Perpendicular lines — product of slopes = −1: Visual walkthrough

We assume nothing but "a line has a steepness." Everything else is built here.

Step 1 — What "steepness" means as a picture

Term by term:

$Δ x$ ::: the sideways step (the $Δ$ , a Greek "D", just means "change in").
$Δ y$ ::: the up step taken during that same sideways move.
Their ratio $Δ y /Δ x$ ::: how much you climb per unit you walk right — the steepness.

WHY a ratio and not just the rise? Because a line is equally steep everywhere. If we walked twice as far right ( $2Δ x$ ) we would climb twice as high ( $2Δ y$ ), and the ratio stays the same. The ratio is the fingerprint of the line; the raw rise is not.

Figure — Perpendicular lines — product of slopes = −1

In the picture, the cyan arrow is our walk. The amber dashed lines are the run and the rise. The slope is the amber rise divided by the amber run.

Step 2 — A line is really a direction arrow

Why switch from "line" to "arrow"? Because we are about to turn the line by $90°$ , and turning is something you do to arrows, not to infinite lines. An arrow has a tip you can watch swing around.

The white line is $L_{1}$ .
The cyan arrow $(Δ x, Δ y)$ points along it.
Its slope $m_{1} = \frac{Δ y}{Δ x}$ is written where the run and rise live.

Step 3 — What a $90°$ turn does to the two axis arrows

Take the two basic direction arrows:

$(1, 0)$ — points right, one step east.
$(0, 1)$ — points up, one step north.

Rotate each a quarter turn anticlockwise (this is the standard positive direction — see Rotation of Vectors by 90°):

$(1, 0) 9 0^{\circ} (0, 1), (0, 1) 9 0^{\circ} (- 1, 0) .$

Reading each:

"Right" $(1, 0)$ swings up to become "up" $(0, 1)$ . ✔ makes sense, east turns to north.
"Up" $(0, 1)$ swings left to become "left" $(- 1, 0)$ . ✔ north turns to west.

The amber curved arrows show each unit arrow swinging a quarter turn. Notice the minus sign is born here: when "up" turns, it lands on the negative $x$ -side. That minus is the whole reason the final answer is $- 1$ and not $+ 1$ .

Step 4 — Rotate our actual arrow: the $(- y, x)$ rule

WHY this exact swap-and-flip? Any arrow $(x, y)$ is built from $x$ copies of "right" plus $y$ copies of "up": $(x, y) = x (1, 0) + y (0, 1) .$ Rotating is fair to each piece separately, so replace each basic arrow by its rotated version from Step 3: $x (0, 1) + y (- 1, 0) = (the y went to the left-side - y, the x went up x) .$

Apply it to our direction arrow $(Δ x, Δ y)$ : $(Δ x, Δ y) ⟶ (- Δ y, Δ x) .$

new run $= - Δ y$ ::: the old rise, now pointing (perhaps) left.
new rise $= Δ x$ ::: the old run, now pointing up.

The cyan arrow is the original; the amber arrow is it after the quarter turn. They meet at a perfect square corner — that square corner is exactly perpendicularity. The old rise and run have traded places, and a minus was stamped on one of them.

Step 5 — Read the new slope off the rotated arrow

Term by term:

numerator $Δ x$ ::: the original run is now the vertical part.
denominator $- Δ y$ ::: the original rise is now the horizontal part, carrying the minus from Step 3.

Compare the two slopes side by side: $m_{1} = \frac{Δ y}{Δ x}, m_{2} = - \frac{Δ x}{Δ y} .$

Look what happened: $m_{2}$ is $m_{1}$ upside-down ( $Δ x$ and $Δ y$ swapped) and negated. That is precisely "flip and flick the sign" — the negative reciprocal. The picture earned that phrase.

Step 6 — Multiply, and watch everything cancel

Cancel term by term:

$Δ y$ on top of the first fraction kills $Δ y$ on the bottom of the second.
$Δ x$ on the bottom of the first kills $Δ x$ on the top of the second.
The lone minus sign survives.

$m_{1} m_{2} = - \frac{Δ y Δ x}{Δ x Δ y} = - 1. ■$

WHY is this inevitable? The two arrows are made of the same $Δ x$ and $Δ y$ , just swapped. Swapping guarantees the letters cancel; the minus from the turn guarantees the leftover is $- 1$ , never $+ 1$ . Nothing about the particular line mattered — so it holds for every line.

The picture shows the two fractions overlapping, arrows drawn between the pairs that cancel, and the surviving amber $- 1$ .

Step 7 — The degenerate case: horizontal meets vertical

WHY the formula stalls: slope is rise-over-run, and a vertical line has run $Δ x = 0$ — dividing by zero has no value, so there is no number $m_{2}$ to plug in. The algebra has nothing to chew on.

WHY they are still perpendicular: the geometry never cared about the formula. The $x$ -axis and $y$ -axis obviously cross at a square corner. Our rotation still works perfectly: the arrow $(Δ x, 0)$ rotates to $(0, Δ x)$ — a horizontal arrow becomes a vertical arrow. The $90°$ is right there in the picture; only the ratio refuses to be written down.

The fix: treat horizontal $⊥$ vertical as a special case verified by the picture, not by the product rule.

The cyan horizontal arrow rotates (amber curve) into the amber vertical arrow — clearly a right angle — while the slope formula for the vertical one shows $\frac{Δ x}{0}$ with a big "undefined" flag.

Step 8 — Every quadrant, one rule

A line can slope up (positive $m_{1}$ , arrow into quadrant I) or down (negative $m_{1}$ , arrow into quadrant IV). Rotate each by $90°$ :

Up-line (e.g. $m_{1} = + 2$ ): rotates to a down-line, $m_{2} = - \frac{1}{2}$ . Product $2 \cdot (- \frac{1}{2}) = - 1$ . ✔
Down-line (e.g. $m_{1} = - \frac{3}{4}$ ): rotates to an up-line, $m_{2} = + \frac{4}{3}$ . Product $(- \frac{3}{4}) \cdot (\frac{4}{3}) = - 1$ . ✔

WHY it always lands on the opposite sign: the negative reciprocal of a positive number is negative, and of a negative number is positive — the "flick the sign" step never fails. So one member of a perpendicular pair always climbs while the other descends (unless we are in the horizontal/vertical special case of Step 7).

The four cyan arrows (one per quadrant) each get an amber perpendicular partner; every partner sits a clean $9 0^{\circ}$ away, and the little product label under each pair reads $- 1$ .

The one-picture summary

This final frame stacks the whole story: original arrow $(Δ x, Δ y)$ in cyan → quarter turn (amber curve) → rotated arrow $(- Δ y, Δ x)$ in amber → the two slopes $m_{1} = \frac{Δ y}{Δ x}$ and $m_{2} = - \frac{Δ x}{Δ y}$ → the cancellation → the surviving $- 1$ .

Recall Feynman: tell the whole walk in plain words

Draw a line and put an arrow along it. The arrow's steepness is "how high it climbs for each step right" — rise over run. Now spin that arrow a quarter turn to make the perpendicular line. When anything spins a quarter turn, its "how far right" and "how high" trade jobs, and one of them gets a minus sign stamped on it (because "up" swings to the left side). So the new steepness is the old one flipped upside-down with a minus. When you multiply the old steepness by the new one, the rise-numbers cancel the rise-numbers and the run-numbers cancel the run-numbers — like two puzzle pieces vanishing — and the only thing left standing is that minus one. It works for uphill lines, downhill lines, every direction. The one exception is a flat line meeting a straight-up line: they are perpendicular, but "straight up" has no rise-over-run number, so we just look at the square corner instead of doing the multiplication.

Connections

Perpendicular Lines — Product of Slopes = $-1$ — the parent result this walkthrough proves.
Slope of a Line — the rise-over-run definition every step leans on.
Rotation of Vectors by 90° — the engine of Steps 3–4.
Parallel Lines — equal slopes — the twin rule (same slope, no turn).
Angle Between Two Lines — perpendicular is the $9 0^{\circ}$ special case.
Equation of a Line — point-slope form, Distance Formula, Pythagoras Theorem — used in the parent's other two proofs.