Noesis
2.3.12 · D5Coordinate Geometry

Question bank — Collinearity of three points

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The core object everywhere below is the signed expression This is twice the signed area of triangle (see Area of Triangle using Coordinates). Collinear . Keep that one sentence in your head — most traps here are just it in disguise.

True or false — justify

TF1. If three points are collinear, the value of is exactly .
True — is twice the signed area, and collinear points enclose no region, so the area is ; the sign doesn't matter when the number is .
TF2. can happen even when the three points are all different.
True — three distinct points can still sit on one line (e.g. ); collinearity is about direction, not about points being equal.
TF3. If two of the three points coincide (say ), then .
True — two identical points plus any third always lie on the line joining them, so they're trivially collinear and ; this is a degenerate case you must accept, not an error.
TF4. A negative value of means the points are "more than collinear" or overlapping.
False — a negative just means the area is signed clockwise; any nonzero (positive or negative) means a genuine triangle, so the points are not collinear.
TF5. The slope test works for every set of three points.
False — it breaks when any two -coordinates are equal (division by zero); the vertical-line case must be handled separately or by the area method.
TF6. If , the three points are collinear.
True — both segments start at and rise at the same rate, so and lie on the same line through ; any pair of the three slopes being equal is enough.
TF7. Swapping the labels of two points (say ) changes whether they are collinear.
False — collinearity is a property of the set of points; relabelling can flip the sign of but never its zero-ness, so the conclusion is unchanged.
TF8. If , then vectors and are parallel.
True — is exactly the 2D cross product ; it vanishes precisely when the two vectors are parallel (or one is zero), which is the cross-product condition for collinearity.
TF9. Three points on a vertical line give .
True — vertical means all equal; substituting gives , so the area method handles verticals with no special case.

Spot the error

SE1. "Points are collinear iff , so I keep the absolute value."
Not wrong, just redundant — forces , so the bars add nothing; but writing bars while solving for an unknown can hide a sign you actually need, so drop them.
SE2. "For I used ."
Error in the skip pattern — term one must be , not ; each multiplies the difference of the other two 's, taken cyclically.
SE3. "Slope equals slope , so I cross-multiply to ."
This step is correct — cross-multiplying is exactly how you avoid the division-by-zero trap; the resulting polynomial equation is valid even when a denominator was .
SE4. "The area came out , so the triangle's area is ."
Error — area is a nonnegative magnitude; the signed value is , and the actual area is . The negative sign only tells you the orientation (clockwise), not a negative area.
SE5. " have undefined slope, so I can't test them — assume not collinear."
Error — undefined slope means a vertical line; equal -coordinates for all three make them collinear. Use (which gives ) instead of guessing.
SE6. "I got from measured coordinates, so the points are collinear."
Careful — exact collinearity needs exactly; a tiny nonzero means an extremely thin but real triangle. With rounded/measured data you decide with a tolerance, not by declaring true collinearity.
SE7. "Three points are collinear, and I computed slope; therefore slope must be too, but was vertical."
Contradiction reveals an input error — if has finite slope and the points are truly collinear, cannot be vertical; a vertical forces vertical too, so the data or arithmetic is inconsistent.

Why questions

WHY1. Why does zero area guarantee collinearity, rather than just "small" triangle?
A line is one-dimensional and encloses no 2D region; the only way three points bound zero area is by lying on one line, so "zero area" and "collinear" are logically identical, not approximate.
WHY2. Why do we build the condition from vectors and (both from ) rather than and ?
Sharing the tail makes the cross product measure the parallelogram spanned by the two "spokes"; it's a cleaner picture, but gives the same zero condition — either works.
WHY3. Why can the slope method miss cases the area method catches?
Slope hides a division; when the slope is undefined and the formula is meaningless, whereas is a plain polynomial with no denominators and stays valid everywhere.
WHY4. Why is signed while the triangle's area is not?
equals the 2D cross product, which encodes orientation (counter-clockwise positive, clockwise negative); area throws that sign away with . For collinearity we only need , where sign is irrelevant.
WHY5. Why does the skip pattern use differences of the other two 's?
It's the expansion of a determinant with a column of 's; the cofactor of each is exactly the determinant of the remaining 's, which is that cyclic difference.
WHY6. Why is collinearity the same as and being linearly dependent?
Linearly dependent means one vector is a scalar multiple of the other (same or opposite direction); that's precisely "parallel," which is the geometric meaning of .
WHY7. Why can we also test collinearity with the Section Formula?
If lies on line , then divides in some ratio ; plugging that ratio into the section formula reproduces 's coordinates exactly — consistency there is another face of collinearity.
WHY8. Why does the parametric line give a collinearity test?
Every point on line is reached by some scalar ; three points are collinear iff each satisfies this equation for one , i.e. all lie on the same parametrised line.

Edge cases

EC1. All three points identical, .
Collinear (and every degenerate quantity is ): a single point trivially lies on infinitely many lines, so and there is no triangle at all.
EC2. Two points equal, third different: .
Collinear — the "triangle" collapses onto the segment , area is , ; a valid but degenerate case, never an error.
EC3. All three on a horizontal line, e.g. .
Collinear — equal 's give slope everywhere and ; the slope method works fine here (no division issue since 's differ).
EC4. All three on a vertical line, e.g. .
Collinear — the slope is undefined so slope method fails, but automatically because the common factors out; always fall back to for verticals.
EC5. Points nearly but not exactly on a line, very small and nonzero.
Not collinear in exact arithmetic — a thin sliver triangle of area still exists; only exactly counts.
EC6. Coordinates include a , e.g. origin as one vertex.
No special rule — a zero coordinate just zeroes one term in ; e.g. with , , still tested against .
EC7. Two points share an but the third does not, e.g. .
Slope is undefined but slope is finite, so slopes can't be equal — not collinear; check with to be safe rather than declaring the test "broken."
EC8. Large or negative coordinates, e.g. .
Sign and size don't change the logic; compute exactly. Here , so collinear — the formula is scale- and sign-blind.
Recall One-line safety rule

Which single test never fails, in every case above? ::: The area/determinant condition — it has no denominators, so verticals, zeros, and coincident points all work automatically.

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