Thanks for your help, this is related to this question
The calculation is: $\lambda A+(1−\lambda)B$ for $\lambda \in[0,1]$
With $A = (0,0)$ and $B = (2,4)$
I am no mathematician, so if you could show the steps to the solution, I would be most grateful.
Cheers, Rudy
On
This is a parameterization of the line segment joining $B$ and $A$. The expression can be rearranged into $B+\lambda(A-B)$, which might be a bit easier to interpret: $(A-B)$ is a direction vector pointing from $B$ to $A$, so you could think of this as describing the motion of a particle that starts at $B$ when $\lambda=0$ and moves at a constant speed toward $A$ until reaching it at $\lambda=1$. If you let $\lambda$ range over all of $\mathbb R$ instead, you’ll get a parameterization of the line through $A$ and $B$.
Another way to think of the expression $\lambda A+(1-\lambda)B$ is as a weighted average of $A$ and $B$, with the weight corresponding to the proportion of the distance between them. This might be a bit more obvious if you write it as $${\lambda A+(1-\lambda)B\over\lambda+(1-\lambda)}.$$ If you take $\lambda=1/2$, for instance, the expression becomes $(A+B)/2$, which you might recognize as a formula for calculating the midpoint of $A$ and $B$.
For your specific example, we have $$\lambda A+(1-\lambda)B=\lambda(0,0)+(1-\lambda)(2,4)=(2-2\lambda,4-4\lambda).$$
You can parametrize the segment joining $(0, 0)$ to $(2, 4)$ as $\{(t, 2t) : 0 \leq t \leq 2\}$. Hope this helps!