Coordinate Area Formulas Advanced. This lesson goes beyond core Geometry. You can skip it.

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Learning goals

Explain how signed edge contributions produce the area of a simple polygon
Calculate exact areas from ordered vertices with the shoelace formula
Predict the effect of changing the starting vertex or reversing the traversal
Identify when a crossing makes a vertex order unsuitable for measuring area

Follow the boundary

A simple polygon has a closed boundary made of straight edges, with edges meeting only at the shared endpoints of consecutive edges. It can be concave, with a corner that points inward. A lattice polygon has integer coordinates at every vertex.

Use the usual axes, with positive $x$ to the right and positive $y$ up. As you follow a simple boundary counterclockwise, the interior is locally on your left. Follow it clockwise and the interior is locally on your right. The signed area will be positive for a counterclockwise walk and negative for a clockwise walk.

Give each directed edge a signed contribution

Draw a horizontal reference line $y=k$ below all the vertices. An edge and the vertical drops from its endpoints enclose a trapezoid down to that line. Give that trapezoid positive area for a leftward edge and negative area for a rightward edge. A vertical edge has zero contribution because its horizontal width is zero.

The dashed drops give the two heights. The arrow gives the sign.

Worked example 1 Sign a single trapezoid

Problem. The directed edge goes from $P(6,4)$ to $Q(2,1)$ . Use the reference line $y=-1$ to find its signed trapezoid contribution. Find the contribution of the reverse edge as well.

Solution. For $P\to Q$ , the width is 4 units. The two heights above the line are 5 and 2 units. This direction is leftward, so the contribution is positive.

\begin{aligned} E(P,Q)&=4\cdot\frac{5+2}{2}\\ E(P,Q)&=14. \end{aligned}

For $Q\to P$ , the same trapezoid has a negative sign because the edge goes rightward.

E(Q,P)=-14.

Check your understanding

An edge moves right by 3 units. Its endpoints are 2 and 6 units above the reference line. What is its signed trapezoid contribution?

12

-24

-12

0

Why the directed trapezoids give signed area#

Cut the plane into vertical strips at the distinct vertex $x$ -coordinates. Within the open part of a strip, the straight boundary segments crossing it stay in the same top-to-bottom order because they never meet there. At a vertical line inside the strip, each entry into the polygon is followed by an exit. Thus the region in that strip consists of pieces with paired upper and lower edges. A piece can taper to a point at a strip endpoint. The strips and their interior pieces are finite in number.

The guide lines show part of a strip. Pair the upper and lower edges separately above and below the notch.

For a counterclockwise boundary, an upper edge of a piece has interior below it, so the edge moves left. A lower edge has interior above it, so the edge moves right. The portion of the upper trapezoid in this strip is added and the portion of the lower trapezoid is subtracted. Both reach the same reference line. Their difference is exactly the area between the two edges. If a strip contains two or more pieces, each upper and lower pair contributes its own area.

Every edge trapezoid is accounted for by its portions in these strips. Vertical cut lines have no area. Therefore the signed contributions add to the polygon’s area. In a clockwise walk, all the edge signs switch and the result is the negative of the area.

Turn the trapezoid sum into the shoelace formula

Why only the cross-products remain#

In Example 1, twice the signed contribution is the signed width times the sum of the heights.

2E(P,Q)=(6-2)(5+2).

Expanding and grouping the four products gives a cross-product difference $d_{PQ}$ and an endpoint difference $b_{PQ}$ .

\begin{aligned} d_{PQ}&=6\cdot2-2\cdot5\\ d_{PQ}&=2. \end{aligned}

\begin{aligned} b_{PQ}&=6\cdot5-2\cdot2\\ b_{PQ}&=26. \end{aligned}

Adding the two differences gives twice the signed contribution.

\begin{aligned} 2E(P,Q)&=d_{PQ}+b_{PQ}\\ 2E(P,Q)&=2+26. \end{aligned}

The second group contains the starting endpoint product $6\cdot5$ and the negative ending endpoint product $-2\cdot2$ . On the next edge, $Q$ becomes the starting point, so that edge supplies $+2\cdot2$ .

Write $P_i=(x_i,y_i)$ for each of the $n$ vertices in boundary order. To include the last edge, use $P_{n+1}=P_1$ . Let $h_i$ be the height of $P_i$ above the reference line.

h_i=y_i-k.

The signed width $x_i-x_{i+1}$ is positive on a leftward edge. Multiplying it by the average of the two heights gives the contribution $E_i$ .

2E_i=(x_i-x_{i+1})(h_i+h_{i+1}).

After expansion, the four products can be grouped into a cross-product difference $d_i$ and an endpoint difference $b_i$ .

d_i=x_i h_{i+1}-x_{i+1}h_i.

b_i=x_i h_i-x_{i+1}h_{i+1}.

2E_i=d_i+b_i.

Around a whole polygon, each endpoint product in $b_i$ occurs once with a plus sign on the outgoing edge and once with a minus sign on the incoming edge. They cancel, including the first vertex because the boundary closes.

Now express $d_i$ using the original coordinates. Substituting the heights gives

d_i=x_i(y_{i+1}-k)-x_{i+1}(y_i-k).

The terms involving $k$ combine to $k$ times the horizontal change $\Delta x_i$ .

\Delta x_i=x_{i+1}-x_i.

d_i=x_i y_{i+1}-x_{i+1}y_i+k\,\Delta x_i.

The changes $\Delta x_i$ add to zero because the walk ends where it began. That leaves just the cross-products in the original coordinates.

Write the vertices in two columns and repeat the first row below the last. Each pair of neighboring rows gives two diagonal products. Subtract the product that goes from the upper $y$ to the lower $x$ from the product that goes from the upper $x$ to the lower $y$ . The pairs of diagonals look like crossed laces.

Check your understanding

Why does the term for the last edge have to use the first vertex again?

A It makes the sum independent of the axes' scale. B It adds the edge that closes the boundary and completes the endpoint cancellation. C It ensures that every edge has a positive contribution. D It allows the interior to contain a notch.

Measure a concave lattice polygon

Worked example 2 Calculate and check a notched area

Problem. A lattice polygon has boundary vertices $A(0,0)$ , $B(6,0)$ , $C(6,1)$ , $D(2,2)$ , $E(6,4)$ , $F(6,5)$ , and $G(0,5)$ in that order. Find its signed area and ordinary area by shoelace. Check its ordinary area using a rectangle and a triangle.

Follow the arrows to read the vertices. The dashed segment closes the triangular notch for the area check.

Solution. Include the closing edge $G\to A$ . In the table, the row for $D\to E$ is negative, so keep that sign when you add.

Edge	First product	Second product	Difference
$A\to B$	$0$	$0$	$0$
$B\to C$	$6$	$0$	$6$
$C\to D$	$12$	$2$	$10$
$D\to E$	$8$	$12$	$-4$
$E\to F$	$30$	$24$	$6$
$F\to G$	$30$	$0$	$30$
$G\to A$	$0$	$0$	$0$

2S=0+6+10-4+6+30+0.

S=24.

The signed area is positive and the ordinary area is 24 square units.

For the geometric check, enclose the polygon in a $6$ by $5$ rectangle. The missing triangle $CDE$ has vertical base $CE$ of length 3. Its perpendicular height is the horizontal gap of 4 from $D$ to that line.

\begin{aligned} K&=6\cdot 5-\tfrac12(3)(4)\\ K&=24. \end{aligned}

In coordinate decomposition, you choose pieces with areas you can find easily. A rectangle with one triangular notch is a good fit. Shoelace uses the same rule for every neighboring pair, so it also works when the edges are slanted and no convenient pieces stand out.

Start anywhere or reverse the walk

Changing the first vertex while keeping the same direction moves some terms to the other end of the sum. Every directed edge stays in the list. Reversing the walk swaps the two products for every edge, so each difference changes sign.

Worked example 3 Reverse the lattice walk

Problem. Use the same lattice polygon as in Example 2, with vertices $A(0,0)$ , $B(6,0)$ , $C(6,1)$ , $D(2,2)$ , $E(6,4)$ , $F(6,5)$ , and $G(0,5)$ . Follow $A,G,F,E,D,C,B$ and close at $A$ . Find the signed and ordinary areas by reversing the earlier edge contributions. State the orientation.

Start at A again and follow the reversed arrows along the same edges.

Solution. The old closing edge is now the first edge in reverse. For example, $G\to F$ has first product 0 and second product 30. Its difference is $-30$ .

2S=0-30-6+4-10-6+0.

S=-24.

The order is clockwise. The ordinary area is 24 square units.

Check your understanding

A simple polygon has signed area $-17$ in the order $A,B,C,D,E$ . What is its signed area in the order $C,D,E,A,B$ ?

17

0

-34

-17

Worked example 4 Keep negative coordinates in the products

Problem. A triangle has vertices $H(-3,-2)$ , $J(4,-1)$ , and $L(1,2)$ in boundary order. Find its exact area and identify the orientation.

Two vertices lie below the horizontal axis. Keep their negative coordinates in the calculation.

Solution. The differences for the three directed edges are

\begin{aligned} d_{HJ}&=3-(-8)\\ d_{HJ}&=11\\ d_{JL}&=8-(-1)\\ d_{JL}&=9\\ d_{LH}&=-2-(-6)\\ d_{LH}&=4. \end{aligned}

2S=11+9+4.

S=12.

The order is counterclockwise. The ordinary area is 12 square units.

Check your understanding

The two product totals for a simple polygon in boundary order are $-11$ and $4$ , in the order used in the formula. What is its ordinary area?

15/2

-15/2

7/2

15

Check that the walk is simple

Connecting the same points in a different sequence can change which edges you draw. Even an absolute value can leave the wrong result when edges cross.

Worked example 5 Find where a crossing loses area

Problem. A rectangle has corners $A(0,0)$ , $B(8,0)$ , $C(8,6)$ , and $D(0,6)$ . A walk visits $A,C,B,D$ and closes at $A$ . Find where its edges cross. Compare the shoelace result for that walk with both the total area of the two triangular regions and the area of the rectangle.

The crossed walk encloses two triangular regions. Its left and right sides have opposite directions.

Solution. Edges $AC$ and $BD$ are the diagonals of the rectangle. They meet at their common midpoint $(4,3)$ .

Crossed walk. The differences on $C\to B$ and $B\to D$ are $-48$ and $48$ . The other two differences are zero.

\begin{aligned} 2S&=0-48+48+0\\ 2S&=0. \end{aligned}

Two triangular regions. Each has a vertical base of length 6 and a perpendicular height of 4. The crossed walk follows one triangle clockwise and the other counterclockwise. Their signed contributions cancel even though both have positive ordinary area.

\begin{aligned} K_{\mathrm{regions}}&=2\cdot\tfrac12(6)(4)\\ K_{\mathrm{regions}}&=24. \end{aligned}

Rectangle. The boundary order $A,B,C,D$ encloses the full rectangle.

\begin{aligned} K_{\mathrm{rectangle}}&=8\cdot6\\ K_{\mathrm{rectangle}}&=48. \end{aligned}

Check your understanding

The edges of a four-vertex walk cross, and its shoelace sum is nonzero. What can you conclude about using the ordinary area formula?

A A nonzero sum proves the walk is simple. B The absolute value always gives the total area of the two regions. C The crossing prevents applying the simple-polygon area rule to this walk. D Reversing the walk removes the crossing.

Coordinate Area Formulas Advanced. This lesson goes beyond core Geometry. You can skip it.

Learning goals

Follow the boundary

Give each directed edge a signed contribution

Turn the trapezoid sum into the shoelace formula

Measure a concave lattice polygon

Start anywhere or reverse the walk

Check that the walk is simple

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Multiple Choice Questions (MCQ)

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