Why are lines and more perplexingly, parallelism, preserved? For lines, is it because the transformation is line-ar?
mpotoole
For lines---yes, it is because all of these equations are linear. Why? Well there's a simple proof for this.
Suppose you're given two points $x_1$ and $x_2$. The line passing through $x_1$ and $x_2$ is given by $x(\alpha) = \alpha x_1 + (1-\alpha)x_2$ for some value $\alpha$.
Now let's apply some affine transformation $A$ to the points on this line. This results in $A x(\alpha) = A ((\alpha x_1) + (1-\alpha)x_2) = \alpha (Ax_1) + (1-\alpha)(Ax_2)$. In other words, we get a line that's passing through two points $Ax_1$ and $Ax_2$.
As for parallelism, that may be a little trickier to prove, but there's a simple justification here. If an affine transformation is a composition of shearing, translation, scaling, and rotation operations, and all of these operations preserve parallelism, then an affine transformation must also preserve parallelism.
Why are lines and more perplexingly, parallelism, preserved? For lines, is it because the transformation is line-ar?
For lines---yes, it is because all of these equations are linear. Why? Well there's a simple proof for this.
Suppose you're given two points $x_1$ and $x_2$. The line passing through $x_1$ and $x_2$ is given by $x(\alpha) = \alpha x_1 + (1-\alpha)x_2$ for some value $\alpha$.
Now let's apply some affine transformation $A$ to the points on this line. This results in $A x(\alpha) = A ((\alpha x_1) + (1-\alpha)x_2) = \alpha (Ax_1) + (1-\alpha)(Ax_2)$. In other words, we get a line that's passing through two points $Ax_1$ and $Ax_2$.
As for parallelism, that may be a little trickier to prove, but there's a simple justification here. If an affine transformation is a composition of shearing, translation, scaling, and rotation operations, and all of these operations preserve parallelism, then an affine transformation must also preserve parallelism.