What is vertices out of order?

List the corners consistently around the boundary; a wrong order gives a self-crossing shape and the wrong area.

What is not wrapping back to the first vertex?

The last term must pair the final vertex with the first; omitting it loses a term.

What are sign slips in the products?

Negative coordinates need careful multiplication; one wrong sign throws off the whole sum.

SEAB Original-style practice: Find the area of the triangle with vertices A(1, 1), B(4, 2) and C(2, 5).

Use the shoelace formula, listing the vertices in order and returning to the first: Area = 12| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) |. = 12| 1(2 - 5) + 4(5 - 1) + 2(1 - 2) | = 12| -3 + 16 - 2 | = 12(11) = 5.5. The area is 5.5 square units. Markers reward correct substitution, taking the modulus, and halving to get 5.5.

SingaporeAdditional MathematicsSyllabus dot point

How can we find the area of a polygon directly from the coordinates of its vertices?

Find the area of a triangle or polygon from the coordinates of its vertices using the shoelace determinant method

A focused answer to the O-Level A-Maths outcome on areas from coordinates. The shoelace method for the area of a triangle or polygon, vertex ordering, and using a zero area as a collinearity test.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to compute the area of a triangle or other polygon when you are given only the coordinates of its corners, using the shoelace (determinant) method. The same calculation, when it gives zero, tells you that the points lie on a single straight line, a handy test for collinearity.

The answer

The shoelace idea

The area of a polygon is found by a criss-cross pattern of products of coordinates that looks like the lacing of a shoe. For a triangle $A(x_1, y_1)$ , $B(x_2, y_2)$ , $C(x_3, y_3)$ :

\text{Area} = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|.

The modulus ensures the area is positive whichever way round you list the vertices.

The polygon version

For more vertices, list them in order around the boundary, wrap back to the first, and sum the cross products $x_i y_{i+1} - x_{i+1} y_i$ :

\text{Area} = \frac{1}{2}\left| \sum (x_i y_{i+1} - x_{i+1} y_i) \right|.

A neat layout writes the coordinates in two columns, repeating the first row at the bottom, then multiplies diagonally.

Order matters

The vertices must be taken consistently in order around the figure (all clockwise or all anticlockwise). A scrambled order gives a wrong, often self-overlapping, answer.

The collinearity test

If three points give an area of zero by the shoelace formula, they are collinear: there is no triangle, because they lie on one line. This is a quick alternative to comparing gradients.

Worked example

Find the area of the triangle with vertices $D(-1, 2)$ , $E(3, -2)$ and $F(4, 5)$ .

Step 1: Lay out the coordinates in order

List $D, E, F$ and repeat $D$ at the bottom for the wrap-around.

Step 2: Form the cross products downward and upward

Downward (each $x$ times the next $y$ ): $(-1)(-2) + (3)(5) + (4)(2) = 2 + 15 + 8 = 25$ .

Upward (each $y$ times the next $x$ ): $(2)(3) + (-2)(4) + (5)(-1) = 6 - 8 - 5 = -7$ .

Step 3: Subtract and take the modulus

\left| 25 - (-7) \right| = |32| = 32.

Step 4: Halve

\text{Area} = \frac{1}{2}(32) = 16 \text{ square units}.

Examples in context

Example 1. Land area from survey points. A plot of land marked by surveyed corner coordinates has its area computed by the shoelace method directly from those points, without dividing the plot into triangles by hand.

Example 2. Checking three points lie on a line. To test whether three plotted data points are collinear, compute the triangle area they would enclose; a value of zero confirms they lie on one straight line, linking area to the straight-line topic.

Try this

Q1. Find the area of the triangle with vertices $(0, 0)$ , $(6, 0)$ and $(0, 4)$ . [2 marks]

Cue. Right triangle: $\tfrac{1}{2}(6)(4) = 12$ square units.

Q2. Show that $(1, 2)$ , $(2, 4)$ and $(3, 6)$ are collinear. [2 marks]

Cue. Shoelace area $= \tfrac{1}{2}|1(4 - 6) + 2(6 - 2) + 3(2 - 4)| = \tfrac{1}{2}|{-2 + 8 - 6}| = 0$ , so they are collinear.

Q3. Find the area of the triangle with vertices $(2, 1)$ , $(5, 1)$ and $(5, 7)$ . [3 marks]

Cue. Right triangle with legs $3$ and $6$ : $\tfrac{1}{2}(3)(6) = 9$ square units.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksFind the area of the triangle with vertices

A(1, 1)

B(4, 2)

and

C(2, 5)

Show worked answer →

Use the shoelace formula, listing the vertices in order and returning to the first:

Area $= \dfrac{1}{2}\left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$ .

$= \dfrac{1}{2}\left| 1(2 - 5) + 4(5 - 1) + 2(1 - 2) \right| = \dfrac{1}{2}\left| -3 + 16 - 2 \right| = \dfrac{1}{2}(11) = 5.5$ .

The area is $5.5$ square units.

Markers reward correct substitution, taking the modulus, and halving to get $5.5$ .

Original5 marksThe vertices of a quadrilateral, taken in order, are

P(0, 0)

Q(4, 0)

R(5, 3)

and

S(1, 4)

. Find its area.

Show worked answer →

Apply the shoelace formula for a polygon, listing the vertices in order and wrapping back to $P$ :

Area $= \dfrac{1}{2}\left| (x_P y_Q - x_Q y_P) + (x_Q y_R - x_R y_Q) + (x_R y_S - x_S y_R) + (x_S y_P - x_P y_S) \right|$ .

$= \dfrac{1}{2}\left| (0 - 0) + (12 - 0) + (20 - 3) + (0 - 0) \right| = \dfrac{1}{2}(12 + 17) = \dfrac{29}{2} = 14.5$ .

The area is $14.5$ square units.

Markers reward setting up the determinant in order, the cross-product terms, and halving the modulus.