How can I calculate the slope of a line given two points?

2015 2 21
2 个回答
回答已采纳
更新时间：2019 4 21
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0 个投票

I need a function which can output the slope of a 2D line given two points. Ofcourse, I could just use m = (y2-y1)/(x2-x1), however, I have a long list of points/lines which I need to find the slope of and I don't believe this would be the most efficient method.
Something similar to the pdist2(point_a, point_b) function which I've previously used to find the distance between two points would be ideal.
Thanks in advance.

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 采纳的回答

Image Analyst
Image Analyst 2015-2-21
编辑：Image Analyst 2015-2-22

2 个投票

If you have the endpoints in 4 arrays, use a vectorized divide:
slopes = (y2 - y1) ./ (x2 - x1);
It is efficient.

9 个评论
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John D'Errico
John D'Errico 2015-2-22
Assuming that x1, x2, y1, y2 are all arrays or vectors, it would be necessary to use ./ there for the division.
Image Analyst
Image Analyst 2015-2-22
I know. I thought I put it in but it's not there for some reason. I must have accidentally delete it when I was putting spaces in there. Thanks for catching it. I've put it in now.
Matlaber
Matlaber 2019-2-15
What is the different between:
slopes = (y2 - y1) ./ (x2 - x1);
and
slopes = (y2 - y1) / (x2 - x1);
Both give same result.
Is '.' have special meaning there?
Image Analyst
Image Analyst 2019-2-16
It's an "elment-by-element" divide rather than a matrix divide. If the y area scalars (not vectors) then there is no difference.
sisir regmi
sisir regmi 2019-4-21
what if we have just 2 array of x and y points?
Image Analyst
Image Analyst 2019-4-21
Try
slopes = diff(y) ./ diff(x);
To get the slopes between a point and the point before it.
sisir regmi
sisir regmi 2019-4-21
its days matrix dimension must agree
X Y
%===================
0.0600 13.22
0.1550 179.6
0.1800 438.2
0.2500 687.8
0.2800 981.3
0.4180 1152
0.5540 931
1.7370 91.24
1.8220 80.37
1.8230 37.61
1.8240 12.09
1.8250 70.55
2.0290 96.34
2.0620 125.6
2.0730 73.59
2.0750 57.15
2.0910 88.75
2.1080 80.1
2.1100 111.6
2.1570 258.1
3.1200 40.66
3.2600 42.7
3.3700 60.61
3.4400 236.8
3.5400 239.6
3.5600 212.1
3.5800 128.4
3.9200 46.09
4.0800 40.58
4.1700 47.02
4.1900 90.86
4.3600 174.5
4.3800 244.2
4.3900 316.9
4.4500 482
4.4560 614.4
4.4400 775.3
4.4600 923.3
4.3500 782.3
4.5500 486.9
4.6900 75.65
5.1380 35.83
5.2800 31.34
5.4000 35.01
5.5400 28.84
5.6700 36.18
5.7800 31.63
5.8000 16.6
5.8100 11.26
5.9000 4.97
5.9001 11.38
5.9100 10.54
5.9200 9.34
6.1900 13.16
6.2000 21.86
6.2100 28.86
6.4230 738.6
6.5700 148.5
6.6500 469.1
6.7100 763.6
6.7200 914.6
6.8300 301.9
6.8500 192
6.9200 118.5
7.1100 79.65
7.2500 49.06
];
figure(1)
slope=diff(y)./diff(x);
matrix dimensions must agree.
Error in untitled5 (line 72)
slope=diff(y)./diff(x)
Image Analyst
Image Analyst 2019-4-21
Well, why do your x and y have different sizes? That doesn't make sense, so explain it to me.

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更多回答（1 个）

Neal Young
Neal Young 2015-2-22

0 个投票

Hi, thank you both for the quick reply. I've noted the points below. Each point is connected to the subsequent point via a bar and I've to find the slope of each of these bars. Therefore the slopes = (y2 - y1) ./ (x2 - x1) formula should be most efficient.
A = [135, 86.5]; B = [152.7, 119]; C = [142.9, 114.2]; D = [192.4, 164.7]; E = [151.8, 170.4]; F = [135.4, 190.6]; G = [143.1, 184]; H = [121.4, 208];

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