Basic Problems of the Coordinate Method

What this move does

Coordinates turn points into pairs of numbers. So the three most basic jobs in geometry, measuring a length, finding a dividing point and computing an area, can all be done by straight calculation:

∣ P_{1} P_{2} ∣ = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}, P (\frac{x _{1} + λ x _{2}}{1 + λ}, \frac{y _{1} + λ y _{2}}{1 + λ}), S = \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111

Distance: draw a right triangle with $P_{1} P_{2}$ as hypotenuse. Its legs are the differences of the $x$ - and $y$ -coordinates; apply the Pythagorean theorem.
Dividing point: project the points onto the $x$ -axis and the $y$ -axis. The ratio $λ$ does not change (parallel lines cut segments proportionally), so use the number-line division formula on each axis. $λ = 1$ gives the midpoint.
Area: a $3 \times 3$ determinant divided by $2$ . It has a sign: positive when the vertices go counterclockwise, negative when they go clockwise. It is $0$ exactly when the three points are collinear.

Predict first

$A (- 1, 2)$ , $B (3, - 2)$ , $C (1, 0)$ . What is the area of $△ A B C$ ?

0

2

4

8

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

Work through one

Example: the area of the triangle P₁(−3, −2), P₂(3, −1), P₃(0, 4)

Step 1 / 4

Set up the determinant
The first column holds the $x$ -coordinates, the second the $y$ -coordinates, and the third is all $1$ s. The order of the rows is the order of the vertices, $P_{1} \to P_{2} \to P_{3}$ .

$S = \frac{1}{2} - 3 30 - 2 - 1 4 111$
Expand by cross-multiplying
Divide by 2 and read the sign
Change the order, and the sign flips

Dividing points work the same way; just keep an eye on the sign of $λ$ .

Example: P₁(−4, −2), P₂(0, 0), with different values of λ

Step 1 / 4

λ = 1: the midpoint
With $λ = 1$ we have $P_{1} P = P P_{2}$ , so $P$ is the midpoint $(\frac{- 4 + 0}{2}, \frac{- 2 + 0}{2}) = (- 2, - 1)$ .

$P = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
λ = 3: close to P₂
λ = −3: beyond P₂
λ = −1: no such point

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: Half the determinant is the signed area of the triangle

When the points are not collinear, drag any vertex: the reading is the area of a genuine triangle, and changing the order of the vertices changes the sign.

Practice

Exercise 1Basic

Find the distance between the points $(- 2, 5)$ and $(4, - 3)$ .

Exercise 2Basic

$A (2, - 1)$ , $B (- 4, 5)$ , and the point $P$ on segment $A B$ satisfies $A P = 2 P B$ . Find the coordinates of $P$ (in the form (a,b)).

Exercise 3Advanced

Find the area of the triangle with vertices $A (1, - 2)$ , $B (5, 1)$ , $C (- 1, 3)$ .

Exercise 4Challenge

$A (0, 0)$ , $B (6, 0)$ , $C (2, 4)$ . The point $P$ lies on segment $B C$ , and the area of $△ A B P$ is $\frac{1}{3}$ of the area of $△ A B C$ . Find the coordinates of $P$ (in the form (a/b,c/d)).

What this move does

Work through one

Example: the area of the triangle P₁(−3, −2), P₂(3, −1), P₃(0, 4)

Example: P₁(−4, −2), P₂(0, 0), with different values of λ

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Putting the whole segment in the denominator of λ

Getting a negative area and assuming it is wrong