points lying on a line from a particular distance

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KalMandy
KalMandy 2017-2-18
回答： Matt J 2017-2-18
Hi All, In my matlab program , I find two points from solving the line equation (given a point and the angle) and the distance equation. ex:
syms x y
[solx,soly]=solve(y-m*x==yc-m*xc , (x-xc).^2+(y-yc).^2==d^2)
here (xc,yc) is the point coordinate, m is tand(angle) and d is distance. my angle rotates from 0-360 degrees.
my problem is (solx,soly) gives me two solutions, although I want to calculate only in angle=0 degrees direction. For example, I do not want the solution for angle 0 and 360 at the same time. Do you know how I can just sort out the point lying in angle 0 direction for instance?
  2 个评论
Matt J
Matt J 2017-2-18
I do not want the solution for angle 0 and 360 at the same time
You should be letting the angle run from only 0 to 180. Angles theta and 180+theta will always correspond to the same line.
Even then, though, you will have 2 intersection points.

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采纳的回答

Matt J
Matt J 2017-2-18
A better way to obtain the points might be parametrically
theta=0:359;
x=xc+d*cosd(theta);
y=yc+d*sind(theta);

更多回答（1 个）

Matt J
Matt J 2017-2-18
编辑：Matt J 2017-2-18
You should be restricting the angle to (-90, 90). tan() has redundant or undefined values outside that interval. In particular, you must avoid angles +/-90 deg. where yc-m*xc will be infinite.
This restriction will still give you two solutions, but you can throw away the solution for which x < xc.
  2 个评论
KalMandy
KalMandy 2017-2-18
编辑：KalMandy 2017-2-18
but I need to find the points lying d distance away from (xc,yc) in all 360 angles (However the solution will not be a circle, due to some other conditions). what can i do in this case?
Matt J
Matt J 2017-2-18
编辑：Matt J 2017-2-18
Then keep both solutions returned by solve(). One corresponds to angle theta, the other to theta+180. As I said, though theta=+/-90 has no solution. You will have to compromise on that case.

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