We call our real parameter [tex]t[/tex] and try to save some work by setting [tex]t=x-1[/tex] so
[tex]x=t+1[/tex]
and
[tex]y = 1.52 t - 2[/tex]
or
[tex](x,y)=(1, -2) + t(1, 1.52)
[/tex]
That's enough to answer the question.
[tex](1, 1.52) [/tex] is the parallel vector, often called the direction vector.
To get the perpendicular vector, we swap the coordinates and negate one of them; it doesn't matter which.
[tex](1.52, -1) [/tex] is the perpendicular vector.
[tex](1, -2) [/tex] is a point on the line, given by [tex]t=0.[/tex]
