Parallel, perpendicular, where they cross, and how far apart.
In the last lesson we learned to write a single line five honest ways, all of them carrying the same fact: how steeply it tilts. Now bring a second line onto the page. Almost everything two lines can do is decided by comparing their slopes — equal slopes can only run forever side by side, and slopes that multiply to −1 can only meet at a right angle. When they aren't parallel they cross at exactly one point, which you find by solving their two equations together. And when you need a length — the shortest reach from a point to a line, or the gap between two parallels — one formula does it: d = |Ax₀ + By₀ + C| / √(A² + B²). By the end you'll read two lines the way you read a sentence.
Two lines are parallel when they tilt at the same angle — same inclination, same slope — yet never touch. Picture two cars on a perfectly straight highway, one lane apart: they hold the same heading forever, so they never collide.
If we write the lines as y = k₁x + b₁ and y = k₂x + b₂, "same heading" means k₁ = k₂. But there's a catch. If the intercepts also match, b₁ = b₂, the two equations describe the very same line — every point of one is a point of the other. We call that coincident. So the full rule is:
Parallel: k₁ = k₂ and b₁ ≠ b₂ (same slope, different intercept — they never meet).
Coincident: k₁ = k₂ and b₁ = b₂ (the same line written twice).
The slope form misses vertical lines, though — they have no slope to compare. The cure is the general form from 30.2, Ax + By + C = 0. Two lines A₁x + B₁y + C₁ = 0 and A₂x + B₂y + C₂ = 0 are parallel exactly when their tilts match with no rotation between them:
A₁B₂ − A₂B₁ = 0 (the same condition k₁ = k₂, but it also catches the vertical lines).
Is y = 2x − 1 parallel to 4x − 2y + 6 = 0? Put the second in slope form: 2y = 4x + 6, so y = 2x + 3. Both have k = 2, and the intercepts differ (−1 ≠ 3). Parallel. If the second had been 4x − 2y − 2 = 0 → y = 2x − 1, it would be the same line — coincident.
Now pin two lines into a perfect right angle. Something surprising happens to their slopes: as one climbs, the other must drop at exactly the matching rate. If line 1 goes "run 1, rise k₁," then turning a quarter-turn swaps and flips those steps, giving line 2 the slope −1/k₁. Multiply them and the product is always −1:
k₁ · k₂ = −1, equivalently k₂ = −1/k₁ (the negative reciprocal).
So a line of slope 2 is perpendicular to a line of slope −½, because 2 · (−½) = −1. A line of slope 3⁄4 is perpendicular to one of slope −4⁄3. Flip it over, change its sign — that's the whole trick.
The rule k₁k₂ = −1 needs both slopes to exist. A horizontal line (k = 0) is perpendicular to a vertical line (no slope), but the product rule can't say so — there's no slope to multiply. For that pair, use the general form: A₁A₂ + B₁B₂ = 0 covers every case, including the horizontal/vertical corner.
That general-form test, A₁A₂ + B₁B₂ = 0, is really the dot product of the two lines' normal vectors (a, b) being zero — perpendicular normals mean perpendicular lines. It is the safe, complete version of "negative reciprocal slopes."
If two lines are not parallel, they share exactly one point — the place that lies on both. To be on both lines is to satisfy both equations at once, so finding the crossing point is just solving the two equations together (a 2×2 system, Stage 12's territory).
Cross y = 2x − 1 with y = −x + 5. They share a point where the y-values agree: 2x − 1 = −x + 5 → 3x = 6 → x = 2. Then y = 2·2 − 1 = 3. The lines cross at (2, 3) — substitute into either line to check, and both give 3.
What if you try this with two parallel lines? Setting them equal gives a false statement like 3 = 7 — no x works, so there is no solution: parallel lines never cross. (Coincident lines give 0 = 0, true for every x — infinitely many "crossings," because every point is shared.) The algebra and the geometry tell the same story.
Set k₂ = ½ with b₂ ≠ 1 and the readout says parallel; match the intercept too and it flips to coincident. Set k₂ = −2 and you get perpendicular, because ½ · (−2) = −1. Any other slope gives a single crossing point.
How far is a point P(x₀, y₀) from a line ℓ? "Distance" always means the shortest reach, and the shortest path from a point to a line is the perpendicular drop — the straight segment that hits the line at a right angle. The foot of that perpendicular is the closest point on ℓ.
You could find the foot and use the distance formula, but there's a beautiful shortcut. Write the line in general form Ax + By + C = 0. Then the perpendicular distance is
d = |A·x₀ + B·y₀ + C| / √(A² + B²).
Read it as a recipe: plug the point into the line's left-hand side, take the size of what comes out (the absolute value, since distance can't be negative), and divide by √(A² + B²) — the length of the line's normal vector (A, B). If the point happens to lie on the line, the top is 0 and the distance is 0, exactly as it should be.
Distance from the origin P(0, 0) to 3x + 4y − 10 = 0:
d = |3·0 + 4·0 − 10| / √(3² + 4²) = |−10| / √(9 + 16) = 10 / 5 = 2.
The √ in the bottom is just the Pythagorean length of the normal (3, 4) — a 3-4-5 triangle, so √25 = 5.
Notice the foot is always the closest spot on the line, and the little square confirms the drop really is perpendicular. Park P right on the line (try (0, 3)) and the distance collapses to 0.
Two parallel lines run a constant width apart — like the two edges of a ruler. To measure that gap, pick any point on one line and find its distance to the other with the formula above; because the lines never converge, you get the same answer wherever you stand.
There's an even slicker form. First make the two lines share the same A and B (rescale one if needed) so they read Ax + By + C₁ = 0 and Ax + By + C₂ = 0. Then the gap is just the difference of the constants over the same normal length:
d = |C₁ − C₂| / √(A² + B²).
Distance between 3x + 4y − 1 = 0 and 3x + 4y − 11 = 0. They already share A = 3, B = 4, so
d = |(−1) − (−11)| / √(3² + 4²) = |10| / 5 = 2.
Check it by the long way: (0, ¼) sits on the first line, and its distance to the second is |3·0 + 4·¼ − 11| / 5 = |1 − 11| / 5 = 10/5 = 2 — same answer, as promised.
The shortcut |C₁ − C₂|/√(A²+B²) only works once the lines share identical A and B. If they read 3x + 4y = 1 and 6x + 8y = 5, halve the second to 3x + 4y = 2.5 first — otherwise you'll divide by the wrong normal length and get a wrong gap.
Two lines, four relationships, two distances — all read off the slopes and the general form. Keep this table close:
| Question | Slope form | General form Aᵢx + Bᵢy + Cᵢ = 0 |
|---|---|---|
| Parallel? | k₁ = k₂, b₁ ≠ b₂ | A₁B₂ − A₂B₁ = 0 |
| Same line (coincident)? | k₁ = k₂ and b₁ = b₂ | all of A, B, C proportional |
| Perpendicular? | k₁·k₂ = −1 | A₁A₂ + B₁B₂ = 0 |
| Crossing point? | solve the two equations together (none if parallel) | |
| Point P(x₀,y₀) to a line? | d = |Ax₀ + By₀ + C| / √(A² + B²) | |
| Two parallel lines apart? | d = |C₁ − C₂| / √(A² + B²) (match A, B first) | |
The thread that ties it together: the vector (A, B) points perpendicular to the line, so its length √(A² + B²) is the natural unit for every distance, and matching or perpendicular normals decide parallel vs. perpendicular. Next, in 30.4 · The equation of a circle, this same point-to-line distance d, compared against a radius r, will instantly tell us whether a line misses, grazes, or cuts a circle.
Put the second in slope form: 3y = 2x + 1, so y = ⅔x + ⅓. Both have slope ⅔ but different intercepts (5 ≠ ⅓), so they are parallel (not the same line).
The perpendicular slope is the negative reciprocal of −3, namely ⅓ (since −3 · ⅓ = −1). Point-slope through (6, 1): y − 1 = ⅓(x − 6) → y = ⅓x − 1.
Add the two equations to cancel y: (2x + y) + (x − y) = 7 + 2 → 3x = 9 → x = 3. Then y = x − 2 = 1. They cross at (3, 1). Check: 2·3 + 1 = 7 ✓ and 3 − 1 = 2 ✓.
Plug into the formula with A = 5, B = −12, C = 10:
d = |5·1 − 12·2 + 10| / √(5² + 12²) = |5 − 24 + 10| / √(25 + 144) = |−9| / 13 = 9⁄13 ≈ 0.69. (5-12-13 is a Pythagorean triple, so √169 = 13.)
They share A = 1, B = 1, so d = |C₁ − C₂| / √(A² + B²) = |(−2) − 4| / √(1 + 1) = 6 / √2 = 3√2 ≈ 4.24. (Rationalize: 6/√2 = 6√2/2 = 3√2.)
Yes, a horizontal and a vertical line always meet at a right angle, so they are perpendicular. But the vertical line has no slope, so there is nothing to multiply — k₁k₂ = −1 can't apply. The general form does: write them as 0·x + 1·y − 3 = 0 and 1·x + 0·y + 2 = 0; then A₁A₂ + B₁B₂ = 0·1 + 1·0 = 0, confirming perpendicular.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the analytic relationships between two lines, addressing the high-school Geometry standards HSG-GPE.B.5 (prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems, e.g., find the equation of a line through a point parallel or perpendicular to a given line) and HSG-GPE.B.7 (use coordinates and the distance formula to compute lengths). The crossing-point work is the geometric face of solving a 2×2 linear system, HSA-REI.C.6, and the parallel/coincident/intersecting trichotomy mirrors the no-solution / infinitely-many / unique-solution cases of HSA-REI. Encourage learners to verify each claim two ways — once with slopes, once by substituting back — and to reach for the general form Ax + By + C = 0 whenever a vertical line is in play, since the slope shortcuts quietly break there.