Area of a triangle

A triangle in the coordinate plane with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ has area given by an explicit determinant. The formula is widely used in geometry, in surveying, and in computer graphics. As a corollary, three points are collinear iff this determinant is zero , a simple algebraic test that avoids slope comparisons.

The formula

Given three points $A_1 = (x_1, y_1), A_2 = (x_2, y_2), A_3 = (x_3, y_3)$ in the plane, the area of triangle $A_1 A_2 A_3$ is

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

The absolute value is necessary: a positive determinant means the vertices are listed counterclockwise; a negative determinant means clockwise. Either way, area is positive.

Derivation

Expand the determinant along the third column:

$\det = x_1(y_2 - y_3) - y_1(x_2 - x_3) + (x_2 y_3 - x_3 y_2)$ $= x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$.

Geometrically this is twice the signed area of the triangle, obtained by computing the cross product of two edge vectors $\vec{A_1 A_2}$ and $\vec{A_1 A_3}$. The factor $1/2$ in the formula corrects for this doubling.

Collinearity test

Three points are collinear iff they form a degenerate triangle, i.e., the area is zero. By the formula, this happens iff

$\det\begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} = 0.$

This is often the most efficient way to check collinearity.

Equation of a line through two points

The line through $(x_1, y_1)$ and $(x_2, y_2)$ consists of all points $(x, y)$ collinear with these two. Set up the collinearity determinant:

$\det\begin{pmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{pmatrix} = 0.$

Expanding gives the line equation in two-point form.

Worked examples

Example 1. Find the area of triangle with vertices $(2, 3), (4, 7), (8, 5)$.

$\det\begin{pmatrix} 2 & 3 & 1 \\ 4 & 7 & 1 \\ 8 & 5 & 1 \end{pmatrix} = 2(7 - 5) - 3(4 - 8) + (20 - 56) = 4 + 12 - 36 = -20$.

Area $= |-20|/2 = 10$.

Example 2. Are $(1, 2), (3, 4), (5, 6)$ collinear?

$\det\begin{pmatrix} 1 & 2 & 1 \\ 3 & 4 & 1 \\ 5 & 6 & 1 \end{pmatrix} = 1(4 - 6) - 2(3 - 5) + 1(18 - 20) = -2 + 4 - 2 = 0$. Yes, collinear.

Example 3. Find the equation of the line through $(1, 2)$ and $(3, 4)$.

$\det\begin{pmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 4 & 1 \end{pmatrix} = 0$.

Expand: $x(2 - 4) - y(1 - 3) + (4 - 6) = -2x + 2y - 2 = 0$, so $y = x + 1$.

Example 4. Find the value of $k$ so that the area of the triangle with vertices $(1, 4), (k, 0), (0, k)$ is $4$.

$\det\begin{pmatrix} 1 & 4 & 1 \\ k & 0 & 1 \\ 0 & k & 1 \end{pmatrix} = 1(0 - k) - 4(k - 0) + (k^2 - 0) = k^2 - 5k$.

Area $= |k^2 - 5k|/2 = 4 \Rightarrow k^2 - 5k = \pm 8$.

Solve: $k^2 - 5k - 8 = 0$ gives $k = (5 \pm \sqrt{57})/2$. Or $k^2 - 5k + 8 = 0$ has no real roots ($\Delta = 25 - 32 < 0$). So $k = (5 + \sqrt{57})/2$ or $(5 - \sqrt{57})/2$.

Example 5. Show that the points $(a, b + c), (b, c + a), (c, a + b)$ are collinear.

$\det\begin{pmatrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{pmatrix}$. Add column $1$ to column $2$: $\begin{pmatrix} a & a + b + c & 1 \\ b & a + b + c & 1 \\ c & a + b + c & 1 \end{pmatrix}$. Pull out $a + b + c$: $(a + b + c) \det\begin{pmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{pmatrix} = 0$ (two equal columns).

So area $= 0$; the points are collinear.

Example 6. Find the area of the parallelogram with sides as vectors $(2, 3)$ and $(5, 1)$ starting from origin.

$\text{Area} = |\det\begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix}| = |2 - 15| = 13$.

Try it yourself

1. Find the area of the triangle with vertices $(0, 0), (4, 0), (0, 3)$.
2. Check collinearity of $(2, 3), (4, 5), (6, 7)$.
3. Find the area of the triangle with vertices $(1, 2), (-1, 3), (3, -1)$.
4. Find the equation of the line through $(2, 1)$ and $(4, 5)$ using determinants.
5. Show that the points $(a, 0), (0, b), (a, b)$ are not collinear unless $a = 0$ or $b = 0$.
6. If the area of the triangle with vertices $(x, 4), (2, -6), (5, 4)$ is $35$ sq units, find $x$.
7. Determine $k$ so that the points $(k, 2k), (-2, 6), (3, 1)$ are collinear.
8. Find the area of the parallelogram with vertices $(0, 0), (3, 0), (4, 5), (1, 5)$. Hint: divide into two triangles.
9. Find the equation of the line through $(3, -1)$ that is parallel to the line through $(2, 4)$ and $(6, 4)$.
10. If $(a, b), (c, d), (e, f)$ are collinear, prove $\det\begin{pmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{pmatrix} = 0$.
11. Find the area of the triangle with vertices $(2, 0), (0, 3), (4, 5)$.
12. Are the points $(0, 0), (1, 2), (3, 6)$ collinear?
13. Find the area of the triangle whose vertices are $(1, 1), (3, 2), (5, 1)$.
14. Show that the line $\det\begin{pmatrix} x & y & 1 \\ -1 & 2 & 1 \\ 3 & -4 & 1 \end{pmatrix} = 0$ passes through both $(-1, 2)$ and $(3, -4)$.

Pitfalls / Tricks

• Always take the absolute value at the end; area is positive.
• A determinant of zero means area zero, i.e., collinear points (degenerate triangle).
• The third column of $1$s is what makes the formula work; do not omit it.
• For a parallelogram with vectors $\vec u, \vec v$ from a vertex, the area is the absolute value of $\det(\vec u, \vec v)$, no $1/2$ factor.

Next we systematise minors and cofactors.