If we sum the coefficients of equation \(R\), we get
\[(1-s)(1-t) + (1-s)t + s = (1 - t - s + st) + (t - st) + s = 1\]
Since \(R\) is an arbitrary point inside the triangle \(ABC\), we can generalize the equation to any point \(P\). That is, any point \(P\) inside the triangle \(ABC\) has the form
These coefficients \(α\), \(β\) and \(γ\) are called the barycentric coordinates of \(P\) with respect to the triangle \(ABC\), such that
If \(α = 0\), then \(P\) lies on the edge \(BC\)
If \(β = 0\), then \(P\) lies on the edge \(AC\)
If \(γ = 0\), then \(P\) lies on the edge \(AB\)
The Area of a Triangle
There are two ways to calculate the area of the triangle:
Approach 1
The area of the triangle is $$ Area = \frac{1}{2} \;. \; Base \;. \; Height $$ where \(Height\) is the perpendicular shortest distance from the the base edge to the opposite vertex. For example, the area of triangle \(PBC\) is $$ Area(▲PBC) = \frac{1}{2} \; ||\vec{BC}|| \; d_{⊥} (P,BC) $$
Approach 2
Another easier way to calculate the area of the triangle, is to use the area of the parallelogram. Consider the parallelogram \(ABCD\). The area of the parallelogram \(ABCD\) is $$ Area(▰ABCD) = ||\vec{AB}|| \; ||\vec{AC}|| \; sin(\theta) $$ We can also express the area of the parallelogram \(ABCD\) as the magnitude of the cross product of two vectors that lie on the parallelogram. That is, $$ Area(▰ABCD) = ||\vec{AB} \times \vec{AC}|| $$ The area of the triangle is equal to half the area of the parallelogram. Hence, the area of the triangle \(ABC\) can be expressed as $$ Area(▲ABC) = \frac{||\vec{AB}|| \; ||\vec{AC}|| \; sin(\theta)}{2} = \frac{||\vec{AB} \times \vec{AC}|| }{2} $$
Barycentric Coordinates Equations
The equations for calculating the barycentric coordinates \(α\), \(β\) and \(γ\) of the triangle \(ABC\) are: $$ α = \frac{Area(▲PBC)}{Area(▲ABC)} $$ $$ β = \frac{Area(▲PAC)}{Area(▲ABC)} $$ $$ γ = \frac{Area(▲PAB)}{Area(▲ABC)} $$
Proof
Let \(P\) be a point inside the triangle \(ABC\). We know that $$ P = αA + βB + γC $$ where \(α + β + γ = 1\) and \(α, β, γ \geq 0\).
We want to prove that $$ α = \frac{Area(▲PBC)}{Area(▲ABC)} $$ $$ β = \frac{Area(▲PAC)}{Area(▲ABC)} $$ $$ γ = \frac{Area(▲PAB)}{Area(▲ABC)} $$ Without loss of generality, consider the barycentric coordinate \(α\).
We know that the area of triangle \(PBC\) is $$ Area(▲PBC) = \frac{1}{2} \; ||\vec{BC}|| \; d_P $$ and the area of triangle \(ABC\) is $$ Area(▲ABC) = \frac{1}{2} \; ||\vec{BC}|| \; d_A $$
Let \(d_P\) be the perpendicular distance from \(P\) to \(BC\) and \(d_A\) be the perpendicular distance from \(A\) to \(BC\).
As \(d_P\) increases, the point \(P\) becomes closer to the point \(A\), causing the value of \(α\) to increase. When \(d_P=d_A\), the point \(P\) coincides the point \(A\) and \(α = 1\).
Similarly, as \(d_P\) decreases, the point \(P\) becomes closer to the edge \(BC\) and the value of \(α\) to also decrease. When \(d_P=0\), the point \(P\) lies on the edge \(BC\) and \(α=0\).
This implies that \(α\) is proportional to \(d_P\) $$ α ∝ d_P $$ where the proportionality constant is \(d_A\). That is, $$ α = \frac{d_P}{d_A} $$
Using the equations for the area of triangle \(▲PBC\) and \(▲ABC\), we can rewrite the equation for \(α\) as: $$ α = \frac{d_P}{d_A} = \frac{\frac{2 \; . \; Area(▲PBC)}{||\vec{BC}||}}{\frac{2 \; . \; Area(▲ABC)}{||\vec{BC}||}} $$ $$ α = \frac{Area(▲PBC)}{Area(▲ABC)} $$ By applying the same reasoning to the other two coordinates, we obtain that $$ β = \frac{Area(▲PAC)}{Area(▲ABC)} $$ $$ γ = \frac{Area(▲PAB)}{Area(▲ABC)} $$ Thus, proof is complete.
\(\blacksquare\)
References
Chapter 7.9.1 - Computer Graphics: Principles and Practice by John F. Hughes
Let \(d_P\) be the perpendicular distance from \(P\) to \(BC\) and \(d_A\) be the perpendicular distance from \(A\) to \(BC\).
