[proofplan] We compare the two displayed points by expressing the second origin $o'$ as a translate of the first origin $o$. If $u=o'-o$, then each displacement from $o'$ differs from the corresponding displacement from $o$ by subtracting $u$. The coefficient condition $\sum_i \lambda_i=1$ makes the total translation error exactly cancel. [/proofplan]

[step:Express the change of origin as a vector in the model space] Define the vector $u \in V$ by \begin{align*} u=o'-o. \end{align*} By the defining torsor identities of an [affine space](/page/Affine%20Space) modeled on $V$, this means \begin{align*} o'=o+u. \end{align*} For each index $i \in \{1,\ldots,m\}$, the displacement from $o'$ to $p_i$ satisfies \begin{align*} p_i-o'=(p_i-o)-u. \end{align*} Indeed, both sides are vectors in $V$, and adding the right-hand side to $o'$ gives \begin{align*} o'+((p_i-o)-u)=o+u+(p_i-o-u)=o+(p_i-o)=p_i. \end{align*} By uniqueness of the vector carrying $o'$ to $p_i$, the identity follows. [/step]

[step:Compute the weighted displacement from the new origin]Define the vectors $s,s' \in V$ by \begin{align*} s=\sum_{i=1}^m \lambda_i(p_i-o) \end{align*} and \begin{align*} s'=\sum_{i=1}^m \lambda_i(p_i-o'). \end{align*} Using the identity from the previous step and linearity of finite sums in the [vector space](/page/Vector%20Space) $V$, we obtain \begin{align*} s'=\sum_{i=1}^m \lambda_i((p_i-o)-u). \end{align*} Therefore \begin{align*} s'=\sum_{i=1}^m \lambda_i(p_i-o)-\left(\sum_{i=1}^m \lambda_i\right)u. \end{align*} Since $\sum_{i=1}^m \lambda_i=1$, this becomes \begin{align*} s'=s-u. \end{align*}[/step]

[guided]We now track exactly how the vector part of the affine combination changes when the origin changes. Define \begin{align*} s=\sum_{i=1}^m \lambda_i(p_i-o) \end{align*} and \begin{align*} s'=\sum_{i=1}^m \lambda_i(p_i-o'). \end{align*} Both $s$ and $s'$ are vectors in $V$, because each displacement $p_i-o$ and $p_i-o'$ lies in $V$, and $V$ is closed under scalar multiplication and finite sums. From the previous step, for every $i \in \{1,\ldots,m\}$, \begin{align*} p_i-o'=(p_i-o)-u. \end{align*} Substituting this identity into the definition of $s'$ gives \begin{align*} s'=\sum_{i=1}^m \lambda_i((p_i-o)-u). \end{align*} Now we use distributivity and additivity in the vector space $V$: scalar multiplication distributes over vector subtraction, and finite sums may be regrouped. Hence \begin{align*} s'=\sum_{i=1}^m \lambda_i(p_i-o)-\sum_{i=1}^m \lambda_i u. \end{align*} Since $u$ is independent of $i$, the second sum is \begin{align*} \sum_{i=1}^m \lambda_i u=\left(\sum_{i=1}^m \lambda_i\right)u. \end{align*} The coefficient hypothesis gives $\sum_{i=1}^m \lambda_i=1$, so \begin{align*} s'=s-u. \end{align*} This is the cancellation mechanism: changing the origin from $o$ to $o'=o+u$ subtracts exactly $u$ from the weighted displacement because the weights sum to one.[/guided]

[step:Translate back to the affine space and cancel the origin shift] Using $o'=o+u$ and $s'=s-u$, we compute in the affine action: \begin{align*} o'+s'=(o+u)+(s-u). \end{align*} By associativity of the affine action of $V$ on $A$, \begin{align*} (o+u)+(s-u)=o+(u+s-u). \end{align*} The vector-space cancellation in $V$ gives \begin{align*} u+s-u=s. \end{align*} Therefore \begin{align*} o'+s'=o+s. \end{align*} Substituting the definitions of $s$ and $s'$ gives \begin{align*} o'+\sum_{i=1}^m \lambda_i(p_i-o')=o+\sum_{i=1}^m \lambda_i(p_i-o). \end{align*} This is the desired equality. [/step]