Area of a triangle

Give a surveyor three GPS pins and no tape measure, and she can still tell you exactly how much land they enclose — coordinate geometry turns "find the area of this triangle" into pure arithmetic on the three $(x,y)$ pairs, with no perpendicular height to construct and no diagram required. This single formula, first met in ICSE Class 10, resurfaces throughout the ISC Class 11 Coordinate Geometry unit: it opens the "brief recall" that also includes the distance and section formulas, and it doubles as the fastest test for whether three points even form a triangle at all, since a "triangle" of zero area is really three points sitting on one straight line. This article rebuilds the formula from a rectangle-and-subtraction picture, applies it to a numeric triangle, and then flips it around into the collinearity test — an algebraic shortcut that avoids the surds of the distance-formula method entirely.

The core insight

For vertices $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ , the triangle's area is half the absolute value of $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$ , and this expression equals zero exactly when the three points are collinear.

The core idea

The formula. Let a triangle have vertices $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ . Its area is

\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|

where: $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the three vertices in any order, and the outer bars are absolute value — the raw sum inside can come out negative depending on whether the vertices happen to be listed clockwise or anticlockwise, but area itself is always reported positive.

Why a fixed pattern matters. Each term pairs one vertex's $x$ -coordinate with the difference of the other two vertices' $y$ -coordinates, cycling $1\to(2,3)$ , $2\to(3,1)$ , $3\to(1,2)$ . Swap that pattern and the formula breaks — this is the single most common source of wrong answers on this topic.

The collinearity test. Because area is zero precisely when the three "vertices" actually lie on one line, dropping the absolute value and setting the inner expression to zero gives a direct algebraic test:

x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) = 0

where: the symbols are the same three vertex coordinates as above; this equation holds if and only if $A$ , $B$ , $C$ are collinear. Compared with the earlier distance-formula collinearity test (checking whether one distance equals the sum of the other two), this one stays in exact rational arithmetic whenever the coordinates are rational — no square roots to compare.

Figure

A triangle plotted on coordinate axes with its three vertices labelled and dashed lines showing the enclosing rectangle used to derive the area formula

Generalises further. The same cyclic pattern, continued around any polygon's vertices in order, gives that polygon's area (the shoelace pattern) — but at Class 11 only the three-point triangle case is required.

Why it belongs with the distance and section formulas. All three are "recall" tools built purely from coordinate subtraction — the distance formula subtracts and combines both coordinates under a square root, the section formula subtracts and weights coordinates by a ratio, and the area formula subtracts $y$ 's and multiplies by the opposite $x$ . Keeping the three patterns visually distinct in your head is worth more marks than memorising any one of them in isolation, since exam questions often mix them within a single multi-part problem — for instance, using the section formula to find a fourth point and then the area formula to check it lies where expected.

Derivation ADeriving the coordinate area formula from an enclosing rectangle

The goal is to show that the compact algebraic expression $\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ genuinely equals the triangle's area, by building it up geometrically from a rectangle that encloses the triangle and three right-angled corner triangles that get subtracted away, using only areas of rectangles and right triangles — no trigonometry, no calculus.

Step 1. Enclose the triangle. Draw the smallest axis-aligned rectangle containing $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ , using the extreme $x$ - and $y$ -values among the three points as its sides. The triangle sits inside this rectangle, touching it at its three vertices, and the rectangle's area splits into the triangle plus (at most) three leftover right-angled corner pieces.

Step 2. Write each corner piece as a right triangle or trapezium built from horizontal and vertical differences of coordinates — for example, the piece cut off near $A$ has legs proportional to differences like $(x_2-x_1)$ and $(y_2-y_1)$ , and similarly for the other two corners. Each such area is a simple product-and-halve expression in the coordinates, just like the area of any right triangle is half base times height.

Step 3. Subtract the three corner areas from the rectangle's total area. When this subtraction is carried out algebraically — expanding every product and collecting terms — the rectangle's own area and most of the corner terms cancel, and what survives collapses into exactly $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$ , still without the halving or the absolute value applied yet.

Step 4. Halve and take the absolute value. The leftover expression can come out negative purely because of the order the vertices were listed (clockwise gives one sign, anticlockwise the other), but a physical area cannot be negative, so the final step is $\text{Area}=\tfrac12\left|\,\cdot\,\right|$ . This closes the derivation: the same construction, run for any three points, always reproduces the stated formula.

Worked example

Find the area of the triangle with vertices $A(1, 1)$ , $B(4, 5)$ , $C(6, 2)$ , and then check whether $A$ , $B$ , $C$ are collinear.

Step 1. Label the coordinates: $x_1=1,\ y_1=1,\ x_2=4,\ y_2=5,\ x_3=6,\ y_3=2$ .

Step 2. Substitute into the formula: $\text{Area} = \dfrac12\left| 1(5-2) + 4(2-1) + 6(1-5) \right| = \dfrac12\left| 1(3) + 4(1) + 6(-4) \right|$ .

Step 3. Simplify inside the bars: $= \dfrac12\left| 3 + 4 - 24 \right| = \dfrac12\left| -17 \right| = \dfrac{17}{2}$ .

Step 4. Since the inner expression is $-17 \ne 0$ , the area is nonzero, confirming $A$ , $B$ , $C$ genuinely form a triangle (not collinear). The area is $\dfrac{17}{2} = 8.5$ square units.

Competitive stretch

Show that $P(2, 3)$ , $Q(5, 7)$ , $R(8, 11)$ are collinear, then find $k$ so that $S(k, 15)$ also lies on that line.

Step 1. Test collinearity of $P,Q,R$ : $2(7-11) + 5(11-3) + 8(3-7) = 2(-4)+5(8)+8(-4) = -8+40-32 = 0$ . Since this is $0$ , the three points are collinear.

Step 2. Force $P$ , $Q$ , $S(k,15)$ to also give zero: $2(7-15) + 5(15-3) + k(3-7) = 0 \implies -16+60-4k=0 \implies 44=4k$ , so $k=11$ . $S(11,15)$ lies on the same line as $P$ , $Q$ , $R$ — matching the pattern that each step from $P$ to $Q$ to $R$ adds $(3,4)$ , so the next point is $(8+3,11+4)=(11,15)$ .

Common traps

Dropping the absolute value. The raw expression can be negative purely because of vertex order; always take the modulus before halving, since area cannot be negative.
Mismatching the cyclic subscripts. Each $x_i$ must pair with the difference of the other two $y$ 's in order — writing $x_1(y_2-y_1)$ instead of $x_1(y_2-y_3)$ silently breaks the formula.
Skipping the collinearity check. When a problem gives three points and asks for "the area of triangle ABC," it is worth confirming the expression is not accidentally zero, especially in multi-part problems built on earlier results.
Confusing this with the distance or section formula. All three recall tools use similar-looking subscripts; this one uses differences of $y$ 's multiplied by the opposite $x$ , not differences of both coordinates together.

Ask the tutor: If your area comes out negative or you lose track of which $y$ -difference pairs with which $x$ , ask the AI tutor to re-derive the formula with you against a labelled sketch of your own triangle.

Why do we take the absolute value instead of just being careful about the order of the vertices?

Because the sign of the raw expression genuinely flips depending on whether you list the vertices clockwise or anticlockwise, and there is no universally "correct" order to insist on — either is a valid way to name a triangle's vertices. Rather than memorising which order gives a positive result, it is simpler and safer to always compute the expression, then take its absolute value before halving. That way the answer is correct regardless of how the question happened to list $A$ , $B$ , $C$ .

Is this formula related to the shoelace formula I've seen mentioned online?

Yes — the three-point version here is exactly the triangle case of the more general "shoelace" pattern for polygon area, where you cycle through all the vertices in order, multiplying each $x$ by the difference of its neighbours' $y$ 's, and sum before halving and taking the modulus. For any polygon with more than three vertices you would extend the same pattern around all of them, but ISC Class 11 only requires the three-vertex triangle case, so you never need the general version for this syllabus.

When should I use this instead of the distance formula to check collinearity?

Almost always prefer the area formula when you already have coordinates, because it only involves multiplication, subtraction, and one halving — all exact rational arithmetic if the coordinates are rational. The distance-formula method requires computing three separate distances (often involving square roots) and then checking whether one equals the sum of the other two, which is slower and introduces surds that are easy to mishandle. The area test reaches the same conclusion with far less arithmetic risk.

Area of a triangle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads