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Using Determinants to Calculate Areas
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In this video, we’re going to learn how we can use determinants to calculate the areas of various polygons.
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We’ll discuss how we can use determinants to calculate the areas of triangles given the coordinates of their vertices and how we can calculate the areas of parallelograms given their vertices or vectors defining them.
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We’ll also discuss how we can extend this to more complicated polygons.
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To start discussing this topic, we first need to answer a very important question.
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What does the determinant of a matrix actually mean?
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We know the definition of a determinant, and we know how to calculate this.
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However, we don’t know what this actually means.
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There’s actually a lot of different interpretations about what the determinant of a matrix can mean.
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We’ll only go through one of these.
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We’ll start with what we mean by the determinant of a two-by-two matrix, the matrix 𝑎, 𝑏, 𝑐, 𝑑.
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The determinant of this two-by-two matrix is the signed area of the parallelogram with sides defined by either the rows or columns of this matrix.
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Now, this is a very complicated-sounding statement.
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However, if we break it down piece by piece, it can be a lot easier to understand.
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We’ll start by choosing to define our parallelogram by the rows of our matrix.
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And we’re going to discuss how we create this parallelogram.
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So our matrix has two rows.
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The first row is 𝑎, 𝑏.
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And the first thing we’ll do is define a vector 𝐕 one, which will have horizontal component 𝑎 and vertical component 𝑏.
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We’ll then do the same for our second row.
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This gives us our second vector 𝐕 two.
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And it’s worth pointing out we can also choose the columns of our matrix to define this parallelogram.
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The only difference is we would choose the columns of our matrix and we’d end up with column vectors.
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However, we’ll just go through the example of rows.
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And now, with these two vectors, we can define a parallelogram.
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If we start at a point and arrange our vectors in the following form, we will get a parallelogram.
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Now, our definition of the determinant tells us that the determinant of this matrix is the signed area of this parallelogram.
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In other words, if we call this parallelogram 𝑃, then the area of 𝑃 is equal to either the determinant of this matrix or negative one multiplied by the determinant of this matrix.
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And that’s because we know the determinant can give us a negative value.
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But we don’t want a negative value for our area.
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This gives us a useful formula for calculating the area of any parallelogram.
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And you’ll often see this written in two different ways.
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First, we’ll take the absolute value of the right-hand side of this equation.
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This then removes the need for the positive or negative value.
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However, some people prefer to use this notation for the determinant of a matrix.
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So you can also see this written as positive or negative times the determinant of our matrix.
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And these all mean exactly the same thing.
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We just calculate the determinant of this matrix and take the positive value.
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And there’s two things worth pointing out here.
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First, remember, we could’ve taken column vectors instead of row vectors.
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And this would give us exactly the same result where we look at the columns of our matrix instead.
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The second thing worth noting is what would happen if 𝐕 one and 𝐕 two did not form a parallelogram.
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For example, if 𝐕 one and 𝐕 two point in exactly the same direction or in opposite directions or one of 𝐕 one or 𝐕 two are the zero vector, then we wouldn’t end up with a shape with area.
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It would have zero area.
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And this is exactly what we mean by the determinant being equal to zero.
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And it also turns out we can extend this definition of the determinant to larger matrices.
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The determinant of a three-by-three matrix is the signed volume of the parallelepiped, that’s the three-dimensional version of a parallelogram, with sides defined by the rows or columns of that matrix.
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Now, it’s beyond the scope of this video to know exactly why this is true.
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However, by assuming this is true and using some algebraic manipulation and geometry, we can arrive at a very useful result.
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If we know the coordinates of the vertices of a triangle 𝑇, we can calculate its area.
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Its area is equal to one-half times the absolute value of the determinant of a three-by-three matrix where each row is the coordinate pair and an extra row for the value of one.
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Once again, the proof of this statement is beyond the scope of this video.
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However, we’ll see how to use this in a few examples.
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Find the area of the triangle below using determinants.
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The question wants us to find the area of the triangle given to us in the diagram.
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And it wants us to do this by using determinants.
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So the first thing we’re going to need to do is recall how we find the area of a triangle by using determinants.
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We recall we just need to find the coordinates of its three vertices and then use the following formula.
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The area of 𝑇 is equal to one-half multiplied by the absolute value of the determinant of a three-by-three matrix where each row in our matrix is formed by the coordinates of the vertex and an extra one.
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So the first thing we need to do is find the three coordinates of the vertices of our triangle.
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We get negative one, negative four; zero, four; and four, negative two.
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And there’s something worth pointing out here.
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It doesn’t matter which of these points we label one, two, or three.
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All this would do is change the sign of our determinant.
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But we’re taking the absolute value of this, so it won’t change our answer.
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So we’ll just choose to number our points from left to right.
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This gives us the area of the triangle 𝑇 is equal to one-half multiplied by the absolute value of the determinant of this three-by-three matrix.
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So what we need to do now is calculate this determinant.
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And there’s a few different ways of doing this.
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We’ll use matrix minors.
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And we’re going to expand out the first column because it contains zero.
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First, we’ll start by recalling the signs we need to multiply by.
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We get positive, negative, positive.
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Next, we need to find our matrix minors.
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We’ll start by finding the one by removing the first row and the first column.
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So we get negative one multiplied by the determinant of the two-by-two matrix four, one, negative two, one.
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Next, the second term in our expansion has a coefficient of zero.
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So we don’t need to do this.
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So we’ll just move on to the third term.
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First, we know we need to multiply this by four.
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And we need to find the matrix minor by removing the first column and third row.
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So we get four multiplied by the determinant of the two-by-two matrix negative four, one, four, one.
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Now, all we need to do is evaluate these two-by-two determinants.
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First, we get four times one minus negative two times one.
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Next, we get negative four multiplied by one minus four times one.
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Now, all that’s left to do is evaluate this expression.
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First, negative one multiplied by four times one minus negative two times one is equal to negative six.
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Next, four times negative four times one minus four times one is equal to negative 32.
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So this just simplified to give us one-half times the absolute value of negative six minus 32.
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Negative six minus 32 is negative 38.
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The absolute value of negative 38 is 38.
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Then we multiply this by one-half to get 19.
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But remember, this gives us the area of a triangle, so we can give this units.
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These must be given in square units.
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Therefore, we were able to show the area of the triangle given to us in the diagram is 19 square units.
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It’s also worth pointing out we could’ve found the area of this triangle by using geometry.
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And this would give us a good way of checking our final answer.
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Let’s now go through an example of how we could use determinants to help us find the coordinates of the vertex of a triangle.
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If the area of a triangle whose vertices are ℎ, zero; six, zero; and zero, three is nine square units, then ℎ is equal to (A) zero or 12, (B) zero or negative 12, (C) negative six or six, (D) negative 12 or 12.
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We’re given that a triangle has an area of nine square units.
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And we’re given the three coordinates of its vertices.
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We need to use this information to find the possible values of ℎ.
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It’s worth pointing out here one method to do this is to plot all three of these points on a diagram and then just calculate the possible areas.
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And this definitely works and can be used to get the correct answer.
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However, we’re going to do this by using determinants.
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Recall, we can find the area 𝐴 of a triangle given its vertices by using the following formula.
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𝐴 is equal to one-half multiplied by the absolute value of the determinant of a three-by-three matrix, where each row in our matrix is the coordinate pair with an extra value of one.
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And it doesn’t matter which order we choose our points.
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So we’ll just order them from left to right.
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So by using the vertices given to us in the question, we get the area of our triangle is equal to one-half times the absolute value of the determinant of the following three-by-three matrix.
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The next thing we’re going to need to do is find the value of this determinant.
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There’s a few different ways of doing this.
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We can see that one of the columns of our matrix has two values of zero.
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So we’ll find this determinant by expanding over the second column.
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So we need to do this by finding the matrix minors.
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Remember, our first two values are zero.
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So we don’t need to calculate these.
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We only need to do the final one with the coefficient of three.
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And remember, we multiply this by negative one to the power of three plus two because it’s in row three, column two.
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Now, remember, we need to find the determinant of the matrix minor we get by removing the third row and second column.
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We can see this is the two-by-two matrix ℎ, one, six, one.
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So we’ve shown the area of our triangle is equal to one-half times the absolute value of negative one to the power of three plus two times three multiplied by the determinant of the two-by-two matrix ℎ, one, six, one.
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Now, all we need to do is evaluate this expression.
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First, negative one to the power of three plus two is just negative one.
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Next, if we evaluate the determinant of our two-by-two matrix, we see it’s equal to ℎ minus six.
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So the area of our triangle is one-half times the absolute value of negative three times ℎ minus six.
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And we can simplify this further.
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We can take the negative three outside of our absolute value.
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And remember, this means it becomes positive three.
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So we get three over two times the absolute value of ℎ minus six.
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Remember, in the question, we’re told the area of our triangle is nine.
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So, in fact, this expression must be equal to nine.
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So we just need to solve an equation involving the absolute value symbol.
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We’ll start by dividing both sides of our equation through by three over two.
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Nine divided by three over two is six.
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We get six is equal to the absolute value of ℎ minus six.
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Remember, to solve equations involving the absolute value symbol, we need to consider the positive and negative solution for our absolute value.
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And by taking the positive and negative, we get two equations we need to solve.
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Six is equal to ℎ minus six, and six is equal to negative one times ℎ minus six.
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And these are both linear equations, and we can solve them.
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We get ℎ is equal to 12 or ℎ is equal to zero.
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And this is option (A).
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Therefore, we were able to show that ℎ is equal to zero or 12.
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And it’s also worth pointing out we could check our answer by substituting our values of ℎ into this coordinate pair, plotting these points, and then calculating the area of the triangle and seeing that we get nine.
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And this can be a good way to check that we got the correct answer.
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Let’s now see an alternate method of calculating the area of a triangle by using determinants.
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Use determinants to work out the area of the triangle with vertices two, negative two; four, negative two; and zero, two by viewing the triangle as half a parallelogram.
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We want to find out the area of a triangle by using determinants.
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And we’re given the coordinates of its vertices.
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So we could just do this directly from our formula.
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However, the question doesn’t want us to do this.
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It wants us to do this by viewing the triangle as half of a parallelogram.
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So to do this, we’ll start by sketching our three points to get an idea of what our parallelogram could look like.
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If we plot the three points and connect them, we get a triangle which looks like this.
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The question then becomes, how are we going to turn this into a parallelogram?
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And although it may not seem like it, there’s actually three different ways we could do this.
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There’s a few different ways of seeing this.
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One way to do this is geometrically.
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Let’s draw an exact copy of our triangle.
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And we can consider the shape if we were to glue two of the sides together.
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If we were to do this, we would get a shape which looks like this.
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And we could find the coordinate of this vertex by using what we know about vectors.
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This is a parallelogram with twice the area of our original triangle because it’s made up of two triangles of equal area.
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But this wasn’t the only choice for the side.
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For example, we could’ve chosen this side.
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If we were to glue these two triangles together along this edge, we would get something which looks like the following diagram.
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And once again, we could find the coordinates of this vertex by using what we know about vectors.
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Once again, the area of this triangle plus the green triangle is still twice the area of our original triangle.
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So it’s a parallelogram with twice the area.
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Finally, we could do exactly the same thing by combining the last two edges.
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And if we did this, we would get a similar story.
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We get a third parallelogram.
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We can find the coordinates of its vertex.
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And this parallelogram is twice the area of our original triangle.
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It doesn’t matter which of these we choose.
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For simplicity, we’ll choose the following example.
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And it’s worth pointing out we didn’t need to do this geometrically.
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You can also do this by choosing two of the sides of the triangle as a vector.
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This would give us the same result.
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The question now wants us to find the area of this parallelogram.
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And it wants us to do this by using determinants.
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Recall, we know how to find the area of a parallelogram by using determinants.
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If we define our parallelogram by the vectors of its sides, 𝐕 one and 𝐕 two, then its area 𝐴 is equal to the absolute value of the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑, where our row vectors are the rows of our matrix.
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And we could’ve also used column vectors and used the columns of our matrix instead.
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It wouldn’t change our answer.
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So to answer this question, we need to find the vectors 𝐕 one and 𝐕 two.
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There’s a few different ways of doing this.
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Let’s start by finding 𝐕 one.
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In 𝐕 one, our 𝑥-component starts at two and ends at zero.
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So the horizontal component or change in 𝑥 of 𝐕 one is zero minus two, which is of course just negative two.
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We can do the same for the vertical component.
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We end at a 𝑦-coordinate of two and begin at negative two.
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So we get two minus negative two, which is four.
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So 𝐕 one is the vector negative two, four.
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We can then do exactly the same to find vector 𝐕 two.
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𝐕 two is two, zero.
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We’re now ready to use our formula.
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The area of our parallelogram is the absolute value of the determinant of the two-by-two matrix negative two, four, two, zero, where negative two, four is 𝐕 one and two, zero is 𝐕 two.
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And we can evaluate this expression.
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We get the absolute value of negative two times zero minus four times two, which is just equal to eight.
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But remember, this is the area of our parallelogram.
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This is twice the area of our triangle.
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So we need to divide this value by two.
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So we divide through by two.
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And we get the area of our triangle is equal to four.
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And one thing worth pointing out here is we never needed to calculate the fourth coordinate of our parallelogram.
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This is because the area of a parallelogram is entirely defined by the vectors which make up the parallelogram.
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Let’s now see an example of how we would calculate the area of a more complicated polygon.
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Consider the quadrilateral with vertices 𝐴 one, three; 𝐵 four, two; 𝐶 4.5, five; and 𝐷 two, six.
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By breaking it into two triangles as shown, calculate the area of the quadrilateral using determinants.
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We’re given a quadrilateral determined by four vertices which we’re given.
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We want to calculate its area by using determinants.
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We can see from the diagram this is not a parallelogram.
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So, instead, we’re going to break this up into triangles and calculate the area of each triangle by using determinants.
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So let’s start by recalling how we calculate the area of a triangle by using determinants.
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We recall, if we know the coordinates of the three vertices of our triangle, then its area is equal to one-half times the absolute value of the determinant of the three-by-three matrix, where each row in this matrix is the coordinate pair of a vertex and then an extra coordinate of one.
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And we’re told to split our quadrilateral into two triangles: triangle 𝐴𝐷𝐶 and triangle 𝐴𝐵𝐶.
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So we’ll need to apply this formula to both of these.
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So let’s start by finding the area of triangle 𝐴𝐵𝐶, which we’ll just write as 𝐴𝐵𝐶.
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From our formula, we need to know the coordinates of these three vertices, which we’re given in the question.
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So we write these into our matrix and then add an extra column of one.
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This gives us the following expression for the area of triangle 𝐴𝐵𝐶.
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Now, to evaluate this expression, we’re going to need to calculate its determinant.
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We’ll do this by expanding the third column.
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First, we’re going to want to find the sign we need to multiply by each of these columns.
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We get positive, negative, positive.
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So our three coefficients are positive one, negative one, and positive one.
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Next, we need to find the determinants of the three matrix minors.
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Removing the first row and third column, we get the two-by-two matrix four, two, 4.5, five.
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We then need to subtract the determinant of our second matrix minor.
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Then, finally, we need to add the determinant of our third matrix minor.
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This gives us the following expression, which is equal to the area of triangle 𝐴𝐵𝐶.
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All we need to do now is evaluate each of these three determinants.
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Let’s start with our first determinant.
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We get four times five minus 4.5 times two.
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That’s 20 minus nine, which is equal to 11.
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Calculating the second determinant and simplifying, we get five minus three times 4.5, which is negative 17 over two.
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But remember, we’re subtracting this value.
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Finally, we need to calculate the third determinant.
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It’s equal to two times one minus three times four.
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That’s two minus 12, which is equal to negative 10.
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So this gives us the area of 𝐴𝐵𝐶 is one-half times the absolute value of 11 minus negative 17 over two plus negative 10.
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And if we evaluate this expression, it’s equal to 19 over four.
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So let’s keep track of this and now do exactly the same thing to find the area of 𝐴𝐷𝐶.
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This time, our three vertices are 𝐴, 𝐶, and 𝐷.
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Remember, to use our formula, we add these coordinate pairs into our matrix and then add an extra coordinate of one.
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This gives us the area of 𝐴𝐷𝐶 is the following expression.
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And we’ll evaluate this in the same way.
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We’ll expand over our third column.
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Of course, the third column is still going to go positive, negative, positive.
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So we need to start by adding the determinant of our first matrix minor by removing the first row and third column.
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That’s the two-by-two matrix 4.5, five, two, six.
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Then we need to subtract the determinant of our second matrix minor by removing the second row and third column.
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That’s the two-by-two matrix one, three, two, six.
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Then we add on the determinant of our third matrix minor, giving us the following expression for the area of 𝐴𝐷𝐶.
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Now, all we need to do is evaluate these determinants.
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Our first determinant is six times 4.5 minus two times five, which is 17.
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Our second determinant is six times one minus two times three, which we can calculate is zero.
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And calculating our third determinant, we get negative 17 over two.
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Finally, we can just evaluate this expression.
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It’s one-half times the absolute value of 17 minus 17 over two, which is equal to 17 over four.
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Remember, the area of our quadrilateral will be the sum of these two values.
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And if we add these two values together and simplify, we get the area of our quadrilateral is nine.
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And there is one interesting thing worth noting about this formula.
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What would happen if the area of our triangle was zero?
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Well, for the area to be zero, it must not be a triangle at all.
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And since a triangle is three points which don’t lie on the same line, that means our three points must lie on the same line.
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In other words, they’re collinear.
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But now look at our formula.
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The only part of our formula which can be equal to zero is the determinant of this matrix.
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This gives us a test for collinearity of three points.
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Let’s now go over the key points of this video.
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First, we showed we can find the area of a parallelogram if its sides are the vectors 𝑎, 𝑏, 𝑐, 𝑑 as the absolute value of the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑.
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We also saw we can instead use column vectors.
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We also saw how to find the area of a triangle if we’re given the coordinates of its vertices by using determinants.
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We also know if a triangle’s area is zero, its vertices must be collinear.
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So this gives us a test for checking if three points are collinear.
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Finally, we saw we can find the area of more complicated polygons if we can split these into triangles.