Q.Find the area of the triangle with vertices $A(5,4)$, $B(2,-6)$ and $C(-2,4)$.

Area From Coordinates: The Shoelace Formula

Imagine you have a polygon drawn on a graph — maybe an irregular pentagon whose vertices you know as coordinates. How do you find its area without splitting it into triangles and measuring each one? There is a neat, mechanical method called the shoelace formula (or the surveyor's formula).

The Intuition

Think of walking around the polygon in order, from vertex to vertex. At each step, you are going to compute a small signed area — the area of a trapezoid under the edge you just walked. If you walk clockwise, some of these trapezoids will have positive area and some negative. When you add them all up, the negatives cancel the parts outside the polygon, and you are left with exactly the area inside.

The Precise Statement

For a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ listed in order (clockwise or anticlockwise), the area is:

where $(x_{n+1}, y_{n+1})$ is taken as $(x_1, y_1)$ — you close the loop.

The absolute value ensures the area is positive regardless of the direction you went around.

How It Works: A Simple Example

Take a triangle with vertices $A(1, 1)$, $B(4, 2)$, $C(2, 5)$. List them in order:

Sum of $x_i y_{i+1}$: $2 + 20 + 2 = 24$

Sum of $x_{i+1} y_i$: $4 + 4 + 5 = 13$

Difference: $24 - 13 = 11$

Area: $\frac{1}{2} \times |11| = 5.5$ square units.

Why It Works (Briefly)

Each term $x_i y_{i+1} - x_{i+1} y_i$ is twice the signed area of the triangle formed by the origin and the edge from $(x_i, y_i)$ to $(x_{i+1}, y_{i+1})$. Summing these over all edges and taking half gives the total signed area of the polygon. The absolute value removes the sign.

When to Use It

• Any polygon with known coordinates — triangles, quadrilaterals, irregular shapes.
• In coordinate geometry problems where vertices are given and you need area quickly.
• In surveying, where land boundaries are marked by GPS coordinates.

A Quick Check

For a rectangle with vertices $(0,0)$, $(a,0)$, $(a,b)$, $(0,b)$:

Sum of $x_i y_{i+1}$: $0\cdot0 + a\cdot b + a\cdot b + 0\cdot0 = 2ab$

Sum of $x_{i+1} y_i$: $a\cdot0 + a\cdot0 + 0\cdot b + 0\cdot b = 0$

Difference: $2ab$

Area: $\frac{1}{2} \times 2ab = ab$ — correct.

The formula is a compact, reliable tool. Once you see the cross-multiplication pattern, you will never need to split polygons into triangles again.