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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.