 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Area With Determinants - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Determinants can be used to find the area of a parallelogram, but they can also be used to find the area of a triangle. Let’s take a look. I’ve got four points to find here and I want to find the area of two triangles. First let’s find the area of triangle SHA.

It doesn’t matter what the triangle looks like, let me draw a quick picture. Let’s say that this is what the triangle looks like. This is point S, this is point H and this is point A. If you look at the two vectors, SA, and SH, those vectors describe not just this triangle but also this parallelogram. And the formula that we’ve studied so far, for the area of a parallelogram, has us take the components of these two vectors and set up a determinant. It gives us the area of the entire parallelogram. Well it turns out that we can always do this.

We can always take two vectors, the vectors for two sides of a triangle, find the area of the associated parallelogram, and then half of that will be the area of the triangle. So that’s what I’m going to do here.

I first look at vector SA. From S to point A, and that’s going to be -1 minus -2, -1 plus 2, is 1. And then 6 minus 0 is 6. And then SH; that’s going to be from S to H, so -5 minus -2, -5 plus 2, -3, and 9 minus 0, 9. So the area of the triangle is going to be ½ of the absolute value of the determinant 1 6, -3 9.

And again this part in here is area of this parallelogram. So I take ½ so I can just get the area of the triangle. So ½, and this determinant is 9 minus -18. So 9 plus 18, 27, and the absolute value of 27 is 27. So 27 over 2 is 13.5. That’s the area of this little triangle.

Now let’s do the same thing to triangle HAN. Here I’m going to look at vectors HA and vectors HN. Usually I like to pick one point and make the two vectors that start from that point; so H and A, H and N. I need to find the components of these two guys. HA goes from H to point A, so -1 minus -5, -1 plus 5, is 4. And 6 minus 9 is -3. HN goes from H to point N; 7 minus -5, 7 plus 5 is 12, 0 minus 9, -9.

And similarly, the area of this triangle will be ½ the absolute value of the determinant, 4 -3, 12 -9. So that’s ½ of the absolute value of 4 times -9, is -36, minus 12 times -3 is also -36, so plus 36. -36 plus 36 is 0. Absolute value of 0 is 0, times ½ is 0.

What does it mean that the area of this triangle is 0? It means that these three points are collinear, H, A, N are collinear. What’s interesting about this? We didn’t get an area, we don’t actually have a proper triangle here, but we discovered a really good test for collinearity. If you want to find out if three points are collinear, try to form a triangle out of them and calculate their area. If the area turns out to be zero, then yes they are collinear, if not they’re not. They form a nice triangle that has an area, they’re not collinear.