Collinearity of three points: Visual walkthrough

We start with three dots. That's all we assume.

Step 1 — Three points, and the one question we care about

WHAT. We drop three points on a flat sheet of paper. Call them $A$ , $B$ , $C$ . Each has an address: two numbers telling you how far right and how far up it sits. We write $A (x_{1}, y_{1})$ — meaning " $A$ is $x_{1}$ steps to the right and $y_{1}$ steps up from the corner." Same for $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ .

WHY. Before any formula, we must fix what the symbols mean. $x_{1}$ is a horizontal distance; $y_{1}$ is a vertical distance. Nothing more. If you've seen this in Slope of a Line, it's the same idea: an address on a grid.

PICTURE. Look at the figure. Two points always sit on some line — you can lay a ruler across any two dots. The real question is only ever about the third dot: does $C$ fall on that ruler, or off to one side?

Step 2 — Turn "on the same line?" into "how much space?"

WHAT. Instead of squinting at a ruler, we connect the three dots with three segments: $A \to B \to C \to A$ . This makes a triangle.

WHY. A ruler test is a yes/no with no number behind it. But a triangle has an area — a single number we can compute from the addresses. And here is the magic swap:

If $C$ is off the line → the triangle is fat → area is a positive number.
If $C$ slides onto the line → the triangle flattens → area shrinks toward $0$ .
If $C$ is exactly on the line → the triangle is a squashed sliver of zero width → area is exactly $0$ .

So "collinear?" becomes "is the area zero?" — and area is something we can build a formula for.

PICTURE. Watch the third dot slide down toward the line. The blue triangle deflates like a balloon until it's a flat line — area $0$ .

Step 3 — Two arrows from A: the vectors AB and AC

WHAT. Pin your finger on $A$ . Draw an arrow from $A$ to $B$ , and another arrow from $A$ to $C$ . An arrow (a vector) is just "how far right, how far up do I travel to get there."

$A B = (rightward travel x_{2} - x_{1}, upward travel y_{2} - y_{1})$

$A C = (rightward travel x_{3} - x_{1}, upward travel y_{3} - y_{1})$

WHY subtract? To travel from $A$ to $B$ , you go from horizontal position $x_{1}$ to horizontal position $x_{2}$ — that's a change of $x_{2} - x_{1}$ . Subtraction always means "end minus start." That's why the parent note wrote $A B = (x_{2} - x_{1}, y_{2} - y_{1})$ .

WHY two arrows from the same point? Because a triangle's area is completely decided by two of its sides sharing a corner. Pin one corner ( $A$ ), point two arrows out, and the triangle is locked in. See Vectors and Cross Product for where this idea leads.

PICTURE. Two arrows fanning out from $A$ . The area we want is the area of the slanted parallelogram they span, cut in half.

Step 4 — Why two arrows enclose a parallelogram (and half of it is our triangle)

WHAT. Copy $A B$ and slide it to the tip of $A C$ ; copy $A C$ and slide it to the tip of $A B$ . The four arrows close up into a parallelogram. Our triangle $A B C$ is exactly half of it (draw the diagonal $B C$ — it cuts the parallelogram into two identical triangles).

WHY a parallelogram first? Because the area of the parallelogram built on two vectors has a beautifully simple formula — and once we have it, dividing by $2$ gives the triangle. This is the reason the parent formula carries that $\frac{1}{2}$ out front.

PICTURE. The shaded parallelogram, with the diagonal splitting it into two matching triangles — one of them is $A B C$ .

Step 5 — The area of that parallelogram: the "cross" of the two arrows

WHAT. Write the two arrows with short names to keep the algebra clean: $A B = (u_{1}, u_{2}), A C = (v_{1}, v_{2}) .$ The area of the parallelogram they span is $Parallelogram = ∣ u_{1} v_{2} - u_{2} v_{1} ∣.$

WHY this exact combination — "right of one times up of the other, minus the swap"? Think about what makes a parallelogram fat. If both arrows point almost the same way, the parallelogram is a thin sliver — small area. If they point at right angles, it's a big fat rectangle — maximum area. The quantity $u_{1} v_{2} - u_{2} v_{1}$ measures precisely how much the two arrows disagree in direction. This is called the cross product in 2D, and it's the same object as a $2 \times 2$ determinant: $u_{1} v_{2} - u_{2} v_{1} = u_{1} v_{1} u_{2} v_{2} .$

Why the absolute value bars $∣ \cdot ∣$ ? Because if the two arrows are in "clockwise" order the expression comes out negative, but area is never negative. The bars throw away the sign.

Now plug the real coordinates back in. Replace $u_{1} = x_{2} - x_{1}$ , $u_{2} = y_{2} - y_{1}$ , $v_{1} = x_{3} - x_{1}$ , $v_{2} = y_{3} - y_{1}$ , and halve it:

$Area = \frac{1}{2} u_{1} (x_{2} - x_{1}) v_{2} (y_{3} - y_{1}) - v_{1} (x_{3} - x_{1}) u_{2} (y_{2} - y_{1}) .$

PICTURE. Two cases side by side: arrows nearly parallel (thin, tiny area) vs. arrows spread wide (fat, big area). The number $u_{1} v_{2} - u_{2} v_{1}$ tracks the fatness.

Step 6 — Multiply out and collect: the symmetric formula appears

WHAT. Expand the bracket $(x_{2} - x_{1}) (y_{3} - y_{1}) - (x_{3} - x_{1}) (y_{2} - y_{1})$ term by term:

$x_{2} y_{3} - x_{2} y_{1} - x_{1} y_{3} + x_{1} y_{1} - (x_{3} y_{2} - x_{3} y_{1} - x_{1} y_{2} + x_{1} y_{1}) .$

WHY. Right now the formula "prefers" point $A$ — everything is measured from $A$ . But collinearity treats all three points equally. Expanding lets the lopsidedness cancel. Notice the two $x_{1} y_{1}$ terms are opposite in sign — they annihilate:

$x_{2} y_{3} - x_{2} y_{1} - x_{1} y_{3} - x_{3} y_{2} + x_{3} y_{1} + x_{1} y_{2} .$

Now gather the terms by which $x$ they contain:

$terms with x_{1} x_{1} (y_{2} - y_{3}) + terms with x_{2} x_{2} (y_{3} - y_{1}) + terms with x_{3} x_{3} (y_{1} - y_{2}) .$

Every point now appears in the same role — a clean, cyclic pattern.

PICTURE. The "1-2-3 skip" wheel: each $x$ skips its own $y$ and grabs the other two $y$ 's difference, marching around the circle $1 \to 2 \to 3 \to 1$ .

Step 7 — Set the area to zero: the collinearity condition

WHAT. From Step 2, "collinear" means area $= 0$ . So set the formula to zero. And since $∣ E ∣ = 0$ happens only when $E = 0$ itself, the absolute-value bars simply vanish:

$\frac{1}{2} E = 0 ⟺ E = 0.$

WHY the bars disappear (and this is a real trap). The bars only matter when you want a positive size. Zero has no sign to worry about. Keeping them isn't wrong — it's just clutter. (See the parent note's Mistake 1.)

PICTURE. The triangle collapsed flat: two arrows now point along the same line, the parallelogram has zero width, $u_{1} v_{2} - u_{2} v_{1} = 0$ .

Step 8 — The degenerate cases the slope shortcut cannot see

WHAT. People love the slope shortcut: "collinear if $\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{y _{3} - y _{2}}{x _{3} - x _{2}}$ ." But watch three danger zones.

Case A — vertical line ( $x_{1} = x_{2} = x_{3}$ ). All three sit above each other. The slope is $\frac{something}{0}$ — division by zero, undefined. The slope test breaks, yet the points are obviously collinear. The area formula sails through: plug $x_{1} = x_{2} = x_{3} = c$ and every term cancels to $0$ .

Case B — two points coincide (e.g. $A = B$ ). One arrow has zero length. There's no genuine triangle; the formula returns $0$ , correctly reporting "no enclosed area." The three points are collinear (two of them are the same dot on any line through the third).

Case C — all three coincide. A single dot. Area is trivially $0$ . Collinear (degenerately).

WHY handle these at all? Because the contract is: the reader must never meet a case you skipped. The area method never divides, so it survives all of them; the slope method divides, so it dies at vertical lines. That's why the parent note says "master the area method — it handles 100% of cases."

PICTURE. Three panels: the vertical stack (slope undefined, area still zero), two dots merged, all three merged.

The one-picture summary

Here is the entire derivation compressed: three dots → two arrows from $A$ → their parallelogram → half of it is the triangle → its area is $\frac{1}{2} ∣ u_{1} v_{2} - u_{2} v_{1} ∣$ → expand and collect into the cyclic $x_{1} (y_{2} - y_{3}) + \dots$ → set it to $0$ → collinear.

Recall Feynman retelling — say it back in plain words

Any two dots already share a line — lay a ruler on them. The only question is whether the third dot lands on that ruler. So I connect all three into a triangle and ask: how much space does it fence in? If the third dot is on the ruler, the triangle is a squashed flat sliver with zero area.

To measure that space, I pin one corner and shoot two arrows to the other corners. Those arrows span a parallelogram; my triangle is half of it. A parallelogram's area is "right-of-one times up-of-the-other, minus the swap" — a number that's big when the arrows point different ways and tiny when they point the same way. When the arrows point the same way, area is zero — which is exactly "all three dots on one line."

Multiplying that out and tidying up gives a pretty pattern where each $x$ skips its own $y$ and multiplies the difference of the other two $y$ 's, cycling $1 \to 2 \to 3$ . Setting it to zero is the collinearity test. And because this recipe never divides, it works even for vertical stacks and repeated dots — where the slope shortcut would choke on a divide-by-zero.

Recall Rebuild it yourself

The area formula's front factor is ::: $\frac{1}{2}$ (a triangle is half the parallelogram spanned by the two arrows) The parallelogram area from arrows $(u_{1}, u_{2}), (v_{1}, v_{2})$ is ::: $∣ u_{1} v_{2} - u_{2} v_{1} ∣$ Why the absolute-value bars vanish in the collinearity test ::: $∣ E ∣ = 0$ happens only when $E = 0$ , and zero has no sign to protect The one method that survives vertical lines ::: the area method — it never divides by anything

Connections

Area of Triangle using Coordinates — the very formula we just built, before setting it to zero.
Vectors and Cross Product — Steps 3–5 are the 2D cross product in disguise.
Determinants — $u_{1} v_{2} - u_{2} v_{1}$ is a $2 \times 2$ determinant.
Slope of a Line — the shortcut that fails at Step 8's vertical case.
Section Formula · Parametric Equations · Linear Dependence — other lenses on "same line."