Exercise 4.2 · Q2: Show $A(a,b+c)$⁠, $B(b,c+a)$⁠, $C(c,a+b)$ are collinear

Basics you need

Solution

1. Why zero area means collinear

   If the three points are collinear, the area is 0: this is Remark (iii) of Section 4.3.

   The converse is what we actually use. Suppose the points are not collinear. Then no two of them coincide (two equal points and any third point always lie on one line), so they form a genuine triangle. Take $AB$ as base: its length is positive, and $C$ is not on the line $AB$⁠, so the height from $C$ is positive too. Hence the area $\tfrac12\times\text{base}\times\text{height}$ is positive. Section 4.3 says that this same area is $|\Delta|$ (the formula from earlier classes, written as a determinant), so $|\Delta| > 0$⁠, that is, $\Delta \ne 0$⁠.

   Turned round: if $\Delta = 0$⁠, the points cannot be non-collinear, so they are collinear. Together with Remark (iii): $\Delta = 0$ exactly when the points are collinear. It is therefore enough to show $\Delta = 0$⁠.

   The question asks for a property of the points (lying on a line), and the tool we have measures area. This step is the bridge between the two; without it, "the determinant is 0" would not yet prove anything about a line. The direction we need is "$\Delta = 0$ gives collinear", and we prove it by showing that non-collinear points always give $\Delta \ne 0$⁠.
   Section 4.3, Remark (iii). Basics: determinant as area

2. Write the area determinant

   One row per point, $x$⁠, then $y$⁠, then 1:

   $$\Delta = \frac12\begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}.$$

   Since $\Delta$ is half the determinant, $\Delta = 0$ exactly when the determinant is 0, so from now on we work with the determinant alone.

   The factor $\tfrac12$ cannot turn a nonzero number into 0, so it plays no part in showing that something is zero.

3. Choose the third column

   No entry is 0, so look for the line whose entries are simplest. Column 3 is $1, 1, 1$⁠, with signs $+, -, +$ (positions (1, 3), (2, 3), (3, 3)):

   $$\begin{aligned} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} &= 1\cdot M_{13} - 1\cdot M_{23}\\ &\quad + 1\cdot M_{33}. \end{aligned}$$
   Every entry of column 3 is 1, so no term carries an extra letter in front of its minor, and each minor uses only the first two columns. Expanding along $R_1$ instead would also work, but would multiply minors by $a$ and by $b + c$ and give longer expressions.
   Signs $(-1)^{1+3} = +$⁠, $(-1)^{2+3} = -$⁠, $(-1)^{3+3} = +$⁠. Lesson: expanding a 3 × 3 determinant

4. The three minors

   Delete column 3 and, in turn, row 1, row 2 and row 3:

   $$\begin{aligned} M_{13} &= \begin{vmatrix} b & c+a \\ c & a+b \end{vmatrix}\\ &= b(a+b) - (c+a)c\\ &= ab + b^2 - c^2 - ac, \end{aligned}$$ $$\begin{aligned} M_{23} &= \begin{vmatrix} a & b+c \\ c & a+b \end{vmatrix}\\ &= a(a+b) - (b+c)c\\ &= a^2 + ab - bc - c^2, \end{aligned}$$ $$\begin{aligned} M_{33} &= \begin{vmatrix} a & b+c \\ b & c+a \end{vmatrix}\\ &= a(c+a) - (b+c)b\\ &= ac + a^2 - b^2 - bc. \end{aligned}$$
   Each minor is a $2\times2$ determinant: main-diagonal product minus other-diagonal product. Expanding the brackets fully, rather than leaving them, is what lets us see the cancellation in the next step.
   $2\times2$ rule, Section 4.2.2. Basics: where the determinant comes from

5. Add with the signs: everything cancels

   The middle minor enters with a minus sign, which flips each of its four terms:

   $$\begin{aligned} &M_{13} - M_{23} + M_{33}\\ &= (ab + b^2 - c^2 - ac)\\ &\quad - a^2 - ab + bc + c^2\\ &\quad + ac + a^2 - b^2 - bc. \end{aligned}$$

   Now pair the terms:

   Term	From	Cancels with
   $+ab$	$M_{13}$	$-ab$ from $-M_{23}$
   $+b^2$	$M_{13}$	$-b^2$ from $M_{33}$
   $-c^2$	$M_{13}$	$+c^2$ from $-M_{23}$
   $-ac$	$M_{13}$	$+ac$ from $M_{33}$
   $-a^2$	$-M_{23}$	$+a^2$ from $M_{33}$
   $+bc$	$-M_{23}$	$-bc$ from $M_{33}$

   All twelve terms cancel in pairs, so the determinant is 0, and $\Delta = \tfrac12\cdot0 = 0$⁠.

   We never gave $a$⁠, $b$⁠, $c$ any particular values, so the cancellation holds for every choice. That is what makes this a proof for all points of this form, not a check of one example.
   A minus sign in front of a bracket changes the sign of every term inside. Basics: algebra the chapter leans on

6. Conclude, and find the line

   The area of triangle $ABC$ is $|\Delta| = 0$⁠, so by Step 1, $A$⁠, $B$⁠, $C$ are collinear.

   The coordinates show which line it is. In each point, $x + y$ is the same number:

   $$\begin{aligned} A:&\ \ a + (b + c) = a + b + c,\\ B:&\ \ b + (c + a) = a + b + c,\\ C:&\ \ c + (a + b) = a + b + c. \end{aligned}$$

   So all three points lie on the straight line $x + y = a + b + c$⁠.

   The determinant proof and the observation $x + y = a + b + c$ are two views of the same fact. The determinant method has the advantage that it works without spotting the line first.

Every fact used, and where it comes from