Affine coordinates and barycentric reasoning

Try

The centroid of a triangle with A=(1,1), B=(7,1), C=(1,4) is G=(3,2). Draw the midpoint M of BC: M=(4,5/2). Then G=A+(2/3)(M−A), so G lies two thirds of the way along the median from A.

This is an affine fact. Stretching the picture horizontally changes lengths and angles but leaves the same division parameter and median concurrence. Affine geometry retains straight-line structure without selecting a dot product.

Predict

Predict which expressions still represent the same geometric point after moving the coordinate origin. The average (A+B+C)/3 does; A+B by itself generally does not. The sum of coefficients is the key.

Predict whether an invertible affine map must preserve perpendicular medians or a circumcircle. Those are metric properties and are not guaranteed. The centroid argument needs neither.

Experiment

Track the coefficient sums and meanings.

Expression	Coefficient sum	Interpretation
(A+B)/2	1	Midpoint point
(A+B+C)/3	1	Centroid point
B−A	0	Displacement vector
2A+B	3	Origin-dependent point formula

Translate the coordinate origin by v. Each point coordinate changes by −v. A coefficient-one combination changes by exactly −v, as a point should; a coefficient-zero combination does not change, as a displacement should. Verify the rows symbolically before using any particular triangle.

Understand

An affine map is F(x)=Lx+b, with L invertible when we want an affine automorphism. For a line P+tv, its image is LP+b+tLv. Since Lv≠0, the image is a line. Parallel directions remain proportional, and invertibility prevents distinct lines from collapsing.

More generally, if Σλᵢ=1, then F(ΣλᵢPᵢ)=ΣλᵢF(Pᵢ). The translation terms match precisely because of the coefficient sum. Hence directed division parameters and barycentric combinations are preserved.

For triangle ABC, let G=(A+B+C)/3. If M=(B+C)/2, then G=A+(2/3)(M−A). Cyclic repetition puts G on all three medians, each with vertex-to-centroid versus centroid-to-midpoint ratio 2:1. Two distinct medians cannot have two common points, because their joining line would then be the same; nondegeneracy rules out their coincidence. Thus the concurrence point is unique.

Apply

Repair “Affine maps preserve ratios of any two lengths.” They preserve a division parameter along a single line, not arbitrary cross-direction length ratios. The map (x,y)↦(2x,y) doubles horizontal distances but leaves vertical ones unchanged.

Barycentric coordinates relative to a nondegenerate triangle are unique. Write P=A+s(B−A)+t(C−A), possible uniquely because the two edge vectors form a basis. Then the weights are 1−s−t,s,t. Nonnegative weights characterize the closed triangle; strictly positive weights characterize its interior. These statements connect affine algebra to the order geometry of convex combinations.

For optional depth, the same coefficient-one rule works in affine spaces over other fields, but inequalities and “inside” require an ordered field. Algebraic barycenters alone do not supply order.

Continue

Use affine coordinates when a proof concerns concurrence, collinearity or ratios on lines, and reserve metric formulas for genuinely metric conclusions. Continue to projective completion to see how adding directions as points changes the parallel story.