### 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 Heronβs formula require us to know or calculate the lengths
of the sides of the triangle. Letβs 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 Heronβs 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, weβre 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.

Letβs 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; πΆ 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.

Letβs start with π sub one,
which is triangle π΄πΆπ·. 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 πΆ 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΅πΆ 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.

Letβs 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; πΆ three, 10. Option (B) π΄ negative 10,
negative four; π΅ negative eight, negative two; πΆ negative five, one. Option (C) π΄ negative three,
six; π΅ eight, negative seven; πΆ negative three, negative eight. Or is it option (D) π΄ negative
10, negative six; π΅ negative two, one; πΆ 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.

Letβs 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.