Slope of a line
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Elie
el 16 de Feb. de 2014
Respondida: Paul
el 16 de Feb. de 2014
I'm trying to find the slope of this line
line([newBoxPolygon(1, 1) newBoxPolygon(4, 1)],[newBoxPolygon(1, 2) newBoxPolygon(4, 2)],'Color','G');
%Slope
X = (max([50 50])-min([50 50]));
Y=(max([1 10000])-min([1 10000]));
Slope_Reference=Y/X;
disp('Slope_Reference:');disp(Slope_Reference);
Is the slope i'm obtaining correct ?
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Mischa Kim
el 16 de Feb. de 2014
Editada: Mischa Kim
el 16 de Feb. de 2014
In general, your equation is correct, k = dY/dX, however, in
X = (max([50 50])-min([50 50])); % 50 - 50 = 0
Y = (max([1 10000])-min([1 10000]));
X = 0 resulting in a slope of Inf. So I am wondering what you are trying to compute the slope of.
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Paul
el 16 de Feb. de 2014
(max([50 50])-min([50 50]))
returns 0, since the minimum and maximum of the vector [50 50] is 50. This means that X=0 and thus Y/X = Y/0 = inf.
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