In this explainer, we will learn how to use determinants to calculate areas of triangles and parallelograms given the coordinates of their vertices.

There are a lot of useful properties of matrices we can use to solve problems. We can use the determinant of matrices to help us calculate the area of a polygon given its vertices. To do this, we will start with the formula for the area of a triangle using determinants.

### Theorem: Area of a Triangle Using Determinants

The area of a triangle with vertices , , and is given by

We take the absolute value of this determinant to ensure the area is nonnegative.

There are other methods of finding the area of a triangle. For example, we know that the area of a triangle is given by half the length of the base times the height. However, this formula requires us to know these lengths rather than just the coordinates of the vertices.

Let’s see an example of using this formula to evaluate the area of a triangle from the coordinates of its vertices.

### Example 1: Finding the Area of a Triangle on the Cartesian Coordinate Using Determinants

Find the area of the triangle below using determinants.

### Answer

In this question, we could find the area of this triangle in many different ways. For example, we could use geometry. However, we are tasked with calculating the area of a triangle by using determinants.

To do this, we will need to use the fact that the area of a triangle with vertices , , and is given by

So, we need to find the vertices of our triangle; we can do this using our sketch.

It is worth pointing out that the order we label the vertices in does not matter, since this would only result in switching the rows of our matrix around, which only changes the sign of the determinant.

Therefore, the area of our triangle is given by

Expanding over the first column, we get giving us that the area of our triangle is 18 square units.

We can check our answer by calculating the area of this triangle using a different method. For example, the area of a triangle is half the length of the base times the height, and we can find both of the values from our sketch.

Taking the horizontal side as the base, we get that the length of the base is 4 and the height of the triangle is 9. So, we can use these to calculate the area of the triangle:

This confirms our answer that the area of our triangle is 18 square units.

We can use the formula for the area of a triangle by using determinants to find the possible coordinates of a vertex of a triangle with a given area, as we will see in our next example.

### Example 2: Finding Information about the Vertices of a Triangle given Its Area

Fill in the blank: If the area of a triangle whose vertices are , , and is 9 square units, then .

- 0 or
- 0 or 12
- or 6
- 12 or

### Answer

In this question, we are given the area of a triangle and the coordinates of two of its vertices, and we need to use this to find the coordinates of the third vertex. We could find an expression for the area of our triangle by using half the length of the base times the height. This would then give us an equation we could solve for . However, let us work out this example by using determinants.

We can find the area of the triangle by using the coordinates of its vertices. A triangle with vertices , , and has an area given by the following:

Substituting in the coordinates of the vertices of this triangle gives us

This area is equal to 9, and we can evaluate the determinant by expanding over the second column:

Therefore, rearranging this equation gives

This gives us two options, either or

We can solve both of these equations to get or , which is option B.

Thus far, we have discussed finding the area of triangles by using determinants. It is possible to extend this idea to polygons with any number of sides. We begin by finding a formula for the area of a parallelogram. There are two different ways we can do this.

The first way we can do this is by viewing the parallelogram as two congruent triangles. If we choose any three vertices of the parallelogram, we have a triangle.

It does not matter which three vertices we choose, we split he parallelogram into two triangles. The side lengths of each of the triangles is the same, so they are congruent and have the same area. We can then find the area of this triangle using determinants:

We can summarize this as follows.

### Formula: Area of a Parallelogram Using Determinants

The area of a parallelogram with any three vertices at , , and is given by

There is a second way we can find the area of a parallelogram by using determinants. Since translating a parallelogram does not alter its area, we can translate any parallelogram to have one of its vertices at the origin. Thus, we only need to determine the area of such a parallelogram. Consider a parallelogram with vertices , , , and , as shown in the following figure.

We can find the area of this parallelogram by splitting it into triangles in two different ways, and both methods will give the same area of the parallelogram. For example, we can split the parallelogram in half along the line segment between and .

We can see that the diagonal line splits the parallelogram into two triangles. These two triangles are congruent because they share the same side lengths. Hence, the area of the parallelogram is twice the area of the triangle pictured below.

We can find the area of this triangle by using determinants:

Expanding over the first row, we get

Since the area of the parallelogram is twice this value, we have

We could also have split the parallelogram along the line segment between the origin and as shown below.

Once again, this splits the triangle into two congruent triangles, and we can calculate the area of one of these triangles as

The area of the parallelogram is twice this value:

In either case, the area of the parallelogram is the absolute value of the determinant of the matrix with the rows as the coordinates of any two of its vertices not at the origin. We summarize this result as follows.

### Theorem: Area of a Parallelogram

If a parallelogram has one vertex at the origin and two of its other vertices at and , then its area is given by

Let’s see an example of how we can apply this formula to determine the area of a parallelogram from the coordinates of its vertices.

### Example 3: Computing the Area of a Parallelogram Using Matrices

Use determinants to calculate the area of the parallelogram with vertices , , , and .

### Answer

Let’s start by recalling how we find the area of a parallelogram by using determinants. The area of a parallelogram with any three vertices at , , and is given by

We can choose any three of the given vertices to calculate the area of this parallelogram. For example, if we choose the first three points, then

Expanding over the first row gives us

Therefore, the area of this parallelogram is 23 square units.

We could also use the fact that if a parallelogram has one vertex at the origin and any two of its other vertices at and , then its area is given by

To use this formula, we need to translate the parallelogram so that one of its vertices is at the origin. Since one of the vertices is the point , we will do this by translating the parallelogram one unit left and one unit down. This gives us the following coordinates for its vertices:

We can actually use any two of the vertices not at the origin to determine the area of this parallelogram. Hence,

Therefore, the area of this parallelogram is 23 square units.

We were able to find the area of a parallelogram by splitting it into two congruent triangles. Similarly, we can find the area of a triangle by considering it as half of a parallelogram, as we will see in our next example.

### Example 4: Computing the Area of a Triangle Using Matrices

Use determinants to work out the area of the triangle with vertices , , and by viewing the triangle as half of a parallelogram.

### Answer

First, we want to construct our parallelogram by using two of the same triangles given to us in the question. This means there will be three different ways to create this parallelogram, since we can combine the two triangles on any side. We can see this in the following three diagrams.

All three of these parallelograms have the same area since they are formed by the same two congruent triangles. However, we do not need the coordinates of the fourth point to find the area of a parallelogram by using determinants. Recall that if a parallelogram has one vertex at the origin and two other vertices at and , then its area is given by

We can use this to determine the area of the parallelogram by translating the shape so that one of its vertices lies at the origin. We translate the point to the origin by translating each of the vertices down two units; this gives us

We use the coordinates of the latter two points to find the area of the parallelogram:

Finally, we remember that the area of our triangle is half of this value, giving us that the area of the triangle with vertices at , , and is 4 square units.

If we can calculate the area of a triangle using determinants, then we can calculate the area of any polygon by splitting it into triangles (called triangulation). Let’s see an example where we are tasked with calculating the area of a quadrilateral by using determinants.

### Example 5: Computing the Area of a Quadrilateral Using Determinants of Matrices

Consider the quadrilateral with vertices , , , and .

By breaking it into two triangles as shown, calculate the area of this quadrilateral using determinants.

### Answer

We want to find the area of this quadrilateral by splitting it up into the triangles as shown. This means we need to calculate the area of these two triangles by using determinants and then add the results together. We have two options for finding the area of a triangle by using determinants: We could treat the triangles as half a parallelogram and use the determinant of a matrix to find the area of this parallelogram, or we could use our formula for the area of a triangle by using the determinant of a matrix. Since we have a diagram with the vertices given, we will use the formula for finding the areas of the triangles directly.

Let’s start with triangle .

We can see from the diagram that , , and . We recall that the area of a triangle with vertices , , and is given by

So, we can find the area of this triangle by using our determinant formula:

We expand this determinant along the first column to get

Similarly, the area of triangle is given by

Summing the areas of these two triangles together, we see that the area of the quadrilateral is 9 square units.

There is another useful property that these formulae give us. Since tells us the signed area of a parallelogram with three vertices at , , and , if this determinant is 0, the triangle with these points as vertices must also have zero area. The area of this triangle can only be zero if the points are not distinct or if the points all lie on the same line (i.e., they are collinear).

### Theorem: Test for Collinear Points

If we have three distinct points , , and , where , then the points are collinear.

Let’s see an example of how to apply this.

### Example 6: Determining If a Set of Points Are Collinear or Not Using Determinants

By using determinants, determine which of the following sets of points are collinear.

- , ,
- , ,
- , ,
- , ,

### Answer

We first recall that three distinct points , , and are collinear if

We note that each given triplet of points is a set of three distinct points. So, we can calculate the determinant of this matrix for each given triplet of points to determine their collinearity. We compute the determinants of all four matrices by expanding over the first row.

Option A would be

Since this is nonzero, the area of the triangle with these points as vertices in also nonzero. Hence, these points are not collinear.

Option B would be

Since this is equal to zero, the area of the triangle with these points as vertices is 0. Hence, these points must be collinear.

Option C would be

Since, this is nonzero, the area of the triangle with these points as vertices in also nonzero. Hence, these points are not collinear.

Option D would be

Since, this is nonzero, the area of the triangle with these points as vertices in also nonzero. Hence, these points are not collinear.

Hence, the points , , and are collinear, which is option B.

Let us finish by recapping a few of the important concepts of this explainer.

### Key Points

- The area of a triangle with vertices , , and is given by
- The area of a parallelogram with any three vertices at , , and is given by
- If a parallelogram has one vertex at the origin and two other vertices at and , then its area is given by
- If we have three distinct points , , and , where , then the points are collinear.