Lesson Video: Using Determinants to Calculate Areas Mathematics • 10th Grade

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

19:03

Video Transcript

Using Determinants to Calculate Areas

In this video, we are going to learn how we can use determinants to calculate the areas of polygons given the coordinates of their vertices. In particular, we will discuss how to calculate the areas of triangles in the plane given the coordinates of their vertices and how to calculate the areas of parallelograms in the plane given the coordinates of three of its vertices. We will also discuss how we can extend this method to find the area of more complicated polygons in the plane by using triangulation.

To begin, we can find the area of a triangle with vertices at π₯ sub one, π¦ sub one; π₯ sub two, π¦ sub two; and π₯ sub three, π¦ sub three using determinants. Its area is given by one-half the absolute value of the determinant of the three-by-three matrix π₯ sub one, π¦ sub one, one, π₯ sub two, π¦ sub two, one, π₯ sub three, π¦ sub three, one. We take the absolute value of the determinant to make sure that our answer is nonnegative.

It is also worth noting that we can label the vertices in any order, since this will only swap the rows of this matrix. This will have the effect of switching the sign, but we take the absolute value. We can prove that this formula holds true by expanding the determinant and rearranging to show that we get an expression for the area of triangle π. However, this is a long process and is beyond the scope of this video.

Instead, we can note that this formula is useful since it only requires us to know the vertices of the triangle, whereas methods such as Heronβs formula require us to know or calculate the lengths of the sides of the triangle. Letβs see an example of applying this method to find the area of a triangle in the coordinate plane.

Find the area of the triangle below using determinants.

In this question, we are given a diagram of a triangle in the coordinate plane and asked to find its area by using determinants. We could directly find the area of this triangle using the formula a half the length of the base times the perpendicular height or by applying Heronβs formula. However, we are told to do this by using determinants.

So we can recall that a triangle with vertices at π₯ sub one, π¦ sub one; π₯ sub two, π¦ sub two; and π₯ sub three, π¦ sub three has area given by one-half the absolute value of the determinant of the three-by-three matrix π₯ sub one, π¦ sub one, one, π₯ sub two, π¦ sub two, one, π₯ sub three, π¦ sub three, one. Therefore, to find the area of this triangle using determinants, we need to find the coordinates of its vertices. We can do this using the given diagram. The vertices are at zero, five; four, five; and three, negative four.

We can now find an expression for the area by substituting the coordinates of the vertices into the formula, where we note that we can label each point in any order since this will only change the sign of the determinant. We have that the area is one-half the absolute value of the determinant of the three-by-three matrix zero, five, one, four, five, one, three, negative four, one.

There are many ways we can evaluate the determinant. One way is to expand over the first row to obtain the following expression. We can evaluate this expression to get one-half the absolute value of negative 36, which is 18 square units. It is worth noting that we can check our answer by finding the length of the base and a perpendicular height from the diagram. In this case, we see that the horizontal side has length four and the height of this triangle is nine.

We can then find the area as one-half the length of the base times the perpendicular height, which is one-half times four multiplied by nine, which we can calculate is 18 square units. Hence, this verifies that the area of the triangle given in the diagram is 18 square units.

In our next example, we will see how we can determine a missing coordinate of the vertex of a triangle using its area and determinants.

Fill in the blank. If the area of a triangle whose vertices are β, zero; six, zero; and zero, three is nine square units, then β equals what. Is it option (A) zero or negative 12? Option (B) zero or 12. Option (C) negative six or six. Or is it option (D) 12 or negative 12?

In this question, weβre given the coordinates of the vertices of a triangle which contains a single unknown coordinate and the area of a triangle. We want to use this to find the possible values of the unknown. We could do this by using any method for finding the area of a triangle. However, since we are given the vertices, we will do this by using determinants.

We recall that the area of a triangle can be calculated by using the formula one-half times the absolute value of the determinant of the three-by-three matrix π₯ sub one, π¦ sub one, one, π₯ sub two, π¦ sub two, one, π₯ sub three, π¦ sub three, one, where π₯ sub π and π¦ sub π are the coordinates of each vertex of the triangle for each value of π.

We can substitute the coordinates of the triangle and the area being equal to nine into this formula to obtain the following equation. We can solve for β by expanding and isolating β on one side of the equation. We can evaluate the determinant in many ways. The easiest is to note that the second column contains two zeros. So expanding over this column will only have a single nonzero term.

We obtain one-half the absolute value of negative three times β minus six. We can simplify this further by taking the factor of negative three out of the absolute value. This becomes a factor of positive three to give the following. This is equal to the area of the triangle, which is nine. We can solve for β by rearranging. We divide both sides of the equation by three over two. This yields that six is equal to the absolute value of β minus six.

We can then solve this absolute value equation by constructing two equations. Either six equals β minus six or six equals negative one times β minus six. Solving each equation gives us that β is equal to 12 or β equals zero. We can then see that this is given by option (B).

Thus far, we have only dealt with finding the area of a triangle using determinants. However, it is possible to extend this idea to any polygon in the plane. A very useful example of this is its application to parallelograms.

Letβs say that we have a parallelogram where we know the coordinates of any three of its vertices. We can split the parallelogram into two congruent triangles along its diagonal. We can then use our formula for the area of a triangle using determinants to find an expression for the area of the parallelogram.

First, we note that the area of the parallelogram π is twice the area of one of the congruent triangles π, which make up the parallelogram. Next, we find the area of the triangle with known vertices using our formula for the determinants to obtain the following expression. We can cancel one-half times two to give one. We then have a formula for the area of a parallelogram from the coordinates of any three vertices.

We can actually simplify this formula and the formula for the area of a triangle even further by considering properties of the determinants and the area of congruent shapes. For instance, we can translate the coordinates of one of the vertices of the parallelogram to the origin. In this case, we would translate horizontally π₯ sub one units and vertically π¦ sub one units, with directions given by the signs of these coordinates. We can find the exact coordinates of the shown points by applying the transformations. For instance, π will be equal to π₯ sub three minus π₯ sub one. However, we will call these points π, π and π, π as shown.

We can now apply our formula for the area of a parallelogram with three known vertices to obtain the following expression for the area. However, since we translated one vertex to be the origin, we can see that a row of this matrix is zero, zero, one. We can expand the determinant over this row to simplify the expression. We find that the area is equal to the absolute value of the determinant of the two-by-two matrix π, π, π, π. We can use either of these formulas to find the area of a parallelogram in the plane from its vertices, as we will see in our next example.

Use determinants to calculate the area of the parallelogram with vertices one, one; negative four, five; negative two, eight; and three, four.

In this question, we are given the coordinates of all four vertices of a parallelogram in the plane and asked to use determinants to calculate the area of this parallelogram. There are two methods we could use, and we will go through both of these.

We can start by recalling that the area of a parallelogram can be calculated as the absolute value of the determinant of the following matrix, whose components of each row are given by the coordinates of a vertex and a final entry of one. We can choose any three vertices to calculate the area of this parallelogram. It is usually a good idea to take vertices with coordinates of zero to make the expansion easier. We have the following expression for the area. We can then evaluate this expression by expanding over the first row and calculating. Doing this gives us an area of 23 square units.

A second method we can use is to translate one vertex of the parallelogram to be at the origin. For instance, we can translate each vertex one unit left and one unit down, though we only need to do this for three vertices. We can then use the fact that the area of a parallelogram with a vertex at the origin and two vertices at π, π and π, π is given by the absolute value of the determinant of the matrix π, π, π, π to find the area of this parallelogram. We substitute the coordinates of the translated points into this formula to obtain the following expression, which we can evaluate. We take the absolute value of the difference in the product of the diagonals to get 23 square units.

In our next example, we will see how we can extend this process to any polygon in the plane using triangulation.

Consider the quadrilateral with vertices π΄ one, three; π΅ four, two; πΆ 4.5, five; and π· two, six. By breaking it into two triangles as shown, calculate the area of this quadrilateral using determinants.

We want to find the area of this quadrilateral using the areas of the two triangles and determinants. We can start by recalling that the area of a triangle is given by half the absolute value of the determinant of the matrix whose rows are given by each coordinate of a vertex and a final component of one. We can apply this to each triangle separately.

Letβs start with π sub one, which is triangle π΄πΆπ·. We substitute the given coordinates of these points into the formula to obtain the following expression. We can then expand over the first row of the matrix to obtain the following expression, which we can calculate is equal to 4.25 square units.

We need to apply this same process for the other triangle. We can do this by substituting the coordinates of π΄, π΅, and πΆ into the formula in any order to obtain the following expression. We can then expand over the first row and evaluate to find that the area of triangle π΄π΅πΆ is 4.75 square units. Finally, we can find the area of the quadrilateral by adding the areas of the two triangles. We find that it has an area of nine square units.

We can apply this process to any quadrilateral with known vertices by first splitting it into triangles. This is called triangulation.

There is one final useful property we can consider by using our formulas for areas using determinants.

Consider a case where the determinant of the following matrix is equal to zero. We know that this determinant tells us the signed area of a parallelogram with three vertices at these coordinates, or twice the signed area of a triangle with these vertices. For this determinant to be equal to zero, the shapes themselves must have zero area. This is only possible if the three points lie on a straight line. Hence, this gives us a method of checking if three points are collinear. It is also worth noting that this property works in reverse. If the three points are collinear, then the triangle with these vertices has zero area. So the determinant of this matrix must be zero.

Letβs see one final example of using this property to check the collinearity of three given points.

By using determinants, determine which of the following sets of points are collinear. Option (A) π΄ negative six, four; π΅ negative eight, four; πΆ three, 10. Option (B) π΄ negative 10, negative four; π΅ negative eight, negative two; πΆ negative five, one. Option (C) π΄ negative three, six; π΅ eight, negative seven; πΆ negative three, negative eight. Or is it option (D) π΄ negative 10, negative six; π΅ negative two, one; πΆ zero, negative nine?

We first recall that we can check the collinearity of a triplet of points in the coordinate plane by using determinants. In particular, if the determinant of this matrix is zero, then the three points will be collinear. And if the three points are collinear, then the determinant of this matrix will be zero.

Therefore, we can determine the collinearity of each triplet by evaluating the determinant of each matrix generated from the three points. Substituting the coordinates given in option (A) into this matrix gives us the determinant of the following three-by-three matrix. We can evaluate the determinant of this matrix by expanding over the first row and evaluating to obtain negative 12. This is nonzero, so the three points are not collinear.

We can follow the same process for the triplet in option (B). We substitute their coordinates into the matrix, expand on the first row, and evaluate to get zero. Since the determinant of this matrix is zero, we can conclude that the points must be collinear.

For due diligence, we can also check the final two options in the same way. For the triplet in option (C), we find that the determinant of this matrix is negative 154. So the three points are not collinear. Similarly, in option (D), we find that the determinant of the generated matrix is negative 94. Hence, only the points in option (B) are collinear.

Letβs now go over the key points in this lesson. First, we saw that we can calculate the area of a triangle from the coordinates of its vertices by using a determinant. Its area is one-half the absolute value of the determinant of the matrix whose rows are each coordinate of each vertex of the triangle and the final component of one. We can also split any polygon into triangles and then use this formula to find the area of each triangle using the coordinates of the vertices. This allows us to find the area of any polygon using determinants from the coordinates of its vertices.

We also saw that we can find the area of a parallelogram from the coordinates of any three of its vertices using determinants. We can simplify both of these determinant formulas to two-by-two matrices by translating the shapes to have a vertex at the origin.

Finally, we saw that we can check the collinearity of points in the plane by using determinants. In particular, if three points in the plane are collinear, then the triangle with these points as vertices has area equal to zero. So the area formula must also give zero. This is equivalent to saying that the determinant of the matrix is zero. The same result is true in reverse. If this determinant is zero, then the three points must be collinear.