# Area of a Triangle (and the beginning of barycentric coordinates)

Let the dimension be $$n=2$$ and $$T$$ a Triangle with positive area $$|T|$$ and the corners $$P_1, P_2, P_3 \in \mathbb{R}^2$$. How do I show:

$$2|T|=\det \begin{pmatrix} 1 & 1 & 1 \\ P_1(1) & P_2(1) & P_3(1) \\ P_1(2) & P_2(2) & P_3(2) \end{pmatrix}$$

With $$P_k(j)$$ being the $$j$$th component of the vectors $$P_k\in \mathbb{R}^2$$

• Here’s one way to think about this: the determinant is equal to the volume of a certain paralellepiped in $\mathbb R^3$. Can you find the triangle in it and relate the triangle’s area to this volume? – amd Jun 11 at 20:07
• thank you: I found this answer toppr.com/guides/maths/determinants/… and I think it is just brilliant – KingDingeling Jun 11 at 20:13
• Good! Though working it out for yourself would’ve been better :) This determinant is also equal to $\det\begin{bmatrix}P_1(1)-P_3(1)&P_2(1)-P_3(1)\\P_1(2)-P_3(2)&P_2(2)-P_3(2)\end{bmatrix}$, which is the area of a paralellogram that’s intimately related to the triangle. – amd Jun 11 at 20:24
• BTW, I don’t really like that linked derivation per se: the $3\times3$ determinant appears magically out of nowhere and almost coincidentally has the same expansion as the area formula derived earlier. By placing the triangle on the plane $x=1$ in $\mathbb R^3$ and examining the pyramid formed by it and the origin, this determinant arises naturally. Perhaps one of the linked videos on that page goes into this. – amd Jun 11 at 20:34

Place the triangle $$\triangle{ABC}$$ on the plane $$x=1$$ in $$\mathbb R^3$$ and consider the pyramid with this triangle as its base and the origin for its apex:
This pyramid’s altitude is $$1$$, so its volume is equal to the area of its base, $$\triangle{ABC}$$. If we glue a reflected copy of this pyramid to its base, we get a parallelepiped whose volume is given by the triple product $$\overrightarrow{OA}\times\overrightarrow{OB}\cdot\overrightarrow{OC} = \det\begin{bmatrix}\overrightarrow{OA} & \overrightarrow{OB} & \overrightarrow{OC}\end{bmatrix} = \det\begin{bmatrix}1&1&1 \\ A(1)&B(1)&C(1) \\ A(2)&B(2)&C(2) \end{bmatrix}.$$ The volume of the pyramid, and hence also the area of the triangle, is half of this value.
We can also construct this geometrically in $$\mathbb R^2$$, but it requires some extra work to pass to a $$3\times3$$ determinant. Similarly to the above construction, we adjoin a reflected version of the triangle to form a parallelogram:
The area of this paralellogram is equal to the value of the determinant $$\det\begin{bmatrix}\overrightarrow{CA}&\overrightarrow{CB}\end{bmatrix} = \det\begin{bmatrix}A(1)-C(1) & B(1)-C(1) \\ A(2)-C(2) & B(2)-C(2)\end{bmatrix}.$$ Embedding this into a $$3\times3$$ matrix and using some properties of determinants, \begin{align} \begin{vmatrix}A(1)-C(1) & B(1)-C(1) \\ A(2)-C(2) & B(2)-C(2)\end{vmatrix} &= \begin{vmatrix}0&0&1 \\ A(1)-C(1) & B(1)-C(1) & 0 \\ A(2)-C(2) & B(2)-C(2) & 0\end{vmatrix} \\ &= \begin{vmatrix}0&0&1 \\ A(1)-C(1) & B(1)-C(1) & C(1) \\ A(2)-C(2) & B(2)-C(2) & C(2)\end{vmatrix} \\ &= \begin{vmatrix}1&1&1 \\ A(1) & B(1) & C(1) \\ A(2) & B(2) & C(2)\end{vmatrix} \end{align}. The second step is valid because if we expand the determinant along the top row, only the lower-left $$2\times2$$ submatrix contributes to its value, so the entries below the $$1$$ in the last column can be anything. I prefer the derivation as a volume, though, because it doesn’t require any algebraic trickery.