Variable acceleration and vector question

I would like some help with question 12 c
I thought that if the particle is moving in the direction of -2i+j then this would mean:
-2(1-2t)i = (2t^2 +t -13)
but in the mark scheme it is -2(2t^2 +t -13)=1-2t
could someone explain why please (photo of question is below)

The velocity components have the same ratio (i/j) so
-2/1 = (1-2t)/(2t^2+t-13)
...
The direction will be the same if the (velocity) components make the same angle with the i axis (for instance), so its just two similar right triangles with i being the adjacent and j being the opposite.