Video: Centroid of a Triangle in the Complex Plane

A triangle has vertices at points π‘Ž, 𝑏, and 𝑐 in the complex plane. Find an expression for the centroid of the triangle in terms of π‘Ž, 𝑏, and 𝑐. You can use the fact that the centroid divides the median in the ratio 2 : 1.

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Video Transcript

A triangle has vertices at points π‘Ž, 𝑏, and 𝑐 in the complex plane. Find an expression for the centroid of the triangle in terms of π‘Ž, 𝑏, and 𝑐. You can use the fact that the centroid divides the median in the ratio two to one.

Let’s draw an arbitrary triangle in the complex plane with vertices π‘Ž, 𝑏, and 𝑐. We’re looking for the centroid of that triangle. And we use the fact that it divides any median of the triangle in the ratio two to one. So what is a median of the triangle? It’s the line segment between a vertex of the triangle and the midpoint of the side opposite that vertex. So if we take the vertex π‘Ž, then we need to find the midpoint of the side opposite. So I think it’s that here. And connecting the two points, we get a median. To find the centroid, we use the fact that it divides any median in the ratio two to one. So the centroid is about here. It’s twice as far from the vertex as it is from the midpoint of the opposite side.

Now, we want to find an expression for this centroid in terms of π‘Ž, 𝑏, and 𝑐. How we’re going to do that? Well, we know that the midpoint of the complex numbers 𝑏 and 𝑐 is just their arithmetic mean, 𝑏 plus 𝑐 over two. Let’s call this π‘š for simplicity. And the centroid divides the line segment from π‘Ž to π‘š in the ratio two-thirds to one-third. So we can get to it by going to π‘š and then going a third of the way from π‘š to π‘Ž. Using what we know about π‘š, we can write the centroid in terms of π‘Ž, 𝑏, and 𝑐. Now, we just need to simplify.

We multiply the numerator and denominator of the second fraction by two to get the fraction with denominator six. And so we write the first fraction over six as well. We can then combine the fractions. And in doing so, we collect some like terms in the numerator. Finally, we cancel the common factor of two in the numerator and denominator and rearrange some of the terms in the numerator to get that the centroid is π‘Ž plus 𝑏 plus 𝑐 over three. Using complex numbers then with simple methods, we get this elegant result.

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