# How to draw orthogonal lines ?

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SUSHMA MB on 14 Aug 2015
Commented: Image Analyst on 15 Aug 2015
I have two points, one is start and the other is end point. I have connected these two points by a straight line. Now i want to draw orthogonal lines over the line. How can i draw it? I have attached a picture for reference. As shown in the picture, i want eight orthogonal planes placed in equidistant. Please help me with this problem.

blaat on 14 Aug 2015
The line between your two points can be described by
y = a (x - x1) + b,
where
a = (y2 - y1)/(x2 - x1)
b = y1,
if we call your two points (x1, y1) and (x2, y2). Lines perpendicular to the original line will have a slope of -1/a and can be expressed as:
y = -1/a (x - x0) + y0,
where (x0, y0) is the point on the original line where it intersects the orthogonal line.
Equidistant points on the line can be easily computed using linspace():
num_orth = 8;
x_orth = linspace(x1, x2, num_orth);
y_orth = linspace(y1, y2, num_orth);

#### 1 Comment

SUSHMA MB on 14 Aug 2015
How to draw the orthogonal lines ?

Mike Garrity on 14 Aug 2015
Edited: Mike Garrity on 14 Aug 2015
Let's say we have the following line:
pt1 = 10*randn(1,2);
pt2 = 10*randn(1,2);
line([pt1(1), pt2(1)],[pt1(2),pt2(2)])
The points where four equally spaced orthogonal lines cross it are:
n = 4;
t = linspace(0,1,n+2); % evenly spaced parameters
t = t(2:(end-1)); % we don't need the start and end points
v = pt2 - pt1;
x = pt1(1) + t*v(1); % p(t) = p1 + t*(p2-p1)
y = pt1(2) + t*v(2);
h = line(x,y);
h.LineStyle = 'none';
h.Marker = 'o';
Next we need to normalize that vector:
delete(h)
v = v / norm(v);
And then we rotate it by 90 degrees. That's just swapping the X & Y components of v, and changing the sign of one:
for i=1:n
line([x(i)+v(2), x(i)-v(2)],[y(i)-v(1), y(i)+v(1)]);
end
The one catch at this point is that the axes might be using different scale factors for the X & Y. This will make the lines look like they're not orthogonal, even if they are mathematically. We can fix this by calling:
axis equal

Mike Garrity on 14 Aug 2015
Things are a bit more interesting in 3D. I discussed that case on the MATLAB Graphics blog in these two posts ( post1 , post2 ).
SUSHMA MB on 14 Aug 2015
Hi...Thank you for the answer. It works. But my points are positive, but why am i getting negative axis. Can i adjust the axis, like i want only positive axis. Like i have points: start_point = [5.95 37.55]; goal_point = [35.62 5.73]; and i want to define the length of the orthogonal lines, i.e., 5 units towards both the side from the point of intersection. Is it possible?

Mike Garrity on 14 Aug 2015
I'm afraid I don't understand the question "why am i getting negative axis".
The length of the orthogonal lines goes at the point where I divided v by norm(v). That was to get them to be unit length. Just multiply by the length you want:
start_point = [5.95 37.55]
goal_point = [35.62 5.73]
line([start_point(1), goal_point(1)],[start_point(2), goal_point(2)],'Marker','o')
n = 8;
t = linspace(0,1,n+2);
t = t(2:(end-1));
v = goal_point - start_point;
x = start_point(1) + t*v(1);
y = start_point(2) + t*v(2);
v = 5*v / norm(v);
for i=1:n
line([x(i)+v(2), x(i)-v(2)],[y(i)-v(1), y(i)+v(1)]);
end
axis equal

SUSHMA MB on 15 Aug 2015
Thank you so much...
SUSHMA MB on 15 Aug 2015
And how to further divide each of theses orthogonal lines in an equal spacing. I mean i want to further divide the line into an equal spacing of 8, but along the orthogonal lines. And this time i only want the points, not necessary to visualize the points on the orthogonal lines. I hope i made myself clear. Thank you in advance.
Image Analyst on 15 Aug 2015
It's just simple 10th grade algebra. But if you want the MATLAB way, just use linspace():
xEquallySpaced = linspace(x(1), x(2), numPoints);
yEquallySpaced = linspace(y(1), y(2), numPoints);