Segment a line to sections and add to it perpendicular line

clc;

Based on this superb question and answers https://www.mathworks.com/matlabcentral/answers/484422-how-to-find-the-coordinates-of-a-point-perpendicular-to-a-line-knowing-the-distance I add a new variable that will help me to segment the line into sections (numOfSegment).

When I set the numOfSegment to 2 it works well but when I add a bigger number (e.g., 4) it plots the perpendicular line out of the line. Why? and how can I plot all the perpendicular lines and store their coordinates?

Code:

clc;
clear all;
close all;
 
A = [0 100]; %[x,y]
B = [12,180]; %[x,y]
 
Clen = 25; % distance off the line AB that C will lie
 
% Call AB the vector that points in the direction
% from A to B.
AB = B - A;
 
% Normalize AB to have unit length
AB = AB/norm(AB);
 
% compute the perpendicular vector to the line
% because AB had unit norm, so will ABperp
ABperp = AB*[0 -1;1 0];
 
 
%%%%%% numOfSegment is set to 2;
numOfSegment = 2;
ABmid = (A + B)/numOfSegment;
 
% Compute new points C and D, each at a ditance
% Clen off the line. Note that since ABperp is
% a vector with unit eEuclidean norm, if I
% multiply it by Clen, then it has length Clen.
C = ABmid + Clen*ABperp;
D = ABmid - Clen*ABperp;
 
% plot them all
subplot(1,2,1)
plot([A(1);B(1)],[A(2);B(2)],'-bo',[C(1);D(1)],[C(2);D(2)],'-rs')
axis equal
 
%%%%%% numOfSegment is set to 4
numOfSegment = 4;
ABmid = (A + B)/numOfSegment;
 
% Compute new points C and D, each at a ditance
% Clen off the line. Note that since ABperp is
% a vector with unit eEuclidean norm, if I
% multiply it by Clen, then it has length Clen.
C = ABmid + Clen*ABperp;
D = ABmid - Clen*ABperp;
 
% plot them all
subplot(1,2,2)
plot([A(1);B(1)],[A(2);B(2)],'-bo',[C(1);D(1)],[C(2);D(2)],'-rs')
axis equal