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Lesson Explainer: Using Determinants to Calculate Areas Mathematics • First Year of Secondary School

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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 (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=12||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

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 (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=12||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

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 areadet=12||||0514513βˆ’41||||.

Expanding over the first column, we get areadetdetdet=12||0Γ—ο€Ό51βˆ’41οˆβˆ’4Γ—ο€Ό51βˆ’41+3Γ—ο€Ό5151||=12|βˆ’4(5Γ—1βˆ’(βˆ’4)Γ—1)+3(5Γ—1βˆ’5Γ—1)|=12|βˆ’36|=18, 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: areabaseheight=12Γ—Γ—=12Γ—4Γ—9=18.

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 (β„Ž,0), (6,0), and (0,3) is 9 square units, then β„Ž=.

  1. 0 or βˆ’12
  2. 0 or 12
  3. βˆ’6 or 6
  4. 12 or βˆ’12

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 β„Ž. 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 (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) has an area given by the following: areadet=12||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

Substituting in the coordinates of the vertices of this triangle gives us areadet=12||||οβ„Ž01601031||||.

This area is equal to 9, and we can evaluate the determinant by expanding over the second column: 9=12||||οβ„Ž01601031||||=12||βˆ’0Γ—ο€Ό6101+0Γ—ο€Όβ„Ž101οˆβˆ’3Γ—ο€Όβ„Ž161||=12|βˆ’3(β„Žβˆ’6)|=32|β„Žβˆ’6|.detdetdetdet

Therefore, rearranging this equation gives 6=|β„Žβˆ’6|.

This gives us two options, either 6=β„Žβˆ’6 or βˆ’6=β„Žβˆ’6.

We can solve both of these equations to get β„Ž=0 or β„Ž=12, 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: areaparallelogramareatriangledetdet()=2()=2Γ—12||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||=||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

We can summarize this as follows.

Formula: Area of a Parallelogram Using Determinants

The area of a parallelogram with any three vertices at (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

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 (0,0), (π‘Ž,𝑏), (𝑐,𝑑), and (𝑒,𝑓), 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 (π‘Ž,𝑏) and (𝑐,𝑑).

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: areatriangledet()=12||||001π‘Žπ‘1𝑐𝑑1||||.

Expanding over the first row, we get areatriangledetdetdetdet()=12|||0×𝑏1𝑑1οˆβˆ’0Γ—ο€Όπ‘Ž1𝑐1+1Γ—ο€Όπ‘Žπ‘π‘π‘‘οˆ|||=12|||ο€Όπ‘Žπ‘π‘π‘‘οˆ|||.

Since the area of the parallelogram is twice this value, we have areaparallelogramdet()=|||ο€Όπ‘Žπ‘π‘π‘‘οˆ|||.

We could also have split the parallelogram along the line segment between the origin and (𝑒,𝑓) 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 12||||001𝑒𝑓1𝑐𝑑1||||=12|||𝑒𝑓𝑐𝑑|||.detdet

The area of the parallelogram is twice this value: areaparallelogramdet()=|||𝑒𝑓𝑐𝑑|||.

In either case, the area of the parallelogram is the absolute value of the determinant of the 2Γ—2 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 (π‘Ž,𝑏) and (𝑐,𝑑), then its area is given by areadet=|||ο€Όπ‘Žπ‘π‘π‘‘οˆ|||.

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 (1,1), (βˆ’4,5), (βˆ’2,8), and (3,4).

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 (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

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 areadet=||||111βˆ’451βˆ’281||||.

Expanding over the first row gives us areadetdetdetsquareunits=||1Γ—ο€Ό5181οˆβˆ’1Γ—ο€Όβˆ’41βˆ’21+1Γ—ο€Όβˆ’45βˆ’28||=|(5βˆ’8)βˆ’(βˆ’4+2)+(βˆ’32+10)|=23.

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 (π‘Ž,𝑏) and (𝑐,𝑑), then its area is given by areadet=|||ο€Όπ‘Žπ‘π‘π‘‘οˆ|||.

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 (1,1), we will do this by translating the parallelogram one unit left and one unit down. This gives us the following coordinates for its vertices: (1βˆ’1,1βˆ’1)=(0,0),(βˆ’4βˆ’1,5βˆ’1)=(βˆ’5,4),(βˆ’2βˆ’1,8βˆ’1)=(βˆ’3,7),(3βˆ’1,4βˆ’1)=(2,3).

We can actually use any two of the vertices not at the origin to determine the area of this parallelogram. Hence, areadet=||ο€Όβˆ’54βˆ’37||=|βˆ’35βˆ’(βˆ’12)|=|βˆ’23|=23.

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 (2,βˆ’2), (4,βˆ’2), and (0,2) 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 (π‘Ž,𝑏) and (𝑐,𝑑), then its area is given by areadet=|||ο€Όπ‘Žπ‘π‘π‘‘οˆ|||.

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 (0,2) to the origin by translating each of the vertices down two units; this gives us (0,2βˆ’2)=(0,0),(2,βˆ’2βˆ’2)=(2,βˆ’4),(4,βˆ’2βˆ’2)=(4,βˆ’4).

We use the coordinates of the latter two points to find the area of the parallelogram: areaparallelogramdet()=||ο€Ό2βˆ’44βˆ’4||=|βˆ’8+16|=8.

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 (2,βˆ’2), (4,βˆ’2), and (0,2) 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 𝐴(1,3), 𝐡(4,2), 𝐢(4.5,5), and 𝐷(2,6).

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 2Γ—2 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 3Γ—3 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 𝐴.

We can see from the diagram that 𝐴(1,3), 𝐡(4,2), and 𝐢(4.5,5). We recall that the area of a triangle with vertices (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=12||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.

So, we can find the area of this triangle by using our determinant formula: areadet(𝐴𝐡𝐢)=12||||1314214.551||||.

We expand this determinant along the first column to get areadetdetdetsquareunits(𝐴𝐡𝐢)=12||1Γ—ο€Ό2151οˆβˆ’4Γ—ο€Ό3151+4.5Γ—ο€Ό3121||=12|1(2βˆ’5)βˆ’4(3βˆ’5)+4.5(3βˆ’2)|=12|βˆ’3+8+4.5|=4.75.

Similarly, the area of triangle 𝐴𝐢𝐷 is given by areadetdetdetdetsquareunits(𝐴𝐢𝐷)=12||||1314.551261||||=12||1Γ—ο€Ό5161οˆβˆ’4.5Γ—ο€Ό3161+2Γ—ο€Ό3151||=12|1(5βˆ’6)βˆ’4.5(3βˆ’6)+2(3βˆ’5)|=12|βˆ’1+13.5βˆ’4|=4.25.

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 detπ‘₯𝑦1π‘₯𝑦1π‘₯𝑦1 tells us the signed area of a parallelogram with three vertices at (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦), 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 (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦), where detπ‘₯𝑦1π‘₯𝑦1π‘₯𝑦1=0, 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.

  1. 𝐴(βˆ’6,4), 𝐡(βˆ’8,4), 𝐢(3,10)
  2. 𝐴(βˆ’10,βˆ’4), 𝐡(βˆ’8,βˆ’2), 𝐢(βˆ’5,1)
  3. 𝐴(βˆ’3,6), 𝐡(8,βˆ’7), 𝐢(βˆ’3,βˆ’8)
  4. 𝐴(βˆ’10,βˆ’6), 𝐡(βˆ’2,1), 𝐢(0,βˆ’9)

Answer

We first recall that three distinct points (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) are collinear if detπ‘₯𝑦1π‘₯𝑦1π‘₯𝑦1=0.

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 detο€βˆ’641βˆ’8413101=βˆ’6(4βˆ’10)βˆ’4(βˆ’8βˆ’3)+(βˆ’80βˆ’12)=βˆ’12.

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 detο€βˆ’10βˆ’41βˆ’8βˆ’21βˆ’511=βˆ’10(βˆ’2βˆ’1)+4(βˆ’8+5)+(βˆ’8βˆ’10)=0.

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 detο€βˆ’3618βˆ’71βˆ’3βˆ’81=βˆ’3(βˆ’7+8)βˆ’6(8+3)+(βˆ’64βˆ’21)=βˆ’154.

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 detο€βˆ’10βˆ’61βˆ’2110βˆ’91=βˆ’10(1+9)+6(βˆ’2βˆ’0)+(18βˆ’0)=βˆ’94.

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 𝐴(βˆ’10,βˆ’4), 𝐡(βˆ’8,βˆ’2), and 𝐢(βˆ’5,1) 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 (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=12||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.
  • The area of a parallelogram with any three vertices at (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦) is given by areadet=||||π‘₯𝑦1π‘₯𝑦1π‘₯𝑦1||||.
  • If a parallelogram has one vertex at the origin and two other vertices at (π‘Ž,𝑏) and (𝑐,𝑑), then its area is given by areadet=|||ο€Όπ‘Žπ‘π‘π‘‘οˆ|||.
  • If we have three distinct points (π‘₯,𝑦), (π‘₯,𝑦), and (π‘₯,𝑦), where detπ‘₯𝑦1π‘₯𝑦1π‘₯𝑦1=0, then the points are collinear.

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