Worked examples — Collinearity of three points

This page drills Collinearity of three points until no case can surprise you. We first lay out every kind of collinearity problem on one table, then solve one example per cell. Before each solution you make a guess — that "Forecast" habit is where the intuition grows.

Two tools do all the work here. Let us name them in plain words before using them.

Why keep both? $S$ never divides, so it survives vertical lines and repeated points. $m$ is faster for a mental check when no denominator is zero. We'll show exactly where each one shines.

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The scenario matrix

Every collinearity question is one of these cells. Each row is a case class you must be able to handle without hesitation.

Cell	Case class	What makes it tricky	Example
C1	All positive coords, generic	none — the "warm-up"	Ex 1
C2	Mixed signs / crosses quadrants	negative differences flip signs	Ex 2
C3	Non-collinear → find the area	$S\ne 0$, must report area	Ex 3
C4	Solve for an unknown coordinate	$S=0$ becomes a linear equation	Ex 4
C5	Vertical line (equal $x$) — slope trap	slope $\to \infty$, must use $S$	Ex 5
C6	Degenerate: two points coincide	is a "triangle" of 2 points collinear?	Ex 6
C7	Real-world word problem	translating words to coords	Ex 7
C8	Exam twist: collinear ⇒ ratio ($k$ in both terms) / quadratic	two unknowns or two answers	Ex 8

Do all eight and you have touched every sign, every degenerate input, and the limiting vertical case.

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Cell C1 — generic, all positive

Forecast: Each step across is $+2$ right and $+2$ up. Feels like a perfect 45° staircase — guess collinear.

1. Write $S=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$. Why this step? $S=0$ is our exact collinearity test; substituting is the whole job.
2. Substitute: $S=1(4-6)+3(6-2)+5(2-4)$. Why this step? Plug the addresses straight in — "1 skips itself, takes 4 and 6", etc.
3. Compute: $S=1(-2)+3(4)+5(-2)=-2+12-10=0$. Why this step? A zero means zero area means one straight line.

Look at the figure: the three dots sit on one teal line, no bend at $B$.

Verify: slope $AB=\frac{4-2}{3-1}=1$, slope $BC=\frac{6-4}{5-3}=1$. Equal ⇒ collinear. ✓

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Cell C2 — mixed signs, crossing quadrants

Forecast: From $A$ to $B$ we go $+2,+4$; from $B$ to $C$ again $+2,+4$. Same step ⇒ guess collinear. The point of this cell is to prove the negatives don't break $S$.

1. $S=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$. Why this step? Same test — signs are handled automatically by subtraction.
2. Substitute carefully: $S=(-2)(1-5)+0(5-(-3))+2((-3)-1)$. Why this step? Watch the double-negatives; $5-(-3)=8$ is where mistakes happen.
3. Compute: $S=(-2)(-4)+0(8)+2(-4)=8+0-8=0$. Why this step? The two $\pm 8$ cancel — geometry of a straight line, not luck.

Verify: slope $AB=\frac{1-(-3)}{0-(-2)}=\frac{4}{2}=2$; slope $BC=\frac{5-1}{2-0}=\frac{4}{2}=2$. Equal ⇒ collinear. ✓

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Cell C3 — not collinear, report the area

Forecast: $P\to Q$ climbs steeply ($+2,+3$), $Q\to R$ climbs gently ($+2,+2$). Direction changed ⇒ guess not collinear, small area.

1. $S=0(3-5)+2(5-0)+4(0-3)$. Why this step? Substitute; the $x_1=0$ term vanishes, easing the arithmetic.
2. Compute: $S=0+10-12=-2$. Why this step? $S\ne 0$, so a real triangle exists — the sign tells orientation, magnitude tells size.
3. Area $=\frac{1}{2}|S|=\frac{1}{2}|-2|=1$. Why this step? $|S|$ is twice the area; halving and dropping the sign gives the physical area.

Verify: slopes differ — $PQ=\frac{3}{2}=1.5$, $QR=\frac{2}{2}=1$. Unequal ⇒ correctly not collinear; area $1$ square unit. ✓

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Cell C4 — solve for an unknown coordinate

Forecast: $x$ jumps evenly $2\to 4\to 6$. For a straight line $y$ must too: $3,?,7$ ⇒ midpoint value $k=5$. Guess $k=5$.

1. Set $S=0$: $2(k-7)+4(7-3)+6(3-k)=0$. Why this step? Collinearity is $S=0$; unknown $k$ makes it a linear equation to solve.
2. Expand: $2k-14+16+18-6k=0$. Why this step? Distribute so like terms in $k$ can be gathered.
3. Collect: $-4k+20=0\Rightarrow k=5$. Why this step? One equation, one unknown — pure algebra now.

Verify: slope $AB=\frac{5-3}{4-2}=1$, slope $BC=\frac{7-5}{6-4}=1$. Equal ⇒ $k=5$ correct. ✓

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Cell C5 — vertical line, the slope trap

Forecast: Every $x$ is $3$ — a straight vertical wall. Guess collinear, and guess the slope method will misbehave.

1. Try slope $AB=\frac{4-1}{3-3}=\frac{3}{0}$. Why this step? To feel the trap: dividing by zero is undefined — the slope method cannot even start.
2. Use $S$ instead: $S=3(4-9)+3(9-1)+3(1-4)$. Why this step? $S$ never divides, so vertical lines are no obstacle at all.
3. Compute: $S=3(-5)+3(8)+3(-3)=-15+24-9=0$. Why this step? Zero ⇒ collinear, confirming the vertical wall with no division needed.

Verify: all three $x$-coordinates equal ($=3$) is itself the definition of a vertical line — every such trio is collinear. $S=0$ agrees. ✓

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Cell C6 — degenerate: two points coincide

Forecast: $A$ and $B$ are literally the same dot. Two distinct points always sit on some line, and the third joins it. Guess collinear (degenerately).

1. $S=2(5-(-1))+2((-1)-5)+9(5-5)$. Why this step? Substitute even though two points match — $S$ handles it gracefully.
2. Compute: $S=2(6)+2(-6)+9(0)=12-12+0=0$. Why this step? Zero, because you can't enclose area with only two genuinely different points.

Verify: Only $A(=B)$ and $C$ are distinct; any two points define a line, and the duplicate lies on it trivially. $S=0$ confirms — collinear (degenerate case). ✓

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Cell C7 — real-world word problem

Forecast: $P\to Q$ is $+3,+6$ (slope $2$); $Q\to R$ is $+2,+4$ (slope $2$). Same ratio ⇒ guess straight (collinear).

1. $S=1(8-12)+4(12-2)+6(2-8)$. Why this step? Translate the words: the posts are points; "straight fence" means collinear, i.e. $S=0$.
2. Compute: $S=1(-4)+4(10)+6(-6)=-4+40-36=0$. Why this step? Zero ⇒ the three posts share one line — the fence is straight.

Verify (units + slope): metres cancel in the ratio; slope $PQ=\frac{6}{3}=2=\frac{4}{2}=$ slope $QR$. Fence is straight. ✓

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Cell C8 — exam twist: unknown gives a quadratic

Forecast: Here $t$ appears in an $x$ and a $y$, so multiplying them may create a $t^2$ term — expect possibly two answers.

1. $S=0$: $t(t-4)+3(4-2)+6(2-t)=0$. Why this step? Same test; but $t$ sits in $x_1$ and $y_2$, so the first term is $t\cdot (t-4)$ — a square appears.
2. Expand: $t^2-4t+6+12-6t=0\Rightarrow t^2-10t+18=0$. Why this step? Collect into standard quadratic form $a{t}^2+bt+c=0$.
3. Solve with the quadratic formula $t=\frac{10\pm \sqrt{100-72}}{2}=\frac{10\pm \sqrt{28}}{2}=5\pm \sqrt{7}$. Why this step? A quadratic can have two roots — two distinct lines through $B,C$ each admit an $A$.

Verify: for $t=5+\sqrt{7}$: substitute back, $S=(5+\sqrt{7})(1+\sqrt{7})+6+12-6(5+\sqrt{7})$. Since $(5+\sqrt{7})(1+\sqrt{7})=5+6\sqrt{7}+7=12+6\sqrt{7}$, we get $12+6\sqrt{7}+18-30-6\sqrt{7}=0$. ✓ (Both roots check symmetrically.)

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Recall Which tool for which cell?

Slope method fails on cells ::: C5 (vertical line) and C6 (repeated point) — any zero denominator. The one tool that solves every cell ::: the signed-area expression $S$, because it never divides. A non-zero $S$ gives you ::: the actual triangle area, $\frac{1}{2}|S|$ (cell C3). Why can C8 have two answers ::: the unknown appears in both an $x$ and a $y$, producing a quadratic.

Connections

• Area of Triangle using Coordinates — where $S$ and the $\frac{1}{2}|S|$ area come from.
• Slope of a Line — the quick mental check, and exactly when it breaks (C5, C6).
• Determinants — $S$ is a $3\times 3$ determinant in disguise.
• Vectors and Cross Product — $S$ equals the 2D cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
• Section Formula — locating a collinear midpoint like $k=5$ in Ex 4.
• Parametric Equations — every collinear trio satisfies $\vec{r}=\vec{a}+t(\vec{b}-\vec{a})$.
• Linear Dependence — collinear direction vectors are linearly dependent.