Review: [[../02 Areas/Math/Barycentric Coordinates|Barycentric Coordinates]] Input: $P, Q, R$ (ordered) [[Vertex|Vertices]] of a [[../02 Areas/Computer Science/Non-Degenerate|Non-Degenerate]] [[Triangle]]. Any point $I$ in the plane of $\triangle PQR$ can be written uniquely in the form: $\large I = \lambda P + \mu Q + \nu R $ where $\large \lambda + \mu + \nu = 1 $ If $v_{p}, v_{q},v_{r}$ are attribute values assigned to [[Vertex|Vertices]] $P,Q,R$ then the linear interpolation at $I$ is: $\large v_{I} = \lambda v_{p} + \mu v_{Q}+\nu v_{R} $ --- Example: $ \large \begin{matrix} &P =(4,7) &Q=(94,47)& R=(24, 67) \\ &v_{P} =-7 &v_{q}=63 &v_{r}=-17 \end{matrix} $ The [[../02 Areas/Math/Barycentric Coordinates|Barycentric Coordinate Functions]]: $\large \begin{align} \mu&= \frac{3x-y-5}{230} \\ \nu&= \frac{3x-y-5}{230} \\ \lambda &= 1-\mu - \nu \end{align} $ Find the [[Linear Interpolation|linearly interpolated value]] at $I=(55,45)$ $\large\begin{matrix} &\mu(I) &= \frac{3(55)-45-5}{230} = &0.5\\ &\nu(I) &= \frac{-4(55)+9(45)-5}{460} = &0.3\\ &\lambda(I) &= 1-0.5-0.3 =&0.2 \end{matrix}$ Therefore, $\begin{align} v_{I} &= \lambda_{I}v_{P}+\mu_{I}v_{Q}+\nu_{I}v_{R}\\ &= (0.2)(-7)+(0.5)(163)+(0.3)(-17) \end{align}$