MATLAB: Projecting a point into a line

projection

How can I project a point (let's say (50,0)) to a line (y = 5.6x – 7.1)?

Thank you.

Best Answer

You need the equation of both perpendicular lines. You already have the equation for the first line. In your line y = 5.6x - 7.1 the slope is 5.6.

The slope of the second line will just be the perpendicular slope of your first line.

m = 5.6; 
b = -7.1; 
x = 50; 
y = 0; 
perpSlope = -1/m;

To get the y intercept of the 2nd line you just need to solve for y=mx+b using your point (x,y)

yInt = -perpSlope * x + y;

Now you've got the two linear equations and you need to find out where they intersect. Here we find the x coordinate of the intersection. m and b are the slope and intercept of line 1, perSlope and yInt are the slope and intercept of line 2.

xIntersection = (yInt - b) / (m - perpSlope);

To get the y coordinate of the intersection, we just plug the x coordinate into one of the equations.

yIntersection = perpSlope * xIntersection + yInt;

Now we can plot it out to make sure it looks rights

figure
plot(x,y, 'rx')
hold on
plot(xIntersection, yIntersection, 'ro')
refline(m, b)
refline(perpSlope, yInt)
axis equal

Related Solutions

MATLAB: How to draw a perpendicular line

The slope of a line perpendicular to another has a slope that is -1 over the slope of the other line. so

coefficients = polyfit(xKnown, yKnown, 1);
slope = coefficients(1);

Now get the min and max x where you want to draw the line.

[x1p, index1] = min(xKnown);
[x2p, index2] = max(xKnown);
% Get y at those endpoints
y1p = yKnown(index1);
y2p = yKnown(index2);
% Compute the slope
perpSlope = -1 / slope;
% Equation of line going through the left most x (at x1p) is
% (y - y1p) = perpSlope * (x - x1p)
y1 = perpSlope * (x1p - xKnown(index1)) + y1p
y2 = perpSlope * (x2p - xKnown(index1)) + y1p
line([x1p, x2p], [y1, y2], 'Color', 'r')

It's untested code, so no guarantees.

MATLAB: Plot continuous line based on user input

"How would I plot a line with a slope of one that goes infinitely until it hits the y axis"

I'm interpreting this as a line that extends from point (xClic,yClic) to the point that crosses the vertical line at x=0, with a slope of 1.

If that interpretation is correct, all you need to so is compute the y intercept.

% Point where user clicked
xClic = -2; 
yClic= -5; 
% Plot that point
plot(xClic, yClic,'bo')
xlim([-8 8])
ylim([-8,8])
hold on
% Compute y intercept for slope of 1
b = yClic-1*xClic;
% Plot the line between the click point and the y intercept
plot([xClic,0],[yClic,b],'b-')

A similar approach is to use refline() but that extends to your axis limits.

h = refline(1,b)
h.LineStyle = ':'

Best Answer

Related Solutions

MATLAB: How to draw a perpendicular line

MATLAB: Plot continuous line based on user input

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