First, note that a linear transformation is not the same as a function with a linear graph, that is, a function in the form of $f(x)=mx+b$.

We deduce a few properties about $T$. First, we see that $T(0)=T(0x)=0T(x)=0$. Also, $T(x)=T(1x)=xT(1)$. If we let $m=T(1)$, then we see that $T(x)=mx$ for some $m$. It can be easily proven that the first property $T(x+y)=T(x)+T(y)$ still holds.

Exercise: show that when $T$ only satisfies the first condition, then it does not need to be in the form $T(x)=mx$. Would this solution still be correct?

Therefore, we simply need to check if all points are on the same graph $T(x)=mx$ for some $m$. The implementation is not very difficult, but do be wary of the corner case where $x=0$.

Failing to resolve the $x=0$ corner case resulted in $25\mathrm{\%}$ of the points, and a slow all-pairs comparison of slope was awarded $50\mathrm{\%}$ of the points.

Time Complexity: $\mathcal{O}(N)$