Question Video: Computing Area of Quadrilateral Using Determinants of Matrices Mathematics

Consider the quadrilateral with vertices 𝐴 (1, 3), 𝐵 (4, 2), 𝐶 (4.5, 5), and 𝐷 (2, 6). By breaking it into two triangles as shown, calculate the area of the quadrilateral using determinants.

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Video Transcript

Consider the quadrilateral with vertices 𝐴 one, three; 𝐵 four, two; 𝐶 4.5, five; and 𝐷 two, six. By breaking it into two triangles as shown, calculate the area of the quadrilateral using determinants.

We’re given a quadrilateral determined by four vertices which we’re given. We want to calculate its area by using determinants. We can see from the diagram this is not a parallelogram. So, instead, we’re going to break this up into triangles and calculate the area of each triangle by using determinants. So let’s start by recalling how we calculate the area of a triangle by using determinants.

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. And we’re told to split our quadrilateral into two triangles: triangle 𝐴𝐷𝐶 and triangle 𝐴𝐵𝐶. So we’ll need to apply this formula to both of these.

So let’s start by finding the area of triangle 𝐴𝐵𝐶, which we’ll just write as 𝐴𝐵𝐶. From our formula, we need to know the coordinates of these three vertices, which we’re given in the question. So we write these into our matrix and then add an extra column of one. This gives us the following expression for the area of triangle 𝐴𝐵𝐶.

Now, to evaluate this expression, we’re going to need to calculate its determinant. We’ll do this by expanding the third column. First, we’re going to want to find the sign we need to multiply by each of these columns. We get positive, negative, positive. So our three coefficients are positive one, negative one, and positive one. Next, we need to find the determinants of the three matrix minors. Removing the first row and third column, we get the two-by-two matrix four, two, 4.5, five. We then need to subtract the determinant of our second matrix minor. Then, finally, we need to add the determinant of our third matrix minor. This gives us the following expression, which is equal to the area of triangle 𝐴𝐵𝐶.

All we need to do now is evaluate each of these three determinants. Let’s start with our first determinant. We get four times five minus 4.5 times two. That’s 20 minus nine, which is equal to 11. Calculating the second determinant and simplifying, we get five minus three times 4.5, which is negative 17 over two. But remember, we’re subtracting this value. Finally, we need to calculate the third determinant. It’s equal to two times one minus three times four. That’s two minus 12, which is equal to negative 10. 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. And if we evaluate this expression, it’s equal to 19 over four.

So let’s keep track of this and now do exactly the same thing to find the area of 𝐴𝐷𝐶. This time, our three vertices are 𝐴, 𝐶, and 𝐷. Remember, to use our formula, we add these coordinate pairs into our matrix and then add an extra coordinate of one. This gives us the area of 𝐴𝐷𝐶 is the following expression. And we’ll evaluate this in the same way. We’ll expand over our third column. Of course, the third column is still going to go positive, negative, positive. So we need to start by adding the determinant of our first matrix minor by removing the first row and third column. That’s the two-by-two matrix 4.5, five, two, six. Then we need to subtract the determinant of our second matrix minor by removing the second row and third column. That’s the two-by-two matrix one, three, two, six. Then we add on the determinant of our third matrix minor, giving us the following expression for the area of 𝐴𝐷𝐶.

Now, all we need to do is evaluate these determinants. Our first determinant is six times 4.5 minus two times five, which is 17. Our second determinant is six times one minus two times three, which we can calculate is zero. And calculating our third determinant, we get negative 17 over two. Finally, we can just evaluate this expression. It’s one-half times the absolute value of 17 minus 17 over two, which is equal to 17 over four.

Remember, the area of our quadrilateral will be the sum of these two values. And if we add these two values together and simplify, we get the area of our quadrilateral is nine. And there is one interesting thing worth noting about this formula. What would happen if the area of our triangle was zero?

Well, for the area to be zero, it must not be a triangle at all. 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. In other words, they’re collinear. But now look at our formula. The only part of our formula which can be equal to zero is the determinant of this matrix. This gives us a test for collinearity of three points.

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