I want to generate a fixed percent of random points on a straight line joining two points.

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I want to generate 5 percent of the randomly generated numbers in a fixed straight line equation. For example, i have two points (x1, y1) and (x2, y2), and now i have drawn a straight line between the two points, with the help of straight line equation. Now i want 100 random numbers to be generated out of which 5 percent of those random numbers should fall in the straight line that is joined between the two points.
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SUSHMA MB
SUSHMA MB 2016 年 8 月 3 日
let the boundary be a square of sides 10m side length.
SUSHMA MB
SUSHMA MB 2016 年 8 月 3 日
Let the
XY_BOUNDARY = [0,70,0,50};
position = [XY_BOUNDARY(2) - XY_BOUNDARY(1); XY_BOUNDARY(4) - XY_BOUNDARY(3)] .* rand(2,1) ...
+ [XY_BOUNDARY(1);XY_BOUNDARY(3)];
In the above code i get a random number within the specified boundary. Now i want the random point position in a specified boundary as well as on the line connecting two points (x1,y1), and (x2,y2)...Please tell me how to do this

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Torsten
Torsten 2016 年 8 月 3 日
Instead of generating 100 random numbers, use "rand" to generate 5 random numbers on [0:1] and set the points to
lambda=rand(5,1);
x=x1+lambda*(x2-x1);
y=y1+lambda*(y2-y1);
Best wishes
Torsten.
  5 件のコメント
SUSHMA MB
SUSHMA MB 2016 年 8 月 3 日
But can you plz tell me that, if 100 points are their, then how can i make atleast 5 points to appear it on a line
SUSHMA MB
SUSHMA MB 2016 年 8 月 3 日
編集済み: SUSHMA MB 2016 年 8 月 3 日
Let the
XY_BOUNDARY = [0,70,0,50};
position = [XY_BOUNDARY(2) - XY_BOUNDARY(1); XY_BOUNDARY(4) - XY_BOUNDARY(3)] .* rand(2,1) ...
+ [XY_BOUNDARY(1);XY_BOUNDARY(3)];
In the above code i get a random number within the specified boundary. Now i want the random point position in a specified boundary as well as on the line connecting two points (x1,y1), and (x2,y2)...Please tell me how to do this

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