Finding the Centroid of a Triangle Using Coordinates

[mathmari]

Hey!

We have a triangle $ABC$ with $A(a_1, a_2)$, $B(b_1, b_2)$ and $C(c_1,c_2)$. I want to show that the coordinates of the centroid S is $\left (\frac{1}{3}(a_1+b_1+c_1),\frac{1}{3}(a_2+b_2+c_2) \right )$.

$S$ is the intersection point of the midpoints of AB, BC and CA.

We have that \begin{align*}&M_{AB}=\left (\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}\right ),&M_{BC}=\left (\frac{b_1+c_1}{2}, \frac{b_2+c_2}{2}\right ),&M_{CA}=\left (\frac{c_1+a_1}{2}, \frac{c_2+a_2}{2}\right )\end{align*}
How can we find then the intersection point?

Do we have to write the line segments $AM_{BC}, BM_{CA}, CM_{AB}$in the form $y=mx+n$ ? (Wondering)

 

[HallsofIvy]

Hey!

We have a triangle $ABC$ with $A(a_1, a_2)$, $B(b_1, b_2)$ and $C(c_1,c_2)$. I want to show that the coordinates of the centroid S is $\left (\frac{1}{3}(a_1+b_1+c_1),\frac{1}{3}(a_2+b_2+c_2) \right )$.

$S$ is the intersection point of the midpoints of AB, BC and CA.

This doesn't quite make sense. "Points" don't intersect! You mean the mutual point of intersection of the three lines, AB, BC, CA.

We have that \begin{align*}&M_{AB}=\left (\frac{a_1+b_1}{2}, \frac{a_2+b_2}{2}\right ),&M_{BC}=\left (\frac{b_1+c_1}{2}, \frac{b_2+c_2}{2}\right ),&M_{CA}=\left (\frac{c_1+a_1}{2}, \frac{c_2+a_2}{2}\right )\end{align*}
How can we find then the intersection point?

Do we have to write the line segments $AM_{BC}, BM_{CA}, CM_{AB}$in the form $y=mx+n$ ? (Wondering)

Yes, to find the point of intersection of three lines, it is a good idea to write them in that form. It is probably simplest to use the "two point form": the line through $A(a_1, a_2)$ and $B(b_1, b_2)$ is $y= \frac{b_1- b_2}{a_1- a_2}(x- a_1)+ b_1$.

 

[mathmari]

This doesn't quite make sense. "Points" don't intersect! You mean the mutual point of intersection of the three lines, AB, BC, CA.

Oh yes, I meant the intersection point of the lines $AM_{BC}, CM_{AB}, BM_{AC}$.

Yes, to find the point of intersection of three lines, it is a good idea to write them in that form. It is probably simplest to use the "two point form": the line through $A(a_1, a_2)$ and $B(b_1, b_2)$ is $y= \frac{b_1- b_2}{a_1- a_2}(x- a_1)+ b_1$.

The line $AM_{BC}$ is: \begin{equation*}y=\frac{y_M-a_2}{x_M-a_1}(x-a_1)+a_2 \Rightarrow y=\frac{\frac{b_2+c_2}{2}-a_2}{\frac{b_1+c_1}{2}-a_1}(x-a_1)+a_2 \Rightarrow y=\frac{\frac{b_2+c_2-2a_2}{2}}{\frac{b_1+c_1-2a_1}{2}}(x-a_1)+a_2 \Rightarrow y=\frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}(x-a_1)+a_2 \Rightarrow y=\frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}x-a_1\cdot \frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}+a_2\end{equation*}

And the line $CM_{AB}$ is: \begin{equation*}y=\frac{y_M-c_2}{x_M-c_1}(x-c_1)+c_2 \Rightarrow y=\frac{\frac{a_2+b_2}{2}-c_2}{\frac{a_1+b_1}{2}-c_1}(x-c_1)+c_2 \Rightarrow y=\frac{\frac{a_2+b_2-2c_2}{2}}{\frac{a_1+b_1-2c_1}{2}}(x-c_1)+c_2 \Rightarrow y=\frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}(x-c_1)+c_2 \Rightarrow y=\frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}x-c_1\cdot \frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}+c_2\end{equation*}

Since we want to calculate the intersection point of these two line segments, we do the following:
\begin{align*}&\frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}x-a_1\cdot \frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}+a_2=\frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}x-c_1\cdot \frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}+c_2 \\ & \Rightarrow \left (\frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}-\frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}\right )x=-c_1\cdot \frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}+c_2+a_1\cdot \frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}-a_2 \\ & \Rightarrow x=\frac{1}{\frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}-\frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}}\cdot \left (-c_1\cdot \frac{a_2+b_2-2c_2}{a_1+b_1-2c_1}+c_2+a_1\cdot \frac{b_2+c_2-2a_2}{b_1+c_1-2a_1}-a_2\right )\end{align*}

Is everything correct so far? How could we continue? (Wondering)

 

[I like Serena]

It should work, but we can see that it's getting quite messy in a hurry.
I didn't check for mistakes, but logically the next step would be to add/subtract and simplify the fractions.

Alternatively, we can prove that the centroid divides the medians in a ratio 2:1 with similarity of triangles (or we can just take it as a property of the centroid).
See the proof here at the bottom.
And then we can calculate $\overrightarrow A + \frac 23\overrightarrow{AM_{BC}}$ to find the coordinates of the centroid. (Thinking)

 

[mathmari]

It should work, but we can see that it's getting quite messy in a hurry.
I didn't check for mistakes, but logically the next step would be to add/subtract and simplify the fractions.

Alternatively, we can prove that the centroid divides the medians in a ratio 2:1 with similarity of triangles (or we can just take it as a property of the centroid).
See the proof here at the bottom.
And then we can calculate $\overrightarrow A + \frac 23\overrightarrow{AM_{BC}}$ to find the coordinates of the centroid. (Thinking)

Ah ok! I will try it! (Nerd)

I had now the following idea:

We have that:
$\vec{AM_{BC}}=\vec{AC}+\vec{CM_{BC}}=\vec{AC}+\frac{\vec{CB}}{2}=\vec{AC}+\frac{\vec{AC}+\vec{AB}}{2}=\frac{3}{2}\vec{AC}+\frac{1}{2}\vec{AB}$
$\vec{BM_{AC}}=\vec{AB}+\vec{AM_{AC}}=\vec{AB}+\frac{\vec{AC}}{2}=\vec{AB}+\frac{1}{2}\vec{AC}$

It holds that $\vec{AS}+\vec{AB}=\vec{SB}$.

Since $S$ is on the vector $\vec{AM_{BC}}$ we have that $\vec{AS}=k\vec{AM_{BC}}$ and since $S$ is also on the vector $\vec{M_{AC}B}$ we have that $SB=n\vec{M_{AC}B}$.

Therefore we get the following:
$$\vec{AS}+\vec{AB}=\vec{SB} \\ \Rightarrow k\vec{AM_{BC}}+\vec{AB}=n\vec{M_{AC}B} \\ \Rightarrow k\left (\frac{3}{2}\vec{AC}+\frac{1}{2}\vec{AB}\right )+\vec{AB}=n\left (\vec{AB}+\frac{1}{2}\vec{AC}\right ) \\ \Rightarrow \frac{3k}{2}\vec{AC}+\frac{k}{2}\vec{AB}+\vec{AB}=n\vec{AB}+\frac{n}{2}\vec{AC} \\ \Rightarrow \left (\frac{3k}{2}-\frac{n}{2}\right )\vec{AC}+\left (\frac{k}{2}+1-n\right )\vec{AB}=0$$

Is everything correct so far? Can we use here linear independency? (Wondering)

 

[I like Serena]

It holds that $\vec{AS}+\vec{AB}=\vec{SB}$.

Shouldn't that be $\vec{AS}+\vec{SB}=\vec{AB}$? (Wondering)

Can we use here linear independency?

I think so yes.

 

[mathmari]

Shouldn't that be $\vec{AS}+\vec{SB}=\vec{AB}$? (Wondering)

Oh yes. I think I have some more typos. Should it be as follows?

We have that:
$\vec{AM_{BC}}=\vec{AC}+\vec{CM_{BC}}=\vec{AC}+\frac{\vec{CB}}{2}=\vec{AC}+\frac{\vec{AC}+\vec{AB}}{2}=\frac{3}{2}\vec{AC}+\frac{1}{2}\vec{AB}$
$\vec{BM_{AC}}=\vec{BA}+\vec{AM_{AC}}=-\vec{AB}+\frac{\vec{AC}}{2}=-\vec{AB}+\frac{1}{2}\vec{AC}$

It holds that $\vec{AS}+\vec{SB}=\vec{AB}$.

Since $S$ is on the vector $\vec{AM_{BC}}$ we have that $\vec{AS}=k\vec{AM_{BC}}$ and since $S$ is also on the vector $\vec{BM_{AC}}$ we have that $SB=n\vec{BM_{AC}}$.

Therefore we get the following:
$$\vec{AS}+\vec{SB}=\vec{AB} \\ \Rightarrow k\vec{AM_{BC}}+n\vec{BM_{AC}}=\vec{AB} \\ \Rightarrow k\left (\frac{3}{2}\vec{AC}+\frac{1}{2}\vec{AB}\right )+n\left (-\vec{AB}+\frac{1}{2}\vec{AC}\right )=\vec{AB} \\ \Rightarrow \frac{3k}{2}\vec{AC}+\frac{k}{2}\vec{AB}-n\vec{AB}+\frac{n}{2}\vec{AC}=\vec{AB} \\ \Rightarrow \left (\frac{3k}{2}+\frac{n}{2}\right )\vec{AC}+\left (\frac{k}{2}-1-n\right )\vec{AB}=0$$ Since $\vec{AC}$ and $\vec{AB}$ are linear independent (because they start at the same point? (Wondering) ) it must hold the following:
$$\frac{3k}{2}+\frac{n}{2}=\frac{k}{2}-1-n=0 \\ \Rightarrow \frac{3k}{2}=-\frac{n}{2} \Rightarrow 3k=-n \ \text{ and } \ \frac{k}{2}=1+n \Rightarrow k=2+2n$$

Therefore we get $3(2+2n)=-n \Rightarrow 6+6n=-n \Rightarrow 7n=-6$.

But shouldn't we get here $\frac{1}{3}$, as the way you suggested above? So have I done something wrong? Or can we not use that way? (Wondering)

 

[I like Serena]

Oh yes. I think I have some more typos. Should it be as follows?

We have that:
$\vec{AM_{BC}}=\vec{AC}+\vec{CM_{BC}}=\vec{AC}+\frac{\vec{CB}}{2}=\vec{AC}+\frac{\vec{AC}+\vec{AB}}{2}=\frac{3}{2}\vec{AC}+\frac{1}{2}\vec{AB}$

Shouldn't we have $\vec{CB} = \vec{AB} - \vec{AC}$?
Generally a vector can be written as a vector to its end point minus a vector to its starting point. (Nerd)

 

[mathmari]

Shouldn't we have $\vec{CB} = \vec{AB} - \vec{AC}$?
Generally a vector can be written as a vector to its end point minus a vector to its starting point. (Nerd)

Ah ok! So, we have the following:

$\vec{AM_{BC}}=\frac{1}{2}\vec{AC}+\frac{1}{2}\vec{AB}$ and $\vec{BM_{AC}}=-\vec{AB}+\frac{1}{2}\vec{AC}$, right? (Wondering)

Therefore we get the following:
$$\vec{AS}+\vec{SB}=\vec{AB} \\ \Rightarrow k\vec{AM_{BC}}+n\vec{BM_{AC}}=\vec{AB} \\ \Rightarrow k\left (\frac{1}{2}\vec{AC}+\frac{1}{2}\vec{AB}\right )+n\left (-\vec{AB}+\frac{1}{2}\vec{AC}\right )=\vec{AB} \\ \Rightarrow \frac{k}{2}\vec{AC}+\frac{k}{2}\vec{AB}-n\vec{AB}+\frac{n}{2}\vec{AC}=\vec{AB} \\ \Rightarrow \left (\frac{k}{2}+\frac{n}{2}\right )\vec{AC}+\left (\frac{k}{2}-1-n\right )\vec{AB}=0$$ From linear independecy we get the following:
$$\frac{k}{2}+\frac{n}{2}=\frac{k}{2}-1-n=0 \\ \Rightarrow \frac{k}{2}=-\frac{n}{2} \Rightarrow k=-n \ \text{ and } \ \frac{k}{2}=1+n \Rightarrow k=2+2n$$

Therefore we get $2+2n=-n \Rightarrow 3n=-2\Rightarrow n=-\frac{2}{3}$ and so $k=\frac{2}{3}$.

Is everything correct? Are $\vec{AC}$ and $\vec{AB}$ linear independent because they start at the same point?
(Wondering)

 

[I like Serena]

Is everything correct?

Looks all correct to me. (Happy)

Are $\vec{AC}$ and $\vec{AB}$ linear independent because they start at the same point?
(Wondering)

Let's consider $\overrightarrow{AC}$ and $\overrightarrow{AM_{AC}}$.
They start from the same point.
Are they independent?

Suppose $\vec{AC}$ and $\vec{AB}$ were not linear independent.
Then one would be a multiple of the other, making them parallel.
But parallel vectors don't span a triangle do they?

In other words, $\vec{AC}$ and $\vec{AB}$ are independent because they span a triangle.
It's not because they start from the same point, because that does not guarantee independence. (Nerd)

 

[mathmari]

Looks all correct to me. (Happy)

So, we have that $$\vec{AS}=\frac{2}{3}\vec{AM_{BC}}=\frac{2}{3}\left (\frac{b_1+c_1}{2}-a_1,\frac{b_2+c_2}{2}-a_2\right ) \Rightarrow (s_1-a_1, s_2-a_2)=\vec{AS}=\frac{2}{3}\left (\frac{b_1+c_1}{2}-a_1,\frac{b_2+c_2}{2}-a_2\right ) $$


So, we get the following relations:
$$s_1-a_1=\frac{2}{3}\left (\frac{b_1+c_1}{2}-a_1\right )\Rightarrow s_1= \frac{2}{3}\frac{b_1+c_1}{2}-\frac{2}{3}a_1+a_1 \Rightarrow s_1= \frac{1}{3}(b_1+c_1)+\frac{1}{3}a_1=\frac{1}{3}(a_1+b_1+c_1)$$
and
$$s_2-a_2=\frac{2}{3}\left (\frac{b_2+c_2}{2}-a_2\right ) \Rightarrow s_2= \frac{2}{3}\frac{b_2+c_2}{2}-\frac{2}{3}a_2+a_2 \Rightarrow s_2= \frac{1}{3}(b_2+c_2)+\frac{1}{3}a_2=\frac{1}{3}(a_2+b_2+c_2)$$ Is everything correct? Could I improve something? (Wondering)

Let's consider $\overrightarrow{AC}$ and $\overrightarrow{AM_{AC}}$.
They start from the same point.
Are they independent?

Suppose $\vec{AC}$ and $\vec{AB}$ were not linear independent.
Then one would be a multiple of the other, making them parallel.
But parallel vectors don't span a triangle do they?

In other words, $\vec{AC}$ and $\vec{AB}$ are independent because they span a triangle.
It's not because they start from the same point, because that does not guarantee independence. (Nerd)

So, since $AB$ and $AC$ are not colinear we have that the vector product of $AB$ and $AC$ spans a parallelogram and the half of it is a triangle, right? (Wondering)

 

[I like Serena]

Yep. All correct. (Smile)

 

[mathmari]

Yep. All correct. (Smile)

Thank you very much! (Mmm)