Perpendicular on a line from a given point

Question

How can I draw a perpendicular on a line segment from a given point? My line segment is defined as (x1, y1), (x2, y2), If I draw a perpendicular from a point (x3,y3) and it

Answer 1

I agree with peter.murray.rust, vectors make the solution clearer:

// first convert line to normalized unit vector
double dx = x2 - x1;
double dy = y2 - y1;
double mag = sqrt(dx*dx + dy*dy);
dx /= mag;
dy /= mag;

// translate the point and get the dot product
double lambda = (dx * (x3 - x1)) + (dy * (y3 - y1));
x4 = (dx * lambda) + x1;
y4 = (dy * lambda) + y1;

Answer 2

Find out the slopes for both the lines, say slopes are m1 and m2 then m1*m2=-1 is the condition for perpendicularity.

Answer 3

You know both the point and the slope, so the equation for the new line is:

y-y3=m*(x-x3)

Since the line is perpendicular, the slope is the negative reciprocal. You now have two equations and can solve for their intersection.

y-y3=-(1/m)*(x-x3)
y-y1=m*(x-x1)

Answer 4

Matlab function code for the following problem

function Pr=getSpPoint(Line,Point)
% getSpPoint(): find Perpendicular on a line segment from a given point
x1=Line(1,1);
y1=Line(1,2);
x2=Line(2,1);
y2=Line(2,1);
x3=Point(1,1);
y3=Point(1,2);

px = x2-x1;
py = y2-y1;
dAB = px*px + py*py;

u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
x = x1 + u * px;
y = y1 + u * py;

Pr=[x,y];

end

Answer 5

This is mostly a duplicate of Arnkrishn's answer. I just wanted to complete his section with a complete Mathematica code snippet:

m = (y2 - y1)/(x2 - x1)
eqn1 = y - y3 == -(1/m)*(x - x3)
eqn2 = y - y1 == m*(x - x1)
Solve[eqn1 && eqn2, {x, y}]

Answer 6

First, calculate the linear function determined by the points (x1,y2),(x2,y2).

We get:

y1 = mx+b1 where m and b1 are constants.

This step is easy to calculate by the formula of linear function between two points.

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Then, calculate the linear function y that goes through (x3,y3).

The function slope is -m, where m is the slope of y1.

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Then calculate the const b2 by the coordinates of the point (x3,y3).

We get y2 = -mx+b2 where m and b2 are constants.

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The last thing to do is to find the intersection of y1, y2. You can find x by solving the equation: -mx+b2 = mx+b1, then place x in one of the equations to find y.