Question Video: Using Determinants to Calculate the Area of a Parallelogram | Nagwa Question Video: Using Determinants to Calculate the Area of a Parallelogram | Nagwa

# Question Video: Using Determinants to Calculate the Area of a Parallelogram Mathematics • First Year of Secondary School

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Consider a parallelogram in which (1, 3), (3, 0), and (−1, 2) are three vertices. Complete the following: The area of this parallelogram equals ＿ square units.

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### Video Transcript

Consider a parallelogram in which one, three; three, zero; and negative one, two are three vertices. Complete the following: The area of this parallelogram equals what square units.

In this question, we’re asked to determine the area of a parallelogram, where we’re given the coordinates of three of its vertices. At first this might seem difficult because a parallelogram has four vertices, and we’re only given three of these vertices. However, there are in fact many different ways we could answer this question. For example, we could sketch the three points on a diagram and then use the fact that the area of a parallelogram is the base times the perpendicular height. And while this would work and give us the correct answer, we’re instead going to go through two different methods of answering this by using determinants.

First, we can recall the general formula for finding the area of a parallelogram by using determinants. If we’re given any three vertices of a parallelogram — let’s say 𝑥 sub one, 𝑦 sub one; 𝑥 sub two, 𝑦 sub two; and 𝑥 sub three, 𝑦 sub three — then its area is the absolute value of the determinant of the three-by-three matrix 𝑥 sub one, 𝑦 sub one, one, 𝑥 sub two, 𝑦 sub two, one, 𝑥 sub three, 𝑦 sub three, one. Since we’re given the coordinates of three vertices of our parallelogram, we can use this formula to determine its area. We substitute the coordinates of the three points into the formula. The area is the absolute value of the determinant of the three-by-three matrix one, three, one, three, zero, one, negative one, two, one.

And it’s worth pointing out here it doesn’t matter how we label our three points. We can write these in any order into our formula since all this will do is change the sign of the determinant. And we’re taking the absolute value of this determinant, so it won’t affect the area. Now, all we need to do is evaluate this expression. And to do this, we need to evaluate the determinant of a three-by-three matrix. And there’s many different ways of doing this, and we can use any we prefer. In this video, we’re going to expand over the second column because it contains zero.

Before we expand over this, we should determine the parity terms of these three expansions. It will be negative, positive, then negative, where we get a negative sign when the row number plus the column number is odd and a positive sign when the row number plus the column number is even. We’re now ready to expand over this column. Expanding over the first term in this column, we get negative three multiplied by the determinant of the two-by-two matrix three, one, negative one, one. We would then expand over the second term in this column. However, it has a factor of zero, so this term would just be equal to zero.

So instead, we’ll move on to the third term. We get negative two multiplied by the determinant of the two-by-two matrix one, one, three, one. And then, don’t forget we’re calculating an area, so we need to take the absolute value of this expression. Now, all that’s left to do is evaluate this expression. Remember, to evaluate the determinant of a two-by-two matrix, we need to find a difference in the product of the diagonals. For example, the determinant of three, one, negative one, one is three times one minus one times negative one, which is just three plus one. And the determinant of our second matrix is one minus three.

This gives us the absolute value of negative three times three plus one minus two times one minus three, which simplifies to give us the absolute value of negative 12 plus four, which is the absolute value of negative eight, which we can calculate is equal to eight.

And this is enough to answer our question. The area of this parallelogram will be eight square units. However, there’s a second method we can use to evaluate this area by using determinants. Instead, we could have also recalled that the area of a parallelogram with one vertex at the origin and two other vertices at the points 𝑎, 𝑏 and 𝑐, 𝑑 is given by the absolute value of the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑. And this is quite a useful result. It’s much easier to calculate the determinant of a two-by-two matrix. However, we need one of the vertices to be at the origin. And in this question, none of the three vertices we’re given are at the origin. So at first, we might not think we can apply this property.

However, we know translating a parallelogram will not affect its area. In fact, translating any shape won’t affect its area. Therefore, if we take a look at one of the points, say, three, zero, we can see this is three units right of the origin. This means if we translate all of the vertices of this parallelogram three units left, then one of the vertices will be at the origin, and we can apply this formula. Therefore, we’ll find the area of this parallelogram by first translating the vertices three units left. First, the point with coordinates one, three will be moved to the point with coordinates one minus three, three, which we can calculate is negative two, three.

Next, we don’t need to perform this to the point three, zero since we already know this will be translated to the origin. So instead, let’s translate the third vertex we’re given — the point negative one, two — three units to the left. We need to subtract three from its 𝑥-coordinate. This gives us the point negative four, two. Now we have a parallelogram with a vertex at the origin. And we know two of its other vertices, the points negative two, three and negative four, two. These will be our coordinates 𝑎, 𝑏 and 𝑐, 𝑑 in our formula. And just like our previous formula, it’s worth pointing out we can label these two vertices in any order we want. It won’t change the value of the area. And in fact, we could have also translated any of these vertices to be at the origin.

In this case, we get the area is equal to the absolute value of the determinant of the two-by-two matrix negative two, three, negative four, two. Now, all that’s left to do is evaluate this expression. To evaluate the determinant of a two-by-two matrix, we find the difference in the products of the diagonals. That’s negative two times two minus three times negative four, which is negative four plus 12. So this simplifies to give us the absolute value of negative four plus 12, which is the absolute value of eight, which is just equal to eight, which agrees with our previous answer. Therefore, we were able to show if a parallelogram has three vertices at the points one, three; three, zero; and negative one, two, then its area is eight square units.

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