Area of a triangle and collinear points

What you will be able to do

• find the area of a triangle from the coordinates of its vertices;
• test whether three points lie on one line, and find an unknown coordinate;
• write the equation of the line through two points as a determinant;
• use the determinant of a matrix as an area scale factor.

The area of a triangle

The triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$ and $C(x_3, y_3)$ has area

The outer bars mean the absolute value, since an area is never negative. Expanding along the third column gives the familiar form

For $A(1, 1)$, $B(4, 2)$ and $C(2, 5)$,

so the area is $\tfrac{11}{2} = 5.5$ square units.

Why does it work? Take the vertex $A$ as the origin. Then the sides $AB$ and $AC$ are two arrows, and the triangle is half of the parallelogram they make. The determinant of the two arrows is the area of the parallelogram.

Three points on a line

If the three points are on one line, the triangle is flat and its area is $0$. So the points are collinear exactly when the determinant is $0$.

Are $(1, 2)$, $(3, 8)$ and $(5, 14)$ on one line? The determinant is $1(8 - 14) - 2(3 - 5) + (42 - 40) = -6 + 4 + 2 = 0$. Yes. Each step across $2$ raises $y$ by $6$.

To find a missing coordinate, set the determinant equal to $0$. For $(2, 3)$, $(4, k)$ and $(6, 9)$ to be collinear:

The line through two points

A point $(x, y)$ is on the line through $(x_1, y_1)$ and $(x_2, y_2)$ exactly when the three points are collinear. So the line is

For $(1, 2)$ and $(3, 8)$, the determinant is $x(2 - 8) - y(1 - 3) + (8 - 6) = -6x + 2y + 2$. Setting it to $0$ and dividing by $-2$ gives $3x - y - 1 = 0$, that is, $y = 3x - 1$.

A matrix as an area scale factor

A $2 \times 2$ matrix $T$ moves every point $v$ to $Tv$. It turns the unit square, with area $1$, into a parallelogram on the columns of $T$. So the area of the image is $|\det T|$. Every shape has its area multiplied by the same number, $|\det T|$. If $\det T$ is negative, the picture is also flipped over.

Take the matrix

Then $\det T = 5$. Any triangle sent through $T$ becomes a triangle with five times the area. This idea is used in A-Level Further Maths and IB Higher Level, where a determinant of $0$ means the picture is flattened to a line.

A worked example

Find the area of the triangle with vertices $(2, 3)$, $(-1, 0)$ and $(5, 1)$, and decide whether $(2, 3)$, $(-1, 0)$ and $(5, 6)$ are collinear.

1. The first determinant. $2(0 - 1) - 3(-1 - 5) + (-1 - 0) = -2 + 18 - 1 = 15$.
2. The area. $\tfrac{1}{2} \cdot 15 = 7.5$ square units.
3. The second set. $2(0 - 6) - 3(-1 - 5) + (-6 - 0) = -12 + 18 - 6 = 0$. The three points are collinear.

Practise with new numbers

Try it in Python

The program finds the determinant with rows $(x, y, 1)$ for two sets of points. It prints the area and tells you whether the points are collinear.

Common mistakes

• Forgetting the half. The determinant is twice the area of the triangle.
• Giving a negative area. Take the absolute value.
• Leaving out the column of ones. The rows are $(x_i, y_i, 1)$.
• Using only two points for collinearity. Any two points are on a line. The test needs a third.
• Taking a determinant of $0$ for "no triangle" in a problem about area. It means the points are collinear.

Remember these

• Area $= \tfrac{1}{2}\left|\det\right|$, with rows $(x_i, y_i, 1)$.
• Three points are collinear exactly when this determinant is $0$.
• The line through two points: replace one row by $(x, y, 1)$ and set the determinant to $0$.
• A matrix multiplies every area by $|\det T|$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the area of the triangle with vertices $(0, 0)$, $(4, 0)$ and $(0, 6)$ using a determinant.

2. Are $(1, 1)$, $(2, 3)$ and $(3, 5)$ collinear?

3. For which $k$ are $(1, 2)$, $(2, k)$ and $(4, 8)$ collinear?

4. Write the equation of the line through $(0, 1)$ and $(2, 5)$ as a determinant, and simplify.

= 0$ along the first row.

5. A matrix has determinant $-3$. It maps a triangle of area $4$ to another triangle. What is the area of the image?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 4 (Determinants), the section on the area of a triangle.
• OpenStax, Precalculus 2e, Section 9.8 (Solving Systems with Cramer's Rule), the subsection on the area of a triangle.