 \cong \angle 5$ by the corresponding angle postulate, and $\angle 5 \cong \angle 7$ because vertical angles are congruent, therefore $\angle 1 \cong \angle 7$ by substitution of $\angle 7$ for $\angle 5$ in the first expression. Therefore **alternate exterior angles are congruent**.

Using the numbered diagram above, we see that **∠1 ≅∠5** by the corresponding angle postulate, and **∠4** is supplementary to **∠1** (obvious from the diagram). That means **∠4 supp. ∠5** because an angle supplementary to a second angle must be supplementary to any other angle of the same measure.

Using the same numbered diagram, we see that **∠1 ≅∠5** by the corresponding angle postulate, and **∠5 supp. ∠8** (obvious from the diagram), so **∠1 supp. ∠8** because an angle supplementary to a second angle must be supplementary to any other angle of the same measure.

Now if you take a look back at all six proofs, you'll notice something interesting:

**congruent** or **supplementary**.

The parallel postulate says that

Through any given point, there is one and only one line parallel to another specific line.

Take a look at this figure. We can draw only one line through point **P** that is parallel to line **L**. This is one of the five postulates in Euclid's original writing on geometry. It's a **postulate** or **axiom**, because it can't actually be proved.

One way to think about this is just to consider line **M || L**. Now move **M** closer and closer to point **P** until **P** is on **M**. We already said that **M || L**, so it still is, and if we *rotate* it around point **P** the slightest bit, it will no longer be parallel.

This figure shows how we can use parallel lines to prove that the angles of a triangle sum to 180˚ (or π radians).

First we draw the line through the top angle of the triangle that is parallel to the base. There is only one such line by the parallel postulate. Then we use the complementarity of angles e, b and d, along with alternate interior angles to prove that the sum of angles a, b and c must be the same as the sum of angles e, b and d, which is 180˚ because those angles are obviously complimentary.

We know that an equation like

$$y = mx + b$$

is the equation of a line with slope **m** and that crosses the y-axis at **y = b**.

Any two lines can cross, at most, at one point. We can find that point by solving the two equations simultaneously. For example, consider these two lines:

$$ \begin{align} y &= 2x + 1, \: \text{and}\\[5pt] y &= -x + 2 \end{align}$$

These lines have different slope, therefore they must intersect at some point. We can find it by setting the two equations equal to each other using the transitive property: If $2x + 1 = y$ and $2 - x = y,$ then it must be true that $2x + 1 = 2 - x.$ Now we solve for **x**:

$$x = \frac{1}{3}$$

We can find the y that goes with that x by plugging ⅓ into either of the linear equations to get **y = 5/3**. So our two lines intersect at the point **(1/3, 5/3)**, and nowhere else.

Now let's repeat the process for two lines that are parallel, say

$$ \begin{align} y &= 2x + 1, \: \text{and}\\[5pt] y &= 2x - 5 \end{align}$$

We know that these lines are parallel because they lie in the same plane (we assume that in Euclidean geometry) and they have the same slope. Let's try to find a point of intersection (though we know there can't be one). Just like in the last example, it must be true that

$$2x + 1 = 2x - 5$$

Now if we gather terms containing x on the left side and constants on the right, we get

$$ \begin{align} 2x - 2x &= -5 - 1, \: \text{or}\\[5pt] 0 &= -6, \end{align}$$

which is never true, so the system of equations has no solution, which is consistent with the idea that these two lines have no point in common.

Later when you solve for the intersection(s) of other kinds of curves in the plane (like circles or parabolas), you'll see find the same kind of result for curves that don't cross.

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