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Slope of a line

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Elie
Elie 2014-2-16
回答： Paul 2014-2-16
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
Mischa Kim 2014-2-16
编辑：Mischa Kim 2014-2-16
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.

更多回答（1 个）

Paul
Paul 2014-2-16
(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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