Hello everyone
I am facing the following geometry problem, and since am not so good at it, i would ask your help:
We have a straight line that goes throw 3 points, A(X1,Y1) B(X2,Y2) and M(X0,Y0).
We know X1,Y1,X2,Y2 and we know d, the distance between A and M, and we have to define the two possible positions of the pair X0,Y0...
Any Help would be highly appreciated, and thanks in advance...
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Here is
O(1)solution with some inaccuracy permitted:You know
You also know slope of line
So, the two possible values of Xm are X1 ± d * cos(z)
Similarly, Ym = Y1 ± d * sin(z)
You can explain little more pls.....
from the image you can say that sin(z)= (y0-y1) / d from this equation you can say that sin(z)*d = y0-y1 so: y0 = y1 + d*sin(z) but y0 may also be in the other direction so y0=y1(+or-)d*sin(z) also you can notice that cos(z)=(x0-x1)/d from this you can say that x0=x1(+or-)d*cos(z) and you can calculate z since that tan(z)=slope which equals to (y2-y1) / (x2-x1) you can make tan inverse to get z
thanks
.