Choosing a Good Coordinate System

In one sentence

The whole point of analytic geometry is to compute geometry with algebra, and the coordinate system is something you put there yourself. For the same figure, where you put the origin and which lines the axes follow does not change the conclusion, but it can change the amount of work a lot. Rules of thumb:

Make a key point of the figure the origin: both its coordinates are $0$ .
Make a key line a coordinate axis: every point on it loses a coordinate.
If the figure has symmetry, make the axis of symmetry a coordinate axis and the center of symmetry the origin: coordinates come in pairs that differ only by a sign.

One theorem, four coordinate systems: AB² + AC² = 2(AM² + BM²)

Step 1 / 4

Placed anywhere: 6 letters
Drop the triangle anywhere in the plane and its three vertices need $6$ letters. Both sides of the equation expand into a dozen or so terms that you then compare. It can be done, but it is tiring.

$A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$
Origin at B, x-axis along BC
Origin at the midpoint M: use the symmetry
The coordinates change, the lengths do not

Predict first

To find the orthocenter $H$ of a triangle (where the three altitudes meet), which of the four placements is the least work?

Placed anywhereOrigin at

B

Origin at the midpoint of

B C

Origin at the foot of an altitude

Choose an answer first; then you can see the result.

Stop 1 · The median of a triangle

The relation $A B^{2} + A C^{2} = 2 (A M^{2} + B M^{2})$ proved above is the median length theorem (Apollonius's theorem). It lets you compute a median from the three sides alone:

A M^{2} = \frac{A B ^{2} + A C ^{2}}{2} - \frac{B C ^{2}}{4} .

The key point is the midpoint $M$ , so put the origin at $M$ and set $B (- c, 0)$ , $C (c, 0)$ symmetrically.

Stop 2 · The three altitudes meet at one point

To prove that the three altitudes are concurrent, show that the intersection of two of them also lies on the third.

Proving the altitudes concurrent in the “foot” coordinate system

Step 1 / 4

Set up the axes
Take $B C$ as the $x$ -axis and the altitude $A D$ as the $y$ -axis, with $A (0, a)$ , $B (b, 0)$ , $C (c, 0)$ and $a \neq = 0$ . The altitude $A D$ is the $y$ -axis: $x = 0$ .

$A (0, a), B (b, 0), C (c, 0)$
The altitude through B
The altitude through C
Conclusion

Stop 3 · A right angle in a square

One problem, several solutions

In the square $A B C D$ , $E$ is the midpoint of $C D$ , and $F$ lies on $B C$ with $C F = \frac{1}{4} B C$ . Prove that $A E ⊥ E F$ .

Pure geometry

Let the side be $4$ , so $D E = E C = 2$ and $C F = 1$ . In $△ A D E$ and $△ E C F$ , $∠ D = ∠ C = 9 0^{\circ}$ and $\frac{A D}{D E} = \frac{4}{2} = \frac{E C}{C F}$ , so the triangles are similar and $∠ D A E = ∠ C E F$ .

Hence $∠ A E D + ∠ C E F = ∠ A E D + ∠ D A E = 9 0^{\circ}$ , so $∠ A E F = 9 0^{\circ}$ .

CostYou first have to spot a pair of similar triangles; if you do not, you are stuck.

Coordinate method

A square has two perpendicular sides, so let them be the axes: $A (0, 0)$ , $B (4, 0)$ , $C (4, 4)$ , $D (0, 4)$ . Then $E (2, 4)$ and $F (4, 3)$ .

$E A = (- 2, - 4)$ and $E F = (2, - 1)$ , with dot product $(- 2) \cdot 2 + (- 4) \cdot (- 1) = 0$ , so $A E ⊥ E F$ . (Slopes work too: $k_{A E} = 2$ , $k_{E F} = - \frac{1}{2}$ , product $- 1$ .)

CostNo inspiration needed, just a well-placed square; the price is a few lines of computation.

This placement uses rule 2: when the figure contains two perpendicular lines (two adjacent sides of the square), make them the axes.

Practice

Exercise 1Basic

The square $A B C D$ has side $6$ . Put the origin at its center, with the axes parallel to the two pairs of opposite sides, and $A$ in the third quadrant. What are the coordinates of $C$ ? (In the form (a,b).)

Exercise 2Basic

Take $B C$ as the $x$ -axis and the altitude to $B C$ as the $y$ -axis, with $A (0, 4)$ , $B (- 2, 0)$ , $C (6, 0)$ . Find the orthocenter $H$ . (In the form (a,b).)

Exercise 3Advanced

In $△ A B C$ , $A B = 5$ , $A C = 7$ , $B C = 8$ , and $M$ is the midpoint of $B C$ . Find $A M^{2}$ .

Exercise 4Challenge

Take $A (0, a)$ , $B (b, 0)$ , $C (c, 0)$ with $a \neq = 0$ . Find the circumcenter $O$ and the centroid $G$ of $△ A B C$ , and prove that $O$ , $G$ , $H$ are collinear with $O G : G H = 1 : 2$ (this line is called the Euler line).

In one sentence

One theorem, four coordinate systems: AB² + AC² = 2(AM² + BM²)

Stop 1 · The median of a triangle

Stop 2 · The three altitudes meet at one point

Proving the altitudes concurrent in the “foot” coordinate system

Stop 3 · A right angle in a square

Pure geometry

Coordinate method

Practice

Mistake Museum · Exhibits from this stop

Choosing coordinates “too well” and losing the general case