proof of concept second approach, normalization of orthogonal not necessary for dot product to become 0.

UPDATE: Nota bene: only additions no divisions, which makes the code fast and secure from division by 0 problems.

($p1,$p2,$p3)=([2,2],[3,4],[4,6]);
 
sub vector {
  my ($x1,$y1) = @{$_[0]};
  my ($x2,$y2) = @{$_[1]};
  my @v=( $x2-$x1, $y2-$y1 );
  my @o=( -$v[1] , $v[0]   );
  return [@v],[@o];
}
 
sub dot {
  my ($x1,$y1) = @{$_[0]};
  my ($x2,$y2) = @{$_[1]};
  return $x1*$x2 + $y1*$y2;
}
 
my ($v12,$o12)= vector($p1,$p2);
 
# the test
my ($v)= vector($p1,$p3);
print dot ($o12,$v);        #  0 because p3 is collinear with p1 ,p2
[download]