How to Find the Area of a Triangle from Its Vertex Coordinates

When you know a triangle’s three vertices in the coordinate plane, you can calculate its area directly from the coordinates. You do not need to find side lengths, angles, or a height first.

The standard formula is half the absolute value of a determinant. It is also a three-point version of the shoelace formula. The absolute value matters because changing the order of the vertices can change the sign of the determinant, but not the triangle’s area.

The coordinate formula

Let the vertices be $A=(x_1,y_1)$, $B=(x_2,y_2)$, and $C=(x_3,y_3)$. Then

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

An equivalent determinant form is

$$\text{Area} = \frac{1}{2}\left|\begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}\right|.$$

Expanding the determinant gives the coordinate formula above. You may also write the same calculation in shoelace form:

$$\text{Area} = \frac{1}{2}\left|x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3\right|.$$

All three expressions give the same result. Use whichever is easiest to substitute into without losing track of the signs.

Why the determinant gives the area

Take one vertex, say $A$, as a reference point. The two vectors from $A$ to the other vertices are

$$\overrightarrow{AB}=(x_2-x_1,\,y_2-y_1), \qquad \overrightarrow{AC}=(x_3-x_1,\,y_3-y_1).$$

The absolute value of their $2\times2$ determinant is the area of the parallelogram they span:

$$\left|(x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1)\right|.$$

A triangle formed by the same two vectors has half that parallelogram’s area. Dividing by $2$ gives the triangle formula. The determinant is signed: its sign records the orientation of the point order, not whether the area itself is negative.

Worked example

Find the area of the triangle with vertices $A=(1,2)$, $B=(5,4)$, and $C=(-1,6)$.

Substitute into the coordinate formula:

$$\begin{aligned} \text{Area} &= \frac{1}{2}\left|1(4-6)+5(6-2)+(-1)(2-4)\right| \\ &= \frac{1}{2}\left|-2+20+2\right| \\ &= \frac{1}{2}(20) \\ &= 10. \end{aligned}$$

The area is 10 square units. The units are squared because area measures a two-dimensional region. For instance, if the coordinates are measured in centimeters, the answer is $10$ square centimeters.

You can check the result using vectors. Here, $\overrightarrow{AB}=(4,2)$ and $\overrightarrow{AC}=(-2,4)$, so the determinant is $4(4)-2(-2)=20$. Half its absolute value is again $10$.

A quick collinearity check

If the three points lie on one straight line, they do not enclose a triangle with positive area. The determinant is then zero, so the formula returns zero. For example, the points $(0,0)$, $(2,2)$, and $(4,4)$ give

$$\frac{1}{2}|0(2-4)+2(4-0)+4(0-2)|=\frac{1}{2}|0|=0.$$

A zero result can therefore indicate that the vertices are collinear. It can also result from a substitution or arithmetic error, so check the coordinates if zero was unexpected.

Common mistakes

• Forgetting the factor of one-half: The determinant gives the area of a parallelogram. A triangle with the same two side vectors has half that area.
• Leaving off the absolute value: A negative determinant indicates the order of the vertices, not negative area.
• Mixing up coordinate pairs: Keep each point’s $x$- and $y$-coordinates together, and use the same vertex order throughout the calculation.
• Reporting the wrong units: If coordinates are in length units, area is in square units.
• Assuming the points must be listed around the triangle: Any order works if you use it consistently. A different order may change the determinant’s sign, but the absolute value keeps the area unchanged.

For a parallelogram, the same determinant idea gives its full area without the factor of one-half.